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/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #BBC_BASIC | BBC BASIC | REM >multifact
FOR i% = 1 TO 5
PRINT "Degree "; i%; ":";
FOR j% = 1 TO 10
PRINT " ";FNmultifact(j%, i%);
NEXT
PRINT
NEXT
END
:
DEF FNmultifact(n%, degree%)
LOCAL i%, mfact%
mfact% = 1
FOR i% = n% TO 1 STEP -degree%
mfact% = mfact% * i%
NEXT
= mfact% |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #C | C | #include <stdio.h>
#include <X11/Xlib.h>
int main()
{
Display *d;
Window inwin; /* root window the pointer is in */
Window inchildwin; /* child win the pointer is in */
int rootx, rooty; /* relative to the "root" window; we are not interested in these,
but we can't pass NULL */
int childx, childy; /* the values we are interested in */
Atom atom_type_prop; /* not interested */
int actual_format; /* should be 32 after the call */
unsigned int mask; /* status of the buttons */
unsigned long n_items, bytes_after_ret;
Window *props; /* since we are interested just in the first value, which is
a Window id */
/* default DISPLAY */
d = XOpenDisplay(NULL);
/* ask for active window (no error check); the client must be freedesktop
compliant */
(void)XGetWindowProperty(d, DefaultRootWindow(d),
XInternAtom(d, "_NET_ACTIVE_WINDOW", True),
0, 1, False, AnyPropertyType,
&atom_type_prop, &actual_format,
&n_items, &bytes_after_ret, (unsigned char**)&props);
XQueryPointer(d, props[0], &inwin, &inchildwin,
&rootx, &rooty, &childx, &childy, &mask);
printf("relative to active window: %d,%d\n", childx, childy);
XFree(props); /* free mem */
(void)XCloseDisplay(d); /* and close the display */
return 0;
} |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #REXX | REXX | /*REXX pgm finds primes whose values in other bases (2──►36) have the most diff. bases. */
parse arg widths . /*obtain optional argument from the CL.*/
if widths=='' | widths=="," then widths= 5 /*Not specified? Then use the default.*/
call genP /*build array of semaphores for primes.*/
names= 'one two three four five six seven eight' /*names for some low decimal numbers. */
$.=
do j=1 for # /*only use primes that are within range*/
do b=36 by -1 for 35; n= base(@.j, b) /*use different bases for each prime. */
L= length(n); if L>widths then iterate /*obtain length; Length too big? Skip.*/
if L==1 then $.L.n= b $.L.n /*Length = unity? Prepend the base.*/
else $.L.n= $.L.n b /* " ¬= " Append " " */
end /*b*/
end /*j*/
/*display info for each of the widths. */
do w=1 for widths; cnt= 0 /*show for each width: cnt,number,bases*/
bot= left(1, w, 0); top= left(9, w, 9) /*calculate range for DO. */
do n=bot to top; y= words($.w.n) /*find the sets of numbers for a width.*/
if y>cnt then do; mxn=n; cnt= max(cnt, y); end /*found a max? Remember it*/
end /*n*/
say
say; say center(' 'word(names, w)"─character numbers that are" ,
'prime in the most bases: ('cnt "bases) ", 101, '─')
do n=bot to top; y= words($.w.n) /*search again for maximums.*/
if y==cnt then say n '──►' strip($.w.n) /*display ───a─── maximum. */
end /*n*/
end /*w*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
base: procedure; parse arg x,r,,z; @= '0123456789abcdefghijklmnopqrsruvwxyz'
do j=1; _= r**j; if _>x then leave
end /*j*/
do k=j-1 to 1 by -1; _= r**k; z= z || substr(@, (x % _) + 1, 1); x= x // _
end /*k*/; return z || substr(@, x+1, 1)
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: @.1=2; @.2=3; @.3=5; @.4=7; @.5=11 /*define some low primes. */
#= 5; sq.#= @.# ** 2 /*number primes so far; prime squared.*/
do j=@.#+2 by 2 to 2 * 36 * 10**widths /*find odd primes from here on. */
parse var j '' -1 _; if _==5 then iterate /*J is ÷ by 5? (right dig).*/
if j//3==0 then iterate; if j//7==0 then iterate /*" " " " 3?; ÷ by 7? */
do k=5 while sq.k<=j /* [↓] divide by the known odd primes.*/
if j//@.k==0 then iterate j /*Is J ÷ X? Then not prime. ___ */
end /*k*/ /* [↑] only process numbers ≤ √ J */
#= # + 1; @.#= j; sq.#= j*j /*bump # Ps; assign next P; P squared.*/
end /*j*/; return |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Elixir | Elixir | defmodule RC do
def queen(n, display \\ true) do
solve(n, [], [], [], display)
end
defp solve(n, row, _, _, display) when n==length(row) do
if display, do: print(n,row)
1
end
defp solve(n, row, add_list, sub_list, display) do
Enum.map(Enum.to_list(0..n-1) -- row, fn x ->
add = x + length(row) # \ diagonal check
sub = x - length(row) # / diagonal check
if (add in add_list) or (sub in sub_list) do
0
else
solve(n, [x|row], [add | add_list], [sub | sub_list], display)
end
end) |> Enum.sum # total of the solution
end
defp print(n, row) do
IO.puts frame = "+" <> String.duplicate("-", 2*n+1) <> "+"
Enum.each(row, fn x ->
line = Enum.map_join(0..n-1, fn i -> if x==i, do: "Q ", else: ". " end)
IO.puts "| #{line}|"
end)
IO.puts frame
end
end
Enum.each(1..6, fn n ->
IO.puts " #{n} Queen : #{RC.queen(n)}"
end)
Enum.each(7..12, fn n ->
IO.puts " #{n} Queen : #{RC.queen(n, false)}" # no display
end) |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Python | Python | def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
def lcm(a, b):
return (a*b) / gcd(a, b)
def isPrime(p):
return (p > 1) and all(f == p for f,e in factored(p))
primeList = [2,3,5,7]
def primes():
for p in primeList:
yield p
while 1:
p += 2
while not isPrime(p):
p += 2
primeList.append(p)
yield p
def factored( a):
for p in primes():
j = 0
while a%p == 0:
a /= p
j += 1
if j > 0:
yield (p,j)
if a < p*p: break
if a > 1:
yield (a,1)
def multOrdr1(a,(p,e) ):
m = p**e
t = (p-1)*(p**(e-1)) # = Phi(p**e) where p prime
qs = [1,]
for f in factored(t):
qs = [ q * f[0]**j for j in range(1+f[1]) for q in qs ]
qs.sort()
for q in qs:
if pow( a, q, m )==1: break
return q
def multOrder(a,m):
assert gcd(a,m) == 1
mofs = (multOrdr1(a,r) for r in factored(m))
return reduce(lcm, mofs, 1)
if __name__ == "__main__":
print multOrder(37, 1000) # 100
b = 10**20-1
print multOrder(2, b) # 3748806900
print multOrder(17,b) # 1499522760
b = 100001
print multOrder(54,b)
print pow( 54, multOrder(54,b),b)
if any( (1==pow(54,r, b)) for r in range(1,multOrder(54,b))):
print 'Exists a power r < 9090 where pow(54,r,b)==1'
else:
print 'Everything checks.' |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Raku | Raku | sub nth-root ($n, $A, $p=1e-9)
{
my $x0 = $A / $n;
loop {
my $x1 = (($n-1) * $x0 + $A / ($x0 ** ($n-1))) / $n;
return $x1 if abs($x1-$x0) < abs($x0 * $p);
$x0 = $x1;
}
}
say nth-root(3,8); |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Liberty_BASIC | Liberty BASIC |
call printImages 0, 25
call printImages 250, 265
call printImages 1000, 1025
end
sub printImages loLim, hiLim
loLim = int(loLim)
hiLIm = int(hiLim)
for i = loLim to hiLim
print str$(i) + suffix$(i) + " ";
next i
print
end sub
function suffix$(n)
n = int(n)
nMod10 = n mod 10
nMod100 = n mod 100
if (nMod10 = 1) and (nMod100 <> 11) then
suffix$ = "st"
else
if (nMod10 = 2) and (nMod100 <> 12) then
suffix$ = "nd"
else
if (NMod10 = 3) and (NMod100 <> 13) then
suffix$ = "rd"
else
suffix$ = "th"
end if
end if
end if
end function
|
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #VBScript | VBScript | Function Narcissist(n)
i = 0
j = 0
Do Until j = n
sum = 0
For k = 1 To Len(i)
sum = sum + CInt(Mid(i,k,1)) ^ Len(i)
Next
If i = sum Then
Narcissist = Narcissist & i & ", "
j = j + 1
End If
i = i + 1
Loop
End Function
WScript.StdOut.Write Narcissist(25) |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Lambdatalk | Lambdatalk |
{def munch
{lambda {:w}
{= :w {+ {S.map {{lambda {:w :i}
{pow {W.get :i :w} {W.get :i :w}}} :w}
{S.serie 0 {- {W.length :w} 1}}}}} }}
-> munch
{S.map {lambda {:i} {if {munch :i} then :i else}}
{S.serie 1 5000}}
->
1
3435
|
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #FOCAL | FOCAL | 01.01 C--PRINT F(0..15) AND M(0..15)
01.10 T "F(0..15)"
01.20 F X=0,15;S N=X;D 4;T %1,N
01.30 T !"M(0..15)"
01.40 F X=0,15;S N=X;D 5;T %1,N
01.50 T !
01.60 Q
04.01 C--N = F(N)
04.10 I (N(D)),4.11,4.2
04.11 S N(D)=1;R
04.20 S D=D+1;S N(D)=N(D-1)-1;D 4;D 5
04.30 S D=D-1;S N(D)=N(D)-N(D+1)
05.01 C--N = M(N)
05.10 I (N(D)),5.11,5.2
05.11 R
05.20 S D=D+1;S N(D)=N(D-1)-1;D 5;D 4
05.30 S D=D-1;S N(D)=N(D)-N(D+1) |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Latitude | Latitude | arr := n times to (Array) map { Object clone. }. |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Logtalk | Logtalk |
| ?- create_protocol(statep, [], [public(state/1)]),
findall(
Id,
(integer::between(1, 10, N),
create_object(Id, [implements(statep)], [], [state(N)])),
Ids
).
Ids = [o1, o2, o3, o4, o5, o6, o7, o8, o9, o10].
|
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Lua | Lua | -- This concept is relevant to tables in Lua
local table1 = {1,2,3}
-- The following will create a table of references to table1
local refTab = {}
for i = 1, 10 do refTab[i] = table1 end
-- Instead, tables should be copied using a function like this
function copy (t)
local new = {}
for k, v in pairs(t) do new[k] = v end
return new
end
-- Now we can create a table of independent copies of table1
local copyTab = {}
for i = 1, 10 do copyTab[i] = copy(table1) end |
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #C.2B.2B | C++ | #include <cstdint>
#include <iomanip>
#include <iostream>
bool is_prime(uint64_t n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
for (uint64_t p = 5; p * p <= n; p += 4) {
if (n % p == 0)
return false;
p += 2;
if (n % p == 0)
return false;
}
return true;
}
class motzkin_generator {
public:
uint64_t next();
private:
uint64_t n = 0;
uint64_t m0 = 1;
uint64_t m1 = 1;
};
uint64_t motzkin_generator::next() {
uint64_t m = n > 1 ? (m1 * (2 * n + 1) + m0 * (3 * n - 3)) / (n + 2) : 1;
++n;
m0 = m1;
m1 = m;
return m;
}
int main() {
std::cout.imbue(std::locale(""));
std::cout << " n M(n) Prime?\n";
std::cout << "-----------------------------------\n";
std::cout << std::boolalpha;
motzkin_generator mgen;
for (int i = 0; i < 42; ++i) {
uint64_t n = mgen.next();
std::cout << std::setw(2) << i << std::setw(25) << n << " "
<< is_prime(n) << '\n';
}
} |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #C.2B.2B | C++ |
#include <iostream>
#include <iterator>
#include <sstream>
#include <vector>
using namespace std;
class MTF
{
public:
string encode( string str )
{
fillSymbolTable();
vector<int> output;
for( string::iterator it = str.begin(); it != str.end(); it++ )
{
for( int i = 0; i < 26; i++ )
{
if( *it == symbolTable[i] )
{
output.push_back( i );
moveToFront( i );
break;
}
}
}
string r;
for( vector<int>::iterator it = output.begin(); it != output.end(); it++ )
{
ostringstream ss;
ss << *it;
r += ss.str() + " ";
}
return r;
}
string decode( string str )
{
fillSymbolTable();
istringstream iss( str ); vector<int> output;
copy( istream_iterator<int>( iss ), istream_iterator<int>(), back_inserter<vector<int> >( output ) );
string r;
for( vector<int>::iterator it = output.begin(); it != output.end(); it++ )
{
r.append( 1, symbolTable[*it] );
moveToFront( *it );
}
return r;
}
private:
void moveToFront( int i )
{
char t = symbolTable[i];
for( int z = i - 1; z >= 0; z-- )
symbolTable[z + 1] = symbolTable[z];
symbolTable[0] = t;
}
void fillSymbolTable()
{
for( int x = 0; x < 26; x++ )
symbolTable[x] = x + 'a';
}
char symbolTable[26];
};
int main()
{
MTF mtf;
string a, str[] = { "broood", "bananaaa", "hiphophiphop" };
for( int x = 0; x < 3; x++ )
{
a = str[x];
cout << a << " -> encoded = ";
a = mtf.encode( a );
cout << a << "; decoded = " << mtf.decode( a ) << endl;
}
return 0;
}
|
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Fortran | Fortran | DIMENSION A(5,4,3,2) !Declares a (real) array A of four dimensions, storage permitting.
