task_url stringlengths 30 116 | task_name stringlengths 2 86 | task_description stringlengths 0 14.4k | language_url stringlengths 2 53 | language_name stringlengths 1 52 | code stringlengths 0 61.9k |
|---|---|---|---|---|---|
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #PureBasic | PureBasic | EnableExplicit
DisableDebugger
Procedure.d maxXY(a.d,b.d,c.d,d.d)
If a<b : Swap a,b : EndIf
If a<c : Swap a,c : EndIf
If a<d : Swap a,d : EndIf
ProcedureReturn a
EndProcedure
Procedure.d minXY(a.d,b.d,c.d,d.d)
If a>b : Swap a,b : EndIf
If a>c : Swap a,c : EndIf
If a>d : Swap a,d : EndIf
ProcedureReturn a
EndProcedure
Procedure Ptree(x1.d, y1.d, x2.d, y2.d, d.i=0)
If d>10 : ProcedureReturn : EndIf
Define dx.d=x2-x1,
dy.d=y1-y2,
x3.d=x2-dy,
y3.d=y2-dx,
x4.d=x1-dy,
y4.d=y1-dx,
x5.d=x4+(dx-dy)/2.0,
y5.d=y4-(dx+dy)/2.0,
p1.d=(maxXY(x1,x2,x3,x4)+minXY(x1,x2,x3,x4))/2.0,
p2.d=(maxXY(y1,y2,y3,y4)+minXY(y1,y2,y3,y4))/2.0,
p3.d=(maxXY(x1,x2,x3,x4)-minXY(x1,x2,x3,x4))
FrontColor(RGB(Random(125,1),Random(255,125),Random(125,1)))
LineXY(x1,y1,x2,y2)
LineXY(x2,y2,x3,y3)
LineXY(x3,y3,x4,y4)
LineXY(x4,y4,x1,y1)
BoxedGradient(minXY(x1,x2,x3,x4),minXY(y1,y2,y3,y4),p3,p3)
FillArea(p1,p2,-1)
Ptree(x4,y4,x5,y5,d+1)
Ptree(x5,y5,x3,y3,d+1)
EndProcedure
Define w1.i=800,
h1.i=w1*11/16,
w2.i=w1/2,
di.i=w1/12
If OpenWindow(0,#PB_Ignore,#PB_Ignore,w1,h1,"Pythagoras tree")
If CreateImage(0,w1,h1,24,0) And StartDrawing(ImageOutput(0))
DrawingMode(#PB_2DDrawing_Gradient)
BackColor($000000)
Ptree(w2-di,h1-10,w2+di,h1-10)
StopDrawing()
EndIf
ImageGadget(0,0,0,0,0,ImageID(0))
Repeat : Until WaitWindowEvent(50)=#PB_Event_CloseWindow
EndIf
End |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #C.23 | C# | using System;
using MathNet.Numerics.LinearAlgebra;
using MathNet.Numerics.LinearAlgebra.Double;
class Program
{
static void Main(string[] args)
{
Matrix<double> A = DenseMatrix.OfArray(new double[,]
{
{ 12, -51, 4 },
{ 6, 167, -68 },
{ -4, 24, -41 }
});
Console.WriteLine("A:");
Console.WriteLine(A);
var qr = A.QR();
Console.WriteLine();
Console.WriteLine("Q:");
Console.WriteLine(qr.Q);
Console.WriteLine();
Console.WriteLine("R:");
Console.WriteLine(qr.R);
}
} |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #68000_Assembly | 68000 Assembly | LEA $000200,A3
JSR PrintString ;(my print routine is 255-terminated and there just so happens to be an FF after the name of the game.) |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #AArch64_Assembly | AArch64 Assembly | sp+0 = argc
sp+8 = argv[0]
sp+16 = argv[1] ... |
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Common_Lisp | Common Lisp | (defun mmul (a b)
(loop for x in a collect
(loop for y in x
for z in b sum (* y z))))
(defun count-tri (lim &aux (prim 0) (cnt 0))
(labels ((count1 (tr &aux (peri (reduce #'+ tr)))
(when (<= peri lim)
(incf prim)
(incf cnt (truncate lim peri))
(count1 (mmul '(( 1 -2 2) ( 2 -1 2) ( 2 -2 3)) tr))
(count1 (mmul '(( 1 2 2) ( 2 1 2) ( 2 2 3)) tr))
(count1 (mmul '((-1 2 2) (-2 1 2) (-2 2 3)) tr)))))
(count1 '(3 4 5))
(format t "~a: ~a prim, ~a all~%" lim prim cnt)))
(loop for p from 2 do (count-tri (expt 10 p))) |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Scheme | Scheme | ((lambda (s) (display (list s (list (quote quote) s)))) (quote (lambda (s) (display (list s (list (quote quote) s)))))) |
http://rosettacode.org/wiki/Pseudo-random_numbers/Combined_recursive_generator_MRG32k3a | Pseudo-random numbers/Combined recursive generator MRG32k3a | MRG32k3a Combined recursive generator (pseudo-code)
/* Constants */
/* First generator */
a1 = [0, 1403580, -810728]
m1 = 2**32 - 209
/* Second Generator */
a2 = [527612, 0, -1370589]
m2 = 2**32 - 22853
d = m1 + 1
class MRG32k3a
x1 = [0, 0, 0] /* list of three last values of gen #1 */
x2 = [0, 0, 0] /* list of three last values of gen #2 */
method seed(u64 seed_state)
assert seed_state in range >0 and < d
x1 = [seed_state, 0, 0]
x2 = [seed_state, 0, 0]
end method
method next_int()
x1i = (a1[0]*x1[0] + a1[1]*x1[1] + a1[2]*x1[2]) mod m1
x2i = (a2[0]*x2[0] + a2[1]*x2[1] + a2[2]*x2[2]) mod m2
x1 = [x1i, x1[0], x1[1]] /* Keep last three */
x2 = [x2i, x2[0], x2[1]] /* Keep last three */
z = (x1i - x2i) % m1
answer = (z + 1)
return answer
end method
method next_float():
return float next_int() / d
end method
end class
MRG32k3a Use:
random_gen = instance MRG32k3a
random_gen.seed(1234567)
print(random_gen.next_int()) /* 1459213977 */
print(random_gen.next_int()) /* 2827710106 */
print(random_gen.next_int()) /* 4245671317 */
print(random_gen.next_int()) /* 3877608661 */
print(random_gen.next_int()) /* 2595287583 */
Task
Generate a class/set of functions that generates pseudo-random
numbers as shown above.
Show that the first five integers generated with the seed `1234567`
are as shown above
Show that for an initial seed of '987654321' the counts of 100_000
repetitions of
floor(random_gen.next_float() * 5)
Is as follows:
0: 20002, 1: 20060, 2: 19948, 3: 20059, 4: 19931
Show your output here, on this page.
| #Sidef | Sidef | class MRG32k3a(seed) {
define(
m1 = (2**32 - 209)
m2 = (2**32 - 22853)
)
define(
a1 = %n< 0 1403580 -810728>
a2 = %n<527612 0 -1370589>
)
has x1 = [seed, 0, 0]
has x2 = x1.clone
method next_int {
x1.unshift(a1.map_kv {|k,v| v * x1[k] }.sum % m1); x1.pop
x2.unshift(a2.map_kv {|k,v| v * x2[k] }.sum % m2); x2.pop
(x1[0] - x2[0]) % (m1 + 1)
}
method next_float { self.next_int / (m1 + 1) -> float }
}
say "Seed: 1234567, first 5 values:"
var rng = MRG32k3a(seed: 1234567)
5.of { rng.next_int }.each { .say }
say "\nSeed: 987654321, values histogram:";
var rng = MRG32k3a(seed: 987654321)
var freq = 100_000.of { rng.next_float * 5 -> int }.freq
freq.sort.each_2d {|k,v| say "#{k} #{v}" } |
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Ring | Ring | # Project : Pythagorean quadruples
limit = 2200
pq = list(limit)
for n = 1 to limit
for m = 1 to limit
for p = 1 to limit
for x = 1 to limit
if pow(x,2) = pow(n,2) + pow(m,2) + pow(p,2)
pq[x] = 1
ok
next
next
next
next
pqstr = ""
for d = 1 to limit
if pq[d] = 0
pqstr = pqstr + d + " "
ok
next
see pqstr + nl
|
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Ruby | Ruby | n = 2200
l_add, l = {}, {}
1.step(n) do |x|
x2 = x*x
x.step(n) {|y| l_add[x2 + y*y] = true}
end
s = 3
1.step(n) do |x|
s1 = s
s += 2
s2 = s
(x+1).step(n) do |y|
l[y] = true if l_add[s1]
s1 += s2
s2 += 2
end
end
puts (1..n).reject{|x| l[x]}.join(" ")
|
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Python | Python | from turtle import goto, pu, pd, color, done
def level(ax, ay, bx, by, depth=0):
if depth > 0:
dx,dy = bx-ax, ay-by
x3,y3 = bx-dy, by-dx
x4,y4 = ax-dy, ay-dx
x5,y5 = x4 + (dx - dy)/2, y4 - (dx + dy)/2
goto(ax, ay), pd()
for x, y in ((bx, by), (x3, y3), (x4, y4), (ax, ay)):
goto(x, y)
pu()
level(x4,y4, x5,y5, depth - 1)
level(x5,y5, x3,y3, depth - 1)
if __name__ == '__main__':
color('red', 'yellow')
pu()
level(-100, 500, 100, 500, depth=8)
done() |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #11l | 11l | I problem
exit(1) |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #C.2B.2B | C++ | /*
* g++ -O3 -Wall --std=c++11 qr_standalone.cpp -o qr_standalone
*/
#include <cstdio>
#include <cstdlib>
#include <cstring> // for memset
#include <limits>
#include <iostream>
#include <vector>
#include <math.h>
class Vector;
class Matrix {
public:
// default constructor (don't allocate)
Matrix() : m(0), n(0), data(nullptr) {}
// constructor with memory allocation, initialized to zero
Matrix(int m_, int n_) : Matrix() {
m = m_;
n = n_;
allocate(m_,n_);
}
// copy constructor
Matrix(const Matrix& mat) : Matrix(mat.m,mat.n) {
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
(*this)(i,j) = mat(i,j);
}
// constructor from array
template<int rows, int cols>
Matrix(double (&a)[rows][cols]) : Matrix(rows,cols) {
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
(*this)(i,j) = a[i][j];
}
// destructor
~Matrix() {
deallocate();
}
// access data operators
double& operator() (int i, int j) {
return data[i+m*j]; }
double operator() (int i, int j) const {
return data[i+m*j]; }
// operator assignment
Matrix& operator=(const Matrix& source) {
// self-assignment check
if (this != &source) {
if ( (m*n) != (source.m * source.n) ) { // storage cannot be reused
allocate(source.m,source.n); // re-allocate storage
}
// storage can be used, copy data
std::copy(source.data, source.data + source.m*source.n, data);
}
return *this;
}
// compute minor
void compute_minor(const Matrix& mat, int d) {
allocate(mat.m, mat.n);
for (int i = 0; i < d; i++)
(*this)(i,i) = 1.0;
for (int i = d; i < mat.m; i++)
for (int j = d; j < mat.n; j++)
(*this)(i,j) = mat(i,j);
}
// Matrix multiplication
// c = a * b
// c will be re-allocated here
void mult(const Matrix& a, const Matrix& b) {
if (a.n != b.m) {
std::cerr << "Matrix multiplication not possible, sizes don't match !\n";
return;
}
// reallocate ourself if necessary i.e. current Matrix has not valid sizes
if (a.m != m or b.n != n)
allocate(a.m, b.n);
memset(data,0,m*n*sizeof(double));
for (int i = 0; i < a.m; i++)
for (int j = 0; j < b.n; j++)
for (int k = 0; k < a.n; k++)
(*this)(i,j) += a(i,k) * b(k,j);
}
void transpose() {
for (int i = 0; i < m; i++) {
for (int j = 0; j < i; j++) {
double t = (*this)(i,j);
(*this)(i,j) = (*this)(j,i);
(*this)(j,i) = t;
}
}
}
// take c-th column of m, put in v
void extract_column(Vector& v, int c);
// memory allocation
void allocate(int m_, int n_) {
// if already allocated, memory is freed
deallocate();
// new sizes
m = m_;
n = n_;
data = new double[m_*n_];
memset(data,0,m_*n_*sizeof(double));
} // allocate
// memory free
void deallocate() {
if (data)
delete[] data;
data = nullptr;
}
int m, n;
private:
double* data;
}; // struct Matrix
// column vector
class Vector {
public:
// default constructor (don't allocate)
Vector() : size(0), data(nullptr) {}
// constructor with memory allocation, initialized to zero
Vector(int size_) : Vector() {
size = size_;
allocate(size_);
}
// destructor
~Vector() {
deallocate();
}
// access data operators
double& operator() (int i) {
return data[i]; }
double operator() (int i) const {
return data[i]; }
// operator assignment
Vector& operator=(const Vector& source) {
// self-assignment check
if (this != &source) {
if ( size != (source.size) ) { // storage cannot be reused
allocate(source.size); // re-allocate storage
}
// storage can be used, copy data
std::copy(source.data, source.data + source.size, data);
}
return *this;
}
// memory allocation
void allocate(int size_) {
deallocate();
// new sizes
size = size_;
data = new double[size_];
memset(data,0,size_*sizeof(double));
} // allocate
// memory free
void deallocate() {
if (data)
delete[] data;
data = nullptr;
}
// ||x||
double norm() {
double sum = 0;
for (int i = 0; i < size; i++) sum += (*this)(i) * (*this)(i);
return sqrt(sum);
}
// divide data by factor
void rescale(double factor) {
for (int i = 0; i < size; i++) (*this)(i) /= factor;
}
void rescale_unit() {
double factor = norm();
rescale(factor);
}
int size;
private:
double* data;
}; // class Vector
// c = a + b * s
void vmadd(const Vector& a, const Vector& b, double s, Vector& c)
{
if (c.size != a.size or c.size != b.size) {
std::cerr << "[vmadd]: vector sizes don't match\n";
return;
}
for (int i = 0; i < c.size; i++)
c(i) = a(i) + s * b(i);
}
// mat = I - 2*v*v^T
// !!! m is allocated here !!!
