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/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Octave | Octave | function p = montepi(samples)
in_circle = 0;
for samp = 1:samples
v = [ unifrnd(-1,1), unifrnd(-1,1) ];
if ( v*v.' <= 1.0 )
in_circle++;
endif
endfor
p = 4*in_circle/samples;
endfunction
l = 1e4;
while (l < 1e7)
disp(montepi(l));
l *= 10;
endwhile |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #PARI.2FGP | PARI/GP | MonteCarloPi(tests)=4.*sum(i=1,tests,norml2([random(1.),random(1.)])<1)/tests; |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Liberty_BASIC | Liberty BASIC | 'The following code relies on the Windows API
Input "Input the text to translate to Morse Code... "; string$
Print PlayMorse$(string$)
End
Function PlayMorse$(string$)
'LetterGap = (3 * BaseTime)
'WordGap = (7 * BaseTime)
BaseTime = 50
freq = 1250
PlayMorse$ = TranslateToMorse$(string$)
morseCode$ = "./-"
For i = 1 To Len(PlayMorse$)
Scan
dwDuration = (Instr(morseCode$, Mid$(PlayMorse$, i, 1)) * BaseTime)
If (Mid$(PlayMorse$, i, 1) <> " ") Then
CallDLL #kernel32, "Beep", freq As ulong, dwDuration As ulong, ret As long
CallDLL #kernel32, "Sleep", BaseTime As long, ret As void
End If
If (Mid$(PlayMorse$, i, 1) <> " ") Then
sleep = (3 * BaseTime)
Else
sleep = (7 * BaseTime)
End If
CallDLL #kernel32, "Sleep", sleep As long, ret As void
Next i
End Function
Function TranslateToMorse$(string$)
string$ = Upper$(string$)
For i = 1 To Len(string$)
While desc$ <> "End"
Read desc$, value$
If desc$ = "" Then desc$ = chr$(34)
If desc$ = Mid$(string$, i, 1) Then
If Mid$(string$, i, 1) <> " " Then value$ = " " + value$
TranslateToMorse$ = TranslateToMorse$ + value$
Exit While
End If
Wend
If desc$ = "End" Then Notice Mid$(string$, i, 1) + " is not accounted for in the Morse Code Table."
Restore
Next i
TranslateToMorse$ = Trim$(TranslateToMorse$)
Data "A", ".-", "B", "-...", "C", "-.-.", "D", "-..", "E", ".", "F", "..-.", "G", "--."
Data "H", "....", "I", "..", "J", ".---", "K", "-.-", "L", ".-..", "M", "--", "N", "-."
Data "O", "---", "P", ".--.", "Q", "--.-", "R", ".-.", "S", "...", "T", "-", "U", "..-"
Data "V", "...-", "W", ".--", "X", "-..-", "Y", "-.--", "Z", "--..", "Á", "--.-", "Ä", ".-.-"
Data "É", "..-..", "Ñ", "--.--", "Ö", "---.", "Ü", "..--", "1", ".----", "2", "..---"
Data "3", "...--", "4", "....-", "5", ".....", "6", "-....", "7", "--...", "8", "---.."
Data "9", "----.", "0", "-----", ",", "--..--", ".", ".-.-.-", "?", "..--..", ";", "-.-.-"
Data ":", "---...", "/", "-..-.", "-", "-....-", "'", ".----.", "+", ".-.-.", "", ".-..-."
Data "@", ".--.-.", "(", "-.--.", ")", "-.--.-", "_", "..--.-", "$", "...-..-", "&", ".-..."
Data "=", "-...-", "!", "..--.", " ", " ", "End", ""
End Function |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Io | Io | keepWins := 0
switchWins := 0
doors := 3
times := 100000
pickDoor := method(excludeA, excludeB,
door := excludeA
while(door == excludeA or door == excludeB,
door = (Random value() * doors) floor
)
door
)
times repeat(
playerChoice := pickDoor()
carDoor := pickDoor()
shownDoor := pickDoor(carDoor, playerChoice)
switchDoor := pickDoor(playerChoice, shownDoor)
(playerChoice == carDoor) ifTrue(keepWins = keepWins + 1)
(switchDoor == carDoor) ifTrue(switchWins = switchWins + 1)
)
("Switching to the other door won #{switchWins} times.\n"\
.. "Keeping the same door won #{keepWins} times.\n"\
.. "Game played #{times} times with #{doors} doors.") interpolate println
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #PicoLisp | PicoLisp | (de modinv (A B)
(let (B0 B X0 0 X1 1 Q 0 T1 0)
(while (< 1 A)
(setq
Q (/ A B)
T1 B
B (% A B)
A T1
T1 X0
X0 (- X1 (* Q X0))
X1 T1 ) )
(if (lt0 X1) (+ X1 B0) X1) ) )
(println
(modinv 42 2017) )
(bye) |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #PL.2FI | PL/I | *process source attributes xref or(!);
/*--------------------------------------------------------------------
* 13.07.2015 Walter Pachl
*-------------------------------------------------------------------*/
minv: Proc Options(main);
Dcl (x,y) Bin Fixed(31);
x=42;
y=2017;
Put Edit('modular inverse of',x,' by ',y,' ---> ',modinv(x,y))
(Skip,3(a,f(4)));
modinv: Proc(a,b) Returns(Bin Fixed(31));
Dcl (a,b,ob,ox,d,t) Bin Fixed(31);
ob=b;
ox=0;
d=1;
If b=1 Then;
Else Do;
Do While(a>1);
q=a/b;
r=mod(a,b);
a=b;
b=r;
t=ox;
ox=d-q*ox;
d=t;
End;
End;
If d<0 Then
d=d+ob;
Return(d);
End;
End; |
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.
| #EchoLisp | EchoLisp |
(lib 'matrix)
(define (mtable i j)
(cond
((and (zero? i) (zero? j)) "😅")
((= i 0) j)
((= j 0) i)
((>= j i ) (* i j ))
(else " ")))
(array-print (build-array 13 13 mtable))
|
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.
| #Seed7 | Seed7 | $ include "seed7_05.s7i";
const func integer: multiFact (in var integer: num, in integer: degree) is func
result
var integer: multiFact is 1;
begin
while num > 1 do
multiFact *:= num;
num -:= degree;
end while;
end func;
const proc: main is func
local
var integer: degree is 0;
var integer: num is 0;
begin
for degree range 1 to 5 do
write("Degree " <& degree <& ": ");
for num range 1 to 10 do
write(multiFact(num, degree) <& " ");
end for;
writeln;
end for;
end func; |
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.
| #Sidef | Sidef | func mfact(s, n) {
n > 0 ? (n * mfact(s, n-s)) : 1
}
{ |s|
say "step=#{s}: #{{|n| mfact(s, n)}.map(1..10).join(' ')}"
} << 1..10 |
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
| #Phix | Phix | with javascript_semantics
--
-- demo\rosetta\n_queens.exw
-- =========================
--
sequence co, -- columns occupied
-- (ro is implicit)
fd, -- forward diagonals
bd, -- backward diagonals
board
atom count
procedure solve(integer row, integer N, integer show)
for col=1 to N do
if not co[col] then
integer fdi = col+row-1,
bdi = row-col+N
if not fd[fdi]
and not bd[bdi] then
board[row][col] = 'Q'
co[col] = 1
fd[fdi] = 1
bd[bdi] = 1
if row=N then
if show then
puts(1,join(board,"\n")&"\n")
puts(1,repeat('=',N)&"\n")
end if
count += 1
else
solve(row+1,N,show)
end if
board[row][col] = '.'
co[col] = 0
fd[fdi] = 0
bd[bdi] = 0
end if
end if
end for
end procedure
procedure n_queens(integer N=8, integer show=1)
co = repeat(0,N)
fd = repeat(0,N*2-1)
bd = repeat(0,N*2-1)
board = repeat(repeat('.',N),N)
count = 0
solve(1,N,show)
printf(1,"%d queens: %d solutions\n",{N,count})
end procedure
for N=1 to iff(platform()=JS?12:14) do
n_queens(N,N<5)
end for
|
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.
| #TypeScript | TypeScript |
// N'th
function suffix(n: number): string {
var nMod10: number = n % 10;
var nMod100: number = n % 100;
if (nMod10 == 1 && nMod100 != 11)
return "st";
else if (nMod10 == 2 && nMod100 != 12)
return "nd";
else if (nMod10 == 3 && nMod100 != 13)
return "rd";
else
return "th";
}
function printImages(loLim: number, hiLim: number) {
for (i = loLim; i <= hiLim; i++)
process.stdout.write(`${i}` + suffix(i) + " ");
process.stdout.write("\n");
}
printImages( 0, 25);
printImages( 250, 265);
printImages(1000, 1025);
|
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).