X = 3*A(2,I,1,K) !Extracts a certain element, multiplies its value by three, result to X.
A(1,2,3,4) = X + 1 !Places a value (the result of the expression X + 1) ... somewhere... |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Emacs_Lisp | Emacs Lisp | (let ((x1 '(0 1 2 3 4 5 6 7 8 9 10))
(x2 '(0 1 1 3 3 7 6 7 3 9 8))
(y '(1 6 17 34 57 86 121 162 209 262 321)))
(apply #'calc-eval "fit(a*X1+b*X2+c,[X1,X2],[a,b,c],[$1 $2 $3])" nil
(mapcar (lambda (items) (cons 'vec items)) (list x1 x2 y)))) |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #C | C |
/* Include statements and constant definitions */
#include <stdio.h>
#define HIGHEST_DEGREE 5
#define LARGEST_NUMBER 10
/* Recursive implementation of multifactorial function */
int multifact(int n, int deg){
return n <= deg ? n : n * multifact(n - deg, deg);
}
/* Iterative implementation of multifactorial function */
int multifact_i(int n, int deg){
int result = n;
while (n >= deg + 1){
result *= (n - deg);
n -= deg;
}
return result;
}
/* Test function to print out multifactorials */
int main(void){
int i, j;
for (i = 1; i <= HIGHEST_DEGREE; i++){
printf("\nDegree %d: ", i);
for (j = 1; j <= LARGEST_NUMBER; j++){
printf("%d ", multifact(j, i));
}
}
}
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #c_sharp | c_sharp |
using System;
using System.Windows.Forms;
static class Program
{
[STAThread]
static void Main()
{
Console.WriteLine(Control.MousePosition.X);
Console.WriteLine(Control.MousePosition.Y);
}
}
|
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Rust | Rust | // [dependencies]
// primal = "0.3"
fn digits(mut n: u32, dig: &mut [u32]) {
for i in 0..dig.len() {
dig[i] = n % 10;
n /= 10;
}
}
fn evalpoly(x: u64, p: &[u32]) -> u64 {
let mut result = 0;
for y in p.iter().rev() {
result *= x;
result += *y as u64;
}
result
}
fn max_prime_bases(ndig: u32, maxbase: u32) {
let mut maxlen = 0;
let mut maxprimebases = Vec::new();
let limit = 10u32.pow(ndig);
let mut dig = vec![0; ndig as usize];
for n in limit / 10..limit {
digits(n, &mut dig);
let bases: Vec<u32> = (2..=maxbase)
.filter(|&x| dig.iter().all(|&y| y < x) && primal::is_prime(evalpoly(x as u64, &dig)))
.collect();
if bases.len() > maxlen {
maxlen = bases.len();
maxprimebases.clear();
}
if bases.len() == maxlen {
maxprimebases.push((n, bases));
}
}
println!(
"{} character numeric strings that are prime in maximum bases: {}",
ndig, maxlen
);
for (n, bases) in maxprimebases {
println!("{} => {:?}", n, bases);
}
println!();
}
fn main() {
for n in 1..=6 {
max_prime_bases(n, 36);
}
} |
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Sidef | Sidef | func max_prime_bases(ndig, maxbase=36) {
var maxprimebases = [[]]
var nwithbases = [0]
var maxprime = (10**ndig - 1)
for p in (idiv(maxprime + 1, 10) .. maxprime) {
var dig = p.digits
var bases = (2..maxbase -> grep {|b| dig.all { _ < b } && dig.digits2num(b).is_prime })
if (bases.len > maxprimebases.first.len) {
maxprimebases = [bases]
nwithbases = [p]
}
elsif (bases.len == maxprimebases.first.len) {
maxprimebases << bases
nwithbases << p
}
}
var (alen, vlen) = (maxprimebases.first.len, maxprimebases.len)
say("\nThe maximum number of prime valued bases for base 10 numeric strings of length ",
ndig, " is #{alen}. The base 10 value list of ", vlen > 1 ? "these" : "this", " is:")
maxprimebases.each_kv {|k,v| say(nwithbases[k], " => ", v) }
}
for n in (1..5) {
max_prime_bases(n)
} |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Erlang | Erlang |
-module( n_queens ).
-export( [display/1, solve/1, task/0] ).
display( Board ) ->
%% Queens are in the positions in the Board list.
%% Top left corner is {1, 1}, Bottom right is {N, N}. There is a queen in the max column.
N = lists:max( [X || {X, _Y} <- Board] ),
[display_row(Y, N, Board) || Y <- lists:seq(1, N)].
solve( N ) ->
Positions = [{X, Y} || X <- lists:seq(1, N), Y <- lists:seq(1, N)],
try
bt( N, Positions, [] )
catch
_:{ok, Board} -> Board
end.
task() ->
task( 4 ),
task( 8 ).
bt( N, Positions, Board ) -> bt_reject( is_not_allowed_queen_placement(N, Board), N, Positions, Board ).
bt_accept( true, _N, _Positions, Board ) -> erlang:throw( {ok, Board} );
bt_accept( false, N, Positions, Board ) -> bt_loop( N, Positions, [], Board ).
bt_loop( _N, [], _Rejects, _Board ) -> failed;
bt_loop( N, [Position | T], Rejects, Board ) ->
bt( N, T ++ Rejects, [Position | Board] ),
bt_loop( N, T, [Position | Rejects], Board ).
bt_reject( true, _N, _Positions, _Board ) -> backtrack;
bt_reject( false, N, Positions, Board ) -> bt_accept( is_all_queens(N, Board), N, Positions, Board ).
diagonals( N, {X, Y} ) ->
D1 = diagonals( N, X + 1, fun diagonals_add1/1, Y + 1, fun diagonals_add1/1 ),
D2 = diagonals( N, X + 1, fun diagonals_add1/1, Y - 1, fun diagonals_subtract1/1 ),
D3 = diagonals( N, X - 1, fun diagonals_subtract1/1, Y + 1, fun diagonals_add1/1 ),
D4 = diagonals( N, X - 1, fun diagonals_subtract1/1, Y - 1, fun diagonals_subtract1/1 ),
D1 ++ D2 ++ D3 ++ D4.
diagonals( _N, 0, _Change_x, _Y, _Change_y ) -> [];
diagonals( _N, _X, _Change_x, 0, _Change_y ) -> [];
diagonals( N, X, _Change_x, _Y, _Change_y ) when X > N -> [];
diagonals( N, _X, _Change_x, Y, _Change_y ) when Y > N -> [];
diagonals( N, X, Change_x, Y, Change_y ) -> [{X, Y} | diagonals( N, Change_x(X), Change_x, Change_y(Y), Change_y )].
diagonals_add1( N ) -> N + 1.
diagonals_subtract1( N ) -> N - 1.
display_row( Row, N, Board ) ->
[io:fwrite("~s", [display_queen(X, Row, Board)]) || X <- lists:seq(1, N)],
io:nl().
display_queen( X, Y, Board ) -> display_queen( lists:member({X, Y}, Board) ).
display_queen( true ) -> " Q";
display_queen( false ) -> " .".
is_all_queens( N, Board ) -> N =:= erlang:length( Board ).
is_diagonal( _N, [] ) -> false;
is_diagonal( N, [Position | T] ) ->
Diagonals = diagonals( N, Position ),
T =/= (T -- Diagonals)
orelse is_diagonal( N, T ).
is_not_allowed_queen_placement( N, Board ) ->
Pieces = erlang:length( Board ),
{Xs, Ys} = lists:unzip( Board ),
Pieces =/= erlang:length( lists:usort(Xs) )
orelse Pieces =/= erlang:length( lists:usort(Ys) )
orelse is_diagonal( N, Board ).
task( N ) ->
io:fwrite( "N = ~p. One solution.~n", [N] ),
Board = solve( N ),
display( Board ).