void compute_householder_factor(Matrix& mat, const Vector& v)
{
int n = v.size;
mat.allocate(n,n);
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
mat(i,j) = -2 * v(i) * v(j);
for (int i = 0; i < n; i++)
mat(i,i) += 1;
}
// take c-th column of a matrix, put results in Vector v
void Matrix::extract_column(Vector& v, int c) {
if (m != v.size) {
std::cerr << "[Matrix::extract_column]: Matrix and Vector sizes don't match\n";
return;
}
for (int i = 0; i < m; i++)
v(i) = (*this)(i,c);
}
void matrix_show(const Matrix& m, const std::string& str="")
{
std::cout << str << "\n";
for(int i = 0; i < m.m; i++) {
for (int j = 0; j < m.n; j++) {
printf(" %8.3f", m(i,j));
}
printf("\n");
}
printf("\n");
}
// L2-norm ||A-B||^2
double matrix_compare(const Matrix& A, const Matrix& B) {
// matrices must have same size
if (A.m != B.m or A.n != B.n)
return std::numeric_limits<double>::max();
double res=0;
for(int i = 0; i < A.m; i++) {
for (int j = 0; j < A.n; j++) {
res += (A(i,j)-B(i,j)) * (A(i,j)-B(i,j));
}
}
res /= A.m*A.n;
return res;
}
void householder(Matrix& mat,
Matrix& R,
Matrix& Q)
{
int m = mat.m;
int n = mat.n;
// array of factor Q1, Q2, ... Qm
std::vector<Matrix> qv(m);
// temp array
Matrix z(mat);
Matrix z1;
for (int k = 0; k < n && k < m - 1; k++) {
Vector e(m), x(m);
double a;
// compute minor
z1.compute_minor(z, k);
// extract k-th column into x
z1.extract_column(x, k);
a = x.norm();
if (mat(k,k) > 0) a = -a;
for (int i = 0; i < e.size; i++)
e(i) = (i == k) ? 1 : 0;
// e = x + a*e
vmadd(x, e, a, e);
// e = e / ||e||
e.rescale_unit();
// qv[k] = I - 2 *e*e^T
compute_householder_factor(qv[k], e);
// z = qv[k] * z1
z.mult(qv[k], z1);
}
Q = qv[0];
// after this loop, we will obtain Q (up to a transpose operation)
for (int i = 1; i < n && i < m - 1; i++) {
z1.mult(qv[i], Q);
Q = z1;
}
R.mult(Q, mat);
Q.transpose();
}
double in[][3] = {
{ 12, -51, 4},
{ 6, 167, -68},
{ -4, 24, -41},
{ -1, 1, 0},
{ 2, 0, 3},
};
int main()
{
Matrix A(in);
Matrix Q, R;
matrix_show(A,"A");
// compute QR decompostion
householder(A, R, Q);
matrix_show(Q,"Q");
matrix_show(R,"R");
// compare Q*R to the original matrix A
Matrix A_check;
A_check.mult(Q, R);
// compute L2 norm ||A-A_check||^2
double l2 = matrix_compare(A,A_check);
// display Q*R
matrix_show(A_check, l2 < 1e-12 ? "A == Q * R ? yes" : "A == Q * R ? no");
return EXIT_SUCCESS;
}
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Ada | Ada | with Ada.Command_Line, Ada.Text_IO;
procedure Command_Name is
begin
Ada.Text_IO.Put_Line(Ada.Command_Line.Command_Name);
end Command_Name; |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Aime | Aime | o_text(argv(0));
o_byte('\n'); |
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Crystal | Crystal | class PythagoranTriplesCounter
def initialize(limit = 0)
@limit = limit
@total = 0
@primitives = 0
generate_triples(3, 4, 5)
end
def total; @total end
def primitives; @primitives end
private def generate_triples(a, b, c)
perim = a + b + c
return if perim > @limit
@primitives += 1
@total += @limit // perim
generate_triples( a-2*b+2*c, 2*a-b+2*c, 2*a-2*b+3*c )
generate_triples( a+2*b+2*c, 2*a+b+2*c, 2*a+2*b+3*c )
generate_triples(-a+2*b+2*c,-2*a+b+2*c,-2*a+2*b+3*c )
end
end
perim = 10
while perim <= 100_000_000
c = PythagoranTriplesCounter.new perim
p [perim, c.total, c.primitives]
perim *= 10
end |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
const array string: prog is [](
"$ include \"seed7_05.s7i\";",
"const array string: prog is [](",
"const proc: main is func",
" local var integer: number is 0;",
" begin",
" for number range 1 to 2 do writeln(prog[number]); end for;",
" for number range 1 to 11 do",
" writeln(literal(prog[number]) <& \",\");",
" end for;",
" writeln(literal(prog[12]) <& \");\");",
" for number range 3 to 12 do writeln(prog[number]); end for;",
" end func;");
const proc: main is func
local var integer: number is 0;
begin
for number range 1 to 2 do writeln(prog[number]); end for;
for number range 1 to 11 do
writeln(literal(prog[number]) <& ",");
end for;
writeln(literal(prog[12]) <& ");");
for number range 3 to 12 do writeln(prog[number]); end for;
end func; |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Shale | Shale | i var i "i var i %c%s%c = 34 i 34 i printf" = 34 i 34 i printf |
http://rosettacode.org/wiki/Pseudo-random_numbers/Combined_recursive_generator_MRG32k3a | Pseudo-random numbers/Combined recursive generator MRG32k3a | MRG32k3a Combined recursive generator (pseudo-code)
/* Constants */
/* First generator */
a1 = [0, 1403580, -810728]
m1 = 2**32 - 209
/* Second Generator */
a2 = [527612, 0, -1370589]
m2 = 2**32 - 22853
d = m1 + 1
class MRG32k3a
x1 = [0, 0, 0] /* list of three last values of gen #1 */
x2 = [0, 0, 0] /* list of three last values of gen #2 */
method seed(u64 seed_state)
assert seed_state in range >0 and < d
x1 = [seed_state, 0, 0]
x2 = [seed_state, 0, 0]
end method
method next_int()
x1i = (a1[0]*x1[0] + a1[1]*x1[1] + a1[2]*x1[2]) mod m1
x2i = (a2[0]*x2[0] + a2[1]*x2[1] + a2[2]*x2[2]) mod m2
x1 = [x1i, x1[0], x1[1]] /* Keep last three */
x2 = [x2i, x2[0], x2[1]] /* Keep last three */
z = (x1i - x2i) % m1
answer = (z + 1)
return answer
end method
method next_float():
return float next_int() / d
end method
end class
MRG32k3a Use:
random_gen = instance MRG32k3a
random_gen.seed(1234567)
print(random_gen.next_int()) /* 1459213977 */
print(random_gen.next_int()) /* 2827710106 */
print(random_gen.next_int()) /* 4245671317 */
print(random_gen.next_int()) /* 3877608661 */
print(random_gen.next_int()) /* 2595287583 */
Task
Generate a class/set of functions that generates pseudo-random
numbers as shown above.
Show that the first five integers generated with the seed `1234567`
are as shown above
Show that for an initial seed of '987654321' the counts of 100_000
repetitions of
floor(random_gen.next_float() * 5)
Is as follows:
0: 20002, 1: 20060, 2: 19948, 3: 20059, 4: 19931
Show your output here, on this page.
| #uBasic.2F4tH | uBasic/4tH | @(0) = 0 ' First generator
@(1) = 1403580
@(2) = -810728
m = SHL(1, 32) - 209
@(3) = 527612 ' Second generator
@(4) = 0
@(5) = -1370589
n = SHL(1, 32) - 22853
d = SHL(1, 32) - 209 + 1 ' m + 1
Proc _Seed(1234567)
Print FUNC(_NextInt)
Print FUNC(_NextInt)
Print FUNC(_NextInt)
Print FUNC(_NextInt)
Print FUNC(_NextInt)
Print
End
_Mod Param(2)
Local(1)
c@ = a@ % b@
If c@ < 0 Then
If b@ < 0 Then
Return (c@-b@)
Else
Return (c@+b@)
Endif
EndIf
Return (c@)
_Seed Param(1) ' seed the PRNG
@(6) = a@
@(7) = 0
@(8) = 0
@(9) = a@
@(10) = 0
@(11) = 0
Return
_NextInt ' get the next random integer value
Local(3)
a@ = FUNC(_Mod((@(0) * @(6) + @(1) * @(7) + @(2) * @(8)), m))
b@ = FUNC(_Mod((@(3) * @(9) + @(4) * @(10) + @(5) * @(11)), n))
c@ = FUNC(_Mod(a@ - b@, m))
' keep last three values of the first generator
@(8) = @(7)
@(7) = @(6)
@(6) = a@
' keep last three values of the second generator
@(11) = @(10)
@(10) = @(9)
@(9) = b@
Return (c@ + 1) |
http://rosettacode.org/wiki/Pseudo-random_numbers/Combined_recursive_generator_MRG32k3a | Pseudo-random numbers/Combined recursive generator MRG32k3a | MRG32k3a Combined recursive generator (pseudo-code)
/* Constants */
/* First generator */
a1 = [0, 1403580, -810728]
m1 = 2**32 - 209
/* Second Generator */
a2 = [527612, 0, -1370589]
m2 = 2**32 - 22853
d = m1 + 1
class MRG32k3a
x1 = [0, 0, 0] /* list of three last values of gen #1 */
x2 = [0, 0, 0] /* list of three last values of gen #2 */
method seed(u64 seed_state)
assert seed_state in range >0 and < d
x1 = [seed_state, 0, 0]
x2 = [seed_state, 0, 0]
end method
method next_int()
x1i = (a1[0]*x1[0] + a1[1]*x1[1] + a1[2]*x1[2]) mod m1
x2i = (a2[0]*x2[0] + a2[1]*x2[1] + a2[2]*x2[2]) mod m2
x1 = [x1i, x1[0], x1[1]] /* Keep last three */
x2 = [x2i, x2[0], x2[1]] /* Keep last three */
z = (x1i - x2i) % m1
answer = (z + 1)
return answer
end method
method next_float():
return float next_int() / d
end method
end class
MRG32k3a Use:
random_gen = instance MRG32k3a
random_gen.seed(1234567)
print(random_gen.next_int()) /* 1459213977 */
print(random_gen.next_int()) /* 2827710106 */
print(random_gen.next_int()) /* 4245671317 */
print(random_gen.next_int()) /* 3877608661 */
print(random_gen.next_int()) /* 2595287583 */
Task
Generate a class/set of functions that generates pseudo-random
numbers as shown above.
Show that the first five integers generated with the seed `1234567`
are as shown above
Show that for an initial seed of '987654321' the counts of 100_000
repetitions of
floor(random_gen.next_float() * 5)
Is as follows:
0: 20002, 1: 20060, 2: 19948, 3: 20059, 4: 19931
Show your output here, on this page.
| #Wren | Wren | // constants
var A1 = [0, 1403580, -810728]
var M1 = 2.pow(32) - 209
var A2 = [527612, 0, -1370589]
var M2 = 2.pow(32) - 22853
var D = M1 + 1
// Python style modulus
var Mod = Fn.new { |x, y|
var m = x % y
return (m < 0) ? m + y.abs : m
}
class MRG32k3a {
construct new() {
_x1 = [0, 0, 0]
_x2 = [0, 0, 0]
}
seed(seedState) {
if (seedState <= 0 || seedState >= D) {
Fiber.abort("Argument must be in the range [0, %(D)].")
}
_x1 = [seedState, 0, 0]
_x2 = [seedState, 0, 0]
}
nextInt {
var x1i = Mod.call(A1[0]*_x1[0] + A1[1]*_x1[1] + A1[2]*_x1[2], M1)
var x2i = Mod.call(A2[0]*_x2[0] + A2[1]*_x2[1] + A2[2]*_x2[2], M2)
_x1 = [x1i, _x1[0], _x1[1]] /* keep last three */
_x2 = [x2i, _x2[0], _x2[1]] /* keep last three */
return Mod.call(x1i - x2i, M1) + 1
}
nextFloat { nextInt / D }
}
var randomGen = MRG32k3a.new()
randomGen.seed(1234567)
for (i in 0..4) System.print(randomGen.nextInt)
var counts = List.filled(5, 0)
randomGen.seed(987654321)
for (i in 1..1e5) {
var i = (randomGen.nextFloat * 5).floor
counts[i] = counts[i] + 1
}
System.print("\nThe counts for 100,000 repetitions are:")
for (i in 0..4) System.print(" %(i) : %(counts[i])") |
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Rust | Rust |
use std::collections::BinaryHeap;
fn a094958_iter() -> Vec<u16> {
(0..12)
.map(|n| vec![1 << n, 5 * (1 << n)])
.flatten()
.filter(|x| x < &2200)
.collect::<BinaryHeap<u16>>()
.into_sorted_vec()
}
fn a094958_filter() -> Vec<u16> {
(1..2200) // ported from Sidef
.filter(|n| ((n & (n - 1) == 0) || (n % 5 == 0 && ((n / 5) & (n / 5 - 1) == 0))))
.collect()
}
fn a094958_loop() -> Vec<u16> {
let mut v = vec![];
for n in 0..12 {
v.push(1 << n);
if 5 * (1 << n) < 2200 {
v.push(5 * (1 << n));
}
}
v.sort();
return v;
}
fn main() {
println!("{:?}", a094958_iter());
println!("{:?}", a094958_loop());
println!("{:?}", a094958_filter());
}
#[cfg(test)]
mod tests {
use super::*;
static HAPPY: &str = "[1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048]";
#[test]
fn test_a094958_iter() {
assert!(format!("{:?}", a094958_iter()) == HAPPY);
}
#[test]
fn test_a094958_loop() {
assert!(format!("{:?}", a094958_loop()) == HAPPY);
}
#[test]
fn test_a094958_filter() {
assert!(format!("{:?}", a094958_filter()) == HAPPY);
}
}
|
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Scala | Scala | object PythagoreanQuadruple extends App {
val MAX = 2200
val MAX2: Int = MAX * MAX * 2
val found = Array.ofDim[Boolean](MAX + 1)
val a2b2 = Array.ofDim[Boolean](MAX2 + 1)
var s = 3
for (a <- 1 to MAX) {
val a2 = a * a
for (b <- a to MAX) a2b2(a2 + b * b) = true
}
for (c <- 1 to MAX) {
var s1 = s
s += 2
var s2 = s
for (d <- (c + 1) to MAX) {
if (a2b2(s1)) found(d) = true
s1 += s2
s2 += 2
}
}
println(f"The values of d <= ${MAX}%d which can't be represented:")
val notRepresented = (1 to MAX).filterNot(d => found(d) )
println(notRepresented.mkString(" "))
} |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #R | R | ## Recursive PT plotting
pythtree <- function(ax,ay,bx,by,d) {
if(d<0) {return()}; clr="darkgreen";
dx=bx-ax; dy=ay-by;
x3=bx-dy; y3=by-dx;
x4=ax-dy; y4=ay-dx;
x5=x4+(dx-dy)/2; y5=y4-(dx+dy)/2;
segments(ax,-ay,bx,-by, col=clr);
segments(bx,-by,x3,-y3, col=clr);
segments(x3,-y3,x4,-y4, col=clr);
segments(x4,-y4,ax,-ay, col=clr);
pythtree(x4,y4,x5,y5,d-1);
pythtree(x5,y5,x3,y3,d-1);
}
## Plotting Pythagoras Tree. aev 3/27/17
## x1,y1,x2,y2 - starting position
## ord - order/depth, fn - file name, ttl - plot title.
pPythagorasT <- function(x1, y1,x2, y2, ord, fn="", ttl="") {
cat(" *** START PYTHT:", date(), "\n");
m=640; i=j=k=m1=m-2; x=y=d=dm=0;
if(fn=="") {pf=paste0("PYTHTR", ord, ".png")} else {pf=paste0(fn, ".png")};
if(ttl=="") {ttl=paste0("Pythagoras tree, order - ", ord)};
cat(" *** Plot file -", pf, "title:", ttl, "\n");
plot(NA, xlim=c(0,m), ylim=c(-m,0), xlab="", ylab="", main=ttl);
pythtree(x1,y1, x2,y2, ord);
dev.copy(png, filename=pf, width=m, height=m);
dev.off(); graphics.off();
cat(" *** END PYTHT:",date(),"\n");
}
## Executing:
pPythagorasT(275,500,375,500,9)
pPythagorasT(275,500,375,500,7) |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #QB64 | QB64 | _Title "Pythagoras Tree"
Dim As Integer sw, sh
sw = 640
sh = 480
Screen _NewImage(sw, sh, 32)
Call pythTree(sw / 2 - sw / 12, sh - 30, sw / 2 + sw / 12, sh - 30, 0)
Sleep
System
Sub pythTree (ax As Integer, ay As Integer, bx As Integer, by As Integer, depth As Integer)
Dim As Single cx, cy, dx, dy, ex, ey
Dim As Integer c
cx = ax - ay + by
cy = ax + ay - bx
dx = bx + by - ay
dy = ax - bx + by
ex = (cx - cy + dx + dy) * 0.5
ey = (cx + cy - dx + dy) * 0.5
c = depth * 15
Color _RGB(c Mod 256, Abs((255 - c)) Mod 256, (144 + c) Mod 256)
Line (cx, cy)-(ax, ay)
Line (ax, ay)-(bx, by)
Line (bx, by)-(dx, dy)
Line (dx, dy)-(cx, cy)
Line (cx, cy)-(ex, ey)
Line (ex, ey)-(dx, dy)
If depth < 12 Then
Call pythTree(cx, cy, ex, ey, depth + 1)
Call pythTree(ex, ey, dx, dy, depth + 1)
End If
End Sub |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #6502_Assembly | 6502 Assembly | ;assuming this is not a subroutine and runs inline.