| #Ol | Ol |
(letrec ((F (lambda (n)
(if (= n 0) 1
(- n (M (F (- n 1)))))))
(M (lambda (n)
(if (= n 0) 0
(- n (F (M (- n 1))))))))
(print (F 19)))
; produces 12
|
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Pascal | Pascal | Program MonteCarlo(output);
uses
Math;
function MC_Pi(expo: integer): real;
var
x, y: real;
i, hits, samples: longint;
begin
samples := 10**expo;
hits := 0;
randomize;
for i := 1 to samples do
begin
x := random;
y := random;
if sqrt(x*x + y*y) < 1.0 then
inc(hits);
end;
MC_Pi := 4.0 * hits / samples;
end;
var
i: integer;
begin
for i := 4 to 8 do
writeln (10**i, ' samples give ', MC_Pi(i):7:5, ' as pi.');
end.
|
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Lua | Lua | local M = {}
-- module-local variables
local BUZZER = pio.PB_10
local dit_length, dah_length, word_length
-- module-local functions
local buzz, dah, dit, init, inter_element_gap, medium_gap, pause, sequence, short_gap
buzz = function(duration)
pio.pin.output(BUZZER)
pio.pin.setlow(BUZZER)
tmr.delay(tmr.SYS_TIMER, duration)
pio.pin.sethigh(BUZZER)
pio.pin.input(BUZZER)
end
dah = function()
buzz(dah_length)
end
dit = function()
buzz(dit_length)
end
init = function(baseline)
dit_length = baseline
dah_length = 2 * baseline
word_length = 4 * baseline
end
inter_element_gap = function()
pause(dit_length)
end
medium_gap = function()
pause(word_length)
end
pause = function(duration)
tmr.delay(tmr.SYS_TIMER, duration)
end
sequence = function(codes)
if codes then
for _,f in ipairs(codes) do
f()
inter_element_gap()
end
short_gap()
end
end
short_gap = function()
pause(dah_length)
end
local morse = {
a = { dit, dah }, b = { dah, dit, dit, dit }, c = { dah, dit, dah, dit },
d = { dah, dit, dit }, e = { dit }, f = { dit, dit, dah, dit },
g = { dah, dah, dit }, h = { dit, dit, dit ,dit }, i = { dit, dit },
j = { dit, dah, dah, dah }, k = { dah, dit, dah }, l = { dit, dah, dit, dit },
m = { dah, dah }, n = { dah, dit }, o = { dah, dah, dah },
p = { dit, dah, dah, dit }, q = { dah, dah, dit, dah }, r = { dit, dah, dit },
s = { dit, dit, dit }, t = { dah }, u = { dit, dit, dah },
v = { dit, dit, dit, dah }, w = { dit, dah, dah }, x = { dah, dit, dit, dah },
y = { dah, dit, dah, dah }, z = { dah, dah, dit, dit },
["0"] = { dah, dah, dah, dah, dah }, ["1"] = { dit, dah, dah, dah, dah },
["2"] = { dit, dit, dah, dah, dah }, ["3"] = { dit, dit, dit, dah, dah },
["4"] = { dit, dit, dit, dit, dah }, ["5"] = { dit, dit, dit, dit, dit },
["6"] = { dah, dit, dit, dit, dit }, ["7"] = { dah, dah, dit, dit, dit },
["8"] = { dah, dah, dah, dit, dit }, ["9"] = { dah, dah, dah, dah, dit },
[" "] = { medium_gap }
}
-- public interface
M.beep = function(message)
message = message:lower()
for _,ch in ipairs { message:byte(1, #message) } do
sequence(morse[string.char(ch)])
end
end
M.set_dit = function(duration)
init(duration)
end
-- initialization code
init(50000)
return M |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #J | J | pick=: {~ ?@# |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Java | Java | import java.util.Random;
public class Monty{
public static void main(String[] args){
int switchWins = 0;
int stayWins = 0;
Random gen = new Random();
for(int plays = 0;plays < 32768;plays++ ){
int[] doors = {0,0,0};//0 is a goat, 1 is a car
doors[gen.nextInt(3)] = 1;//put a winner in a random door
int choice = gen.nextInt(3); //pick a door, any door
int shown; //the shown door
do{
shown = gen.nextInt(3);
//don't show the winner or the choice
}while(doors[shown] == 1 || shown == choice);
stayWins += doors[choice];//if you won by staying, count it
//the switched (last remaining) door is (3 - choice - shown), because 0+1+2=3
switchWins += doors[3 - choice - shown];
}
System.out.println("Switching wins " + switchWins + " times.");
System.out.println("Staying wins " + stayWins + " times.");
}
} |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #PowerShell | PowerShell | function invmod($a,$n){
if ([int]$n -lt 0) {$n = -$n}
if ([int]$a -lt 0) {$a = $n - ((-$a) % $n)}
$t = 0
$nt = 1
$r = $n
$nr = $a % $n
while ($nr -ne 0) {
$q = [Math]::truncate($r/$nr)
$tmp = $nt
$nt = $t - $q*$nt
$t = $tmp
$tmp = $nr
$nr = $r - $q*$nr
$r = $tmp
}
if ($r -gt 1) {return -1}
if ($t -lt 0) {$t += $n}
return $t
}
invmod 42 2017 |
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.
| #Elixir | Elixir | defmodule RC do
def multiplication_tables(n) do
IO.write " X |"
Enum.each(1..n, fn i -> :io.fwrite("~4B", [i]) end)
IO.puts "\n---+" <> String.duplicate("----", n)
Enum.each(1..n, fn j ->
:io.fwrite("~2B |", [j])
Enum.each(1..n, fn i ->
if i<j, do: (IO.write " "), else: :io.fwrite("~4B", [i*j])
end)
IO.puts ""
end)
end
end
RC.multiplication_tables(12) |
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.
| #Swift | Swift | func multiFactorial(_ n: Int, k: Int) -> Int {
return stride(from: n, to: 0, by: -k).reduce(1, *)
}
let multis = (1...5).map({degree in
(1...10).map({member in
multiFactorial(member, k: degree)
})
})
for (i, degree) in multis.enumerated() {
print("Degree \(i + 1): \(degree)")
} |
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.
| #Tcl | Tcl | package require Tcl 8.6
proc mfact {n m} {
set mm [expr {-$m}]
for {set r $n} {[incr n $mm] > 1} {set r [expr {$r * $n}]} {}
return $r
}
foreach n {1 2 3 4 5 6 7 8 9 10} {
puts $n:[join [lmap m {1 2 3 4 5 6 7 8 9 10} {mfact $m $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
| #PHP | PHP |
<html>
<head>
<title>
n x n Queen solving program
</title>
</head>
<body>
<?php
echo "<h1>n x n Queen solving program</h1>";
//Get the size of the board
$boardX = $_POST['boardX'];
$boardY = $_POST['boardX'];
// Function to rotate a board 90 degrees
function rotateBoard($p, $boardX) {
$a=0;
while ($a < count($p)) {
$b = strlen(decbin($p[$a]))-1;
$tmp[$b] = 1 << ($boardX - $a - 1);
++$a;
}
ksort($tmp);
return $tmp;
}
// This function will find rotations of a solution
function findRotation($p, $boardX,$solutions){
$tmp = rotateBoard($p,$boardX);
// Rotated 90
if (in_array($tmp,$solutions)) {}
else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX);
// Rotated 180
if (in_array($tmp,$solutions)){}
else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX);
// Rotated 270
if (in_array($tmp,$solutions)){}
else {$solutions[] = $tmp;}
// Reflected
$tmp = array_reverse($p);
if (in_array($tmp,$solutions)){}
else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX);
// Reflected and Rotated 90
if (in_array($tmp,$solutions)){}
else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX);
// Reflected and Rotated 180
if (in_array($tmp,$solutions)){}
else {$solutions[] = $tmp;}
$tmp = rotateBoard($tmp,$boardX);
// Reflected and Rotated 270
if (in_array($tmp,$solutions)){}
else {$solutions[] = $tmp;}
return $solutions;
}
// This is a function which will render the board
function renderBoard($p,$boardX) {
$img = 'data:image/png;base64,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';
echo "<table border=1 cellspacing=0 style='text-align:center;display:inline'>";
for ($y = 0; $y < $boardX; ++$y) {
echo '<tr>';
for ($x = 0; $x < $boardX; ++$x){
if (($x+$y) & 1) { $cellCol = '#9C661F';}
else {$cellCol = '#FCE6C9';}
if ($p[$y] == 1 << $x) { echo "<td bgcolor=".$cellCol."><img width=30 height=30 src='".$img."'></td>";}
else { echo "<td bgcolor=".$cellCol."> </td>";}
}
echo '<tr>';
}
echo '<tr></tr></table> ';
}
//This function allows me to generate the next order of rows.
function pc_next_permutation($p) {
$size = count($p) - 1;
// slide down the array looking for where we're smaller than the next guy
for ($i = $size - 1; $p[$i] >= $p[$i+1]; --$i) { }
// if this doesn't occur, we've finished our permutations
// the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
if ($i == -1) { return false; }
// slide down the array looking for a bigger number than what we found before
for ($j = $size; $p[$j] <= $p[$i]; --$j) { }
// swap them
$tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp;
// now reverse the elements in between by swapping the ends
for (++$i, $j = $size; $i < $j; ++$i, --$j)
{ $tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp; }
return $p;
}
//This function needs to check the current state to see if there are any
function checkBoard($p,$boardX) {
$a = 0; //this is the row being checked
while ($a < count($p)) {
$b = 1;
while ($b < ($boardX - $a)){
$x = $p[$a+$b] << $b;
$y = $p[$a+$b] >> $b;
if ($p[$a] == $x | $p[$a] == $y) {
return false;
}
++$b;
}
++$a;
}
return true;
}
if (isset($_POST['process']) && isset($_POST['boardX']))
{
//Within here is the code that needs to be run if process is clicked.