|
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Racket | Racket |
#lang racket
(require math)
(define (order a n)
(unless (coprime? a n) (error 'order "arguments must be coprime"))
(for/fold ([o 1]) ([r (factorize n)])
(lcm o (order1 a r))))
(define (order1 a p&e)
(match-define (list p e) p&e)
(define m (expt p e))
(define t (* (- p 1) (expt p (- e 1))))
(define qs
(for/fold ([qs '(1)]) ([f (factorize t)])
(match f [(list f0 f1)
(for*/list ([q qs] [j (in-range (+ 1 f1))])
(* q (expt f0 j)))])))
(for/or ([q (sort qs <)] #:when (= (modular-expt a q m) 1)) q))
(order 37 1000)
(order (+ (expt 10 100) 1) 7919)
(order (+ (expt 10 1000) 1) 15485863)
(order (- (expt 10 10000) 1) 22801763489)
(order 13 (+ 1 (expt 10 80)))
|
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Raku | Raku | use Prime::Factor;
sub mo-prime($a, $p, $e) {
my $m = $p ** $e;
my $t = ($p - 1) * ($p ** ($e - 1)); # = Phi($p**$e) where $p prime
my @qs = 1;
for prime-factors($t).Bag -> $f {
@qs = flat @qs.map(-> $q { (0..$f.value).map(-> $j { $q * $f.key ** $j }) });
}
@qs.sort.first: -> $q { expmod( $a, $q, $m ) == 1 };
}
sub mo($a, $m) {
$a gcd $m == 1 or die "$a and $m are not relatively prime";
[lcm] flat 1, prime-factors($m).Bag.map: { mo-prime($a, .key, .value) };
}
multi MAIN('test') {
use Test;
for (10, 21, 25, 150, 1231, 123141, 34131) -> $n {
is ([*] prime-factors($n).Bag.map( { .key ** .value } )), $n, "$n factors correctly";
}
is mo(37, 1000), 100, 'mo(37,1000) == 100';
my $b = 10**20-1;
is mo(2, $b), 3748806900, 'mo(2,10**20-1) == 3748806900';
is mo(17, $b), 1499522760, 'mo(17,10**20-1) == 1499522760';
$b = 100001;
is mo(54, $b), 9090, 'mo(54,100001) == 9090';
} |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #RATFOR | RATFOR |
program nth
#
integer root
real number, precision
real temp0, temp1
1 format('Enter the base number: ')
2 format('Enter the desired root: ')
3 format('Enter the desired precision: ')
4 format(F12.6)
5 format(I6)
write(6,1)
read(5,4)number
write(6,2)
read(5,5)root
write(6,3)
read(5,4)precision
temp0 = number
temp1 = number/root
while ( abs(temp0 - temp1) > precision )
{
temp0 = temp1
temp1 = ((root - 1.0) * temp1 + number / temp1 ** (root - 1.0)) / root
}
6 format(' number root precision')
write(6,6)
7 format(f12.6,i6,f12.6)
write (6,7)number,root,precision
8 format('The root is: ',F12.6)
write (6,8)temp1
end
|
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #REXX | REXX | /*REXX program calculates the Nth root of X, with DIGS (decimal digits) accuracy. */
parse arg x root digs . /*obtain optional arguments from the CL*/
if x=='' | x=="," then x= 2 /*Not specified? Then use the default.*/
if root=='' | root=="," then root= 2 /* " " " " " " */
if digs=='' | digs=="," then digs=65 /* " " " " " " */
numeric digits digs /*set the decimal digits to DIGS. */
say ' x = ' x /*echo the value of X. */
say ' root = ' root /* " " " " ROOT. */
say ' digits = ' digs /* " " " " DIGS. */
say ' answer = ' root(x, root) /*show the value of ANSWER. */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
root: procedure; parse arg x 1 Ox, r 1 Or /*arg1 ──► x & Ox, 2nd ──► r & Or*/
if r=='' then r=2 /*Was root specified? Assume √. */
if r=0 then return '[n/a]' /*oops-ay! Can't do zeroth root.*/
complex= x<0 & R//2==0 /*will the result be complex? */
oDigs=digits() /*get the current number of digs.*/
if x=0 | r=1 then return x/1 /*handle couple of special cases.*/
dm=oDigs+5 /*we need a little guard room. */
r=abs(r); x=abs(x) /*the absolute values of R and X.*/
rm=r-1 /*just a fast version of ROOT -1*/
numeric form /*take a good guess at the root─┐*/
parse value format(x,2,1,,0) 'E0' with ? 'E' _ . /* ◄────────────────────────────┘*/
g= (? / r'E'_ % r) + (x>1) /*kinda uses a crude "logarithm".*/
d=5 /*start with five decimal digits.*/
do until d==dm; d=min(d+d,dm) /*each time, precision doubles. */
numeric digits d /*tell REXX to use D digits. */
old=-1 /*assume some kind of old guess. */
do until old=g; old=g /*where da rubber meets da road─┐*/
g=format((rm*g**r+x)/r/g**rm,, d-2) /* ◄────── the root computation─┘*/
end /*until old=g*/ /*maybe until the cows come home.*/
end /*until d==dm*/ /*and wait for more cows to come.*/
if g=0 then return 0 /*in case the jillionth root = 0.*/
if Or<0 then g=1/g /*root < 0 ? Reciprocal it is! */
if \complex then g=g*sign(Ox) /*adjust the sign (maybe). */
numeric digits oDigs /*reinstate the original digits. */
return (g/1) || left('j', complex) /*normalize # to digs, append j ?*/ |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Lua | Lua | function getSuffix (n)
local lastTwo, lastOne = n % 100, n % 10
if lastTwo > 3 and lastTwo < 21 then return "th" end
if lastOne == 1 then return "st" end
if lastOne == 2 then return "nd" end
if lastOne == 3 then return "rd" end
return "th"
end
function Nth (n) return n .. "'" .. getSuffix(n) end
for i = 0, 25 do print(Nth(i), Nth(i + 250), Nth(i + 1000)) end |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #Wren | Wren | var narc = Fn.new { |n|
var power = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
var limit = 10
var result = []
var x = 0
while (result.count < n) {
if (x >= limit) {
for (i in 0..9) power[i] = power[i] * i
limit = limit * 10
}
var sum = 0
var xx = x
while (xx > 0) {
sum = sum + power[xx%10]
xx = (xx/10).floor
}
if (sum == x) result.add(x)
x = x + 1
}
return result
}
System.print(narc.call(25)) |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #langur | langur | # sum power of digits
val .spod = f(.n) fold f{+}, map(f (.x-'0') ^ (.x-'0'), s2cp toString .n)
# Munchausen
writeln "Answers: ", where f(.n) .n == .spod(.n), series 0..5000 |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Forth | Forth | defer m
: f ( n -- n )
dup 0= if 1+ exit then
dup 1- recurse m - ;
:noname ( n -- n )
dup 0= if exit then
dup 1- recurse f - ;
is m
: test ( xt n -- ) cr 0 do i over execute . loop drop ;
' m defer@ 20 test \ 0 0 1 2 2 3 4 4 5 6 6 7 7 8 9 9 10 11 11 12
' f 20 test \ 1 1 2 2 3 3 4 5 5 6 6 7 8 8 9 9 10 11 11 12 |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #M2000_Interpreter | M2000 Interpreter |
Module CheckIt {
Form 60, 40
Foo=Lambda Id=1 (m)->{
class Alfa {
x, id
Class:
Module Alfa(.x, .id) {}
}
=Alfa(m, id)
id++
}
Dim A(10)<<Foo(20)
\\ for each arrayitem call Foo(20)
TestThis()
\\ call once foo(20) and result copy to each array item
Dim A(10)=Foo(20)
TestThis()
Bar=Lambda Foo (m)->{
->Foo(m)
}
\\ Not only the same id, but the same group
\\ each item is pointer to group
Dim A(10)=Bar(20)
TestThis()
Sub TestThis()
Local i
For i=0 to 9 {
For A(i){
.x++
Print .id , .x
}
}
Print
End Sub
}
Checkit
|
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | {x, x, x, x} /. x -> Random[] |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Maxima | Maxima | a: [1, 2]$
b: makelist(copy(a), 3);
[[1,2],[1,2],[1,2]]
b[1][2]: 1000$
b;
[[1,1000],[1,2],[1,2]] |
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #F.23 | F# |
// Motzkin numbers. Nigel Galloway: September 10th., 2021
let M=let rec fN g=seq{yield List.item 1 g; yield! fN(0L::(g|>List.windowed 3|>List.map(List.sum))@[0L;0L])} in fN [0L;1L;0L;0L]
M|>Seq.take 42|>Seq.iter(printfn "%d")
|
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #Factor | Factor | USING: combinators formatting io kernel math math.primes
tools.memory.private ;
MEMO: motzkin ( m -- n )
dup 2 < [
drop 1
] [
{
[ 2 * 1 + ]
[ 1 - motzkin * ]
[ 3 * 3 - ]
[ 2 - motzkin * + ]
[ 2 + /i ]
} cleave
] if ;
" n motzkin(n)\n" print
42 [
dup motzkin [ commas ] keep prime? "prime" "" ?
"%2d %24s %s\n" printf
] each-integer |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #Clojure | Clojure | (def lowercase (map char (range (int \a) (inc (int \z)))))
(defn move-to-front [x xs]
(cons x (remove #{x} xs)))
(defn encode [text table & {:keys [acc] :or {acc []}}]
(let [c (first text)
idx (.indexOf table c)]
(if (empty? text) acc (recur (drop 1 text) (move-to-front c table) {:acc (conj acc idx)}))))
(defn decode [indices table & {:keys [acc] :or {acc []}}]
(if (empty? indices) (apply str acc)
(let [n (first indices)
c (nth table n)]
(recur (drop 1 indices) (move-to-front c table) {:acc (conj acc c)}))))
(doseq [word ["broood" "bananaaa" "hiphophiphop"]]
(let [encoded (encode word lowercase)
decoded (decode encoded lowercase)]
(println (format "%s encodes to %s which decodes back to %s."
word encoded decoded)))) |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Go | Go | package main
import "fmt"
type md struct {
dim []int
ele []float64
}
func newMD(dim ...int) *md {
n := 1
for _, d := range dim {
n *= d
}
return &md{append([]int{}, dim...), make([]float64, n)}
}
func (m *md) index(i ...int) (x int) {
for d, dx := range m.dim {
x = x*dx + i[d]
}
return
}
func (m *md) at(i ...int) float64 {
return m.ele[m.index(i...)]
}
func (m *md) set(x float64, i ...int) {
m.ele[m.index(i...)] = x
}
func (m *md) show(i ...int) {
fmt.Printf("m%d = %g\n", i, m.at(i...))
}
func main() {
m := newMD(5, 4, 3, 2)
m.show(4, 3, 2, 1)
m.set(87, 4, 3, 2, 1)
m.show(4, 3, 2, 1)
for i := 0; i < m.dim[0]; i++ {
for j := 0; j < m.dim[1]; j++ {
for k := 0; k < m.dim[2]; k++ {
for l := 0; l < m.dim[3]; l++ {
x := m.index(i, j, k, l)
m.set(float64(x)+.1, i, j, k, l)
}
}
}
}
fmt.Println(m.ele[:10])
fmt.Println(m.ele[len(m.ele)-10:])
m.show(4, 3, 2, 1)
} |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #ERRE | ERRE | PROGRAM MULTIPLE_REGRESSION
!$DOUBLE
CONST N=14,M=2,Q=3 ! number of points and M.R. polynom degree
DIM X[N],Y[N] ! data points
DIM S[N],T[N] ! linear system coefficient
DIM A[M,Q] ! sistem to be solved
BEGIN
DATA(1.47,1.50,1.52,1.55,1.57,1.60,1.63,1.65,1.68,1.70,1.73,1.75,1.78,1.80,1.83)
DATA(52.21,53.12,54.48,55.84,57.20,58.57,59.93,61.29,63.11,64.47,66.28,68.10,69.92,72.19,74.46)
FOR I%=0 TO N DO
READ(X[I%])
END FOR
FOR I%=0 TO N DO
READ(Y[I%])
END FOR
FOR K%=0 TO 2*M DO
S[K%]=0 T[K%]=0
FOR I%=0 TO N DO
S[K%]=S[K%]+X[I%]^K%
IF K%<=M THEN T[K%]=T[K%]+Y[I%]*X[I%]^K% END IF
END FOR
END FOR
! build linear system
FOR ROW%=0 TO M DO
FOR COL%=0 TO M DO
A[ROW%,COL%]=S[ROW%+COL%]
END FOR
A[ROW%,COL%]=T[ROW%]
END FOR
PRINT("LINEAR SYSTEM COEFFICENTS") PRINT
FOR I%=0 TO M DO
FOR J%=0 TO M+1 DO
WRITE(" ######.#";A[I%,J%];)
END FOR
PRINT
END FOR
PRINT
FOR J%=0 TO M DO
FOR I%=J% TO M DO
EXIT IF A[I%,J%]<>0
END FOR
IF I%=M+1 THEN
PRINT("SINGULAR MATRIX !")
!$STOP
END IF
FOR K%=0 TO M+1 DO
SWAP(A[J%,K%],A[I%,K%])
END FOR
Y=1/A[J%,J%]
FOR K%=0 TO M+1 DO
A[J%,K%]=Y*A[J%,K%]
END FOR
FOR I%=0 TO M DO
IF I%<>J% THEN
Y=-A[I%,J%]
FOR K%=0 TO M+1 DO
A[I%,K%]=A[I%,K%]+Y*A[J%,K%]
END FOR
END IF
END FOR
END FOR
PRINT
PRINT("SOLUTIONS") PRINT
FOR I%=0 TO M DO
PRINT("c";I%;"=";)
WRITE("#####.#######";A[I%,M+1])
END FOR
END PROGRAM |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Fortran | Fortran | *-----------------------------------------------------------------------
* MR - multiple regression using the SLATEC library routine DHFTI
*
* Finds the nearest approximation to BETA in the system of linear equations:
*
* X(j,i) . BETA(i) = Y(j)
* where
* 1 ... j ... N
* 1 ... i ... K
* and
* K .LE. N
*
* INPUT ARRAYS ARE DESTROYED!
*
*___Name___________Type_______________In/Out____Description_____________
* X(N,K) Double precision In Predictors
* Y(N) Double precision Both On input: N Observations
* On output: K beta weights
* N Integer In Number of observations
* K Integer In Number of predictor variables
* DWORK(N+2*K) Double precision Neither Workspace
* IWORK(K) Integer Neither Workspace
*-----------------------------------------------------------------------
SUBROUTINE MR (X, Y, N, K, DWORK, IWORK)
IMPLICIT NONE
INTEGER K, N, IWORK
DOUBLE PRECISION X, Y, DWORK
DIMENSION X(N,K), Y(N), DWORK(N+2*K), IWORK(K)
* local variables
INTEGER I, J
DOUBLE PRECISION TAU, TOT
* maximum of all column sums of magnitudes
TAU = 0.