cmp TestValue ;a label for a memory address that contains some value we want to test the accumulator against
beq continue
rts ;unlike the Z80 there is no conditional return so we have to branch around the return instruction.
continue: |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #AArch64_Assembly | AArch64 Assembly |
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program ending64.s */
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
/*******************************************/
/* Initialized data */
/*******************************************/
.data
/*******************************************/
/* code section */
/*******************************************/
.text
.global main
main: // entry of program
OK:
// end program OK
mov x0,0 // return code
b 100f
NONOK:
// if error detected end program no ok
mov x0,1 // return code
100: // standard end of the program
mov x8,EXIT // request to exit program
svc 0 // perform the system call Linux
/********************************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
|
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #Common_Lisp | Common Lisp | (defun sign (x)
(if (zerop x)
x
(/ x (abs x))))
(defun norm (x)
(let ((len (car (array-dimensions x))))
(sqrt (loop for i from 0 to (1- len) sum (expt (aref x i 0) 2)))))
(defun make-unit-vector (dim)
(let ((vec (make-array `(,dim ,1) :initial-element 0.0d0)))
(setf (aref vec 0 0) 1.0d0)
vec))
;; Return a nxn identity matrix.
(defun eye (n)
(let ((I (make-array `(,n ,n) :initial-element 0)))
(loop for j from 0 to (- n 1) do
(setf (aref I j j) 1))
I))
(defun array-range (A ma mb na nb)
(let* ((mm (1+ (- mb ma)))
(nn (1+ (- nb na)))
(B (make-array `(,mm ,nn) :initial-element 0.0d0)))
(loop for i from 0 to (1- mm) do
(loop for j from 0 to (1- nn) do
(setf (aref B i j)
(aref A (+ ma i) (+ na j)))))
B))
(defun rows (A) (car (array-dimensions A)))
(defun cols (A) (cadr (array-dimensions A)))
(defun mcol (A n) (array-range A 0 (1- (rows A)) n n))
(defun mrow (A n) (array-range A n n 0 (1- (cols A))))
(defun array-embed (A B row col)
(let* ((ma (rows A))
(na (cols A))
(mb (rows B))
(nb (cols B))
(C (make-array `(,ma ,na) :initial-element 0.0d0)))
(loop for i from 0 to (1- ma) do
(loop for j from 0 to (1- na) do
(setf (aref C i j) (aref A i j))))
(loop for i from 0 to (1- mb) do
(loop for j from 0 to (1- nb) do
(setf (aref C (+ row i) (+ col j))
(aref B i j))))
C))
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #ALGOL_68 | ALGOL 68 |
BEGIN
print ((program idf, newline))
END
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Amazing_Hopper | Amazing Hopper |
#include <hbasic.h>
Begin
GetParam(name File)
Print("My Program name: ", name File,Newl)
End
|
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #D | D | void main() @safe {
import std.stdio, std.range, std.algorithm, std.typecons, std.numeric;
enum triples = (in uint n) pure nothrow @safe /*@nogc*/ =>
iota(1, n + 1)
.map!(z => iota(1, z + 1)
.map!(x => iota(x, z + 1).map!(y => tuple(x, y, z))))
.joiner.joiner
.filter!(t => t[0] ^^ 2 + t[1] ^^ 2 == t[2] ^^ 2 && t[].only.sum <= n)
.map!(t => tuple(t[0 .. 2].gcd == 1, t[]));
auto xs = triples(100);
writeln("Up to 100 there are ", xs.count, " triples, ",
xs.filter!q{ a[0] }.count, " are primitive.");
} |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Sidef | Sidef | s = %(s = %%(%s); printf(s, s);
); printf(s, s); |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Smalltalk | Smalltalk | [:s| Transcript show: s, s printString; cr ] value: '[:s| Transcript show: s, s printString; cr ] value: '
|
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Sidef | Sidef | # Finds all solutions (a,b) such that: a^2 + b^2 = n^2
func sum_of_two_squares(n) is cached {
n == 0 && return [[0, 0]]
var prod1 = 1
var prod2 = 1
var prime_powers = []
for p,e in (n.factor_exp) {
if (p % 4 == 3) { # p = 3 (mod 4)
e.is_even || return [] # power must be even
prod2 *= p**(e >> 1)
}
elsif (p == 2) { # p = 2
if (e.is_even) { # power is even
prod2 *= p**(e >> 1)
}
else { # power is odd
prod1 *= p
prod2 *= p**((e - 1) >> 1)
prime_powers.append([p, 1])
}
}
else { # p = 1 (mod 4)
prod1 *= p**e
prime_powers.append([p, e])
}
}
prod1 == 1 && return [[prod2, 0]]
prod1 == 2 && return [[prod2, prod2]]
# All the solutions to the congruence: x^2 = -1 (mod prod1)
var square_roots = gather {
gather {
for p,e in (prime_powers) {
var pp = p**e
var r = sqrtmod(-1, pp)
take([[r, pp], [pp - r, pp]])
}
}.cartesian { |*a|
take(Math.chinese(a...))
}
}
var solutions = []
for r in (square_roots) {
var s = r
var q = prod1
while (s*s > prod1) {
(s, q) = (q % s, s)
}
solutions.append([prod2 * s, prod2 * (q % s)])
}
for p,e in (prime_powers) {
for (var i = e%2; i < e; i += 2) {
var sq = p**((e - i) >> 1)
var pp = p**(e - i)
solutions += (
__FUNC__(prod1 / pp).map { |pair|
pair.map {|r| sq * prod2 * r }
}
)
}
}
solutions.map {|pair| pair.sort } \
.uniq_by {|pair| pair[0] } \
.sort_by {|pair| pair[0] }
}
# Finds all solutions (a,b,c) such that: a^2 + b^2 + c^2 = n^2
func sum_of_three_squares(n) {
gather {
for k in (1 .. n//3) {
var t = sum_of_two_squares(n**2 - k**2) || next
take(t.map { [k, _...] }...)
}
}
}
say gather {
for n in (1..2200) {
sum_of_three_squares(n) || take(n)
}
} |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Racket | Racket | #lang racket
(require racket/draw pict)
(define (draw-pythagoras-tree order x0 y0 x1 y1)
(λ (the-dc dx dy)
(define (inr order x0 y0 x1 y1)
(when (positive? order)
(let* ((y0-1 (- y0 y1))
(x1-0 (- x1 x0))
(x2 (+ x1 y0-1))
(y2 (+ y1 x1-0))
(x3 (+ x0 y0-1))
(y3 (+ y0 x1-0))
(x4 (+ x2 x3 (/ (+ x0 x2) -2)))
(y4 (+ y2 y3 (/ (+ y0 y2) -2)))
(path (new dc-path%)))
(send* path [move-to x0 y0]
[line-to x1 y1] [line-to x2 y2] [line-to x3 y3]
[close])
(send the-dc draw-path path dx dy)
(inr (sub1 order) x3 y3 x4 y4)
(inr (sub1 order) x4 y4 x2 y2))))
(define old-brush (send the-dc get-brush))
(define old-pen (send the-dc get-pen))
(send the-dc set-pen (new pen% [width 1] [color "black"]))
(inr (add1 order) x0 y0 x1 y1)
(send the-dc set-brush old-brush)
(send the-dc set-pen old-pen)))
(dc (draw-pythagoras-tree 7 (+ 200 32) 255 (- 200 32) 255) 400 256) |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #Action.21 | Action! | PROC Main()
DO
IF Rand(0)=10 THEN
PrintE("Terminate program by Break() procedure")
Break()
FI
OD
PrintE("This is a dead code")
RETURN |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #Ada | Ada | with Ada.Task_Identification; use Ada.Task_Identification;
procedure Main is
-- Create as many task objects as your program needs
begin
-- whatever logic is required in your Main procedure
if some_condition then
Abort_Task (Current_Task);
end if;
end Main; |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #D | D | import std.stdio, std.math, std.algorithm, std.traits,
std.typecons, std.numeric, std.range, std.conv;
template elementwiseMat(string op) {
T[][] elementwiseMat(T)(in T[][] A, in T B) pure nothrow {
if (A.empty)
return null;
auto R = new typeof(return)(A.length, A[0].length);
foreach (immutable r, const row; A)
R[r][] = mixin("row[] " ~ op ~ "B");
return R;
}
T[][] elementwiseMat(T, U)(in T[][] A, in U[][] B)
pure nothrow if (is(Unqual!T == Unqual!U)) {
assert(A.length == B.length);
if (A.empty)
return null;
auto R = new typeof(return)(A.length, A[0].length);
foreach (immutable r, const row; A) {
assert(row.length == B[r].length);
R[r][] = mixin("row[] " ~ op ~ "B[r][]");
}
return R;
}
}
alias mSum = elementwiseMat!q{ + },
mSub = elementwiseMat!q{ - },
pMul = elementwiseMat!q{ * },
pDiv = elementwiseMat!q{ / };
bool isRectangular(T)(in T[][] mat) pure nothrow {
return mat.all!(r => r.length == mat[0].length);
}
T[][] matMul(T)(in T[][] a, in T[][] b) pure nothrow
in {
assert(a.isRectangular && b.isRectangular &&
a[0].length == b.length);
} body {
auto result = new T[][](a.length, b[0].length);
auto aux = new T[b.length];
foreach (immutable j; 0 .. b[0].length) {
foreach (immutable k; 0 .. b.length)
aux[k] = b[k][j];
foreach (immutable i; 0 .. a.length)
result[i][j] = a[i].dotProduct(aux);
}
return result;
}
Unqual!T[][] transpose(T)(in T[][] m) pure nothrow {
auto r = new Unqual!T[][](m[0].length, m.length);
foreach (immutable nr, row; m)
foreach (immutable nc, immutable c; row)
r[nc][nr] = c;
return r;
}
T norm(T)(in T[][] m) pure nothrow {
return transversal(m, 0).map!q{ a ^^ 2 }.sum.sqrt;
}
Unqual!T[][] makeUnitVector(T)(in size_t dim) pure nothrow {
auto result = new Unqual!T[][](dim, 1);
foreach (row; result)
row[] = 0;
result[0][0] = 1;
return result;
}
/// Return a nxn identity matrix.
Unqual!T[][] matId(T)(in size_t n) pure nothrow {
auto Id = new Unqual!T[][](n, n);
foreach (immutable r, row; Id) {
row[] = 0;
row[r] = 1;
}
return Id;
}
T[][] slice2D(T)(in T[][] A,
in size_t ma, in size_t mb,
in size_t na, in size_t nb) pure nothrow {
auto B = new T[][](mb - ma + 1, nb - na + 1);
foreach (immutable i, brow; B)
brow[] = A[ma + i][na .. na + brow.length];
return B;
}
size_t rows(T)(in T[][] A) pure nothrow { return A.length; }
size_t cols(T)(in T[][] A) pure nothrow {
return A.length ? A[0].length : 0;
}
T[][] mcol(T)(in T[][] A, in size_t n) pure nothrow {
return slice2D(A, 0, A.rows - 1, n, n);
}
T[][] matEmbed(T)(in T[][] A, in T[][] B,
in size_t row, in size_t col) pure nothrow {
auto C = new T[][](rows(A), cols(A));
foreach (immutable i, const arow; A)
C[i][] = arow[]; // Some wasted copies.
foreach (immutable i, const brow; B)
C[row + i][col .. col + brow.length] = brow[];
return C;
}
// Main routines ---------------
T[][] makeHouseholder(T)(in T[][] a) {
immutable m = a.rows;
immutable T s = a[0][0].sgn;
immutable e = makeUnitVector!T(m);
immutable u = mSum(a, pMul(e, a.norm * s));
immutable v = pDiv(u, u[0][0]);
immutable beta = 2.0 / v.transpose.matMul(v)[0][0];
return mSub(matId!T(m), pMul(v.matMul(v.transpose), beta));
}
Tuple!(T[][],"Q", T[][],"R") QRdecomposition(T)(T[][] A) {
immutable m = A.rows;
immutable n = A.cols;
auto Q = matId!T(m);
// Work on n columns of A.
foreach (immutable i; 0 .. (m == n ? n - 1 : n)) {
// Select the i-th submatrix. For i=0 this means the original
// matrix A.
immutable B = slice2D(A, i, m - 1, i, n - 1);
// Take the first column of the current submatrix B.
immutable x = mcol(B, 0);
// Create the Householder matrix for the column and embed it
// into an mxm identity.
immutable H = matEmbed(matId!T(m), x.makeHouseholder, i, i);
// The product of all H matrices from the right hand side is
// the orthogonal matrix Q.
Q = Q.matMul(H);
// The product of all H matrices with A from the LHS is the
// upper triangular matrix R.
A = H.matMul(A);
}
// Return Q and R.
return typeof(return)(Q, A);
}
// Polynomial regression ---------------
/// Solve an upper triangular system by back substitution.
T[][] solveUpperTriangular(T)(in T[][] R, in T[][] b) pure nothrow {
immutable n = R.cols;
auto x = new T[][](n, 1);
foreach_reverse (immutable k; 0 .. n) {
T tot = 0;
foreach (immutable j; k + 1 .. n)
tot += R[k][j] * x[j][0];
x[k][0] = (b[k][0] - tot) / R[k][k];
}
return x;
}
/// Solve a linear least squares problem by QR decomposition.