//First I need to create the different possible rows
for ($x = 0; $x < $boardX; ++$x){
$row[$x] = 1 << $x;
}
//Now I need to create all the possible orders of rows, will be equal to [boardY]!
$solcount = 0;
$solutions = array();
while ($row != false) {
if (checkBoard($row,$boardX)){
if(!in_array($row,$solutions)){
$solutions[] = $row;
renderBoard($row,$boardX);
$solutions = findRotation($row,$boardX,$solutions);
++$solcount;
}
}
$row = pc_next_permutation($row);
}
echo "<br><br>    Rows/Columns: ".$boardX."<br>    Unique Solutions: ".$solcount."<br>    Total Solutions: ".count($solutions)." - Note: This includes symmetrical solutions<br>";
//print_r($solutions);
}
//This code collects the starting parameters
echo <<<_END
<form name="input" action="index.php" method="post">
    Number of columns/rows <select name="boardX" />
<option value="1">One</option>
<option value="2">Two</option>
<option value="3">Three</option>
<option value="4" >Four</option>
<option value="5">Five</option>
<option value="6">Six</option>
<option value="7">Seven</option>
<option value="8" selected="selected">Eight</option>
<option value="9">Nine</option>
<option value="10">Ten</option>
</select>
<input type="hidden" name="process" value="yes" />
 <input type="submit" value="Process" />
</form>
_END;
?>
</body>
</html>
|
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.
| #uBasic.2F4tH | uBasic/4tH | For x = 0 to 25 ' Test range 0..25
Push x : GoSub _PrintOrdinal
Next x : Print
For x = 250 to 265 ' Test range 250..265
Push x : GoSub _PrintOrdinal
Next x : Print
For x = 1000 to 1025 ' Test range 1000..1025
Push x : GoSub _PrintOrdinal
Next x : Print
End ' End test program
' ( n --)
_PrintOrdinal ' Ordinal subroutine
If Tos() > -1 Then ' If within range then
Print Using "____#";Tos();"'"; ' Print the number
' Take care of 11, 12 and 13
If (Tos()%100 > 10) * (Tos()%100 < 14) Then
Gosub (Pop() * 0) + 100 ' Clear stack and print "th"
Return ' We're done here
EndIf
Push Pop() % 10 ' Calculate n mod 10
GoSub 100 + 10 * ((Tos()>0) + (Tos()>1) + (Tos()>2) - (3 * (Pop()>3)))
Else ' And decide which ordinal to use
Print Pop();" is less than zero" ' Otherwise, it is an error
EndIf
Return
' Select and print proper ordinal
100 Print "th"; : Return
110 Print "st"; : Return
120 Print "nd"; : Return
130 Print "rd"; : Return |
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).
| #Order | Order | #include <order/interpreter.h>
#define ORDER_PP_DEF_8f \
ORDER_PP_FN(8fn(8N, \
8if(8is_0(8N), \
1, \
8sub(8N, 8m(8f(8dec(8N)))))))
#define ORDER_PP_DEF_8m \
ORDER_PP_FN(8fn(8N, \
8if(8is_0(8N), \
0, \
8sub(8N, 8f(8m(8dec(8N)))))))
//Test
ORDER_PP(8for_each_in_range(8fn(8N, 8print(8f(8N))), 0, 19))
ORDER_PP(8for_each_in_range(8fn(8N, 8print(8m(8N))), 0, 19)) |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Perl | Perl | sub pi {
my $nthrows = shift;
my $inside = 0;
foreach (1 .. $nthrows) {
my $x = rand() * 2 - 1;
my $y = rand() * 2 - 1;
if (sqrt($x*$x + $y*$y) < 1) {
$inside++;
}
}
return 4 * $inside / $nthrows;
}
printf "%9d: %07f\n", $_, pi($_) for 10**4, 10**6; |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | Dictionary = Join[CharacterRange["a", "z"], CharacterRange["0", "9"]];
mark = 0.1; gap = 0.125; (* gap should be equal to mark. But longer gap makes audio code easier to decode *)
shortgap = 3*gap; medgap = 7*gap;
longmark = 3*mark;
MorseDictionary = {
".-", "-...", "-.-.", "-..",
".", "..-.", "--.", "....", "..",
".---", "-.-", ".-..", "--", "-.",
"---", ".--.", "--.-", ".-.",
"...", "-", "..-", "...-", ".--",
"-..-", "-.--", "--..",
"-----", ".----", "..---", "...--", "....-", ".....",
"-....", "--...", "---..", "----."
};
MorseDictionary = # <> " " & /@ MorseDictionary; (* Force short gap silence after each letter/digit *)
Tones = {
SoundNote[None, medgap],
SoundNote[None, shortgap],
{SoundNote["C", mark, "Clarinet"], SoundNote[None, gap]},
{SoundNote["C", longmark, "Clarinet"], SoundNote[None, gap]},
{SoundNote["F#", mark, "Clarinet"], SoundNote[None, gap]} (* Use F# short mark to denote unrecognized character *)
};
codeRules = MapThread[Rule, {Dictionary, MorseDictionary}];
decodeRules = MapThread[Rule, {MorseDictionary, Dictionary}];
soundRules = MapThread[Rule, {{" ", " ", ".", "-", "?"}, Tones}];
(* The order of the rules here is important. Otherwise medium gaps and short gaps get confounded *)
morseCode[s_String] := StringReplace[ToLowerCase@s, codeRules~Join~{x_ /; FreeQ[Flatten@{Dictionary, " "}, x] -> "? "}]
morseDecode[s_String] := StringReplace[s, decodeRules]
sonicMorse[s_String] := EmitSound@Sound@Flatten[Characters@morseCode@s /. soundRules] |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #JavaScript | JavaScript |
function montyhall(tests, doors) {
'use strict';
tests = tests ? tests : 1000;
doors = doors ? doors : 3;
var prizeDoor, chosenDoor, shownDoor, switchDoor, chosenWins = 0, switchWins = 0;
// randomly pick a door excluding input doors
function pick(excludeA, excludeB) {
var door;
do {
door = Math.floor(Math.random() * doors);
} while (door === excludeA || door === excludeB);
return door;
}
// run tests
for (var i = 0; i < tests; i ++) {
// pick set of doors
prizeDoor = pick();
chosenDoor = pick();
shownDoor = pick(prizeDoor, chosenDoor);
switchDoor = pick(chosenDoor, shownDoor);
// test set for both choices
if (chosenDoor === prizeDoor) {
chosenWins ++;
} else if (switchDoor === prizeDoor) {
switchWins ++;
}
}
// results
return {
stayWins: chosenWins + ' ' + (100 * chosenWins / tests) + '%',
switchWins: switchWins + ' ' + (100 * switchWins / tests) + '%'
};
}
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Prolog | Prolog |
egcd(_, 0, 1, 0) :- !.
egcd(A, B, X, Y) :-
divmod(A, B, Q, R),
egcd(B, R, S, X),
Y is S - Q*X.
modinv(A, B, N) :-
egcd(A, B, X, Y),
A*X + B*Y =:= 1,
N is X mod B.
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #PureBasic | PureBasic | EnableExplicit
Declare main()
Declare.i mi(a.i, b.i)
If OpenConsole("MODULAR-INVERSE")
main() : Input() : End
EndIf
Macro ModularInverse(a, b)
PrintN(~"\tMODULAR-INVERSE(" + RSet(Str(a),5) + "," +
RSet(Str(b),5)+") = " +
RSet(Str(mi(a, b)),5))
EndMacro
Procedure main()
ModularInverse(42, 2017) ; = 1969
ModularInverse(40, 1) ; = 0
ModularInverse(52, -217) ; = 96
ModularInverse(-486, 217) ; = 121
ModularInverse(40, 2018) ; = -1
EndProcedure
Procedure.i mi(a.i, b.i)
Define x.i = 1,
y.i = Int(Abs(b)),
r.i = 0
If y = 1 : ProcedureReturn 0 : EndIf
While x < y
r = (a * x) % b
If r = 1 Or (y + r) = 1
Break
EndIf
x + 1
Wend
If x > y - 1 : x = -1 : EndIf
ProcedureReturn x
EndProcedure |
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.
| #Erlang | Erlang |
-module( multiplication_tables ).