DO J = 1, K
TOT = 0.
DO I = 1, N
TOT = TOT + ABS(X(I,J))
END DO
IF (TOT > TAU) TAU = TOT
END DO
TAU = TAU * EPSILON(TAU) ! tolerance argument
* call function
CALL DHFTI (X, N, N, K, Y, N, 1, TAU,
$ J, DWORK(1), DWORK(N+1), DWORK(N+K+1), IWORK)
IF (J < K) PRINT *, 'mr: solution is rank deficient!'
RETURN
END ! of MR
*-----------------------------------------------------------------------
PROGRAM t_mr ! polynomial regression example
IMPLICIT NONE
INTEGER N, K
PARAMETER (N=15, K=3)
INTEGER IWORK(K), I, J
DOUBLE PRECISION XIN(N), X(N,K), Y(N), DWORK(N+2*K)
DATA XIN / 1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68,
$ 1.70, 1.73, 1.75, 1.78, 1.80, 1.83 /
DATA Y / 52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29,
$ 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46 /
* make coefficient matrix
DO J = 1, K
DO I = 1, N
X(I,J) = XIN(I) **(J-1)
END DO
END DO
* solve
CALL MR (X, Y, N, K, DWORK, IWORK)
* print result
10 FORMAT ('beta: ', $)
20 FORMAT (F12.4, $)
30 FORMAT ()
PRINT 10
DO J = 1, K
PRINT 20, Y(J)
END DO
PRINT 30
STOP 'program complete'
END
|
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #C.23 | C# | namespace RosettaCode.Multifactorial
{
using System;
using System.Linq;
internal static class Program
{
private static void Main()
{
Console.WriteLine(string.Join(Environment.NewLine,
Enumerable.Range(1, 5)
.Select(
degree =>
string.Join(" ",
Enumerable.Range(1, 10)
.Select(
number =>
Multifactorial(number, degree))))));
}
private static int Multifactorial(int number, int degree)
{
if (degree < 1)
{
throw new ArgumentOutOfRangeException("degree");
}
var count = 1 + (number - 1) / degree;
if (count < 1)
{
throw new ArgumentOutOfRangeException("number");
}
return Enumerable.Range(0, count)
.Aggregate(1, (accumulator, index) => accumulator * (number - degree * index));
}
}
} |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Clojure | Clojure | (let [point (.. java.awt.MouseInfo getPointerInfo getLocation)] [(.getX point) (.getY point)]) |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Common_Lisp | Common Lisp |
(ql:quickload "ltk")
(in-package :ltk-user)
(defun motion (event)
(format t "~a x position is ~a~&" event (event-x event)))
(with-ltk ()
;; create a small window. Enter the mouse to see lots of events.
(bind *tk* "<Motion>" #'motion))
|
http://rosettacode.org/wiki/Multi-base_primes | Multi-base primes | Prime numbers are prime no matter what base they are represented in.
A prime number in a base other than 10 may not look prime at first glance.
For instance: 19 base 10 is 25 in base 7.
Several different prime numbers may be expressed as the "same" string when converted to a different base.
107 base 10 converted to base 6 == 255
173 base 10 converted to base 8 == 255
353 base 10 converted to base 12 == 255
467 base 10 converted to base 14 == 255
743 base 10 converted to base 18 == 255
1277 base 10 converted to base 24 == 255
1487 base 10 converted to base 26 == 255
2213 base 10 converted to base 32 == 255
Task
Restricted to bases 2 through 36; find the strings that have the most different bases that evaluate to that string when converting prime numbers to a base.
Find the conversion string, the amount of bases that evaluate a prime to that string and the enumeration of bases that evaluate a prime to that string.
Display here, on this page, the string, the count and the list for all of the: 1 character, 2 character, 3 character, and 4 character strings that have the maximum base count that evaluate to that string.
Should be no surprise, the string '2' has the largest base count for single character strings.
Stretch goal
Do the same for the maximum 5 character string.
| #Wren | Wren | import "/math" for Int, Nums
var maxDepth = 5
var maxBase = 36
var c = Int.primeSieve(maxBase.pow(maxDepth), false)
var digits = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
var maxStrings = []
var mostBases = -1
var process = Fn.new { |indices|
var minBase = 2.max(Nums.max(indices) + 1)
if (maxBase - minBase + 1 < mostBases) return // can't affect results so return
var bases = []
for (b in minBase..maxBase) {
var n = 0
for (i in indices) n = n * b + i
if (!c[n]) bases.add(b)
}
var count = bases.count
if (count > mostBases) {
mostBases = count
maxStrings = [[indices.toList, bases]]
} else if (count == mostBases) {
maxStrings.add([indices.toList, bases])
}
}
var printResults = Fn.new {
System.print("%(maxStrings[0][1].count)")
for (m in maxStrings) {
var s = m[0].reduce("") { |acc, i| acc + digits[i] }
System.print("%(s) -> %(m[1])")
}
}
var nestedFor // recursive
nestedFor = Fn.new { |indices, length, level|
if (level == indices.count) {
process.call(indices)
} else {
indices[level] = (level == 0) ? 1 : 0
while (indices[level] < length) {
nestedFor.call(indices, length, level + 1)
indices[level] = indices[level] + 1
}
}
}
for (depth in 1..maxDepth) {
System.write("%(depth) character strings which are prime in most bases: ")
maxStrings = []
mostBases = -1
var indices = List.filled(depth, 0)
nestedFor.call(indices, maxBase, 0)
printResults.call()
System.print()
} |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #ERRE | ERRE |
!------------------------------------------------
! QUEENS.R : solve queens problem on a NxN board
!------------------------------------------------
PROGRAM QUEENS
DIM COL%[15]
BEGIN
MAXSIZE%=15
PRINT(TAB(25);" --- PROBLEMA DELLE REGINE --- ")
PRINT
PRINT("Board dimension ";)
INPUT(N%)
PRINT
IF (N%<1 OR N%>MAXSIZE%)
THEN
PRINT("Illegal dimension!!")
ELSE
FOR CURCOLNBR%=1 TO N%
COL%[CURCOLNBR%]=0
END FOR
CURCOLNBR%=1
WHILE CURCOLNBR%>0 DO
PLACEDAQUEEN%=FALSE
I%=COL%[CURCOLNBR%]+1
WHILE (I%<=N%) AND NOT PLACEDAQUEEN% DO
PLACEDAQUEEN%=TRUE
J%=1
WHILE PLACEDAQUEEN% AND (J%<CURCOLNBR%) DO
PLACEDAQUEEN%=COL%[J%]<>I%
J%=J%+1
END WHILE
IF PLACEDAQUEEN%
THEN
DIAGNBR%=I%+CURCOLNBR%
J%=1
WHILE PLACEDAQUEEN% AND (J%<CURCOLNBR%) DO
PLACEDAQUEEN%=(COL%[J%]+J%)<>DIAGNBR%
J%=J%+1
END WHILE
ELSE
END IF
IF PLACEDAQUEEN%
THEN
DIAGNBR%=I%-CURCOLNBR%
J%=1
WHILE PLACEDAQUEEN% AND (J%<CURCOLNBR%) DO
PLACEDAQUEEN%=(COL%[J%]-J%)<>DIAGNBR%
J%=J%+1
END WHILE
ELSE
END IF
IF NOT PLACEDAQUEEN%
THEN
I%=I%+1
ELSE
COL%[CURCOLNBR%]=I%
END IF
END WHILE
IF NOT PLACEDAQUEEN%
THEN
COL%[CURCOLNBR%]=0
CURCOLNBR%=CURCOLNBR%-1
ELSE
IF CURCOLNBR%=N%
THEN
NSOL%=NSOL%+1
PRINT("Soluzione";NSOL%;":";)
FOR I%=1 TO N%
PRINT(COL%[I%];)
END FOR
PRINT
ELSE
CURCOLNBR%=CURCOLNBR%+1
END IF
END IF
END WHILE
PRINT("Search completed")
REPEAT
GET(CH$)
UNTIL CH$<>""
END IF
END PROGRAM
|
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #REXX | REXX | /*REXX pgm computes multiplicative order of a minimum integer N such that a^n mod m≡1*/
wa= 0; wm= 0 /* ═a═ ══m══ */ /*maximum widths of the A and M values.*/
@.=.; @.1= 3 10
@.2= 37 1000
@.3= 37 10000
@.4= 37 3343
@.5= 37 3344
@.6= 2 1000
pad= left('', 9)
d= 500 /*use 500 decimal digits for a starter.*/
do w=1 for 2 /*when W≡1, find max widths of A and M.*/
do j=1 while @.j\==.; parse var @.j a . 1 r m , n
if w==1 then do; wa= max(wa, length(a) ); wm= max(wm, length(m) ); iterate
end
if m//a==0 then n= ' [solution not possible]' /*test co─prime for A and B. */
numeric digits d /*start with 100 decimal digits. */
if n=='' then do n= 2; p= r * a /*compute product──may have an exponent*/
parse var p 'E' _ /*try to extract the exponent from P. */
if _\=='' then do; numeric digits _+d /*bump the decimal digs.*/
p=r*a /*recalculate integer P.*/
end
if p//m==1 then leave /*now, perform the nitty─gritty modulo.*/
r= p /*assign product to R for next multiply*/
end /*n*/ /* [↑] // is really ÷ remainder.*/
say pad 'a=' right(a,wa) pad "m=" right(m,wm) pad 'multiplicative order:' n
end /*j*/
end /*w*/ /*stick a fork in it, we're all done. */ |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Ring | Ring |
decimals(12)
see "cube root of 5 is : " + root(3, 5, 0) + nl
func root n, a, d
y = 0 x = a / n
while fabs (x - y) > d
y = ((n - 1)*x + a/pow(x,(n-1))) / n
temp = x
x = y
y = temp
end
return x
|
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Ruby | Ruby | def nthroot(n, a, precision = 1e-5)
x = Float(a)
begin
prev = x
x = ((n - 1) * prev + a / (prev ** (n - 1))) / n
end while (prev - x).abs > precision
x
end
p nthroot(5,34) # => 2.02439745849989 |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Maple | Maple | toOrdinal := proc(n:: nonnegint)
if 1 <= n and n <= 10 then
if n >= 4 then
printf("%ath", n);
elif n = 3 then
printf("%ard", n);
elif n = 2 then
printf("%and", n);
else
printf("%ast", n);
end if:
else
printf(convert(n, 'ordinal'));
end if:
return NULL;
end proc:
a := [[0, 25], [250, 265], [1000, 1025]]:
for i in a do
for j from i[1] to i[2] do
toOrdinal(j);
printf(" ");
end do;
printf("\n\n");
end do; |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #XPL0 | XPL0 | func IPow(A, B); \A^B
int A, B, T, I;
[T:= 1;
for I:= 1 to B do T:= T*A;
return T;
];
int Count, M, N, Sum, T, Dig;
[Text(0, "0 ");
Count:= 1;
for M:= 1 to 9 do
for N:= IPow(10, M-1) to IPow(10, M)-1 do
[Sum:= 0;
T:= N;
while T do
[T:= T/10;
Dig:= rem(0);
Sum:= Sum + IPow(Dig, M);
];
if Sum = N then
[IntOut(0, N); ChOut(0, ^ );
Count:= Count+1;
if Count >= 25 then exit;
];
];
] |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #Lua | Lua | function isMunchausen (n)
local sum, nStr, digit = 0, tostring(n)
for pos = 1, #nStr do
digit = tonumber(nStr:sub(pos, pos))
sum = sum + digit ^ digit
end
return sum == n
end
-- alternative, faster version based on the C version,
-- avoiding string manipulation, for Lua 5.3 or higher
local function isMunchausen (n)
local sum, digit, acc = 0, 0, n
while acc > 0 do
digit = acc % 10.0
sum = sum + digit ^ digit
acc = acc // 10 -- integer div
end
return sum == n
end
for i = 1, 5000 do
if isMunchausen(i) then print(i) end
end |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #Fortran | Fortran | module MutualRec
implicit none
contains
pure recursive function m(n) result(r)
integer :: r
integer, intent(in) :: n
if ( n == 0 ) then
r = 0
return
end if
r = n - f(m(n-1))
end function m
pure recursive function f(n) result(r)
integer :: r
integer, intent(in) :: n
if ( n == 0 ) then
r = 1
return
end if
r = n - m(f(n-1))
end function f
end module |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Modula-3 | Modula-3 | VAR a: ARRAY[1..N] OF T |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #NGS | NGS | { [foo()] * n } |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Nim | Nim | import sequtils, strutils