T[][] lsqr(T)(T[][] A, in T[][] b) pure nothrow {
const qr = A.QRdecomposition;
immutable n = qr.R.cols;
return solveUpperTriangular(
slice2D(qr.R, 0, n - 1, 0, n - 1),
slice2D(qr.Q.transpose.matMul(b), 0, n - 1, 0, 0));
}
T[][] polyFit(T)(in T[][] x, in T[][] y, in size_t n) pure nothrow {
immutable size_t m = x.cols;
auto A = new T[][](m, n + 1);
foreach (immutable i, row; A)
foreach (immutable j, ref item; row)
item = x[0][i] ^^ j;
return lsqr(A, y.transpose);
}
void main() {
// immutable (Q, R) = QRdecomposition([[12.0, -51, 4],
immutable qr = QRdecomposition([[12.0, -51, 4],
[ 6.0, 167, -68],
[-4.0, 24, -41]]);
immutable form = "[%([%(%2.3f, %)]%|,\n %)]\n";
writefln(form, qr.Q);
writefln(form, qr.R);
immutable x = [[0.0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]];
immutable y = [[1.0, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321]];
polyFit(x, y, 2).writeln;
} |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Applesoft_BASIC | Applesoft BASIC |
10 GOSUB 40"GET PROGRAM NAME
20 PRINT N$
30 END
40 REMGET PROGRAM NAME
50 GOSUB 100"GET INPUT BUFFER
60 GOSUB 200"REMOVE RUN PREFIX
70 GOSUB 300"REMOVE , SUFFIXES
80 GOSUB 400"TRIM SPACES
90 RETURN
100 REMGET INPUT BUFFER
110 N$ = ""
120 FOR I = 512 TO 767
130 B = PEEK (I) - 128
140 IF B < 32 THEN RETURN
150 N$ = N$ + CHR$ (B)
160 NEXT I
170 RETURN
200 REMREMOVE RUN PREFIX
210 P = 1
220 FOR I = 1 TO 3
230 FOR J = P TO LEN(N$)
240 C$ = MID$ (N$,J,1)
250 P = P + 1
260 IF C$ = " " THEN NEXT J
270 IF C$ = MID$("RUN",I,1) THEN NEXT I:N$ = MID$(N$,P,LEN(N$)-P+1):RETURN
280 PRINT "YOU NEED TO RUN THIS PROGRAM USING THE RUN COMMAND FROM DOS."
290 END
300 REMREMOVE , SUFFIXES
310 L = LEN (N$)
320 FOR I = 1 TO L
330 C$ = MID$ (N$,I,1)
340 IF C$ = "," THEN N$ = LEFT$(N$,I - 1): RETURN
350 NEXT I
360 RETURN
400 REMTRIM SPACES
410 GOSUB 600
500 REMLEFT TRIM SPACES
510 L = LEN(N$) - 1
520 FOR I = L TO 0 STEP -1
530 IF I < 0 THEN RETURN
540 IF LEFT$ (N$,1) <> " " THEN RETURN
550 IF I THEN N$ = RIGHT$ (N$, I)
560 NEXT I
570 N$ = "
580 RETURN
600 REMRIGHT TRIM SPACES
610 L = LEN(N$) - 1
620 FOR I = L TO 0 STEP -1
630 IF I < 0 THEN RETURN
640 IF RIGHT$ (N$,1) <> " " THEN RETURN
650 IF I THEN N$ = LEFT$ (N$, I)
660 NEXT I
670 N$ = "
680 RETURN
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #ARM_Assembly | ARM Assembly |
/* ARM assembly Raspberry PI */
/* program namepgm.s */
/* Constantes */
.equ STDOUT, 1
.equ WRITE, 4
.equ EXIT, 1
/* Initialized data */
.data
szMessage: .asciz "Program : " @
szRetourLigne: .asciz "\n"
.text
.global main
main:
push {fp,lr} /* save des 2 registres */
add fp,sp,#8 /* fp <- adresse début */
ldr r0, iAdrszMessage @ adresse of message
bl affichageMess @ call function
ldr r0,[fp,#4] @ recup name of program in command line
bl affichageMess @ call function
ldr r0, iAdrszRetourLigne @ adresse of message
bl affichageMess @ call function
/* fin standard du programme */
mov r0, #0 @ return code
pop {fp,lr} @restaur des 2 registres
mov r7, #EXIT @ request to exit program
swi 0 @ perform the system call
iAdrszMessage: .int szMessage
iAdrszRetourLigne: .int szRetourLigne
/******************************************************************/
/* affichage des messages avec calcul longueur */
/******************************************************************/
/* r0 contient l adresse du message */
affichageMess:
push {fp,lr} /* save des 2 registres */
push {r0,r1,r2,r7} /* save des autres registres */
mov r2,#0 /* compteur longueur */
1: /*calcul de la longueur */
ldrb r1,[r0,r2] /* recup octet position debut + indice */
cmp r1,#0 /* si 0 c est fini */
beq 1f
add r2,r2,#1 /* sinon on ajoute 1 */
b 1b
1: /* donc ici r2 contient la longueur du message */
mov r1,r0 /* adresse du message en r1 */
mov r0,#STDOUT /* code pour écrire sur la sortie standard Linux */
mov r7, #WRITE /* code de l appel systeme "write" */
swi #0 /* appel systeme */
pop {r0,r1,r2,r7} /* restaur des autres registres */
pop {fp,lr} /* restaur des 2 registres */
bx lr /* retour procedure */
|
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Delphi | Delphi |
[Pythagorean triples for Rosetta code.
Counts (1) all Pythagorean triples (2) primitive Pythagorean triples,
with perimeter not greater than a given value.
Library subroutine M3, Prints header and is then overwritten.
Here, the last character sets the teleprinter to figures.]
..PZ [simulate blank tape]
PFGKIFAFRDLFUFOFE@A6FG@E8FEZPF
@&*!MAX!PERIM!!!!!TOTAL!!!!!!PRIM@&#.
..PZ
[Library subroutine P7, prints long strictly positive integer;
10 characters, right justified, padded left with spaces.
Closed, even; 35 storage locations; working position 4D.]
T 56 K
GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSFL4F
T4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@
[Subroutine for positive integer division.
Input: 4D = dividend, 6D = divisor.
Output: 4D = remainder, 6D = quotient.
37 locations; working locations 0D, 8D.]
T 100 K
GKA3FT35@A6DU8DTDA4DRDSDG13@T36@ADLDE4@T36@T6DA4DSDG23@
T4DA6DYFYFT6DT36@A8DSDE35@T36@ADRDTDA6DLDT6DE15@EFPF
[Subroutine to return GCD of two non-negative 35-bit integers.
Input: Integers at 4D, 6D.
Output: GCD at 4D; changes 6D.
41 locations; working location 0D.]
T 200 K
GKA3FT39@S4DE37@T40@A4DTDA6DRDSDG15@T40@ADLDE6@T40@A6DSDG20@T6D
T40@A4DSDE29@T40@ADRDTDE16@S6DE39@TDA4DT6DSDT4DE5@A6DT4DEFPF
[************************ ROSETTA CODE TASK *************************
Subroutine to count Pythagorean triples with given maximum perimeter.
Input: 0D = maximum perimeter.
Output: 4D = number of triples, 6D = number of primitive.
0D is changed.
Must be loaded at an even address.
Uses the well-known fact that a primitive Pythagorean triple is of the form
(m^2 - n^2, 2*m*n, m^2 + n^2) where m, n are coprime and of opposite parity.]
T 300 K
G K
A 3 F [make link]
E 16 @ [jump over variables and constants]
[Double values are put here to ensure even address]
[Variables]
[2] P F P F [maximum perimeter]
[4] P F P F [total number of Pythagorean triples]
[6] P F P F [number of primitive Pythagorean triples]
[8] P F P F [m]
[10] P F P F [n]
[Constants]
T12#Z PF T12Z [clears sandwich digit between 12 and 13]
[12] P D P F [double-value 1]
T14#Z PF T14Z [clears sandwich digit between 14 and 15]
[14] P1F P F [double-value 2]
[Continue with code]
[16] T 69 @ [plant link for return]
A D [load maximum perimeter]
T 2#@ [store locally]
T 4#@ [initialize counts of triangles to 0]
T 6#@
A 12#@ [load 1]
T 8#@ [m := 1]
[Next m, inc by 1]
[23] T F [clear acc]
A 8#@ [load m]
A 12#@ [add 1]
T 8#@ [update m]
H 8#@ [mult reg := m]
C 12#@ [acc := m AND 1]
A 12#@ [add 1]
T 10#@ [n := 1 if m even, 2 if m odd]
[Here to count triangles arising from m, n.
It's assumed m and n are known coprime.]
[31] A 31 @ [call the count subroutine,]
G 70 @ [result is in 6D]
S 6 D [load negative count]
G 40 @ [jump if count > 0]
[No triangles found for this n.
If n = 1 or 2 then whole thing is finished.
Else move on to next m.]
T F [clear acc]
A 14#@ [load 2]
S 10#@ [2 - n]
G 23 @ [if n > 2, go to next m]
E 64 @ [if n <= 2, exit]
[Found triangles, count is in 6D]
[40] T F [clear acc]
A 4#@ [load total count]
A 6 D [add count just found]
T 4#@ [update total count]
A 6#@ [load primitive count]
A 12#@ [add 1]
T 6#@ [update primitive count]
[47] T F [clear acc]
A 10#@ [load n]
A 14#@ [add 2]
U 10#@ [update n]
S 8#@ [is n > m?]
E 23 @ [if so, loop back for next m]
[Test whether m and n are coprime.]
T F [clear acc]
A 8#@ [load m]
T 4 D [to 4D for GCD routine]
A 10#@ [load n]
T 6 D [to 6D for GCD routine]
A 58 @ [call GCD routine,]
G 200 F [GCD is returned in 4D]
A 4 D [load GCD]
S 14#@ [is GCD = 1? (test by subtracting 2)]
E 47 @ [no, go straight to next n]
G 31 @ [yes, count triangles, then next n]
[64] T F [exit, clear acc]
A 4#@ [load total number of triples]
T 4 D [return in 4D]
A 6#@ [load number of primitive triples]
T 6 D [return in 6D]
[69] E F
[2nd-level subroutine to count triangles arising from m, n.
Assumes m, n are coprime and of opposite parity,
and m is in the multiplier register.
Result is returned in 6D.]
[70] A 3 F [make and plant link for return]
T 91 @
A 2#@ [acc := maximum perimeter]
T 4 D [to 4D for division routine]
A 8#@ [load m]
A 10#@ [add n]
T D [m + n to 0D]
V D [acc := m*(m + n)]
[Need to shift product 34 left to restore integer scaling.
Since we want 2*m*(m+n), shift 35 left.]
L F [13 left (maximum possible)]
L F [13 more]
L 128 F [9 more]
T 6 D [perimeter to 6D for division routine]
A 4 D [load maximum perimeter]
S 6 D [is perimeter > maximum?]
G 89 @ [quick exit if so]
T F [clear acc]
A 86 @ [call division routine,]
G 100 F [leaves count in 6D]
E 91 @ [jump to exit]
[89] T F [acc := 0]
T 6 D [return count = 0]
[91] E F
[Main routine. Load at an even address.]
T 500 K
G K
[The initial maximum perimeter is repeatedly multiplied by 10]
T#Z PF TZ [clears sandwich digit between 0 and 1]
[0] P50F PF [initial maximum perimeter <---------- EDIT HERE]
[2] P 3 F [number of values to calculate <---------- EDIT HERE]
[3] P D [1]
[4] P F P F [maximum perimeter]
[6] P F P F [total number of triples]
[8] P F P F [number of primitive triples]
[10] P F [negative count of values]
[11] # F [figures shift]
[12] @ F [carriage return]
[13] & F [line feed]
[14] K 4096 F [null char]
[Enter with acc = 0]
[15] S 2 @ [initialize a negative counter]
T 10 @ [(standard EDSAC practice)]
A #@ [initialize maximum perimeter]
T 4#@
[19] T F [clear acc]
A 4#@ [load maximum perimeter]
T D [to 0D for subroutine]
A 22 @ [call subroutine to count triples]
G 300 F
A 4 D [returns total number in 4D]
T 6#@ [save locally]
A 6 D [returns number of primitive in 6D]
T 8#@ [save locally]
[Print the result]
A 4#@ [load maximum perimeter]
T D [to 0D for print subroutine]
A 30 @ [call print subroutine]
G 56 F
A 6#@ [repeat for total number of triples]
T D
A 34 @
G 56 F
A 8#@ [repeat for number of primitive triples]
T D
A 38 @
G 56 F
O 12 @
O 13 @
A 10 @ [load negative count]
A 3 @ [add 1]
E 53 @ [out if reached 0]
T 10 @ [else update count]
A 4#@ [load max perimeter]
U D [temp store]
L 1 F [times 4]
A D [times 5]
L D [times 10]
T 4#@ [update]
E 19 @ [loop back]
[53] O 14 @ [done; print null to flush printer buffer]
Z F [stop]
E 15 Z [define entry point]
P F [acc = 0 on entry]
|
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #SmileBASIC | SmileBASIC | Q$="Q$=%SPRINT FORMAT$(Q$,CHR$(34)+Q$+CHR$(34)+CHR$(10))"
PRINT FORMAT$(Q$,CHR$(34)+Q$+CHR$(34)+CHR$(10)) |
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Swift | Swift | func missingD(upTo n: Int) -> [Int] {
var a2 = 0, s = 3, s1 = 0, s2 = 0
var res = [Int](repeating: 0, count: n + 1)
var ab = [Int](repeating: 0, count: n * n * 2 + 1)
for a in 1...n {
a2 = a * a
for b in a...n {
ab[a2 + b * b] = 1
}
}
for c in 1..<n {
s1 = s
s += 2
s2 = s
for d in c+1...n {
if ab[s1] != 0 {
res[d] = 1
}
s1 += s2
s2 += 2
}
}
return (1...n).filter({ res[$0] == 0 })
}
print(missingD(upTo: 2200)) |
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #VBA | VBA | Const n = 2200
Public Sub pq()
Dim s As Long, s1 As Long, s2 As Long, x As Long, x2 As Long, y As Long: s = 3
Dim l(n) As Boolean, l_add(9680000) As Boolean '9680000=n * n * 2
For x = 1 To n
x2 = x * x
For y = x To n
l_add(x2 + y * y) = True
Next y
Next x
For x = 1 To n
s1 = s
s = s + 2
s2 = s
For y = x + 1 To n
If l_add(s1) Then l(y) = True
s1 = s1 + s2
s2 = s2 + 2
Next
Next
For x = 1 To n
If Not l(x) Then Debug.Print x;
Next
Debug.Print
End Sub |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Raku | Raku | class Square {
has Complex ($.position, $.edge);
method size { $!edge.abs }
method svg-polygon {
qq[<polygon points="{join ' ', map
{ ($!position + $_ * $!edge).reals.join(',') },
0, 1, 1+1i, 1i}" style="fill:lime;stroke=black" />]
}
method left-child {
self.new:
position => $!position + i*$!edge,
edge => sqrt(2)/2*cis(pi/4)*$!edge;
}
method right-child {
self.new:
position => $!position + i*$!edge + self.left-child.edge,
edge => sqrt(2)/2*cis(-pi/4)*$!edge;
}
}
BEGIN say '<svg width="500" height="500">';
END say '</svg>';
sub tree(Square $s, $level = 0) {
return if $level > 8;
say $s.svg-polygon;
tree($s.left-child, $level+1);
tree($s.right-child, $level+1);
}
tree Square.new: :position(250+0i), :edge(60+0i); |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #Aime | Aime | void
f1(integer a)
{
if (a) {
exit(1);
}
}
integer
main(void)
{
f1(3);
return 0;
} |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #ALGOL_68 | ALGOL 68 | IF problem = 1 THEN
stop
FI |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #F.23 | F# |
// QR decomposition. Nigel Galloway: January 11th., 2022
let n=[[12.0;-51.0;4.0];[6.0;167.0;-68.0];[-4.0;24.0;-41.0]]|>MathNet.Numerics.LinearAlgebra.MatrixExtensions.matrix
let g=n|>MathNet.Numerics.LinearAlgebra.Matrix.qr
printfn $"Matrix\n------\n%A{n}\nQ\n-\n%A{g.Q}\nR\n-\n%A{g.R}"
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #AutoHotkey | AutoHotkey |
MsgBox, % A_ScriptName
|
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #EDSAC_order_code | EDSAC order code |
[Pythagorean triples for Rosetta code.