-export( [print_upto/1, task/0, upto/1] ).
print_upto( N ) ->
Upto_tuples = [{X, {Y, Sum}} || {X, Y, Sum} <- upto(N)],
io:fwrite( " " ),
[io:fwrite( "~5B", [X]) || X <- lists:seq(1, N)],
io:nl(),
io:nl(),
[print_upto(X, proplists:get_all_values(X, Upto_tuples)) || X <- lists:seq(1, N)].
task() -> print_upto( 12 ).
upto( N ) -> [{X, Y, X*Y} || X <- lists:seq(1, N), Y <- lists:seq(1, N), Y >= X].
print_upto( N, Uptos ) ->
io:fwrite( "~2B", [N] ),
io:fwrite( "~*s", [5*(N - 1), " "] ),
[io:fwrite("~5B", [Sum]) || {_Y, Sum} <- Uptos],
io:nl().
|
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.
| #uBasic.2F4tH | uBasic/4tH | print "Degree | Multifactorials 1 to 10"
for x = 1 to 53 : print "-"; : next : print
for d = 1 to 5
print d;" ";"| ";
for n = 1 to 10
print FUNC(_multiFact(n, d));" ";
next
print
next
end
_multiFact param (2)
local (2)
c@ = 1
for d@ = a@ to 2 step -b@
c@ = c@ * d@
next
return (c@) |
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.
| #VBScript | VBScript |
Function multifactorial(n,d)
If n = 0 Then
multifactorial = 1
Else
For i = n To 1 Step -d
If i = n Then
multifactorial = n
Else
multifactorial = multifactorial * i
End If
Next
End If
End Function
For j = 1 To 5
WScript.StdOut.Write "Degree " & j & ": "
For k = 1 To 10
If k = 10 Then
WScript.StdOut.Write multifactorial(k,j)
Else
WScript.StdOut.Write multifactorial(k,j) & " "
End If
Next
WScript.StdOut.WriteLine
Next
|
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
| #Solution_with_recursion | Solution with recursion |
<html>
<body>
<pre>
<?php
/*************************************************************************
*
* This algorithm solves the 8 queens problem using backtracking.
* Please try with N<=25 * * * *************************************************************************/
class Queens {
var $size;
var $arr;
var $sol;
function Queens($n = 8) {
$this->size = $n;
$this->arr = array();
$this->sol = 0;
// Inicialiate array;
for($i=0; $i<$n; $i++) {
$this->arr[$i] = 0;
}
}
function isSolution($n) {
for ($i = 0; $i < $n; $i++) {
if ($this->arr[$i] == $this->arr[$n] ||
($this->arr[$i] - $this->arr[$n]) == ($n - $i) ||
($this->arr[$n] - $this->arr[$i]) == ($n - $i))
{
return false;
}
}
return true;
}
function PrintQueens() {
echo("solution #".(++$this->sol)."\n");
// echo("solution #".($this->size)."\n");
for ($i = 0; $i < $this->size; $i++) {
for ($j = 0; $j < $this->size; $j++) {
if ($this->arr[$i] == $j) echo("& ");
else echo(". ");
}
echo("\n");
}
echo("\n");
}
// backtracking Algorithm
function run($n = 0) {
if ($n == $this->size){
$this->PrintQueens();
}
else {
for ($i = 0; $i < $this->size; $i++) {
$this->arr[$n] = $i;
if($this->isSolution($n)){
$this->run($n+1);
}
}
}
}
}
$myprogram = new Queens(8);
$myprogram->run();
?>
</pre>
</body>
</html>
|
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.
| #UNIX_Shell | UNIX Shell | nth() {
local ordinals=(th st nd rd)
local -i n=$1 i
if (( n < 0 )); then
printf '%s%s\n' - "$(nth $(( -n )) )"
return 0
fi
case $(( n % 100 )) in
11|12|13) i=0;;
*) (( i= n%10 < 4 ? n%10 : 0 ));;
esac
printf '%d%s\n' "$n" "${ordinals[i]}"
}
for n in {0..25} {250..265} {1000..1025}; do
nth $n
done | column |
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.
| #VBA | VBA | Private Function ordinals() As Variant
ordinals = [{"th","st","nd","rd"}]
End Function
Private Function Nth(n As Variant, Optional apostrophe As Boolean = False) As String
Dim mod10 As Integer: mod10 = n Mod 10 + 1
If mod10 > 4 Or n Mod 100 = mod10 + 9 Then mod10 = 1
Nth = CStr(n) & String$(Val(-apostrophe), "'") & ordinals()(mod10)
End Function
Public Sub main()
Ranges = [{0,25;250,265;1000,1025}]
For i = 1 To UBound(Ranges)
For j = Ranges(i, 1) To Ranges(i, 2)
If j Mod 10 = 0 Then Debug.Print
Debug.Print Format(Nth(j, i = 2), "@@@@@@@");
Next j
Debug.Print
Next i
End Sub |
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).
| #Oz | Oz | declare
fun {F N}
if N == 0 then 1
elseif N > 0 then N - {M {F N-1}}
end
end
fun {M N}
if N == 0 then 0
elseif N > 0 then N - {F {M N-1}}
end
end
in
{Show {Map {List.number 0 9 1} F}}
{Show {Map {List.number 0 9 1} M}} |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Phix | Phix | with javascript_semantics
integer N = 100
for i=1 to 6 do
integer inside = 0
for n=1 to N do
integer x = rand(N),
y = rand(N)
inside += (x*x+y*y<N*N)
end for
pp({N,4*inside/N})
N *= 10
end for
|
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #PHP | PHP | <?
$loop = 1000000; # loop to 1,000,000
$count = 0;
for ($i=0; $i<$loop; $i++) {
$x = rand() / getrandmax();
$y = rand() / getrandmax();
if(($x*$x) + ($y*$y)<=1) $count++;
}
echo "loop=".number_format($loop).", count=".number_format($count).", pi=".($count/$loop*4);
?> |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #MATLAB | MATLAB | function [morseText,morseSound] = text2morse(string,playSound)
%% Translate AlphaNumeric Text to Morse Text
string = lower(string);
%Defined such that the ascii code of the characters in the string map
%to the indecies of the dictionary.
morseDictionary = {{' ',' '},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'0','-----'},{'1','.----'},{'2','..---'},{'3','...--'},...
{'4','....-'},{'5','.....'},{'6','-....'},{'7','--...'},...
{'8','---..'},{'9','----.'},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},{'',''},{'',''},{'',''},...
{'',''},{'',''},{'',''},...
{'a','.-'},{'b','-...'},{'c','-.-.'},{'d','-..'},...
{'e','.'},{'f','..-.'},{'g','--.'},{'h','....'},...
{'i','..'},{'j','.---'},{'k','-.-'},{'l','.-..'},...
{'m','--'},{'n','-.'},{'o','---'},{'p','.--.'},...
{'q','--.-'},{'r','.-.'},{'s','...'},{'t','-'},...
{'u','..-'},{'v','...-'},{'w','.--'},{'x','-..-'},...
{'y','-.--'},{'z','--..'}};
%Iterates through each letter in the string and converts it to morse
%code
morseText = arrayfun(@(x)[morseDictionary{x}{2} '|'],(string - 31),'UniformOutput',false);
%The output of the previous operation is a cell array, we want it to be
%a string. This line accomplishes that.
morseText = cell2mat(morseText);
morseText(end) = []; %delete extra pipe
%% Translate Morse Text to Morse Audio
%Generate the tones for each element of the code
SamplingFrequency = 8192; %Hz
ditLength = .1; %s
dit = (0:1/SamplingFrequency:ditLength);
dah = (0:1/SamplingFrequency:3*ditLength);
dit = sin(3520*dit);
dah = sin(3520*dah);
silent = zeros(1,length(dit));
%A dictionary of the audio components of each symbol
morseTiming = {{'.',[dit silent]},{'-',[dah silent]},{'|',[silent silent]},{' ',[silent silent]}};
morseSound = [];
for i = (1:length(morseText))
%Iterate through each cell in the morseTiming cell array and
%find which timing sequence corresponds to the current morse
%text symbol.