# Creating a sequence containing sequences of integers.
var s1 = newSeq[seq[int]](5)
for item in s1.mitems: item = @[1]
echo "s1 = ", s1 # @[@[1], @[1], @[1], @[1], @[1]]
s1[0].add 2
echo "s1 = ", s1 # @[@[1, 2], @[1], @[1], @[1], @[1]]
# Using newSeqWith.
var s2 = newSeqWith(5, @[1])
echo "s2 = ", s2 # @[@[1], @[1], @[1], @[1], @[1]]
s2[0].add 2
echo "s2 = ", s2 # @[@[1, 2], @[1], @[1], @[1], @[1]]
# Creating a sequence containing pointers.
proc newInt(n: int): ref int =
new(result)
result[] = n
var s3 = newSeqWith(5, newInt(1))
echo "s3 contains references to ", s3.mapIt(it[]).join(", ") # 1, 1, 1, 1, 1
s3[0][] = 2
echo "s3 contains references to ", s3.mapIt(it[]).join(", ") # 2, 1, 1, 1, 1
# How to create non distinct elements.
let p = newInt(1)
var s4 = newSeqWith(5, p)
echo "s4 contains references to ", s4.mapIt(it[]).join(", ") # 1, 1, 1, 1, 1
s4[0][] = 2
echo "s4 contains references to ", s4.mapIt(it[]).join(", ") # 2, 2, 2, 2, 2 |
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #Fermat | Fermat |
Array m[42];
m[1]:=1;
m[2]:=2;
!!(1,0); {precompute and print m[0] thru m[2]}
!!(1,0);
!!(2,1);
for n=3 to 41 do
m[n]:=(m[n-1]*(2*n+1) + m[n-2]*(3*n-3))/(n+2);
!!(m[n],Isprime(m[n]));
od;
; {built-in Isprime function returns 0 for 1, 1 for primes, and}
; {the smallest prime factor for composites, so this actually gives}
; {slightly more information than requested}
|
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #FreeBASIC | FreeBASIC |
#include "isprime.bas"
dim as ulongint M(0 to 41)
M(0) = 1 : M(1) = 1
print "1" : print "1"
for n as integer = 2 to 41
M(n) = M(n-1)
for i as uinteger = 0 to n-2
M(n) += M(i)*M(n-2-i)
next i
print M(n),
if isprime(M(n)) then print "is a prime" else print
next n
|
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #Go | Go | package main
import (
"fmt"
"rcu"
)
func motzkin(n int) []int {
m := make([]int, n+1)
m[0] = 1
m[1] = 1
for i := 2; i <= n; i++ {
m[i] = (m[i-1]*(2*i+1) + m[i-2]*(3*i-3)) / (i + 2)
}
return m
}
func main() {
fmt.Println(" n M[n] Prime?")
fmt.Println("-----------------------------------")
m := motzkin(41)
for i, e := range m {
fmt.Printf("%2d %23s %t\n", i, rcu.Commatize(e), rcu.IsPrime(e))
}
} |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #Common_Lisp | Common Lisp | (defconstant +lower+ (coerce "abcdefghijklmnopqrstuvwxyz" 'list))
(defun move-to-front (x xs)
(cons x (remove x xs)))
(defun enc (text table)
(map 'list
(lambda (c)
(let ((idx (position c table)))
(setf table (move-to-front c table))
idx))
text))
(defun dec (indices table)
(coerce (mapcar (lambda (idx)
(let ((c (nth idx table)))
(setf table (move-to-front c table))
c))
indices)
'string))
(loop for word in '("broood" "bananaaa" "hiphophiphop")
do (let* ((encoded (enc word +lower+))
(decoded (dec encoded +lower+)))
(assert (string= word decoded))
(format T "~s encodes to ~a which decodes back to ~s.~%"
word encoded decoded))) |
http://rosettacode.org/wiki/Monads/Writer_monad | Monads/Writer monad | The Writer monad is a programming design pattern which makes it possible to compose functions which return their result values paired with a log string. The final result of a composed function yields both a value, and a concatenation of the logs from each component function application.
Demonstrate in your programming language the following:
Construct a Writer monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that monad (or just use what the language already provides)
Write three simple functions: root, addOne, and half
Derive Writer monad versions of each of these functions
Apply a composition of the Writer versions of root, addOne, and half to the integer 5, deriving both a value for the Golden Ratio φ, and a concatenated log of the function applications (starting with the initial value, and followed by the application of root, etc.) | #ALGOL_68 | ALGOL 68 | BEGIN
MODE MWRITER = STRUCT( LONG REAL value
, STRING log
);
PRIO BIND = 9;
OP BIND = ( MWRITER m, PROC( LONG REAL )MWRITER f )MWRITER:
( MWRITER n := f( value OF m );
log OF n := log OF m + log OF n;
n
);
OP LEN = ( STRING s )INT: ( UPB s + 1 ) - LWB s;
PRIO PAD = 9;
OP PAD = ( STRING s, INT width )STRING: IF LEN s >= width THEN s ELSE s + ( width - LEN s ) * " " FI;
PROC unit = ( LONG REAL v, STRING s )MWRITER: ( v, " " + s PAD 17 + ":" + fixed( v, -19, 15 ) + REPR 10 );
PROC root = ( LONG REAL v )MWRITER: unit( long sqrt( v ), "Took square root" );
PROC add one = ( LONG REAL v )MWRITER: unit( v+1, "Added one" );
PROC half = ( LONG REAL v )MWRITER: unit( v/2, "Divided by two" );
MWRITER mw2 := unit( 5, "Initial value" ) BIND root BIND add one BIND half;
print( ( "The Golden Ratio is", fixed( value OF mw2, -19, 15 ), newline ) );
print( ( newline, "This was derived as follows:-", newline ) );
print( ( log OF mw2 ) )
END |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #J | J | A1=:5 4 3 2$0 |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #Java | Java | public class MultiDimensionalArray {
public static void main(String[] args) {
// create a regular 4 dimensional array and initialize successive elements to the values 1 to 120
int m = 1;
int[][][][] a4 = new int[5][4][3][2];
for (int i = 0; i < a4.length; ++i) {
for (int j = 0; j < a4[0].length; ++j) {
for (int k = 0; k < a4[0][0].length; ++k) {
for (int l = 0; l < a4[0][0][0].length; ++l) {
a4[i][j][k][l] = m++;
}
}
}
}
System.out.println("First element = " + a4[0][0][0][0]); // access and print value of first element
a4[0][0][0][0] = 121; // change value of first element
System.out.println();
for (int i = 0; i < a4.length; ++i) {
for (int j = 0; j < a4[0].length; ++j) {
for (int k = 0; k < a4[0][0].length; ++k) {
for (int l = 0; l < a4[0][0][0].length; ++l) {
System.out.printf("%4d", a4[i][j][k][l]);
}
}
}
}
}
} |
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #11l | 11l | V n = 12
L(j) 1..n
print(‘#3’.format(j), end' ‘ ’)
print(‘│’)
L 1..n
print(‘────’, end' ‘’)
print(‘┼───’)
L(i) 1..n
L(j) 1..n
print(I j < i {‘ ’} E ‘#3 ’.format(i * j), end' ‘’)
print(‘│ ’i) |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Go | Go | package main
import (
"fmt"
"github.com/gonum/matrix/mat64"
)
func givens() (x, y *mat64.Dense) {
height := []float64{1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63,
1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83}
weight := []float64{52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93,
61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46}
degree := 2
x = Vandermonde(height, degree)
y = mat64.NewDense(len(weight), 1, weight)
return
}
func Vandermonde(a []float64, degree int) *mat64.Dense {
x := mat64.NewDense(len(a), degree+1, nil)
for i := range a {
for j, p := 0, 1.; j <= degree; j, p = j+1, p*a[i] {
x.Set(i, j, p)
}
}
return x
}
func main() {
x, y := givens()
fmt.Printf("%.4f\n", mat64.Formatted(mat64.QR(x).Solve(y)))
} |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #C.2B.2B | C++ |
#include <algorithm>
#include <iostream>
#include <iterator>
/*Generate multifactorials to 9
Nigel_Galloway
November 14th., 2012.
*/
int main(void) {
for (int g = 1; g < 10; g++) {
int v[11], n=0;
generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;});
std::cout << std::endl;
}
return 0;
}
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Delphi | Delphi | procedure TForm1.FormMouseMove(Sender: TObject; Shift: TShiftState; X,
Y: Integer);
begin
lblMousePosition.Caption := ('X:' + IntToStr(X) + ', Y:' + IntToStr(Y));
end; |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #EasyLang | EasyLang | on mouse_move
clear
text mouse_x & " " & mouse_y
. |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #F.23 | F# |
let rec iterate f value = seq {
yield value
yield! iterate f (f value) }
let up i = i + 1
let right i = i
let down i = i - 1
let noCollisionGivenDir solution number dir =
Seq.forall2 (<>) solution (Seq.skip 1 (iterate dir number))
let goodAddition solution number =
List.forall (noCollisionGivenDir solution number) [ up; right; down ]
let rec extendSolution n ps =
[0..n - 1]
|> List.filter (goodAddition ps)
|> List.map (fun num -> num :: ps)
let allSolutions n =
iterate (List.collect (extendSolution n)) [[]]
// Print one solution for the 8x8 case
let printOneSolution () =
allSolutions 8
|> Seq.item 8
|> Seq.head
|> List.iter (fun rowIndex ->
printf "|"
[0..8] |> List.iter (fun i -> printf (if i = rowIndex then "X|" else " |"))
printfn "")
// Print number of solution for the other cases
let printNumberOfSolutions () =
printfn "Size\tNr of solutions"
[1..11]
|> List.map ((fun i -> Seq.item i (allSolutions i)) >> List.length)
|> List.iteri (fun i cnt -> printfn "%d\t%d" (i+1) cnt)
printOneSolution()
printNumberOfSolutions()
|
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Ruby | Ruby | require 'prime'
def powerMod(b, p, m)
p.to_s(2).each_char.inject(1) do |result, bit|
result = (result * result) % m
bit=='1' ? (result * b) % m : result
end
end
def multOrder_(a, p, k)
pk = p ** k
t = (p - 1) * p ** (k - 1)
r = 1
for q, e in t.prime_division
x = powerMod(a, t / q**e, pk)
while x != 1
r *= q
x = powerMod(x, q, pk)
end
end
r
end
def multOrder(a, m)
m.prime_division.inject(1) do |result, f|
result.lcm(multOrder_(a, *f))
end
end
puts multOrder(37, 1000)
b = 10**20-1
puts multOrder(2, b)
puts multOrder(17,b)
b = 100001
puts multOrder(54,b)
puts powerMod(54, multOrder(54,b), b)
if (1...multOrder(54,b)).any? {|r| powerMod(54, r, b) == 1}
puts 'Exists a power r < 9090 where powerMod(54,r,b)==1'
else
puts 'Everything checks.'