Counts (1) all Pythagorean triples (2) primitive Pythagorean triples,
with perimeter not greater than a given value.
Library subroutine M3, Prints header and is then overwritten.
Here, the last character sets the teleprinter to figures.]
..PZ [simulate blank tape]
PFGKIFAFRDLFUFOFE@A6FG@E8FEZPF
@&*!MAX!PERIM!!!!!TOTAL!!!!!!PRIM@&#.
..PZ
[Library subroutine P7, prints long strictly positive integer;
10 characters, right justified, padded left with spaces.
Closed, even; 35 storage locations; working position 4D.]
T 56 K
GKA3FT26@H28#@NDYFLDT4DS27@TFH8@S8@T1FV4DAFG31@SFLDUFOFFFSFL4F
T4DA1FA27@G11@XFT28#ZPFT27ZP1024FP610D@524D!FO30@SFL8FE22@
[Subroutine for positive integer division.
Input: 4D = dividend, 6D = divisor.
Output: 4D = remainder, 6D = quotient.
37 locations; working locations 0D, 8D.]
T 100 K
GKA3FT35@A6DU8DTDA4DRDSDG13@T36@ADLDE4@T36@T6DA4DSDG23@
T4DA6DYFYFT6DT36@A8DSDE35@T36@ADRDTDA6DLDT6DE15@EFPF
[Subroutine to return GCD of two non-negative 35-bit integers.
Input: Integers at 4D, 6D.
Output: GCD at 4D; changes 6D.
41 locations; working location 0D.]
T 200 K
GKA3FT39@S4DE37@T40@A4DTDA6DRDSDG15@T40@ADLDE6@T40@A6DSDG20@T6D
T40@A4DSDE29@T40@ADRDTDE16@S6DE39@TDA4DT6DSDT4DE5@A6DT4DEFPF
[************************ ROSETTA CODE TASK *************************
Subroutine to count Pythagorean triples with given maximum perimeter.
Input: 0D = maximum perimeter.
Output: 4D = number of triples, 6D = number of primitive.
0D is changed.
Must be loaded at an even address.
Uses the well-known fact that a primitive Pythagorean triple is of the form
(m^2 - n^2, 2*m*n, m^2 + n^2) where m, n are coprime and of opposite parity.]
T 300 K
G K
A 3 F [make link]
E 16 @ [jump over variables and constants]
[Double values are put here to ensure even address]
[Variables]
[2] P F P F [maximum perimeter]
[4] P F P F [total number of Pythagorean triples]
[6] P F P F [number of primitive Pythagorean triples]
[8] P F P F [m]
[10] P F P F [n]
[Constants]
T12#Z PF T12Z [clears sandwich digit between 12 and 13]
[12] P D P F [double-value 1]
T14#Z PF T14Z [clears sandwich digit between 14 and 15]
[14] P1F P F [double-value 2]
[Continue with code]
[16] T 69 @ [plant link for return]
A D [load maximum perimeter]
T 2#@ [store locally]
T 4#@ [initialize counts of triangles to 0]
T 6#@
A 12#@ [load 1]
T 8#@ [m := 1]
[Next m, inc by 1]
[23] T F [clear acc]
A 8#@ [load m]
A 12#@ [add 1]
T 8#@ [update m]
H 8#@ [mult reg := m]
C 12#@ [acc := m AND 1]
A 12#@ [add 1]
T 10#@ [n := 1 if m even, 2 if m odd]
[Here to count triangles arising from m, n.
It's assumed m and n are known coprime.]
[31] A 31 @ [call the count subroutine,]
G 70 @ [result is in 6D]
S 6 D [load negative count]
G 40 @ [jump if count > 0]
[No triangles found for this n.
If n = 1 or 2 then whole thing is finished.
Else move on to next m.]
T F [clear acc]
A 14#@ [load 2]
S 10#@ [2 - n]
G 23 @ [if n > 2, go to next m]
E 64 @ [if n <= 2, exit]
[Found triangles, count is in 6D]
[40] T F [clear acc]
A 4#@ [load total count]
A 6 D [add count just found]
T 4#@ [update total count]
A 6#@ [load primitive count]
A 12#@ [add 1]
T 6#@ [update primitive count]
[47] T F [clear acc]
A 10#@ [load n]
A 14#@ [add 2]
U 10#@ [update n]
S 8#@ [is n > m?]
E 23 @ [if so, loop back for next m]
[Test whether m and n are coprime.]
T F [clear acc]
A 8#@ [load m]
T 4 D [to 4D for GCD routine]
A 10#@ [load n]
T 6 D [to 6D for GCD routine]
A 58 @ [call GCD routine,]
G 200 F [GCD is returned in 4D]
A 4 D [load GCD]
S 14#@ [is GCD = 1? (test by subtracting 2)]
E 47 @ [no, go straight to next n]
G 31 @ [yes, count triangles, then next n]
[64] T F [exit, clear acc]
A 4#@ [load total number of triples]
T 4 D [return in 4D]
A 6#@ [load number of primitive triples]
T 6 D [return in 6D]
[69] E F
[2nd-level subroutine to count triangles arising from m, n.
Assumes m, n are coprime and of opposite parity,
and m is in the multiplier register.
Result is returned in 6D.]
[70] A 3 F [make and plant link for return]
T 91 @
A 2#@ [acc := maximum perimeter]
T 4 D [to 4D for division routine]
A 8#@ [load m]
A 10#@ [add n]
T D [m + n to 0D]
V D [acc := m*(m + n)]
[Need to shift product 34 left to restore integer scaling.
Since we want 2*m*(m+n), shift 35 left.]
L F [13 left (maximum possible)]
L F [13 more]
L 128 F [9 more]
T 6 D [perimeter to 6D for division routine]
A 4 D [load maximum perimeter]
S 6 D [is perimeter > maximum?]
G 89 @ [quick exit if so]
T F [clear acc]
A 86 @ [call division routine,]
G 100 F [leaves count in 6D]
E 91 @ [jump to exit]
[89] T F [acc := 0]
T 6 D [return count = 0]
[91] E F
[Main routine. Load at an even address.]
T 500 K
G K
[The initial maximum perimeter is repeatedly multiplied by 10]
T#Z PF TZ [clears sandwich digit between 0 and 1]
[0] P50F PF [initial maximum perimeter <---------- EDIT HERE]
[2] P 3 F [number of values to calculate <---------- EDIT HERE]
[3] P D [1]
[4] P F P F [maximum perimeter]
[6] P F P F [total number of triples]
[8] P F P F [number of primitive triples]
[10] P F [negative count of values]
[11] # F [figures shift]
[12] @ F [carriage return]
[13] & F [line feed]
[14] K 4096 F [null char]
[Enter with acc = 0]
[15] S 2 @ [initialize a negative counter]
T 10 @ [(standard EDSAC practice)]
A #@ [initialize maximum perimeter]
T 4#@
[19] T F [clear acc]
A 4#@ [load maximum perimeter]
T D [to 0D for subroutine]
A 22 @ [call subroutine to count triples]
G 300 F
A 4 D [returns total number in 4D]
T 6#@ [save locally]
A 6 D [returns number of primitive in 6D]
T 8#@ [save locally]
[Print the result]
A 4#@ [load maximum perimeter]
T D [to 0D for print subroutine]
A 30 @ [call print subroutine]
G 56 F
A 6#@ [repeat for total number of triples]
T D
A 34 @
G 56 F
A 8#@ [repeat for number of primitive triples]
T D
A 38 @
G 56 F
O 12 @
O 13 @
A 10 @ [load negative count]
A 3 @ [add 1]
E 53 @ [out if reached 0]
T 10 @ [else update count]
A 4#@ [load max perimeter]
U D [temp store]
L 1 F [times 4]
A D [times 5]
L D [times 10]
T 4#@ [update]
E 19 @ [loop back]
[53] O 14 @ [done; print null to flush printer buffer]
Z F [stop]
E 15 Z [define entry point]
P F [acc = 0 on entry]
|
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #SNOBOL4 | SNOBOL4 | S = ' OUTPUT = " S = 0" S "0"; OUTPUT = REPLACE(S,+"","0");END'
OUTPUT = " S = '" S ""; OUTPUT = REPLACE(S,+"","'");END |
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Wren | Wren | var N = 2200
var N2 = N * N * 2
var s = 3
var s1 = 0
var s2 = 0
var r = List.filled(N + 1, false)
var ab = List.filled(N2 + 1, false)
for (a in 1..N) {
var a2 = a * a
for (b in a..N) ab[a2 + b*b] = true
}
for (c in 1..N) {
s1 = s
s = s + 2
s2 = s
var d = c + 1
while (d <= N) {
if (ab[s1]) r[d] = true
s1 = s1 + s2
s2 = s2 + 2
d = d + 1
}
}
for (d in 1..N) {
if (!r[d]) System.write("%(d) ")
}
System.print() |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Ring | Ring | # Project : Pythagoras tree
load "guilib.ring"
paint = null
new qapp
{
win1 = new qwidget() {
setwindowtitle("Pythagoras tree")
setgeometry(100,100,800,600)
label1 = new qlabel(win1) {
setgeometry(10,10,800,600)
settext("")
}
new qpushbutton(win1) {
setgeometry(150,500,100,30)
settext("draw")
setclickevent("draw()")
}
show()
}
exec()
}
func draw
p1 = new qpicture()
color = new qcolor() {
setrgb(0,0,255,255)
}
pen = new qpen() {
setcolor(color)
setwidth(1)
}
paint = new qpainter() {
begin(p1)
setpen(pen)
w = 800
h = floor(w*11/16)
w2 = floor(w/2)
diff = floor(w/12)
pythagorastree(w2 - diff,h -10,w2 + diff ,h -10 ,0)
endpaint()
}
label1 { setpicture(p1) show() }
return
func pythagorastree(x1,y1,x2,y2,depth)
if depth > 10
return
ok
dx = x2 - x1
dy = y1 - y2
x3 = x2 - dy
y3 = y2 - dx
x4 = x1 - dy
y4 = y1 - dx
x5 = x4 + floor((dx - dy) / 2)
y5 = y4 - floor((dx + dy) / 2)
paint.drawline(x1,y1,x2,y2)
paint.drawline(x2,y2,x3,y3)
paint.drawline(x4,y4,x1,y1)
pythagorastree(x4, y4, x5, y5, depth +1)
pythagorastree(x5, y5, x3, y3, depth +1) |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #ALGOL_W | ALGOL W | if anErrorOccured then assert( false ); |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #AppleScript | AppleScript | if (someCondition) then error number -128 |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #Fortran | Fortran | program qrtask
implicit none
integer, parameter :: n = 4
real(8) :: durer(n, n) = reshape(dble([ &
16, 5, 9, 4, &
3, 10, 6, 15, &
2, 11, 7, 14, &
13, 8, 12, 1 &
]), [n, n])
real(8) :: q(n, n), r(n, n), qr(n, n), id(n, n), tau(n)
integer, parameter :: lwork = 1024
real(8) :: work(lwork)
integer :: info, i, j
q = durer
call dgeqrf(n, n, q, n, tau, work, lwork, info)
r = 0d0
forall (i = 1:n, j = 1:n, j >= i) r(i, j) = q(i, j)
call dorgqr(n, n, n, q, n, tau, work, lwork, info)
qr = matmul(q, r)
id = matmul(q, transpose(q))
call show(4, durer, "A")
call show(4, q, "Q")
call show(4, r, "R")
call show(4, qr, "Q*R")
call show(4, id, "Q*Q'")
contains
subroutine show(n, a, s)
character(*) :: s
integer :: n, i
real(8) :: a(n, n)
print *, s
do i = 1, n
print 1, a(i, :)
1 format (*(f12.6,:,' '))
end do
end subroutine
end program |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #AWK | AWK |
# syntax: TAWK -f PROGRAM_NAME.AWK
#
# GAWK can provide the invoking program name from ARGV[0] but is unable to
# provide the AWK script name that follows -f. Thompson Automation's TAWK
# version 5.0c, last released in 1998 and no longer commercially available, can
# provide the AWK script name that follows -f from the PROGFN built-in
# variable. It should also provide the invoking program name, E.G. TAWK, from
# ARGV[0] but due to a bug it holds the fully qualified -f name instead.
#
# This example is posted here with hopes the TAWK built-in variables PROGFN
# (PROGram File Name) and PROGLN (PROGram Line Number) be added to GAWK by its
# developers.
#
BEGIN {
printf("%s -f %s\n",ARGV[0],PROGFN)
printf("line number %d\n",PROGLN)
exit(0)
}
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #BASIC | BASIC | appname = COMMAND$(0) |
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Eiffel | Eiffel |
class
APPLICATION
create
make
feature
make
local
perimeter: INTEGER
do
perimeter := 100
from
until
perimeter > 1000000
loop
total := 0
primitive_triples := 0
count_pythagorean_triples (3, 4, 5, perimeter)
io.put_string ("There are " + total.out + " triples, below " + perimeter.out + ". Of which " + primitive_triples.out + " are primitives.%N")
perimeter := perimeter * 10
end
end
count_pythagorean_triples (a, b, c, perimeter: INTEGER)
-- Total count of pythagorean triples and total count of primitve triples below perimeter.
local
p: INTEGER
do
p := a + b + c
if p <= perimeter then
primitive_triples := primitive_triples + 1
total := total + perimeter // p
count_pythagorean_triples (a + 2 * (- b + c), 2 * (a + c) - b, 2 * (a - b + c) + c, perimeter)
count_pythagorean_triples (a + 2 * (b + c), 2 * (a + c) + b, 2 * (a + b + c) + c, perimeter)
count_pythagorean_triples (- a + 2 * (b + c), 2 * (- a + c) + b, 2 * (- a + b + c) + c, perimeter)
end
end
feature {NONE}
primitive_triples: INTEGER
total: INTEGER
end
|
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #SPL | SPL | d='JC5wcmludCgiZD0nIitkKyInOyIrJC5iNjRkZWNvZGUoZCkp';$.print("d='"+d+"';"+$.b64decode(d))
|
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #SPWN | SPWN | d='JC5wcmludCgiZD0nIitkKyInOyIrJC5iNjRkZWNvZGUoZCkp';$.print("d='"+d+"';"+$.b64decode(d))
|
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #Yabasic | Yabasic | limite = 2200
s = 3
dim l(limite)
dim ladd(limite * limite * 2)
for x = 1 to limite
x2 = x * x
for y = x to limite
ladd(x2 + y * y) = 1
next y
next x
for x = 1 to limite
s1 = s
s = s + 2
s2 = s
for y = x +1 to limite
if ladd(s1) = 1 l(y) = 1
s1 = s1 + s2
s2 = s2 + 2
next y
next x
for x = 1 to limite
if l(x) = 0 print str$(x), " ";
next x
print
end |
http://rosettacode.org/wiki/Pythagorean_quadruples | Pythagorean quadruples |
One form of Pythagorean quadruples is (for positive integers a, b, c, and d):
a2 + b2 + c2 = d2
An example:
22 + 32 + 62 = 72
which is:
4 + 9 + 36 = 49
Task
For positive integers up 2,200 (inclusive), for all values of a,
b, c, and d,
find (and show here) those values of d that can't be represented.