cellNum = find(cellfun(@(x)(x{1}==morseText(i)),morseTiming));
morseSound = [morseSound morseTiming{cellNum}{2}];
end
morseSound(end-length(silent):end) = []; %Delete the extra silent tone at the end
if(playSound)
sound(morseSound,SamplingFrequency); %Play sound
end
end %text2morse |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #jq | jq | cat /dev/urandom | tr -cd '0-9' | fold -w 1 | jq -nrf monty-hall.jq
|
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Julia | Julia | using Printf
function play_mh_literal{T<:Integer}(ncur::T=3, ncar::T=1)
ncar < ncur || throw(DomainError())
curtains = shuffle(collect(1:ncur))
cars = curtains[1:ncar]
goats = curtains[(ncar+1):end]
pick = rand(1:ncur)
isstickwin = pick in cars
deleteat!(curtains, findin(curtains, pick))
if !isstickwin
deleteat!(goats, findin(goats, pick))
end
if length(goats) > 0 # reveal goat
deleteat!(curtains, findin(curtains, shuffle(goats)[1]))
else # no goats, so reveal car
deleteat!(curtains, rand(1:(ncur-1)))
end
pick = shuffle(curtains)[1]
isswitchwin = pick in cars
return (isstickwin, isswitchwin)
end
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Python | Python | >>> def extended_gcd(aa, bb):
lastremainder, remainder = abs(aa), abs(bb)
x, lastx, y, lasty = 0, 1, 1, 0
while remainder:
lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
x, lastx = lastx - quotient*x, x
y, lasty = lasty - quotient*y, y
return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)
>>> def modinv(a, m):
g, x, y = extended_gcd(a, m)
if g != 1:
raise ValueError
return x % m
>>> modinv(42, 2017)
1969
>>> |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Quackery | Quackery | [ dup 1 != if
[ tuck 1 0
[ swap temp put
temp put
over 1 > while
tuck /mod swap
temp take tuck *
temp take swap -
again ]
2drop
temp release
temp take
dup 0 < if
[ over + ] ]
nip ] is modinv ( n n --> n )
42 2017 modinv echo |
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.
| #Euphoria | Euphoria | puts(1," x")
for i = 1 to 12 do
printf(1," %3d",i)
end for
puts(1,'\n')
for i = 1 to 12 do
printf(1,"%2d",i)
for j = 1 to 12 do
if j<i then
puts(1," ")
else
printf(1," %3d",i*j)
end if
end for
puts(1,'\n')
end for |
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.
| #Wortel | Wortel | @let {
facd &[d n]?{<= n d n @prod@range[n 1 @-d]}
; tacit implementation
facdt ^(!?(/^> .1 ^(@prod @range ~1jdtShj &^!(@- @id))) @,)
; recursive
facdrec &[n d] ?{<= n d n *n !!facdrec -n d d}
; output
l @to 10
~@each @to 5 &n !console.log "Degree {n}: {@join @s !*\facd n l}"
} |
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.
| #Wren | Wren | import "/fmt" for Fmt
var mf = Fn.new { |n, d|
var prod = 1
while (n > 1) {
prod = prod * n
n = n - d
}
return prod
}
for (d in 1..5) {
System.write("degree %(d): ")
for (n in 1..10) System.write(Fmt.d(8, mf.call(n, d)))
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
| #Picat | Picat | import sat.
% import mip.
queens_sat(N,Q) =>
Q = new_array(N,N),
Q :: 0..1,
foreach (K in 1-N..N-1)
sum([Q[I,J] : I in 1..N, J in 1..N, I-J==K]) #=< 1
end,
foreach (K in 2..2*N)
sum([Q[I,J] : I in 1..N, J in 1..N, I+J==K]) #=< 1
end,
foreach (I in 1..N)
sum([Q[I,J] : J in 1..N]) #= 1
end,
foreach (J in 1..N)
sum([Q[I,J] : I in 1..N]) #= 1
end,
solve([inout],Q). |
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.
| #Vlang | Vlang | fn ord(n int) string {
mut s := "th"
c := n % 10
if c in [1,2,3] {
if n%100/10 == 1 {
return "$n$s"
}
match c {
1 {
s = 'st'
}
2 {
s = 'nd'
}
3 {
s = 'rd'
}
else{}
}
}
return "$n$s"
}
fn main() {
for n := 0; n <= 25; n++ {
print("${ord(n)} ")
}
println('')
for n := 250; n <= 265; n++ {
print("${ord(n)} ")
}
println('')
for n := 1000; n <= 1025; n++ {
print("${ord(n)} ")
}
println('')
} |
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.
| #Wren | Wren | import "/fmt" for Conv
var ranges = [ 0..25, 250..265, 1000..1025 ]
for (r in ranges) {
r.each { |i| System.write("%(Conv.ord(i)) ") }
System.print("\n")
} |
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).
| #PARI.2FGP | PARI/GP | F(n)=if(n,n-M(F(n-1)),1)
M(n)=if(n,n-F(M(n-1)),0) |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Picat | Picat |
sim1(N, F) = C =>
C = 0,
I = 0,
while (I <= N)
C := C + apply(F),
I := I + 1
end. |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #PicoLisp | PicoLisp | (de carloPi (Scl)
(let (Dim (** 10 Scl) Dim2 (* Dim Dim) Pi 0)
(do (* 4 Dim)
(let (X (rand 0 Dim) Y (rand 0 Dim))
(when (>= Dim2 (+ (* X X) (* Y Y)))
(inc 'Pi) ) ) )
(format Pi Scl) ) )
(for N 6
(prinl (carloPi N)) ) |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Modula-2 | Modula-2 | MODULE MorseCode;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteMorseCode(str : ARRAY OF CHAR);
VAR i : CARDINAL;
BEGIN
WriteString(str);
WriteLn;
FOR i:=0 TO HIGH(str) DO
CASE CAP(str[i]) OF
'A': WriteString(".-");
| 'B': WriteString("-...");
| 'C': WriteString("-.-");
| 'D': WriteString("-..");
| 'E': WriteString(".");
| 'F': WriteString("..-.");
| 'G': WriteString("--.");
| 'H': WriteString("....");
| 'I': WriteString("..");
| 'J': WriteString(".---");
| 'K': WriteString("-.-");
| 'L': WriteString(".-..");
| 'M': WriteString("--");
| 'N': WriteString("-.");
| 'O': WriteString("---");
| 'P': WriteString(".--.");
| 'Q': WriteString("--.-");
| 'R': WriteString(".-.");
| 'S': WriteString("...");
| 'T': WriteString("-");
| 'U': WriteString("..-");
| 'V': WriteString("...-");
| 'W': WriteString(".--");
| 'X': WriteString("-..-");
| 'Y': WriteString("-.--");
| 'Z': WriteString("--..");
| '0': WriteString("-----");
| '1': WriteString(".----");
| '2': WriteString("..---");
| '3': WriteString("...--");
| '4': WriteString("....-");
| '5': WriteString(".....");
| '6': WriteString("-....");
| '7': WriteString("--...");
| '8': WriteString("---..");
| '9': WriteString("----.");
| ' ': WriteString(" ");
ELSE
IF (str[i] # 0C) THEN
WriteString("?");
END
END;
WriteString(" ");
END;
END WriteMorseCode;
BEGIN
WriteMorseCode("hello world");
WriteLn;
ReadChar;
END MorseCode. |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Kotlin | Kotlin | // version 1.1.2
import java.util.Random
fun montyHall(games: Int) {
var switchWins = 0
var stayWins = 0
val rnd = Random()
(1..games).forEach {
val doors = IntArray(3) // all zero (goats) by default
doors[rnd.nextInt(3)] = 1 // put car in a random door
val choice = rnd.nextInt(3) // choose a door at random
var shown: Int
do {
shown = rnd.nextInt(3) // the shown door
}
while (doors[shown] == 1 || shown == choice)
stayWins += doors[choice]
switchWins += doors[3 - choice - shown]
}
println("Simulating $games games:")
println("Staying wins $stayWins times")
println("Switching wins $switchWins times")
}
fun main(args: Array<String>) {
montyHall(1_000_000)
} |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Racket | Racket |
(require math)
(modular-inverse 42 2017)
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Raku | Raku | sub inverse($n, :$modulo) {
my ($c, $d, $uc, $vc, $ud, $vd) = ($n % $modulo, $modulo, 1, 0, 0, 1);
my $q;
while $c != 0 {
($q, $c, $d) = ($d div $c, $d % $c, $c);
($uc, $vc, $ud, $vd) = ($ud - $q*$uc, $vd - $q*$vc, $uc, $vc);
}
return $ud % $modulo;
}
say inverse 42, :modulo(2017) |
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.
| #Excel | Excel | FNOVERHALFCARTESIANPRODUCT
=LAMBDA(f,
LAMBDA(n,
LET(
ixs, SEQUENCE(n, n, 1, 1),
REM, "1-based indices.",
x, 1 + MOD(ixs - 1, n),
y, 1 + QUOTIENT(ixs - 1, n),
IF(x >= y,
f(x)(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.
| #XPL0 | XPL0 | code ChOut=8, CrLf=9, IntOut=11;
func MultiFac(N, D); \Return multifactorial of N in degree D
int N, D;
int F;
[F:= 1;
repeat F:= F*N;
N:= N-D;
until N <= 1;
return F;
];
int I, J; \generate table of multifactorials
for J:= 1 to 5 do
[for I:= 1 to 10 do
[IntOut(0, MultiFac(I, J)); ChOut(0, 9\tab\)];
CrLf(0);
] |
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.