end |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Run_BASIC | Run BASIC | print "Root 125th Root of 5643 Precision .001 ";using( "#.###############", NthRoot( 125, 5642, 0.001 ))
print "125th Root of 5643 Precision .001 ";using( "#.###############", NthRoot( 125, 5642, 0.001 ))
print "125th Root of 5643 Precision .00001 ";using( "#.###############", NthRoot( 125, 5642, 0.00001))
print " 3rd Root of 27 Precision .00001 ";using( "#.###############", NthRoot( 3, 27, 0.00001))
print " 2nd Root of 2 Precision .00001 ";using( "#.###############", NthRoot( 2, 2, 0.00001))
print " 10th Root of 1024 Precision .00001 ";using( "#.###############", NthRoot( 10, 1024, 0.00001))
wait
function NthRoot( root, A, precision)
x0 = A
x1 = A /root
while abs( x1 -x0) >precision
x0 = x1
x1 = x1 / 1.0 ' force float
x1 = (( root -1.0) *x1 +A /x1^( root -1.0)) /root
wend
NthRoot =x1
end function
end |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Rust | Rust | // 20210212 Rust programming solution
fn nthRoot(n: f64, A: f64) -> f64 {
let p = 1e-9_f64 ;
let mut x0 = A / n ;
loop {
let mut x1 = ( (n-1.0) * x0 + A / f64::powf(x0, n-1.0) ) / n;
if (x1-x0).abs() < (x0*p).abs() { return x1 };
x0 = x1
}
}
fn main() {
println!("{}", nthRoot(3. , 8. ));
} |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | suffixlist = {"th", "st", "nd", "rd", "th", "th", "th", "th", "th","th"};
addsuffix[n_] := Module[{suffix},
suffix = Which[
Mod[n, 100] <= 10, suffixlist[[Mod[n, 10] + 1]],
Mod[n, 100] > 20, suffixlist[[Mod[n, 10] + 1]],
True, "th"
];
ToString[n] <> suffix
]
addsuffix[#] & /@ Range[0, 25] (* test 1 *)
addsuffix[#] & /@ Range[250, 265] (* test 2 *)
addsuffix[#] & /@ Range[1000, 1025] (* test 3 *) |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #zkl | zkl | fcn isNarcissistic(n){
ns,m := n.split(), ns.len() - 1;
ns.reduce('wrap(s,d){ z:=d; do(m){z*=d} s+z },0) == n
} |
http://rosettacode.org/wiki/Narcissistic_decimal_number | Narcissistic decimal number | A Narcissistic decimal number is a non-negative integer,
n
{\displaystyle n}
, that is equal to the sum of the
m
{\displaystyle m}
-th powers of each of the digits in the decimal representation of
n
{\displaystyle n}
, where
m
{\displaystyle m}
is the number of digits in the decimal representation of
n
{\displaystyle n}
.
Narcissistic (decimal) numbers are sometimes called Armstrong numbers, named after Michael F. Armstrong.
They are also known as Plus Perfect numbers.
An example
if
n
{\displaystyle n}
is 153
then
m
{\displaystyle m}
, (the number of decimal digits) is 3
we have 13 + 53 + 33 = 1 + 125 + 27 = 153
and so 153 is a narcissistic decimal number
Task
Generate and show here the first 25 narcissistic decimal numbers.
Note:
0
1
=
0
{\displaystyle 0^{1}=0}
, the first in the series.
See also
the OEIS entry: Armstrong (or Plus Perfect, or narcissistic) numbers.
MathWorld entry: Narcissistic Number.
Wikipedia entry: Narcissistic number.
| #ZX_Spectrum_Basic | ZX Spectrum Basic | 1 DIM K(10): DIM M(10)
2 FOR Y=0 TO 9: LET M(Y+1)=Y: NEXT Y
3 FOR N=1 TO 7
4 FOR J=N TO 0 STEP -1
5 FOR I=N-J TO 0 STEP -1
6 FOR H=N-J-I TO 0 STEP -1
7 FOR G=N-J-I-H TO 0 STEP -1
8 FOR F=N-J-I-H-G TO 0 STEP -1
9 FOR E=N-J-I-H-G-F TO 0 STEP -1
10 FOR D=N-J-I-H-G-F-E TO 0 STEP -1
11 FOR C=N-J-I-H-G-F-E-D TO 0 STEP -1
12 FOR B=N-J-I-H-G-F-E-D-C TO 0 STEP -1
13 LET A=N-J-I-H-G-F-E-D-C-B
14 LET X=B+C*M(3)+D*M(4)+E*M(5)+F*M(6)+G*M(7)+H*M(8)+I*M(9)+J*M(10)
15 LET S$=STR$ (X)
16 IF LEN (S$)<N THEN GO TO 34
17 IF LEN (S$)<>N THEN GO TO 33
18 FOR Y=1 TO 10: LET K(Y)=0: NEXT Y
19 FOR Y=1 TO N
20 LET Z= CODE (S$(Y))-47
21 LET K(Z)=K(Z)+1
22 NEXT Y
23 IF A<>K(1) THEN GO TO 33
24 IF B<>K(2) THEN GO TO 33
25 IF C<>K(3) THEN GO TO 33
26 IF D<>K(4) THEN GO TO 33
27 IF E<>K(5) THEN GO TO 33
28 IF F<>K(6) THEN GO TO 33
29 IF G<>K(7) THEN GO TO 33
30 IF H<>K(8) THEN GO TO 33
31 IF I<>K(9) THEN GO TO 33
32 IF J=K(10) THEN PRINT X,
33 NEXT B: NEXT C: NEXT D: NEXT E: NEXT F: NEXT G: NEXT H: NEXT I: NEXT J
34 FOR Y=2 TO 9
35 LET M(Y+1)=M(Y+1)*Y
36 NEXT Y
37 NEXT N
38 PRINT
39 PRINT "DONE" |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #M2000_Interpreter | M2000 Interpreter |
Module Munchausen {
Inventory p=0:=0,1:=1
for i=2 to 9 {Append p, i:=i**i}
Munchausen=lambda p (x)-> {
m=0
t=x
do {
m+=p(x mod 10)
x=x div 10
} until x=0
=m=t
}
For i=1 to 5000
If Munchausen(i) then print i,
Next i
Print
}
Munchausen
|
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #FreeBASIC | FreeBASIC | ' FB 1.05.0 Win64
' Need forward declaration of M as it's used
' by F before its defined
Declare Function M(n As Integer) As Integer
Function F(n As Integer) As Integer
If n = 0 Then
Return 1
End If
Return n - M(F(n - 1))
End Function
Function M(n As Integer) As Integer
If n = 0 Then
Return 0
End If
Return n - F(M(n - 1))
End Function
Dim As Integer n = 24
Print "n :";
For i As Integer = 0 to n : Print Using "###"; i; : Next
Print
Print String(78, "-")
Print "F :";
For i As Integer = 0 To n : Print Using "###"; F(i); : Next
Print
Print "M :";
For i As Integer = 0 To n : Print Using "###"; M(i); : Next
Print
Print "Press any key to quit"
Sleep |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #OCaml | OCaml | Array.make n (new foo);;
(* here (new foo) can be any expression that returns a new object,
record, array, or string *) |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Oforth | Oforth | ListBuffer init(10, #[ Float rand ]) println |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #ooRexx | ooRexx | -- get an array of directory objects
array = fillArrayWith(3, .directory)
say "each object will have a different identityHash"
say
loop d over array
say d d~identityHash
end
::routine fillArrayWith
use arg size, class
array = .array~new(size)
loop i = 1 to size
-- Note, this assumes this object class can be created with
-- no arguments
array[i] = class~new
end
return array |
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #Haskell | Haskell | import Control.Monad.Memo (Memo, memo, startEvalMemo)
import Math.NumberTheory.Primes.Testing (isPrime)
import System.Environment (getArgs)
import Text.Printf (printf)
type I = Integer
-- The n'th Motzkin number, where n is assumed to be ≥ 0. We memoize the
-- computations using MonadMemo.
motzkin :: I -> Memo I I I
motzkin 0 = return 1
motzkin 1 = return 1
motzkin n = do
m1 <- memo motzkin (n-1)
m2 <- memo motzkin (n-2)
return $ ((2*n+1)*m1 + (3*n-3)*m2) `div` (n+2)
-- The first n+1 Motzkin numbers, starting at 0.
motzkins :: I -> [I]
motzkins = startEvalMemo . mapM motzkin . enumFromTo 0
-- For i = 0 to n print i, the i'th Motzkin number, and if it's a prime number.
printMotzkins :: I -> IO ()
printMotzkins n = mapM_ prnt $ zip [0 :: I ..] (motzkins n)
where prnt (i, m) = printf "%2d %20d %s\n" i m $ prime m
prime m = if isPrime m then "prime" else ""
main :: IO ()
main = do
[n] <- map read <$> getArgs
printMotzkins n |
http://rosettacode.org/wiki/Move-to-front_algorithm | Move-to-front algorithm | Given a symbol table of a zero-indexed array of all possible input symbols
this algorithm reversibly transforms a sequence
of input symbols into an array of output numbers (indices).
The transform in many cases acts to give frequently repeated input symbols
lower indices which is useful in some compression algorithms.
Encoding algorithm
for each symbol of the input sequence:
output the index of the symbol in the symbol table
move that symbol to the front of the symbol table
Decoding algorithm
# Using the same starting symbol table
for each index of the input sequence:
output the symbol at that index of the symbol table
move that symbol to the front of the symbol table
Example
Encoding the string of character symbols 'broood' using a symbol table of the lowercase characters a-to-z
Input
Output
SymbolTable
broood
1
'abcdefghijklmnopqrstuvwxyz'
broood
1 17
'bacdefghijklmnopqrstuvwxyz'
broood
1 17 15
'rbacdefghijklmnopqstuvwxyz'
broood
1 17 15 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0
'orbacdefghijklmnpqstuvwxyz'
broood
1 17 15 0 0 5
'orbacdefghijklmnpqstuvwxyz'
Decoding the indices back to the original symbol order:
Input
Output
SymbolTable
1 17 15 0 0 5
b
'abcdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
br
'bacdefghijklmnopqrstuvwxyz'
1 17 15 0 0 5
bro
'rbacdefghijklmnopqstuvwxyz'
1 17 15 0 0 5
broo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
brooo
'orbacdefghijklmnpqstuvwxyz'
1 17 15 0 0 5
broood
'orbacdefghijklmnpqstuvwxyz'
Task
Encode and decode the following three strings of characters using the symbol table of the lowercase characters a-to-z as above.
Show the strings and their encoding here.
Add a check to ensure that the decoded string is the same as the original.
The strings are:
broood
bananaaa
hiphophiphop
(Note the misspellings in the above strings.)
| #D | D | import std.stdio, std.string, std.ascii, std.algorithm;
ptrdiff_t[] mtfEncoder(in string data) pure nothrow @safe
in {
assert(data.countchars("a-z") == data.length);
} out(result) {
assert(result.length == data.length);
assert(result.all!(e => e >= 0 && e < lowercase.length));
} body {
ubyte[lowercase.length] order = lowercase.representation;
auto encoded = new typeof(return)(data.length);
size_t i = 0;
foreach (immutable b; data) {
immutable j = encoded[i++] = order[].countUntil(b);
bringToFront(order[0 .. j], order[j .. j + 1]);
}
return encoded;
}
string mtfDecoder(in ptrdiff_t[] encoded) pure nothrow @safe
in {
assert(encoded.all!(e => e >= 0 && e < lowercase.length));
} out(result) {
assert(result.length == encoded.length);
assert(result.countchars("a-z") == result.length);
} body {
ubyte[lowercase.length] order = lowercase.representation;
auto decoded = new char[encoded.length];
size_t i = 0;
foreach (immutable code; encoded) {
decoded[i++] = order[code];
bringToFront(order[0 .. code], order[code .. code + 1]);
}
return decoded;
}
void main() {
foreach (immutable word; ["broood", "bananaaa", "hiphophiphop"]) {
immutable encoded = word.mtfEncoder;
immutable decoded = encoded.mtfDecoder;
writefln("'%s' encodes to %s, which decodes back to '%s'",
word, encoded, decoded);
assert(word == decoded);
}
} |
http://rosettacode.org/wiki/Monads/Writer_monad | Monads/Writer monad | The Writer monad is a programming design pattern which makes it possible to compose functions which return their result values paired with a log string. The final result of a composed function yields both a value, and a concatenation of the logs from each component function application.