Show the values of d on one line of output (optionally with a title).
Related tasks
Euler's sum of powers conjecture.
Pythagorean triples.
Reference
the Wikipedia article: Pythagorean quadruple.
| #zkl | zkl | # find values of d where d^2 =/= a^2 + b^2 + c^2 for any integers a, b, c #
# where d in [1..2200], a, b, c =/= 0 #
# max number to check #
const max_number = 2200;
const max_square = max_number * max_number;
# table of numbers that can be the sum of two squares #
sum_of_two_squares:=Data(max_square+1,Int).fill(0); # 4 meg byte array
foreach a in ([1..max_number]){
a2 := a * a;
foreach b in ([a..max_number]){
sum2 := ( b * b ) + a2;
if(sum2 <= max_square) sum_of_two_squares[ sum2 ] = True; # True-->1
}
}
# now find d such that d^2 - c^2 is in sum of two squares #
solution:=Data(max_number+1,Int).fill(0); # another byte array
foreach d in ([1..max_number]){
d2 := d * d;
foreach c in ([1..d-1]){
diff2 := d2 - ( c * c );
if(sum_of_two_squares[ diff2 ]){ solution[ d ] = True; break; }
}
}
# print the numbers whose squares are not the sum of three squares #
foreach d in ([1..max_number]){
if(not solution[ d ]) print(d, " ");
}
println(); |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Ruby | Ruby |
# frozen_string_literal: true
def setup
sketch_title 'Pythagoras Tree'
background(255)
stroke(0, 255, 0)
tree(width / 2.3, height, width / 1.8, height, 10)
end
def tree(x1, y1, x2, y2, depth)
return if depth <= 0
dx = (x2 - x1)
dy = (y1 - y2)
x3 = (x2 - dy)
y3 = (y2 - dx)
x4 = (x1 - dy)
y4 = (y1 - dx)
x5 = (x4 + 0.5 * (dx - dy))
y5 = (y4 - 0.5 * (dx + dy))
# square
begin_shape
fill(0.0, 255.0 / depth, 0.0)
vertex(x1, y1)
vertex(x2, y2)
vertex(x3, y3)
vertex(x4, y4)
vertex(x1, y1)
end_shape
# triangle
begin_shape
fill(0.0, 255.0 / depth, 0.0)
vertex(x3, y3)
vertex(x4, y4)
vertex(x5, y5)
vertex(x3, y3)
end_shape
tree(x4, y4, x5, y5, depth - 1)
tree(x5, y5, x3, y3, depth - 1)
end
def settings
size(800, 400)
end
|
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #APL | APL |
#!/usr/local/bin/apl --script --
⍝⍝ GNU APL script
⍝⍝ Usage: errout.apl <code>
⍝⍝
⍝⍝ $ echo $? ## to see exit code
⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝⍝
args ← 2↓⎕ARG~'--script' '--' ⍝⍝ strip off script args we don't need
err ← ⍎⊃args[1]
∇main
→(0=err)/ok
error:
'Error! exiting.'
⍎')off 1' ⍝⍝ NOTE: exit code arg was added to )OFF in SVN r1499 Nov 2021
ok:
'No error, continuing...'
⍎')off'
∇
main
|
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #ARM_Assembly | ARM Assembly |
/* ARM assembly Raspberry PI */
/* program ending.s */
/* Constantes */
.equ EXIT, 1 @ Linux syscall
/* Initialized data */
.data
/* code section */
.text
.global main
main: @ entry of program
push {fp,lr} @ saves registers
OK:
@ end program OK
mov r0, #0 @ return code
b 100f
NONOK:
@ if error detected end program no ok
mov r0, #1 @ return code
100: @ standard end of the program
pop {fp,lr} @restaur registers
mov r7, #EXIT @ request to exit program
swi 0 @ perform the system call Linux
|
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #Futhark | Futhark |
import "lib/github.com/diku-dk/linalg/linalg"
module linalg_f64 = mk_linalg f64
let eye (n: i32): [n][n]f64 =
let arr = map (\ind -> let (i,j) = (ind/n,ind%n) in if (i==j) then 1.0 else 0.0) (iota (n*n))
in unflatten n n arr
let norm v = linalg_f64.dotprod v v |> f64.sqrt
let qr [n] [m] (a: [m][n]f64): ([m][m]f64, [m][n]f64) =
let make_householder [d] (x: [d]f64): [d][d]f64 =
let div = if x[0] > 0 then x[0] + norm x else x[0] - norm x
let v = map (/div) x
let v[0] = 1
let fac = 2.0 / linalg_f64.dotprod v v
in map2 (map2 (-)) (eye d) (map (map (*fac)) (linalg_f64.outer v v))
let step ((x,y):([m][m]f64,[m][n]f64)) (i:i32): ([m][m]f64,[m][n]f64) =
let h = eye m
let h[i:m,i:m] = make_householder y[i:m,i]
let q': [m][m]f64 = linalg_f64.matmul x h
let a': [m][n]f64 = linalg_f64.matmul h y
in (q',a')
let q = eye m
in foldl step (q,a) (iota n)
entry main = qr [[12.0, -51.0, 4.0],[6.0, 167.0, -68.0],[-4.0, 24.0, -41.0]]
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Blue | Blue | global _start
: syscall ( num:eax -- result:eax ) syscall ;
: exit ( status:edi -- noret ) 60 syscall ;
: bye ( -- noret ) 0 exit ;
: write ( buf:esi len:edx fd:edi -- ) 1 syscall drop ;
1 const stdout
: print ( buf len -- ) stdout write ;
: newline ( -- ) s" \n" print ;
: println ( buf len -- ) print newline ;
: find0 ( start:rsi -- end:rsi ) lodsb 0 cmp latest xne ;
: cstrlen ( str:rdi -- len:rsi ) dup find0 swap sub dec ;
: cstr>str ( cstr:rdx -- str:rsi len:rdx ) dup cstrlen xchg ;
: print-arg ( arg -- ) cstr>str println ;
: arg0 ( rsp -- rsp ) 8 add @ ; inline
: _start ( rsp -- noret ) arg0 print-arg bye ; |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #C | C | #include <stdio.h>
int main(int argc, char **argv) {
printf("Executable: %s\n", argv[0]);
return 0;
} |
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Elixir | Elixir | defmodule RC do
def count_triples(limit), do: count_triples(limit,3,4,5)
defp count_triples(limit, a, b, c) when limit<(a+b+c), do: {0,0}
defp count_triples(limit, a, b, c) do
{p1, t1} = count_triples(limit, a-2*b+2*c, 2*a-b+2*c, 2*a-2*b+3*c)
{p2, t2} = count_triples(limit, a+2*b+2*c, 2*a+b+2*c, 2*a+2*b+3*c)
{p3, t3} = count_triples(limit,-a+2*b+2*c,-2*a+b+2*c,-2*a+2*b+3*c)
{1+p1+p2+p3, div(limit, a+b+c)+t1+t2+t3}
end
end
list = for n <- 1..8, do: Enum.reduce(1..n, 1, fn(_,acc)->10*acc end)
Enum.each(list, fn n -> IO.inspect {n, RC.count_triples(n)} end) |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Standard_ML | Standard ML | (fn s => print (s ^ "\"" ^ String.toString s ^ "\";\n")) "(fn s => print (s ^ \"\\\"\" ^ String.toString s ^ \"\\\";\\n\")) "; |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Swift | Swift | ({print($0+$0.debugDescription+")")})("({print($0+$0.debugDescription+\")\")})(") |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Rust | Rust | /* add to file Cargo.toml:
[dependencies]
svg = "0.10.0"
*/
use svg::node::element::{Group, Polygon};
fn main() {
let mut doc = svg::Document::new().set("stroke", "white");
let mut base: Vec<[(f64, f64); 2]> = vec![[(-200.0, 0.0), (200.0, 0.0)]];
for lvl in 0..12u8 {
let rg = |step| lvl.wrapping_mul(step).wrapping_add(80 - step * 2);
let mut group = Group::new().set("fill", format!("#{:02X}{:02X}18", rg(20), rg(30))); // level color
let mut next_base = Vec::new();
for [a, b] in base {
let v = (b.0 - a.0, b.1 - a.1);
let c = (a.0 + v.1, a.1 - v.0);
let d = (c.0 + v.0, c.1 + v.1);
let e = (c.0 + 0.5 * (v.0 + v.1), c.1 + 0.5 * (v.1 - v.0));
group = group.add(Polygon::new().set("points", vec![a, c, e, d, c, d, b]));
next_base.extend([[c, e], [e, d]]);
}
base = next_base;
doc = doc.add(group);
}
let (x0, y0) = (base.iter()).fold((0.0, 0.0), |(x0, y0), [(x, y), _]| (x.min(x0), y.min(y0)));
let file = "pythagoras_tree.svg";
match svg::save(file, &doc.set("viewBox", (x0, y0, -x0 * 2.0, -y0))) {
Ok(_) => println!("{file} file written successfully!"),
Err(e) => println!("failed to write {file}: {e}"),
}
} |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #Arturo | Arturo | problem: true
if problem -> exit |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #AutoHotkey | AutoHotkey | If (problem)
ExitApp |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #Go | Go | package main
import (
"fmt"
"math"
"github.com/skelterjohn/go.matrix"
)
func sign(s float64) float64 {
if s > 0 {
return 1
} else if s < 0 {
return -1
}
return 0
}
func unitVector(n int) *matrix.DenseMatrix {
vec := matrix.Zeros(n, 1)
vec.Set(0, 0, 1)
return vec
}
func householder(a *matrix.DenseMatrix) *matrix.DenseMatrix {
m := a.Rows()
s := sign(a.Get(0, 0))
e := unitVector(m)
u := matrix.Sum(a, matrix.Scaled(e, a.TwoNorm()*s))
v := matrix.Scaled(u, 1/u.Get(0, 0))
// (error checking skipped in this solution)
prod, _ := v.Transpose().TimesDense(v)
β := 2 / prod.Get(0, 0)
prod, _ = v.TimesDense(v.Transpose())
return matrix.Difference(matrix.Eye(m), matrix.Scaled(prod, β))
}
func qr(a *matrix.DenseMatrix) (q, r *matrix.DenseMatrix) {
m := a.Rows()
n := a.Cols()
q = matrix.Eye(m)
last := n - 1
if m == n {
last--
}
for i := 0; i <= last; i++ {
// (copy is only for compatibility with an older version of gomatrix)
b := a.GetMatrix(i, i, m-i, n-i).Copy()
x := b.GetColVector(0)
h := matrix.Eye(m)
h.SetMatrix(i, i, householder(x))
q, _ = q.TimesDense(h)
a, _ = h.TimesDense(a)
}
return q, a
}
func main() {
// task 1: show qr decomp of wp example
a := matrix.MakeDenseMatrixStacked([][]float64{
{12, -51, 4},
{6, 167, -68},
{-4, 24, -41}})
q, r := qr(a)
fmt.Println("q:\n", q)
fmt.Println("r:\n", r)
// task 2: use qr decomp for polynomial regression example
x := matrix.MakeDenseMatrixStacked([][]float64{
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}})
y := matrix.MakeDenseMatrixStacked([][]float64{
{1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321}})
fmt.Println("\npolyfit:\n", polyfit(x, y, 2))
}
func polyfit(x, y *matrix.DenseMatrix, n int) *matrix.DenseMatrix {
m := x.Cols()
a := matrix.Zeros(m, n+1)
for i := 0; i < m; i++ {
for j := 0; j <= n; j++ {
a.Set(i, j, math.Pow(x.Get(0, i), float64(j)))
}
}
return lsqr(a, y.Transpose())
}
func lsqr(a, b *matrix.DenseMatrix) *matrix.DenseMatrix {
q, r := qr(a)
n := r.Cols()
prod, _ := q.Transpose().TimesDense(b)
return solveUT(r.GetMatrix(0, 0, n, n), prod.GetMatrix(0, 0, n, 1))
}
func solveUT(r, b *matrix.DenseMatrix) *matrix.DenseMatrix {
n := r.Cols()
x := matrix.Zeros(n, 1)
for k := n - 1; k >= 0; k-- {
sum := 0.
for j := k + 1; j < n; j++ {
sum += r.Get(k, j) * x.Get(j, 0)
}
x.Set(k, 0, (b.Get(k, 0)-sum)/r.Get(k, k))
}
return x
} |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #C.23 | C# | using System;
namespace ProgramName
{
class Program
{
static void Main(string[] args)
{
Console.Write(Environment.CommandLine);
}
}
} |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #C.2B.2B | C++ | #include <iostream>
using namespace std;
int main(int argc, char **argv) {
char *program = argv[0];
cout << "Program: " << program << endl;
} |
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Erlang | Erlang | %%
%% Pythagorian triples in Erlang, J.W. Luiten
%%
-module(triples).
-export([main/1]).
%% Transformations t1, t2 and t3 to generate new triples
t1(A, B, C) ->
{A-2*B+2*C, 2*A-B+2*C, 2*A-2*B+3*C}.
t2(A, B, C) ->
{A+2*B+2*C, 2*A+B+2*C, 2*A+2*B+3*C}.
t3(A, B, C) ->
{2*B+2*C-A, B+2*C-2*A, 2*B+3*C-2*A}.
%% Generation of triples
count_triples(A, B, C, Tot_acc, Cnt_acc, Max_perimeter) when (A+B+C) =< Max_perimeter ->
Tot1 = Tot_acc + Max_perimeter div (A+B+C),
{A1, B1, C1} = t1(A, B, C),
{Tot2, Cnt2} = count_triples(A1, B1, C1, Tot1, Cnt_acc+1, Max_perimeter),
{A2, B2, C2} = t2(A, B, C),
{Tot3, Cnt3} = count_triples(A2, B2, C2, Tot2, Cnt2, Max_perimeter),
{A3, B3, C3} = t3(A, B, C),
{Tot4, Cnt4} = count_triples(A3, B3, C3, Tot3, Cnt3, Max_perimeter),
{Tot4, Cnt4};
count_triples(_A, _B, _C, Tot_acc, Cnt_acc, _Max_perimeter) ->
{Tot_acc, Cnt_acc}.
count_triples(A, B, C, Pow) ->
Max = trunc(math:pow(10, Pow)),
{Tot, Prim} = count_triples(A, B, C, 0, 0, Max),
{Pow, Tot, Prim}.
count_triples(Pow) ->
count_triples(3, 4, 5, Pow).
%% Display a single result.
display_result({Pow, Tot, Prim}) ->
io:format("Up to 10 ** ~w : ~w triples, ~w primitives~n", [Pow, Tot, Prim]).
main(Max) ->
L = lists:seq(1, Max),
Answer = lists:map(fun(X) -> count_triples(X) end, L),
lists:foreach(fun(Result) -> display_result(Result) end, Answer). |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Tcl | Tcl | join { {} A B } any_string
=> any_stringAany_stringB |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Turbo_Pascal | Turbo Pascal | program quine;
const
apos: Char = Chr(39);
comma: Char = Chr(44);
lines: Array[1..17] of String[80] = (
'program quine;',
'',
'const',
' apos: Char = Chr(39);',
' comma: Char = Chr(44);',
' lines: Array[1..17] of String[80] = (',
' );',
'',
'var',
' num: Integer;',
'',
'begin',
' for num := 1 to 6 do writeln(lines[num]);',
' for num := 1 to 16 do writeln(apos, lines[num], apos, comma);',
'% writeln(apos, lines[17], apos);',
' for num := 7 to 17 do writeln(lines[num]);',
'end.'