| #zkl | zkl | fcn mfact(n,m){ [n..1,-m].reduce('*,1) }
foreach m in ([1..5]){ println("%d: %s".fmt(m,[1..10].apply(mfact.fp1(m)))) } |
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
| #PicoLisp | PicoLisp | (load "@lib/simul.l") # for 'permute'
(de queens (N)
(let (R (range 1 N) Cnt 0)
(for L (permute (range 1 N))
(when
(= N # from the Python solution
(length (uniq (mapcar + L R)))
(length (uniq (mapcar - L R))) )
(inc 'Cnt) ) )
Cnt ) ) |
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.
| #XBasic | XBasic |
PROGRAM "nth"
VERSION "0.0002"
DECLARE FUNCTION Entry()
INTERNAL FUNCTION Suffix$(n&&)
INTERNAL FUNCTION PrintImages (loLim&&, hiLim&&)
FUNCTION Entry()
PrintImages( 0, 25)
PrintImages( 250, 265)
PrintImages(1000, 1025)
END FUNCTION
FUNCTION Suffix$(n&&)
nMod10@@ = n&& MOD 10
nMod100@@ = n&& MOD 100
SELECT CASE TRUE
CASE (nMod10@@ = 1) AND (nMod100@@ <> 11):
RETURN ("st")
CASE (nMod10@@ = 2) AND (nMod100@@ <> 12):
RETURN ("nd")
CASE (nMod10@@ = 3) AND (nMod100@@ <> 13):
RETURN ("rd")
CASE ELSE:
RETURN ("th")
END SELECT
END FUNCTION
FUNCTION PrintImages(loLim&&, hiLim&&)
FOR i&& = loLim&& TO hiLim&&
PRINT TRIM$(STRING$(i&&)); Suffix$(i&&); " ";
NEXT
PRINT
END FUNCTION
END 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
(
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).
| #Pascal | Pascal | Program MutualRecursion;
{M definition comes after F which uses it}
function M(n : Integer) : Integer; forward;
function F(n : Integer) : Integer;
begin
if n = 0 then
F := 1
else
F := n - M(F(n-1));
end;
function M(n : Integer) : Integer;
begin
if n = 0 then
M := 0
else
M := n - F(M(n-1));
end;
var
i : Integer;
begin
for i := 0 to 19 do begin
write(F(i) : 4)
end;
writeln;
for i := 0 to 19 do begin
write(M(i) : 4)
end;
writeln;
end. |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #PowerShell | PowerShell | function Get-Pi ($Iterations = 10000) {
$InCircle = 0
for ($i = 0; $i -lt $Iterations; $i++) {
$x = Get-Random 1.0
$y = Get-Random 1.0
if ([Math]::Sqrt($x * $x + $y * $y) -le 1) {
$InCircle++
}
}
$Pi = [decimal] $InCircle / $Iterations * 4
$RealPi = [decimal] "3.141592653589793238462643383280"
$Diff = [Math]::Abs(($Pi - $RealPi) / $RealPi * 100)
New-Object PSObject `
| Add-Member -PassThru NoteProperty Iterations $Iterations `
| Add-Member -PassThru NoteProperty Pi $Pi `
| Add-Member -PassThru NoteProperty "% Difference" $Diff
} |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Nim | Nim | import os, strutils, tables
const Morse = {'A': ".-", 'B': "-...", 'C': "-.-.", 'D': "-..", 'E': ".",
'F': "..-.", 'G': "--.", 'H': "....", 'I': "..", 'J': ".---",
'K': "-.-", 'L': ".-..", 'M': "--", 'N': "-.", 'O': "---",
'P': ".--.", 'Q': "--.-", 'R': ".-.", 'S': "...", 'T': "-",
'U': "..-", 'V': "...-", 'W': ".--", 'X': "-..-", 'Y': "-.--",
'Z': "--..", '0': "-----", '1': ".----", '2': "..---", '3': "...--",
'4': "....-", '5': ".....", '6': "-....", '7': "--...", '8': "---..",
'9': "----.", '.': ".-.-.-", ',': "--..--", '?': "..--..", '\'': ".----.",
'!': "-.-.--", '/': "-..-.", '(': "-.--.", ')': "-.--.-", '&': ".-...",
':': "---...", ';': "-.-.-.", '=': "-...-", '+': ".-.-.", '-': "-....-",
'_': "..--.-", '"': ".-..-.", '$': "...-..-", '@': ".--.-."}.toTable
proc morse(s: string): string =
var r: seq[string]
for c in s:
r.add Morse.getOrDefault(c.toUpperAscii, "")
result = r.join(" ")
var m: seq[string]
for arg in commandLineParams():
m.add morse(arg)
echo m.join(" ") |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Liberty_BASIC | Liberty BASIC |
'adapted from BASIC solution
DIM doors(3) '0 is a goat, 1 is a car
total = 10000 'set desired number of iterations
switchWins = 0
stayWins = 0
FOR plays = 1 TO total
winner = INT(RND(1) * 3) + 1
doors(winner) = 1'put a winner in a random door
choice = INT(RND(1) * 3) + 1'pick a door, any door
DO
shown = INT(RND(1) * 3) + 1
'don't show the winner or the choice
LOOP WHILE doors(shown) = 1 OR shown = choice
if doors(choice) = 1 then
stayWins = stayWins + 1 'if you won by staying, count it
else
switchWins = switchWins + 1'could have switched to win
end if
doors(winner) = 0 'clear the doors for the next test
NEXT
PRINT "Result for ";total;" games."
PRINT "Switching wins "; switchWins; " times."
PRINT "Staying wins "; stayWins; " times."
|
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #REXX | REXX | /*REXX program calculates and displays the modular inverse of an integer X modulo Y.*/
parse arg x y . /*obtain two integers from the C.L. */
if x=='' | x=="," then x= 42 /*Not specified? Then use the default.*/
if y=='' | y=="," then y= 2017 /* " " " " " " */
say 'modular inverse of ' x " by " y ' ───► ' modInv(x,y)
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
modInv: parse arg a,b 1 ob; z= 0 /*B & OB are obtained from the 2nd arg.*/
$= 1 /*initialize modular inverse to unity. */
if b\=1 then do while a>1
parse value a/b a//b b z with q b a t
z= $ - q * z
$= trunc(t)
end /*while*/
if $<0 then $= $ + ob /*Negative? Then add the original B. */
return $ |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Ring | Ring |
see "42 %! 2017 = " + multInv(42, 2017) + nl
func multInv a,b
b0 = b
x0 = 0
multInv = 1
if b = 1 return 0 ok
while a > 1
q = floor(a / b)
t = b
b = a % b
a = t
t = x0
x0 = multInv - q * x0
multInv = t
end
if multInv < 0 multInv = multInv + b0 ok
return multInv
|
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.
| #F.23 | F# |
open System
let multTable () =
Console.Write (" X".PadRight (4))
for i = 1 to 12 do Console.Write ((i.ToString "####").PadLeft 4)
Console.Write "\n ___"
for i = 1 to 12 do Console.Write " ___"
Console.WriteLine ()
for row = 1 to 12 do
Console.Write (row.ToString("###").PadLeft(3).PadRight(4))
for col = 1 to 12 do
if row <= col then Console.Write ((row * col).ToString("###").PadLeft(4))
else
Console.Write ("".PadLeft 4)
Console.WriteLine ()
Console.WriteLine ()
Console.ReadKey () |> ignore
multTable ()
|
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
| #PL.2FI | PL/I |
NQUEENS: PROC OPTIONS (MAIN);
DCL A(35) BIN FIXED(31) EXTERNAL;
DCL COUNT BIN FIXED(31) EXTERNAL;
COUNT = 0;
DECLARE SYSIN FILE;
DCL ABS BUILTIN;
DECLARE SYSPRINT FILE;
DECLARE N BINARY FIXED (31); /* COUNTER */
/* MAIN LOOP STARTS HERE */
GET LIST (N) FILE(SYSIN); /* N QUEENS, N X N BOARD */
PUT SKIP (1) FILE(SYSPRINT);
PUT SKIP LIST('BEGIN N QUEENS PROCESSING *****') FILE(SYSPRINT);
PUT SKIP LIST('SOLUTIONS FOR N: ',N) FILE(SYSPRINT);
PUT SKIP (1) FILE(SYSPRINT);
IF N < 4 THEN DO;
/* LESS THAN 4 MAKES NO SENSE */
PUT SKIP (2) FILE(SYSPRINT);
PUT SKIP LIST (N,' N TOO LOW') FILE (SYSPRINT);
PUT SKIP (2) FILE(SYSPRINT);
RETURN (1);
END;
IF N > 35 THEN DO;
/* WOULD TAKE WEEKS */
PUT SKIP (2) FILE(SYSPRINT);
PUT SKIP LIST (N,' N TOO HIGH') FILE (SYSPRINT);
PUT SKIP (2) FILE(SYSPRINT);
RETURN (1);
END;
CALL QUEEN(N);
PUT SKIP (2) FILE(SYSPRINT);
PUT SKIP LIST (COUNT,' SOLUTIONS FOUND') FILE(SYSPRINT);
PUT SKIP (1) FILE(SYSPRINT);
PUT SKIP LIST ('END OF PROCESSING ****') FILE(SYSPRINT);
RETURN(0);
/* MAIN LOOP ENDS ABOVE */
PLACE: PROCEDURE (PS);
DCL PS BIN FIXED(31);
DCL I BIN FIXED(31) INIT(0);
DCL A(50) BIN FIXED(31) EXTERNAL;
DO I=1 TO PS-1;
IF A(I) = A(PS) THEN RETURN(0);
IF ABS ( A(I) - A(PS) ) = (PS-I) THEN RETURN(0);
END;
RETURN (1);
END PLACE;
QUEEN: PROCEDURE (N);
DCL N BIN FIXED (31);
DCL K BIN FIXED (31);
DCL A(50) BIN FIXED(31) EXTERNAL;
DCL COUNT BIN FIXED(31) EXTERNAL;
K = 1;
A(K) = 0;
DO WHILE (K > 0);
A(K) = A(K) + 1;
DO WHILE ( ( A(K)<= N) & (PLACE(K) =0) );
A(K) = A(K) +1;
END;
IF (A(K) <= N) THEN DO;
IF (K = N ) THEN DO;
COUNT = COUNT + 1;
END;
ELSE DO;
K= K +1;
A(K) = 0;
END; /* OF INSIDE ELSE */
END; /* OF FIRST IF */
ELSE DO;
K = K -1;
END;
END; /* OF EXTERNAL WHILE LOOP */
END QUEEN;
END NQUEENS; |
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.