Demonstrate in your programming language the following:
Construct a Writer monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that monad (or just use what the language already provides)
Write three simple functions: root, addOne, and half
Derive Writer monad versions of each of these functions
Apply a composition of the Writer versions of root, addOne, and half to the integer 5, deriving both a value for the Golden Ratio φ, and a concatenated log of the function applications (starting with the initial value, and followed by the application of root, etc.) | #AppleScript | AppleScript | -- WRITER MONAD FOR APPLESCRIPT
-- How can we compose functions which take simple values as arguments
-- but return an output value which is paired with a log string ?
-- We can prevent functions which expect simple values from choking
-- on log-wrapped output (from nested functions)
-- by writing Unit/Return() and Bind() for the Writer monad in AppleScript
on run {}
-- Derive logging versions of three simple functions, pairing
-- each function with a particular comment string
-- (a -> b) -> (a -> (b, String))
set wRoot to writerVersion(root, "obtained square root")
set wSucc to writerVersion(succ, "added one")
set wHalf to writerVersion(half, "divided by two")
loggingHalfOfRootPlusOne(5)
--> value + log string
end run
-- THREE SIMPLE FUNCTIONS
on root(x)
x ^ (1 / 2)
end root
on succ(x)
x + 1
end succ
on half(x)
x / 2
end half
-- DERIVE A LOGGING VERSION OF A FUNCTION BY COMBINING IT WITH A
-- LOG STRING FOR THAT FUNCTION
-- (SEE 'on run()' handler at top of script)
-- (a -> b) -> String -> (a -> (b, String))
on writerVersion(f, strComment)
script
on call(x)
{value:sReturn(f)'s call(x), comment:strComment}
end call
end script
end writerVersion
-- DEFINE A COMPOSITION OF THE SAFE VERSIONS
on loggingHalfOfRootPlusOne(x)
logCompose([my wHalf, my wSucc, my wRoot], x)
end loggingHalfOfRootPlusOne
-- Monadic UNIT/RETURN and BIND functions for the writer monad
on writerUnit(a)
try
set strValue to ": " & a as string
on error
set strValue to ""
end try
{value:a, comment:"Initial value" & strValue}
end writerUnit
on writerBind(recWriter, wf)
set recB to wf's call(value of recWriter)
set v to value of recB
try
set strV to " -> " & (v as string)
on error
set strV to ""
end try
{value:v, comment:(comment of recWriter) & linefeed & (comment of recB) & strV}
end writerBind
-- THE TWO HIGHER ORDER FUNCTIONS ABOVE ENABLE COMPOSITION OF
-- THE LOGGING VERSIONS OF EACH FUNCTION
on logCompose(lstFunctions, varValue)
reduceRight(lstFunctions, writerBind, writerUnit(varValue))
end logCompose
-- xs: list, f: function, a: initial accumulator value
-- the arguments available to the function f(a, x, i, l) are
-- v: current accumulator value
-- x: current item in list
-- i: [ 1-based index in list ] optional
-- l: [ a reference to the list itself ] optional
on reduceRight(xs, f, a)
set sf to sReturn(f)
repeat with i from length of xs to 1 by -1
set a to sf's call(a, item i of xs, i, xs)
end repeat
end reduceRight
-- Unit/Return and bind for composing handlers in script wrappers
-- lift 2nd class function into 1st class wrapper
-- handler function --> first class script object
on sReturn(f)
script
property call : f
end script
end sReturn
-- return a new script in which function g is composed
-- with the f (call()) of the Mf script
-- Mf -> (f -> Mg) -> Mg
on sBind(mf, g)
script
on call(x)
sReturn(g)'s call(mf's call(x))
end call
end script
end sBind |
http://rosettacode.org/wiki/Monads/List_monad | Monads/List monad | A Monad is a combination of a data-type with two helper functions written for that type.
The data-type can be of any kind which can contain values of some other type – common examples are lists, records, sum-types, even functions or IO streams. The two special functions, mathematically known as eta and mu, but usually given more expressive names like 'pure', 'return', or 'yield' and 'bind', abstract away some boilerplate needed for pipe-lining or enchaining sequences of computations on values held in the containing data-type.
The bind operator in the List monad enchains computations which return their values wrapped in lists. One application of this is the representation of indeterminacy, with returned lists representing a set of possible values. An empty list can be returned to express incomputability, or computational failure.
A sequence of two list monad computations (enchained with the use of bind) can be understood as the computation of a cartesian product.
The natural implementation of bind for the List monad is a composition of concat and map, which, used with a function which returns its value as a (possibly empty) list, provides for filtering in addition to transformation or mapping.
Demonstrate in your programming language the following:
Construct a List Monad by writing the 'bind' function and the 'pure' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)
Make two functions, each which take a number and return a monadic number, e.g. Int -> List Int and Int -> List String
Compose the two functions with bind | #AppleScript | AppleScript | -- MONADIC FUNCTIONS (for list monad) ------------------------------------------
-- Monadic bind for lists is simply ConcatMap
-- which applies a function f directly to each value in the list,
-- and returns the set of results as a concat-flattened list
-- bind :: (a -> [b]) -> [a] -> [b]
on bind(f, xs)
-- concat :: a -> a -> [a]
script concat
on |λ|(a, b)
a & b
end |λ|
end script
foldl(concat, {}, map(f, xs))
end bind
-- Monadic return/unit/inject for lists: just wraps a value in a list
-- a -> [a]
on unit(a)
[a]
end unit
-- TEST ------------------------------------------------------------------------
on run
-- Pythagorean triples drawn from integers in the range [1..n]
-- {(x, y, z) | x <- [1..n], y <- [x+1..n], z <- [y+1..n], (x^2 + y^2 = z^2)}
pythagoreanTriples(25)
--> {{3, 4, 5}, {5, 12, 13}, {6, 8, 10}, {7, 24, 25}, {8, 15, 17},
-- {9, 12, 15}, {12, 16, 20}, {15, 20, 25}}
end run
-- pythagoreanTriples :: Int -> [(Int, Int, Int)]
on pythagoreanTriples(maxInteger)
script X
on |λ|(X)
script Y
on |λ|(Y)
script Z
on |λ|(Z)
if X * X + Y * Y = Z * Z then
unit([X, Y, Z])
else
[]
end if
end |λ|
end script
bind(Z, enumFromTo(1 + Y, maxInteger))
end |λ|
end script
bind(Y, enumFromTo(1 + X, maxInteger))
end |λ|
end script
bind(X, enumFromTo(1, maxInteger))
end pythagoreanTriples
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if n < m then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn |
http://rosettacode.org/wiki/Multi-dimensional_array | Multi-dimensional array | For the purposes of this task, the actual memory layout or access method of this data structure is not mandated.
It is enough to:
State the number and extent of each index to the array.
Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.
Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.
Task
State if the language supports multi-dimensional arrays in its syntax and usual implementation.
State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.
Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array.
The idiomatic method for the language is preferred.
The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, if indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).
State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.
If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.
Show all output here, (but you may judiciously use ellipses to shorten repetitive output text).
| #JavaScript | JavaScript | function array() {
var dimensions= Array.prototype.slice.call(arguments);
var N=1, rank= dimensions.length;
for (var j= 0; j<rank; j++) N*= dimensions[j];
this.dimensions= dimensions;
this.values= new Array(N);
} |
http://rosettacode.org/wiki/Multiplication_tables | Multiplication tables | Task
Produce a formatted 12×12 multiplication table of the kind memorized by rote when in primary (or elementary) school.
Only print the top half triangle of products.
| #360_Assembly | 360 Assembly | * 12*12 multiplication table 14/08/2015
MULTTABL CSECT
USING MULTTABL,R12
LR R12,R15
LA R10,0 buffer pointer
LA R3,BUFFER
MVC 0(4,R3),=C' | '
LA R10,4(R10)
LA R5,12
LA R4,1 i=1
LOOPN LA R3,BUFFER do i=1 to 12
AR R3,R10
XDECO R4,XDEC i
MVC 0(4,R3),XDEC+8 output i
LA R10,4(R10)
LA R4,1(R4)
BCT R5,LOOPN end i
XPRNT BUFFER,52
XPRNT PORT,52 border
LA R5,12
LA R4,1 i=1 (R4)
LOOPI LA R10,0 do i=1 to 12
MVC BUFFER,=CL52' '
LA R3,BUFFER
AR R3,R10
XDECO R4,XDEC
MVC 0(2,R3),XDEC+10
LA R10,2(R10)
LA R3,BUFFER
AR R3,R10
MVC 0(2,R3),=C'| '
LA R10,2(R10)
LA R7,12
LA R6,1 j=1 (R6)
LOOPJ CR R6,R4 do j=1 to 12
BNL MULT
LA R3,BUFFER
AR R3,R10
MVC 0(4,R3),=C' '
LA R10,4(R10)
B NEXTJ
MULT LR R9,R4 i
MR R8,R6 i*j in R8R9
LA R3,BUFFER
AR R3,R10
XDECO R9,XDEC
MVC 0(4,R3),XDEC+8
LA R10,4(R10)
NEXTJ LA R6,1(R6)
BCT R7,LOOPJ end j
ELOOPJ XPRNT BUFFER,52
LA R4,1(R4)
BCT R5,LOOPI end i
ELOOPI XR R15,R15
BR R14
BUFFER DC CL52' '
XDEC DS CL12
PORT DC C'--+-------------------------------------------------'
YREGS
END MULTTABL |
http://rosettacode.org/wiki/Multiple_regression | Multiple regression | Task
Given a set of data vectors in the following format:
y
=
{
y
1
,
y
2
,
.
.
.
,
y
n
}
{\displaystyle y=\{y_{1},y_{2},...,y_{n}\}\,}
X
i
=
{
x
i
1
,
x
i
2
,
.
.
.
,
x
i
n
}
,
i
∈
1..
k
{\displaystyle X_{i}=\{x_{i1},x_{i2},...,x_{in}\},i\in 1..k\,}
Compute the vector
β
=
{
β
1
,
β
2
,
.
.
.
,
β
k
}
{\displaystyle \beta =\{\beta _{1},\beta _{2},...,\beta _{k}\}}
using ordinary least squares regression using the following equation:
y
j
=
Σ
i
β
i
⋅
x
i
j
,
j
∈
1..
n
{\displaystyle y_{j}=\Sigma _{i}\beta _{i}\cdot x_{ij},j\in 1..n}
You can assume y is given to you as a vector (a one-dimensional array), and X is given to you as a two-dimensional array (i.e. matrix).
| #Haskell | Haskell | import Numeric.LinearAlgebra
import Numeric.LinearAlgebra.LAPACK
m :: Matrix Double
m = (3><3)
[7.589183,1.703609,-4.477162,
-4.597851,9.434889,-6.543450,
0.4588202,-6.115153,1.331191]
v :: Matrix Double
v = (3><1)
[1.745005,-4.448092,-4.160842] |
http://rosettacode.org/wiki/Multifactorial | Multifactorial | The factorial of a number, written as
n
!
{\displaystyle n!}
, is defined as
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
.
Multifactorials generalize factorials as follows:
n
!
=
n
(
n
−
1
)
(
n
−
2
)
.
.
.
(
2
)
(
1
)
{\displaystyle n!=n(n-1)(n-2)...(2)(1)}
n
!
!
=
n
(
n
−
2
)
(
n
−
4
)
.
.
.
{\displaystyle n!!=n(n-2)(n-4)...}
n
!
!
!
=
n
(
n
−
3
)
(
n
−
6
)
.
.
.
{\displaystyle n!!!=n(n-3)(n-6)...}
n
!
!
!
!
=
n
(
n
−
4
)
(
n
−
8
)
.
.
.
{\displaystyle n!!!!=n(n-4)(n-8)...}
n
!
!
!
!
!
=
n
(
n
−
5
)
(
n
−
10
)
.
.
.
{\displaystyle n!!!!!=n(n-5)(n-10)...}
In all cases, the terms in the products are positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold:
Write a function that given n and the degree, calculates the multifactorial.
Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
| #Clojure | Clojure | (defn !! [m n]
(->> (iterate #(- % m) n) (take-while pos?) (apply *)))
(doseq [m (range 1 6)]
(prn m (map #(!! m %) (range 1 11)))) |
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #EchoLisp | EchoLisp |
(lib 'plot)
(plot-x-minmax 10) ; set logical dimensions of plotting area
(plot-y-minmax 100)
→ (("x" 0 10) ("y" 0 100))
;; press ESC to see the canvas
;; the mouse position is displayed as , for example, [ x: 5.6 y : 88.7]
;; 0 <= x <= 10, 0 <= y <= 100
|
http://rosettacode.org/wiki/Mouse_position | Mouse position | Task
Get the current location of the mouse cursor relative to the active window.
Please specify if the window may be externally created.
| #Elm | Elm | import Graphics.Element exposing (Element, show)
import Mouse
main : Signal Element
main =
Signal.map show Mouse.position |
http://rosettacode.org/wiki/N-queens_problem | N-queens problem |
Solve the eight queens puzzle.
You can extend the problem to solve the puzzle with a board of size NxN.
For the number of solutions for small values of N, see OEIS: A000170.
Related tasks
A* search algorithm
Solve a Hidato puzzle
Solve a Holy Knight's tour
Knight's tour
Peaceful chess queen armies
Solve a Hopido puzzle
Solve a Numbrix puzzle
Solve the no connection puzzle
| #Factor | Factor | USING: kernel sequences math math.combinatorics formatting io locals ;
IN: queens
: /= ( x y -- ? ) = not ; inline
:: safe? ( board q -- ? )
[let q board nth :> x
q <iota> [
x swap
[ board nth ] keep
q swap -
[ + /= ]
[ - /= ] 3bi and
] all?
] ;
: solution? ( board -- ? )
dup length <iota> [ dupd safe? ] all? nip ;
: queens ( n -- l )
<iota> all-permutations [ solution? ] filter ;
: .queens ( n -- )
queens
[
[ 1 + "%d " printf ] each nl
] each ; |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
include "bigint.s7i";
const type: oneFactor is new struct
var bigInteger: prime is 0_;
var integer: exp is 0;
end struct;
const func oneFactor: oneFactor (in bigInteger: prime, in integer: exp) is func
result
var oneFactor: aFactor is oneFactor.value;
begin
aFactor.prime := prime;
aFactor.exp := exp;
end func;
const func array oneFactor: factor (in var bigInteger: n) is func
result
var array oneFactor: pf is 0 times oneFactor.value;
local
var integer: e is 0;
var bigInteger: d is 0_;
var bigInteger: s is 0_;
begin
e := lowestSetBit(n);
if e > 0 then
n >>:= e;
pf := [] (oneFactor(2_, e));
end if;
s := sqrt(n);
d := 3_;
while n > 1_ do
if d > s then
d := n;
end if;
e := 0;
while n rem d = 0_ do
n := n div d;
incr(e);
end while;
if e > 0 then
pf &:= oneFactor(d, e);
s := sqrt(n);
end if;
d +:= 2_;
end while;
end func;
const func bigInteger: moBachShallit58(in bigInteger: a, in bigInteger: n, in array oneFactor: pf) is func
result
var bigInteger: mo is 0_;
local
var bigInteger: n1 is 0_;
var oneFactor: pe is oneFactor.value;
var bigInteger: x is 0_;
var bigInteger: y is 0_;
var integer: o is 0;
var bigInteger: o1 is 0_;
begin
n1 := n - 1_;
mo := 1_;
for pe range pf do
y := n1 div pe.prime ** pe.exp;
x := modPow(a, y, n);
o := 0;
while x > 1_ do
x := modPow(x, pe.prime, n);
incr(o);
end while;
o1 := pe.prime ** o;
mo *:= o1 div gcd(mo, o1);
end for;
end func;
const func boolean: isProbablyPrime (in bigInteger: primeCandidate, in var integer: count) is func
result
var boolean: isProbablyPrime is TRUE;
local
var bigInteger: aRandomNumber is 0_;
begin
while isProbablyPrime and count > 0 do
aRandomNumber := rand(1_, pred(primeCandidate));
isProbablyPrime := modPow(aRandomNumber, pred(primeCandidate), primeCandidate) = 1_;
decr(count);
end while;
# writeln(count);
end func;
const proc: moTest (in bigInteger: a, in bigInteger: n) is func
begin
if bitLength(a) < 100 then
write("ord(" <& a <& ")");
else
write("ord([big])");
end if;
if bitLength(n) < 100 then
write(" mod " <& n <& " ");
else
write(" mod [big] ");
end if;
if not isProbablyPrime(n, 20) then
writeln("not computed. modulus must be prime for this algorithm.")
else
writeln("= " <& moBachShallit58(a, n, factor(n - 1_)));
end if;
end func;
const proc: main is func
local
var bigInteger: b is 100_;
begin
moTest(37_, 3343_);
moTest(10_ ** 100 + 1_, 7919_);
moTest(10_ ** 1000 + 1_, 15485863_);
moTest(10_ ** 10000 - 1_, 22801763489_);
moTest(1511678068_, 7379191741_);
moTest(3047753288_, 2257683301_);
end func; |
http://rosettacode.org/wiki/Multiplicative_order | Multiplicative order | The multiplicative order of a relative to m is the least positive integer n such that a^n is 1 (modulo m).
Example
The multiplicative order of 37 relative to 1000 is 100 because 37^100 is 1 (modulo 1000), and no number smaller than 100 would do.
One possible algorithm that is efficient also for large numbers is the following: By the Chinese Remainder Theorem, it's enough to calculate the multiplicative order for each prime exponent p^k of m, and
combine the results with the least common multiple operation.
Now the order of a with regard to p^k must divide Φ(p^k). Call this number t, and determine it's factors q^e. Since each multiple of the order will also yield 1 when used as exponent for a, it's enough to find the least d such that (q^d)*(t/(q^e)) yields 1 when used as exponent.
Task
Implement a routine to calculate the multiplicative order along these lines. You may assume that routines to determine the factorization into prime powers are available in some library.
An algorithm for the multiplicative order can be found in Bach & Shallit, Algorithmic Number Theory, Volume I: Efficient Algorithms, The MIT Press, 1996:
Exercise 5.8, page 115:
Suppose you are given a prime p and a complete factorization
of p-1. Show how to compute the order of an
element a in (Z/(p))* using O((lg p)4/(lg lg p)) bit
operations.
Solution, page 337:
Let the prime factorization of p-1 be q1e1q2e2...qkek . We use the following observation:
if x^((p-1)/qifi) = 1 (mod p) ,
and fi=ei or x^((p-1)/qifi+1) != 1 (mod p) , then qiei-fi||ordp x. (This follows by combining Exercises 5.1 and 2.10.)
Hence it suffices to find, for each i , the exponent fi such that the condition above holds.
This can be done as follows: first compute q1e1, q2e2, ... ,
qkek . This can be done using O((lg p)2) bit operations. Next, compute y1=(p-1)/q1e1, ... , yk=(p-1)/qkek .
This can be done using O((lg p)2) bit operations. Now, using the binary method,
compute x1=ay1(mod p), ... , xk=ayk(mod p) .
This can be done using O(k(lg p)3) bit operations, and k=O((lg p)/(lg lg p)) by Theorem 8.8.10.
Finally, for each i , repeatedly raise xi to the qi-th power (mod p) (as many as ei-1 times), checking to see when 1 is obtained.
This can be done using O((lg p)3) steps.
The total cost is dominated by O(k(lg p)3) , which is O((lg p)4/(lg lg p)).
| #Sidef | Sidef | say 37.znorder(1000) #=> 100
say 54.znorder(100001) #=> 9090 |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Sather | Sather | class MATH is
nthroot(n:INT, a:FLT):FLT
pre n > 0
is
x0 ::= a / n.flt;
m ::= n - 1;
loop
x1 ::= (m.flt * x0 + a/(x0^(m.flt))) / n.flt;
if (x1 - x0).abs < (x0 * 1.0e-9).abs then
return x1;
end;
x0 := x1;
end;
end;
end; |
http://rosettacode.org/wiki/Nth_root | Nth root | Task
Implement the algorithm to compute the principal nth root
A
n
{\displaystyle {\sqrt[{n}]{A}}}
of a positive real number A, as explained at the Wikipedia page.
| #Scala | Scala | def nroot(n: Int, a: Double): Double = {
@tailrec
def rec(x0: Double) : Double = {
val x1 = ((n - 1) * x0 + a/math.pow(x0, n-1))/n
if (x0 <= x1) x0 else rec(x1)
}
rec(a)
} |
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #MATLAB | MATLAB | function s = nth(n)
tens = mod(n, 100);
if tens > 9 && tens < 20
suf = 'th';
else
switch mod(n, 10)
case 1
suf = 'st';
case 2
suf = 'nd';
case 3
suf = 'rd';
otherwise
suf = 'th';
end
end
s = sprintf('%d%s', n, suf);
end |
http://rosettacode.org/wiki/Munchausen_numbers | Munchausen numbers | A Munchausen number is a natural number n the sum of whose digits (in base 10), each raised to the power of itself, equals n.
(Munchausen is also spelled: Münchhausen.)
For instance: 3435 = 33 + 44 + 33 + 55
Task
Find all Munchausen numbers between 1 and 5000.
Also see
The OEIS entry: A046253
The Wikipedia entry: Perfect digit-to-digit invariant, redirected from Munchausen Number
| #MAD | MAD | NORMAL MODE IS INTEGER
DIMENSION P(5)
THROUGH CLCPOW, FOR D=0, 1, D.G.5
P(D) = D
THROUGH CLCPOW, FOR X=1, 1, X.GE.D
CLCPOW P(D) = P(D) * D
THROUGH TEST, FOR D1=0, 1, D1.G.5
THROUGH TEST, FOR D2=0, 1, D2.G.5
THROUGH TEST, FOR D3=0, 1, D3.G.5
THROUGH TEST, FOR D4=1, 1, D4.G.5
N = D1*1000 + D2*100 + D3*10 + D4
WHENEVER P(D1)+P(D2)+P(D3)+P(D4) .E. N
PRINT FORMAT FMT,N
TEST END OF CONDITIONAL
VECTOR VALUES FMT = $I4*$
END OF PROGRAM |
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
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{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #F.C5.8Drmul.C3.A6 | Fōrmulæ | package main
import "fmt"
func F(n int) int {
if n == 0 { return 1 }
return n - M(F(n-1))
}
func M(n int) int {
if n == 0 { return 0 }
return n - F(M(n-1))
}
func main() {
for i := 0; i < 20; i++ {
fmt.Printf("%2d ", F(i))
}
fmt.Println()
for i := 0; i < 20; i++ {
fmt.Printf("%2d ", M(i))
}
fmt.Println()
} |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Oz | Oz | declare
Xs = {MakeList 5} %% a list of 5 unbound variables
in
{ForAll Xs OS.rand} %% fill it with random numbers (CORRECT)
{Show Xs} |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Pascal | Pascal | (Foo->new) x $n
# here Foo->new can be any expression that returns a reference representing
# a new object |
http://rosettacode.org/wiki/Multiple_distinct_objects | Multiple distinct objects | Create a sequence (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
By initialized we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to "zero" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of "zero" for that type.
This task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the same mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.
This task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).
See also: Closures/Value capture
| #Perl | Perl | (Foo->new) x $n
# here Foo->new can be any expression that returns a reference representing
# a new object |
http://rosettacode.org/wiki/Motzkin_numbers | Motzkin numbers | Definition
The nth Motzkin number (denoted by M[n]) is the number of different ways of drawing non-intersecting chords between n points on a circle (not necessarily touching every point by a chord).
By convention M[0] = 1.
Task
Compute and show on this page the first 42 Motzkin numbers or, if your language does not support 64 bit integers, as many such numbers as you can. Indicate which of these numbers are prime.
See also
oeis:A001006 Motzkin numbers
| #J | J | nextMotzkin=: , (({:*1+2*#) + _2&{*3*#-1:)%2+# |
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.