);
var
num: Integer;
begin
for num := 1 to 6 do writeln(lines[num]);
for num := 1 to 16 do writeln(apos, lines[num], apos, comma);
writeln(apos, lines[17], apos);
for num := 7 to 17 do writeln(lines[num]);
end. |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Scala | Scala | import java.awt._
import java.awt.geom.Path2D
import javax.swing.{JFrame, JPanel, SwingUtilities, WindowConstants}
object PythagorasTree extends App {
SwingUtilities.invokeLater(() => {
new JFrame {
class PythagorasTree extends JPanel {
setPreferredSize(new Dimension(640, 640))
setBackground(Color.white)
override def paintComponent(g: Graphics): Unit = {
val (depthLimit, hue) = (7, 0.15f)
def drawTree(g: Graphics2D, x1: Float, y1: Float, x2: Float, y2: Float, depth: Int): Unit = {
if (depth == depthLimit) return
val (dx, dy) = (x2 - x1, y1 - y2)
val (x3, y3) = (x2 - dy, y2 - dx)
val (x4, y4) = (x1 - dy, y1 - dx)
val (x5, y5) = (x4 + 0.5F * (dx - dy), y4 - 0.5F * (dx + dy))
val square = new Path2D.Float {
moveTo(x1, y1); lineTo(x2, y2); lineTo(x3, y3); lineTo(x4, y4); closePath()
}
val triangle = new Path2D.Float {
moveTo(x3, y3); lineTo(x4, y4); lineTo(x5, y5); closePath()
}
g.setColor(Color.getHSBColor(hue + depth * 0.02f, 1, 1))
g.fill(square)
g.setColor(Color.lightGray)
g.draw(square)
g.setColor(Color.getHSBColor(hue + depth * 0.035f, 1, 1))
g.fill(triangle)
g.setColor(Color.lightGray)
g.draw(triangle)
drawTree(g, x4, y4, x5, y5, depth + 1)
drawTree(g, x5, y5, x3, y3, depth + 1)
}
super.paintComponent(g)
drawTree(g.asInstanceOf[Graphics2D], 275, 500, 375, 500, 0)
}
}
setDefaultCloseOperation(WindowConstants.EXIT_ON_CLOSE)
setTitle("Pythagoras Tree")
setResizable(false)
add(new PythagorasTree, BorderLayout.CENTER)
pack()
setLocationRelativeTo(null)
setVisible(true)
}
})
} |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #AutoIt | AutoIt | If problem Then
Exit
Endif |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #AWK | AWK | if(problem)exit 1 |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #Haskell | Haskell |
import Data.List
import Text.Printf (printf)
eps = 1e-6 :: Double
-- a matrix is represented as a list of columns
mmult :: Num a => [[a]] -> [[a]] -> [[a]]
nth :: Num a => [[a]] -> Int -> Int -> a
mmult_num :: Num a => [[a]] -> a -> [[a]]
madd :: Num a => [[a]] -> [[a]] -> [[a]]
idMatrix :: Num a => Int -> Int -> [[a]]
adjustWithE :: [[Double]] -> Int -> [[Double]]
mmult a b = [ [ sum $ zipWith (*) ak bj | ak <- (transpose a) ] | bj <- b ]
nth mA i j = (mA !! j) !! i
mmult_num mA n = map (\c -> map (*n) c) mA
madd mA mB = zipWith (\c1 c2 -> zipWith (+) c1 c2) mA mB
idMatrix n m = [ [if (i==j) then 1 else 0 | i <- [1..n]] | j <- [1..m]]
adjustWithE mA n = let lA = length mA in
(idMatrix n (n - lA)) ++ (map (\c -> (take (n - lA) (repeat 0.0)) ++ c ) mA)
-- auxiliary functions
sqsum :: Floating a => [a] -> a
norm :: Floating a => [a] -> a
epsilonize :: [[Double]] -> [[Double]]
sqsum a = foldl (\x y -> x + y*y) 0 a
norm a = sqrt $! sqsum a
epsilonize mA = map (\c -> map (\x -> if abs x <= eps then 0 else x) c) mA
-- Householder transformation; householder A = (Q, R)
uTransform :: [Double] -> [Double]
hMatrix :: [Double] -> Int -> Int -> [[Double]]
householder :: [[Double]] -> ([[Double]], [[Double]])
-- householder_rec Q R A
householder_rec :: [[Double]] -> [[Double]] -> Int -> ([[Double]], [[Double]])
uTransform a = let t = (head a) + (signum (head a))*(norm a) in
1 : map (\x -> x/t) (tail a)
hMatrix a n i = let u = uTransform (drop i a) in
madd
(idMatrix (n-i) (n-i))
(mmult_num
(mmult [u] (transpose [u]))
((/) (-2) (sqsum u)))
householder_rec mQ mR 0 = (mQ, mR)
householder_rec mQ mR n = let mSize = length mR in
let mH = adjustWithE (hMatrix (mR!!(mSize - n)) mSize (mSize - n)) mSize in
householder_rec (mmult mQ mH) (mmult mH mR) (n - 1)
householder mA = let mSize = length mA in
let (mQ, mR) = householder_rec (idMatrix mSize mSize) mA mSize in
(epsilonize mQ, epsilonize mR)
backSubstitution :: [[Double]] -> [Double] -> [Double] -> [Double]
backSubstitution mR [] res = res
backSubstitution mR@(hR:tR) q@(h:t) res =
let x = (h / (head hR)) in
backSubstitution
(map tail tR)
(tail (zipWith (-) q (map (*x) hR)))
(x : res)
showMatrix :: [[Double]] -> String
showMatrix mA =
concat $ intersperse "\n"
(map (\x -> unwords $ printf "%10.4f" <$> (x::[Double])) (transpose mA))
mY = [[12, 6, -4], [-51, 167, 24], [4, -68, -41]] :: [[Double]]
q = [21, 245, 35] :: [Double]
main = let (mQ, mR) = householder mY in
putStrLn ("Q: \n" ++ showMatrix mQ) >>
putStrLn ("R: \n" ++ showMatrix mR) >>
putStrLn ("q: \n" ++ show q) >>
putStrLn ("x: \n" ++ show (backSubstitution (reverse (map reverse mR)) (reverse q) []))
|
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Clojure | Clojure | ":";exec lein exec $0 ${1+"$@"}
":";exit
(ns scriptname
(:gen-class))
(defn -main [& args]
(let [program (first *command-line-args*)]
(println "Program:" program)))
(when (.contains (first *command-line-args*) *source-path*)
(apply -main (rest *command-line-args*))) |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #COBOL | COBOL | identification division.
program-id. sample.
data division.
working-storage section.
01 progname pic x(16).
procedure division.
sample-main.
display 0 upon argument-number
accept progname from argument-value
display "argument-value zero :" progname ":"
display "function module-id :" function module-id ":"
goback.
end program sample. |
http://rosettacode.org/wiki/Primorial_numbers | Primorial numbers | Primorial numbers are those formed by multiplying successive prime numbers.
The primorial number series is:
primorial(0) = 1 (by definition)
primorial(1) = 2 (2)
primorial(2) = 6 (2×3)
primorial(3) = 30 (2×3×5)
primorial(4) = 210 (2×3×5×7)
primorial(5) = 2310 (2×3×5×7×11)
primorial(6) = 30030 (2×3×5×7×11×13)
∙ ∙ ∙
To express this mathematically, primorialn is
the product of the first n (successive) primes:
p
r
i
m
o
r
i
a
l
n
=
∏
k
=
1
n
p
r
i
m
e
k
{\displaystyle primorial_{n}=\prod _{k=1}^{n}prime_{k}}
─── where
p
r
i
m
e
k
{\displaystyle prime_{k}}
is the kth prime number.
In some sense, generating primorial numbers is similar to factorials.
As with factorials, primorial numbers get large quickly.
Task
Show the first ten primorial numbers (0 ──► 9, inclusive).
Show the length of primorial numbers whose index is: 10 100 1,000 10,000 and 100,000.
Show the length of the one millionth primorial number (optional).
Use exact integers, not approximations.
By length (above), it is meant the number of decimal digits in the numbers.
Related tasks
Sequence of primorial primes
Factorial
Fortunate_numbers
See also
the MathWorld webpage: primorial
the Wikipedia webpage: primorial.
the OEIS webpage: A002110.
| #11l | 11l | F get_primes(primes_count)
V limit = 17 * primes_count
V is_prime = [0B] * 2 [+] [1B] * (limit - 1)
L(n) 0 .< Int(limit ^ 0.5 + 1.5)
I is_prime[n]
L(i) (n * n .< limit + 1).step(n)
is_prime[i] = 0B
[Int] primes
L(prime) is_prime
I prime
primes.append(L.index)
I primes.len == primes_count
L.break
R primes
V primes = get_primes(100000)
F primorial(n)
BigInt r = 1
L(i) 0 .< n
r *= :primes[i]
R r
print(‘First ten primorials: ’(0.<10).map(n -> primorial(n)))
L(e) 6
V n = 10 ^ e
print(‘primorial(#.) has #. digits’.format(n, String(primorial(n)).len)) |
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #ERRE | ERRE | PROGRAM PIT
BEGIN
PRINT(CHR$(12);) !CLS
PRINT(TIME$)
FOR POWER=1 TO 7 DO
PLIMIT=10#^POWER
UPPERBOUND=INT(1+PLIMIT^0.5)
PRIMITIVES=0
TRIPLES=0
EXTRAS=0 ! will count the in-range multiples of any primitive
FOR M=2 TO UPPERBOUND DO
FOR N=1+(M MOD 2=1) TO M-1 STEP 2 DO
TERM1=2*M*N
TERM2=M*M-N*N
TERM3=M*M+N*N
PERIMETER=TERM1+TERM2+TERM3
IF PERIMETER<=PLIMIT THEN TRIPLES=TRIPLES+1
A=TERM1
B=TERM2
REPEAT
R=A-B*INT(A/B)
A=B
B=R
UNTIL R<=0
! we've found a primitive triple if a = 1, since hcf =1.
! and it is inside perimeter range. Save it in an array
IF (A=1) AND (PERIMETER<=PLIMIT) THEN
PRIMITIVES=PRIMITIVES+1
!-----------------------------------------------
!swap so in increasing order of side length
!-----------------------------------------------
IF TERM1>TERM2 THEN SWAP(TERM1,TERM2)
!-----------------------------------------------
!we have the primitive & removed any multiples.
!Now calculate ALL the multiples in range.
!-----------------------------------------------
NEX=INT(PLIMIT/PERIMETER)
EXTRAS=EXTRAS+NEX
END IF
!scan
END FOR
END FOR
PRINT("Primit. with perimeter <=";10#^power;"is";primitives;"&";extras;"non-prim.triples.")
PRINT(TIME$)
END FOR
PRINT PRINT("** End **")
END PROGRAM |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #TXR | TXR | @(deffilter me ("ME" "@(bind me "ME") @(output) @@(deffilter me ("ME" "@{me :filter me}")) @{me :filter (me :from_html)} @(end)"))
@(bind me "ME")
@(output)
@@(deffilter me ("ME" "@{me :filter me}"))
@{me :filter (me :from_html)}
@(end) |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #UNIX_Shell | UNIX Shell | #!/bin/sh
cat < "$0" |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Scilab | Scilab | side = 1; //side length of the square
depth = 8; //final number of branch levels
//L-system definition:
//Alphabet: UTDB+-[]
//U: go upwards T: top of the square
//D: go downwards B: bottom of the square
//[: start new branch ]: end current branch
//+: branch to the right -: branch to the left
//Axiom: UTDB
//Rule: T -> [+UTD-UTD]
//L-system sentence generation
sentence = 'UTDB'
rule = '[+UTD-UTD]';
for i=1:depth
sentence = strsubst(sentence,'T',rule);
end
sentence = strsplit(sentence)';
//Empty tree
tree_size = 1.0...
+ length(find(sentence == "U" | sentence == "T" |...
sentence == "D" | sentence == "B"))...
+ 2 * length(find(sentence == "]" | sentence == "-" |...
sentence == "+"));
tree=zeros(tree_size,1);
//Vectorial operation to calculate a new point in the tree
deff('z = new_point(origin,rho,theta)',...
'z = origin + rho * exp(%i*theta)');
//Drawing the tree
curr_angle = %pi/2;
curr_pos = 1;
ratio = 1/sqrt(2);
for ind = 1:size(sentence,'c')
charac = sentence(ind);
select charac
case 'U' then //Draw line upwards
tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
curr_pos = curr_pos + 1;
case 'T' then //Draw top of the square
curr_angle = curr_angle - %pi/2;
tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
curr_pos = curr_pos + 1;
case 'D' then //Draw line downwards
curr_angle = curr_angle - %pi/2;
tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
curr_pos = curr_pos + 1;
case 'B' then //Draw the bottom
curr_angle = curr_angle - %pi/2;
tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
curr_pos = curr_pos + 1;
case '[' then //Start branch
side = side * ratio;
case '+' then //Start going to the left
curr_angle = curr_angle - %pi/4;
// tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
// tree(curr_pos+2) = new_point(tree(curr_pos+1),side,%pi+curr_angle);
// curr_pos = curr_pos + 2;
curr_angle = curr_angle + %pi/2;
case '-' then //Start going to the left
// tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
// tree(curr_pos+2) = new_point(tree(curr_pos+1),side,%pi+curr_angle);
// curr_pos = curr_pos + 2;
curr_angle = curr_angle + %pi/2;
case ']' then
side = side / ratio;
curr_angle = curr_angle - %pi/4;
// tree(curr_pos+1) = new_point(tree(curr_pos),side,curr_angle);
// tree(curr_pos+2) = new_point(tree(curr_pos+1),side,%pi+curr_angle);
// curr_pos = curr_pos + 2;
curr_angle = curr_angle + %pi;
else
error('L-system sentence error');
end
end
scf(); clf();
xname('Pythagoras tree: '+string(depth)+' levels')
plot2d(real(tree),imag(tree),14);
set(gca(),'isoview','on');
set(gca(),'axes_visible',['off','off','off']); |
http://rosettacode.org/wiki/Pythagoras_tree | Pythagoras tree |
The Pythagoras tree is a fractal tree constructed from squares. It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem.
Task
Construct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).