| #XLISP | XLISP | (DEFUN NTH (N)
(COND
((AND (> (MOD N 100) 3) (< (MOD N 100) 21)) `(,N TH))
((= (MOD N 10) 1) `(,N ST))
((= (MOD N 10) 2) `(,N ND))
((= (MOD N 10) 3) `(,N RD))
(T `(,N TH))))
(DEFUN RANGE (X Y)
(IF (<= X Y)
(CONS X (RANGE (+ X 1) Y))))
(DEFUN TEST-NTH ()
(DISPLAY (MAPCAR NTH (RANGE 1 25)))
(NEWLINE)
(DISPLAY (MAPCAR NTH (RANGE 250 265)))
(NEWLINE)
(DISPLAY (MAPCAR NTH (RANGE 1000 1025))))
(TEST-NTH) |
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).
| #Perl | Perl | sub F { my $n = shift; $n ? $n - M(F($n-1)) : 1 }
sub M { my $n = shift; $n ? $n - F(M($n-1)) : 0 }
# Usage:
foreach my $sequence (\&F, \&M) {
print join(' ', map $sequence->($_), 0 .. 19), "\n";
} |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #PureBasic | PureBasic | OpenConsole()
Procedure.d MonteCarloPi(throws.d)
inCircle.d = 0
For i = 1 To throws.d
randX.d = (Random(2147483647)/2147483647)*2-1
randY.d = (Random(2147483647)/2147483647)*2-1
dist.d = Sqr(randX.d*randX.d + randY.d*randY.d)
If dist.d < 1
inCircle = inCircle + 1
EndIf
Next i
pi.d = (4 * inCircle / throws.d)
ProcedureReturn pi.d
EndProcedure
PrintN ("'built-in' #Pi = " + StrD(#PI,20))
PrintN ("MonteCarloPi(10000) = " + StrD(MonteCarloPi(10000),20))
PrintN ("MonteCarloPi(100000) = " + StrD(MonteCarloPi(100000),20))
PrintN ("MonteCarloPi(1000000) = " + StrD(MonteCarloPi(1000000),20))
PrintN ("MonteCarloPi(10000000) = " + StrD(MonteCarloPi(10000000),20))
PrintN("Press any key"): Repeat: Until Inkey() <> ""
|
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #OCaml | OCaml | let codes = [
'a', ".-"; 'b', "-..."; 'c', "-.-.";
'd', "-.."; 'e', "."; 'f', "..-.";
'g', "--."; 'h', "...."; 'i', "..";
'j', ".---"; 'k', "-.-"; 'l', ".-..";
'm', "--"; 'n', "-."; 'o', "---";
'p', ".--."; 'q', "--.-"; 'r', ".-.";
's', "..."; 't', "-"; 'u', "..-";
'v', "...-"; 'w', ".--"; 'x', "-..-";
'y', "-.--"; 'z', "--.."; '0', "-----";
'1', ".----"; '2', "..---"; '3', "...--";
'4', "....-"; '5', "....."; '6', "-....";
'7', "--..."; '8', "---.."; '9', "----.";
]
let oc = open_out "/dev/dsp"
let bip u =
for i = 0 to pred u do
let j = sin(0.6 *. (float i)) in
let k = ((j +. 1.0) /. 2.0) *. 127.0 in
output_byte oc (truncate k)
done
let gap u =
for i = 0 to pred u do
output_byte oc 0
done
let morse =
let u = 1000 in (* length of one unit *)
let u2 = u * 2 in
let u3 = u * 3 in
let u6 = u * 6 in
String.iter (function
| ' ' -> gap u6
| 'a'..'z' | 'A'..'Z' | '0'..'9' as c ->
let s = List.assoc c codes in
String.iter (function
'.' -> bip u; gap u
| '-' -> bip u3; gap u
| _ -> assert false
) s; gap u2
| _ -> prerr_endline "unknown char")
let () = morse "rosettacode morse" |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Lua | Lua | function playgame(player)
local car = math.random(3)
local pchoice = player.choice()
local function neither(a, b) --slow, but it works
local el = math.random(3)
return (el ~= a and el ~= b) and el or neither(a, b)
end
local el = neither(car, pchoice)
if(player.switch) then pchoice = neither(pchoice, el) end
player.wins = player.wins + (pchoice == car and 1 or 0)
end
for _, v in ipairs{true, false} do
player = {choice = function() return math.random(3) end,
wins = 0, switch = v}
for i = 1, 20000 do playgame(player) end
print(player.wins)
end |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Ruby | Ruby | #based on pseudo code from http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#Iterative_method_2 and from translating the python implementation.
def extended_gcd(a, b)
last_remainder, remainder = a.abs, b.abs
x, last_x, y, last_y = 0, 1, 1, 0
while remainder != 0
last_remainder, (quotient, remainder) = remainder, last_remainder.divmod(remainder)
x, last_x = last_x - quotient*x, x
y, last_y = last_y - quotient*y, y
end
return last_remainder, last_x * (a < 0 ? -1 : 1)
end
def invmod(e, et)
g, x = extended_gcd(e, et)
if g != 1
raise 'The maths are broken!'
end
x % et
end |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Run_BASIC | Run BASIC | print multInv(42, 2017)
end
function multInv(a,b)
b0 = b
multInv = 1
if b = 1 then goto [endFun]
while a > 1
q = a / b
t = b
b = a mod b
a = t
t = x0
x0 = multInv - q * x0
multInv = int(t)
wend
if multInv < 0 then multInv = multInv + b0
[endFun]
end function |
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.
| #Factor | Factor | USING: io kernel math math.parser math.ranges sequences ;
IN: multiplication-table
: print-row ( n -- )
[ number>string 2 CHAR: space pad-head write " |" write ]
[ 1 - [ " " write ] times ]
[
dup 12 [a,b]
[ * number>string 4 CHAR: space pad-head write ] with each
] tri nl ;
: print-table ( -- )
" " write
1 12 [a,b] [ number>string 4 CHAR: space pad-head write ] each nl
" +" write
12 [ "----" write ] times nl
1 12 [a,b] [ print-row ] each ; |
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
| #PowerBASIC | PowerBASIC | #COMPILE EXE
#DIM ALL
SUB aux(n AS INTEGER, i AS INTEGER, a() AS INTEGER, _
u() AS INTEGER, v() AS INTEGER, m AS QUAD)
LOCAL j, k, p, q AS INTEGER
IF i > n THEN
INCR m
FOR k = 1 TO n : PRINT a(k); : NEXT : PRINT
ELSE
FOR j = i TO n
k = a(j)
p = i - k + n
q = i + k - 1
IF u(p) AND v(q) THEN
u(p) = 0 : v(q) = 0
a(j) = a(i) : a(i) = k
CALL aux(n, i + 1, a(), u(), v(), m)
u(p) = 1 : v(q) = 1
a(i) = a(j) : a(j) = k
END IF
NEXT
END IF
END SUB
FUNCTION PBMAIN () AS LONG
LOCAL n, i AS INTEGER
LOCAL m AS QUAD
IF COMMAND$(1) <> "" THEN
n = VAL(COMMAND$(1))
REDIM a(1 TO n) AS INTEGER
REDIM u(1 TO 2 * n - 1) AS INTEGER
REDIM v(1 TO 2 * n - 1) AS INTEGER
FOR i = 1 TO n
a(i) = i
NEXT
FOR i = 1 TO 2 * n - 1
u(i) = 1
v(i) = 1
NEXT
m = 0
CALL aux(n, 1, a(), u(), v(), m)
PRINT m
END IF
END FUNCTION |
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.