Related tasks
Fractal tree
| #Sidef | Sidef | require('Imager')
func tree(img, x1, y1, x2, y2, depth) {
depth <= 0 && return()
var dx = (x2 - x1)
var dy = (y1 - y2)
var x3 = (x2 - dy)
var y3 = (y2 - dx)
var x4 = (x1 - dy)
var y4 = (y1 - dx)
var x5 = (x4 + 0.5*(dx - dy))
var y5 = (y4 - 0.5*(dx + dy))
# square
img.polygon(
points => [
[x1, y1],
[x2, y2],
[x3, y3],
[x4, y4],
],
color => [0, 255/depth, 0],
)
# triangle
img.polygon(
points => [
[x3, y3],
[x4, y4],
[x5, y5],
],
color => [0, 255/depth, 0],
)
tree(img, x4, y4, x5, y5, depth - 1)
tree(img, x5, y5, x3, y3, depth - 1)
}
var (width=1920, height=1080)
var img = %O<Imager>.new(xsize => width, ysize => height)
img.box(filled => 1, color => 'white')
tree(img, width/2.3, height, width/1.8, height, 10)
img.write(file => 'pythagoras_tree.png') |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #Axe | Axe | Returnʳ |
http://rosettacode.org/wiki/Program_termination | Program termination |
Task
Show the syntax for a complete stoppage of a program inside a conditional.
This includes all threads/processes which are part of your program.
Explain the cleanup (or lack thereof) caused by the termination (allocated memory, database connections, open files, object finalizers/destructors, run-on-exit hooks, etc.).
Unless otherwise described, no special cleanup outside that provided by the operating system is provided.
| #BASIC | BASIC | IF problem = 1 THEN
END
END IF |
http://rosettacode.org/wiki/QR_decomposition | QR decomposition | Any rectangular
m
×
n
{\displaystyle m\times n}
matrix
A
{\displaystyle {\mathit {A}}}
can be decomposed to a product of an orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
and an upper (right) triangular matrix
R
{\displaystyle {\mathit {R}}}
, as described in QR decomposition.
Task
Demonstrate the QR decomposition on the example matrix from the Wikipedia article:
A
=
(
12
−
51
4
6
167
−
68
−
4
24
−
41
)
{\displaystyle A={\begin{pmatrix}12&-51&4\\6&167&-68\\-4&24&-41\end{pmatrix}}}
and the usage for linear least squares problems on the example from Polynomial regression. The method of Householder reflections should be used:
Method
Multiplying a given vector
a
{\displaystyle {\mathit {a}}}
, for example the first column of matrix
A
{\displaystyle {\mathit {A}}}
, with the Householder matrix
H
{\displaystyle {\mathit {H}}}
, which is given as
H
=
I
−
2
u
T
u
u
u
T
{\displaystyle H=I-{\frac {2}{u^{T}u}}uu^{T}}
reflects
a
{\displaystyle {\mathit {a}}}
about a plane given by its normal vector
u
{\displaystyle {\mathit {u}}}
. When the normal vector of the plane
u
{\displaystyle {\mathit {u}}}
is given as
u
=
a
−
∥
a
∥
2
e
1
{\displaystyle u=a-\|a\|_{2}\;e_{1}}
then the transformation reflects
a
{\displaystyle {\mathit {a}}}
onto the first standard basis vector
e
1
=
[
1
0
0
.
.
.
]
T
{\displaystyle e_{1}=[1\;0\;0\;...]^{T}}
which means that all entries but the first become zero. To avoid numerical cancellation errors, we should take the opposite sign of
a
1
{\displaystyle a_{1}}
:
u
=
a
+
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle u=a+{\textrm {sign}}(a_{1})\|a\|_{2}\;e_{1}}
and normalize with respect to the first element:
v
=
u
u
1
{\displaystyle v={\frac {u}{u_{1}}}}
The equation for
H
{\displaystyle H}
thus becomes:
H
=
I
−
2
v
T
v
v
v
T
{\displaystyle H=I-{\frac {2}{v^{T}v}}vv^{T}}
or, in another form
H
=
I
−
β
v
v
T
{\displaystyle H=I-\beta vv^{T}}
with
β
=
2
v
T
v
{\displaystyle \beta ={\frac {2}{v^{T}v}}}
Applying
H
{\displaystyle {\mathit {H}}}
on
a
{\displaystyle {\mathit {a}}}
then gives
H
a
=
−
sign
(
a
1
)
∥
a
∥
2
e
1
{\displaystyle H\;a=-{\textrm {sign}}(a_{1})\;\|a\|_{2}\;e_{1}}
and applying
H
{\displaystyle {\mathit {H}}}
on the matrix
A
{\displaystyle {\mathit {A}}}
zeroes all subdiagonal elements of the first column:
H
1
A
=
(
r
11
r
12
r
13
0
∗
∗
0
∗
∗
)
{\displaystyle H_{1}\;A={\begin{pmatrix}r_{11}&r_{12}&r_{13}\\0&*&*\\0&*&*\end{pmatrix}}}
In the second step, the second column of
A
{\displaystyle {\mathit {A}}}
, we want to zero all elements but the first two, which means that we have to calculate
H
{\displaystyle {\mathit {H}}}
with the first column of the submatrix (denoted *), not on the whole second column of
A
{\displaystyle {\mathit {A}}}
.
To get
H
2
{\displaystyle H_{2}}
, we then embed the new
H
{\displaystyle {\mathit {H}}}
into an
m
×
n
{\displaystyle m\times n}
identity:
H
2
=
(
1
0
0
0
H
0
)
{\displaystyle H_{2}={\begin{pmatrix}1&0&0\\0&H&\\0&&\end{pmatrix}}}
This is how we can, column by column, remove all subdiagonal elements of
A
{\displaystyle {\mathit {A}}}
and thus transform it into
R
{\displaystyle {\mathit {R}}}
.
H
n
.
.
.
H
3
H
2
H
1
A
=
R
{\displaystyle H_{n}\;...\;H_{3}H_{2}H_{1}A=R}
The product of all the Householder matrices
H
{\displaystyle {\mathit {H}}}
, for every column, in reverse order, will then yield the orthogonal matrix
Q
{\displaystyle {\mathit {Q}}}
.
H
1
H
2
H
3
.
.
.
H
n
=
Q
{\displaystyle H_{1}H_{2}H_{3}\;...\;H_{n}=Q}
The QR decomposition should then be used to solve linear least squares (Multiple regression) problems
A
x
=
b
{\displaystyle {\mathit {A}}x=b}
by solving
R
x
=
Q
T
b
{\displaystyle R\;x=Q^{T}\;b}
When
R
{\displaystyle {\mathit {R}}}
is not square, i.e.
m
>
n
{\displaystyle m>n}
we have to cut off the
m
−
n
{\displaystyle {\mathit {m}}-n}
zero padded bottom rows.
R
=
(
R
1
0
)
{\displaystyle R={\begin{pmatrix}R_{1}\\0\end{pmatrix}}}
and the same for the RHS:
Q
T
b
=
(
q
1
q
2
)
{\displaystyle Q^{T}\;b={\begin{pmatrix}q_{1}\\q_{2}\end{pmatrix}}}
Finally, solve the square upper triangular system by back substitution:
R
1
x
=
q
1
{\displaystyle R_{1}\;x=q_{1}}
| #J | J | QR =: 128!:0 |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #CoffeeScript | CoffeeScript | #!/usr/bin/env coffee
main = () ->
program = __filename
console.log "Program: " + program
if not module.parent then main() |
http://rosettacode.org/wiki/Program_name | Program name | The task is to programmatically obtain the name used to invoke the program. (For example determine whether the user ran "python hello.py", or "python hellocaller.py", a program importing the code from "hello.py".)
Sometimes a multiline shebang is necessary in order to provide the script name to a language's internal ARGV.
See also Command-line arguments
Examples from GitHub.
| #Common_Lisp | Common Lisp | ;;; Play nice with shebangs
(set-dispatch-macro-character #\# #\!
(lambda (stream character n)
(declare (ignore character n))
(read-line stream nil nil t)
nil)) |
http://rosettacode.org/wiki/Primorial_numbers | Primorial numbers | Primorial numbers are those formed by multiplying successive prime numbers.
The primorial number series is:
primorial(0) = 1 (by definition)
primorial(1) = 2 (2)
primorial(2) = 6 (2×3)
primorial(3) = 30 (2×3×5)
primorial(4) = 210 (2×3×5×7)
primorial(5) = 2310 (2×3×5×7×11)
primorial(6) = 30030 (2×3×5×7×11×13)
∙ ∙ ∙
To express this mathematically, primorialn is
the product of the first n (successive) primes:
p
r
i
m
o
r
i
a
l
n
=
∏
k
=
1
n
p
r
i
m
e
k
{\displaystyle primorial_{n}=\prod _{k=1}^{n}prime_{k}}
─── where
p
r
i
m
e
k
{\displaystyle prime_{k}}
is the kth prime number.
In some sense, generating primorial numbers is similar to factorials.
As with factorials, primorial numbers get large quickly.
Task
Show the first ten primorial numbers (0 ──► 9, inclusive).
Show the length of primorial numbers whose index is: 10 100 1,000 10,000 and 100,000.
Show the length of the one millionth primorial number (optional).
Use exact integers, not approximations.
By length (above), it is meant the number of decimal digits in the numbers.
Related tasks
Sequence of primorial primes
Factorial
Fortunate_numbers
See also
the MathWorld webpage: primorial
the Wikipedia webpage: primorial.
the OEIS webpage: A002110.
| #C | C |
#include <inttypes.h>
#include <math.h>
#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>
#include <string.h>
#include <gmp.h>
/* Eratosthenes bit-sieve */
int es_check(uint32_t *sieve, uint64_t n)
{
if ((n != 2 && !(n & 1)) || (n < 2))
return 0;
else
return !(sieve[n >> 6] & (1 << (n >> 1 & 31)));
}
uint32_t *es_sieve(const uint64_t nth, uint64_t *es_size)
{
*es_size = nth * log(nth) + nth * (log(log(nth)) - 0.9385f) + 1;
uint32_t *sieve = calloc((*es_size >> 6) + 1, sizeof(uint32_t));
for (uint64_t i = 3; i < sqrt(*es_size) + 1; i += 2)
if (!(sieve[i >> 6] & (1 << (i >> 1 & 31))))
for (uint64_t j = i * i; j < *es_size; j += (i << 1))
sieve[j >> 6] |= (1 << (j >> 1 & 31));
return sieve;
}
size_t mpz_number_of_digits(const mpz_t op)
{
char *opstr = mpz_get_str(NULL, 10, op);
const size_t oplen = strlen(opstr);
free(opstr);
return oplen;
}
#define PRIMORIAL_LIMIT 1000000
int main(void)
{
/* Construct a sieve of the first 1,000,000 primes */
uint64_t sieve_size;
uint32_t *sieve = es_sieve(PRIMORIAL_LIMIT, &sieve_size);
mpz_t primorial;
mpz_init_set_ui(primorial, 1);
uint64_t prime_count = 0;
int print = 1;
double unused;
for (uint64_t i = 2; i < sieve_size && prime_count <= PRIMORIAL_LIMIT; ++i) {
if (print) {
if (prime_count < 10)
gmp_printf("Primorial(%" PRIu64 ") = %Zd\n", prime_count, primorial);
/* Is the current number a power of 10? */
else if (!modf(log10(prime_count), &unused))
printf("Primorial(%" PRIu64 ") has %zu digits\n", prime_count, mpz_number_of_digits(primorial));
print = 0;
}
if (es_check(sieve, i)) {
mpz_mul_ui(primorial, primorial, i);
prime_count++;
print = 1;
}
}
free(sieve);
mpz_clear(primorial);
return 0;
}
|
http://rosettacode.org/wiki/Pythagorean_triples | Pythagorean triples | A Pythagorean triple is defined as three positive integers
(
a
,
b
,
c
)
{\displaystyle (a,b,c)}
where
a
<
b
<
c
{\displaystyle a<b<c}
, and
a
2
+
b
2
=
c
2
.
{\displaystyle a^{2}+b^{2}=c^{2}.}
They are called primitive triples if
a
,
b
,
c
{\displaystyle a,b,c}
are co-prime, that is, if their pairwise greatest common divisors
g
c
d
(
a
,
b
)
=
g
c
d
(
a
,
c
)
=
g
c
d
(
b
,
c
)
=
1
{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}
.
Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (
g
c
d
(
a
,
b
)
=
1
{\displaystyle {\rm {gcd}}(a,b)=1}
).
Each triple forms the length of the sides of a right triangle, whose perimeter is
P
=
a
+
b
+
c
{\displaystyle P=a+b+c}
.
Task
The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.
Extra credit
Deal with large values. Can your program handle a maximum perimeter of 1,000,000? What about 10,000,000? 100,000,000?
Note: the extra credit is not for you to demonstrate how fast your language is compared to others; you need a proper algorithm to solve them in a timely manner.
Related tasks
Euler's sum of powers conjecture
List comprehensions
Pythagorean quadruples
| #Euphoria | Euphoria | function tri(atom lim, sequence in)
sequence r
atom p
p = in[1] + in[2] + in[3]
if p > lim then
return {0, 0}
end if
r = {1, floor(lim / p)}
r += tri(lim, { in[1]-2*in[2]+2*in[3], 2*in[1]-in[2]+2*in[3], 2*in[1]-2*in[2]+3*in[3]})
r += tri(lim, { in[1]+2*in[2]+2*in[3], 2*in[1]+in[2]+2*in[3], 2*in[1]+2*in[2]+3*in[3]})
r += tri(lim, {-in[1]+2*in[2]+2*in[3], -2*in[1]+in[2]+2*in[3], -2*in[1]+2*in[2]+3*in[3]})
return r
end function
atom max_peri
max_peri = 10
while max_peri <= 100000000 do
printf(1,"%d: ", max_peri)
? tri(max_peri, {3, 4, 5})
max_peri *= 10
end while |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #Unlambda | Unlambda | ``d`.v`.vv```s``si`k`ki``si`k`d`..`.``.d`.``.c`.s`.``.``.s`.``.``.vv``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`.dv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.vv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.vv``s``sc`d`.vv``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`.sv``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`.sv``s``sc`d`.iv``s``sc`d`.`v``s``sc`d`.kv``s``sc`d`.`v``s``sc`d`.kv``s``sc`d`.iv``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`.sv``s``sc`d`.iv``s``sc`d`.`v``s``sc`d`.kv``s``sc`d`.`v``s``sc`d`.dv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.dv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.cv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.sv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.sv``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.`v``s``sc`d`.`v``s``sc`d`..v``s``sc`d`.vvv |
http://rosettacode.org/wiki/Quine | Quine | A quine is a self-referential program that can,
without any external access, output its own source.
A quine (named after Willard Van Orman Quine) is also known as:
self-reproducing automata (1972)
self-replicating program or self-replicating computer program
self-reproducing program or self-reproducing computer program
self-copying program or self-copying computer program
It is named after the philosopher and logician
who studied self-reference and quoting in natural language,
as for example in the paradox "'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation."
"Source" has one of two meanings. It can refer to the text-based program source.
For languages in which program source is represented as a data structure, "source" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.
The usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once quoted in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.
Task
Write a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out. Empty programs producing no output are not allowed.
There are several difficulties that one runs into when writing a quine, mostly dealing with quoting:
Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.
Some languages have a function for getting the "source code representation" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.
Another solution is to construct the quote character from its character code, without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.
Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. "\n"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.
If the language has a way of getting the "source code representation", it usually handles the escaping of characters, so this is not a problem.
Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.
Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.
Next to the Quines presented here, many other versions can be found on the Quine page.
Related task
print itself.
| #V | V | [p [put ' 'put] map ' ' puts]. |
Subsets and Splits
Select Specific Languages Codes
Retrieves specific programming language names and codes from training data, providing basic filtering but limited analytical value beyond identifying these particular languages.