| #XPL0 | XPL0 |
\N'th
code Rem=2, CrLf=9, IntIn=10, IntOut=11, Text=12;
procedure Suf(N, S);
integer N;
character S;
integer NMod10, NMod100;
begin
NMod10:= Rem(N/10);
NMod100:= Rem(N/100);
case of
NMod10 = 1 & NMod100 # 11: S(0):= "st";
NMod10 = 2 & NMod100 # 12: S(0):= "nd";
NMod10 = 3 & NMod100 # 13: S(0):= "rd"
other
S(0):= "th";
end;
procedure PrintImages(LoLim, HiLim);
integer LoLim, HiLim;
integer I;
character S;
begin
for I:= LoLim, HiLim do
begin
Suf(I, addr S);
IntOut(0, I); Text(0, S); Text(0, " ")
end;
CrLf(0)
end;
begin
PrintImages(0, 25);
PrintImages(250, 265);
PrintImages(1000, 1025)
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).
| #Phix | Phix | with javascript_semantics
forward function M(integer n)
function F(integer n)
return iff(n?n-M(F(n-1)):1)
end function
function M(integer n)
return iff(n?n-F(M(n-1)):0)
end function
for i=0 to 20 do
printf(1," %d",F(i))
end for
printf(1,"\n")
for i=0 to 20 do
printf(1," %d",M(i))
end for
|
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Python | Python | >>> import random, math
>>> throws = 1000
>>> 4.0 * sum(math.hypot(*[random.random()*2-1
for q in [0,1]]) < 1
for p in xrange(throws)) / float(throws)
3.1520000000000001
>>> throws = 1000000
>>> 4.0 * sum(math.hypot(*[random.random()*2-1
for q in [0,1]]) < 1
for p in xrange(throws)) / float(throws)
3.1396359999999999
>>> throws = 100000000
>>> 4.0 * sum(math.hypot(*[random.random()*2-1
for q in [0,1]]) < 1
for p in xrange(throws)) / float(throws)
3.1415666400000002 |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #Ol | Ol |
(display "Please, enter the string in lower case bounded by \" sign: ")
(lfor
(list->ff '(
(#\a . ".-" ) (#\b . "-..." ) (#\c . "-.-." )
(#\d . "-.." ) (#\e . "." ) (#\f . "..-." )
(#\g . "--." ) (#\h . "...." ) (#\i . ".." )
(#\j . ".---" ) (#\k . "-.-" ) (#\l . ".-.." )
(#\m . "--" ) (#\n . "-." ) (#\o . "---" )
(#\p . ".--." ) (#\q . "--.-" ) (#\r . ".-." )
(#\s . "..." ) (#\t . "-" ) (#\u . "..-" )
(#\v . "...-" ) (#\w . ".--" ) (#\x . "-..-" )
(#\y . "-.--" ) (#\z . "--.." ) (#\1 . ".----")
(#\2 . "..---") (#\3 . "...--") (#\4 . "....-")
(#\5 . ".....") (#\6 . "-....") (#\7 . "--...")
(#\8 . "---..") (#\9 . "----.") (#\0 . "-----")
(#\space . " ") (#\. . " ")))
(str-iter (read))
(lambda (codes char)
(let ((out (getf codes char)))
(if out (display out)))
codes))
; ==> Please, enter the string in lower case bounded by " sign:
; <== "hello world"
; ==> ......-...-..--- .-----.-..-..-..
|
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Lua.2FTorch | Lua/Torch | function montyHall(n)
local car = torch.LongTensor(n):random(3) -- door with car
local choice = torch.LongTensor(n):random(3) -- player's choice
local opens = torch.LongTensor(n):random(2):csub(1):mul(2):csub(1) -- -1 or +1
local iscarchoice = choice:eq(car)
local nocarchoice = 1-iscarchoice
opens[iscarchoice] = (((opens + choice - 1) % 3):abs() + 1)[iscarchoice]
opens[nocarchoice] = (6 - car - choice)[nocarchoice]
local change = torch.LongTensor(n):bernoulli() -- 0: stay, 1: change
local win = iscarchoice:long():cmul(1-change) + nocarchoice:long():cmul(change)
return car, choice, opens, change, win
end
function montyStats(n)
local car, pchoice, opens, change, win = montyHall(n)
local change_and_win = change [ win:byte()]:sum()/ change :sum()*100
local no_change_and_win = (1-change)[ win:byte()]:sum()/(1-change):sum()*100
local change_and_win_not = change [1-win:byte()]:sum()/ change :sum()*100
local no_change_and_win_not = (1-change)[1-win:byte()]:sum()/(1-change):sum()*100
print(string.format(" %9s %9s" , "no change", "change" ))
print(string.format("win %8.4f%% %8.4f%%", no_change_and_win , change_and_win ))
print(string.format("win not %8.4f%% %8.4f%%", no_change_and_win_not, change_and_win_not))
end
montyStats(1e7) |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Rust | Rust | fn mod_inv(a: isize, module: isize) -> isize {
let mut mn = (module, a);
let mut xy = (0, 1);
while mn.1 != 0 {
xy = (xy.1, xy.0 - (mn.0 / mn.1) * xy.1);
mn = (mn.1, mn.0 % mn.1);
}
while xy.0 < 0 {
xy.0 += module;
}
xy.0
}
fn main() {
println!("{}", mod_inv(42, 2017))
} |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Scala | Scala |
def gcdExt(u: Int, v: Int): (Int, Int, Int) = {
@tailrec
def aux(a: Int, b: Int, x: Int, y: Int, x1: Int, x2: Int, y1: Int, y2: Int): (Int, Int, Int) = {
if(b == 0) (x, y, a) else {
val (q, r) = (a / b, a % b)
aux(b, r, x2 - q * x1, y2 - q * y1, x, x1, y, y1)
}
}
aux(u, v, 1, 0, 0, 1, 1, 0)
}
def modInv(a: Int, m: Int): Option[Int] = {
val (i, j, g) = gcdExt(a, m)
if (g == 1) Option(if (i < 0) i + m else i) else Option.empty
} |
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.
| #FALSE | FALSE | [$100\>[" "]?$10\>[" "]?." "]p:
[$p;! m: 2[$m;\>][" "1+]# [$13\>][$m;*p;!1+]#%"
"]l:
1[$13\>][$l;!1+]#% |
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
| #PowerShell | PowerShell |
function PlaceQueen ( [ref]$Board, $Row, $N )
{
# For the current row, start with the first column
$Board.Value[$Row] = 0
# While haven't exhausted all columns in the current row...
While ( $Board.Value[$Row] -lt $N )
{
# If not the first row, check for conflicts
$Conflict = $Row -and
( (0..($Row-1)).Where{ $Board.Value[$_] -eq $Board.Value[$Row] }.Count -or
(0..($Row-1)).Where{ $Board.Value[$_] -eq $Board.Value[$Row] - $Row + $_ }.Count -or
(0..($Row-1)).Where{ $Board.Value[$_] -eq $Board.Value[$Row] + $Row - $_ }.Count )
# If no conflicts and the current column is a valid column...
If ( -not $Conflict -and $Board.Value[$Row] -lt $N )
{
# If this is the last row
# Board completed successfully
If ( $Row -eq ( $N - 1 ) )
{
return $True
}
# Recurse
# If all nested recursions were successful
# Board completed successfully
If ( PlaceQueen $Board ( $Row + 1 ) $N )
{
return $True
}
}
# Try the next column
$Board.Value[$Row]++
}
# Everything was tried, nothing worked
Return $False
}
function Get-NQueensBoard ( $N )
{
# Start with a default board (array of column positions for each row)
$Board = @( 0 ) * $N
# Place queens on board
# If successful...
If ( PlaceQueen -Board ([ref]$Board) -Row 0 -N $N )
{
# Convert board to strings for display
$Board | ForEach { ( @( "" ) + @(" ") * $_ + "Q" + @(" ") * ( $N - $_ ) ) -join "|" }
}
Else
{
"There is no solution for N = $N"
}
}
|
http://rosettacode.org/wiki/N%27th | N'th | Write a function/method/subroutine/... that when given an integer greater than or equal to zero returns a string of the number followed by an apostrophe then the ordinal suffix.
Example
Returns would include 1'st 2'nd 3'rd 11'th 111'th 1001'st 1012'th
Task
Use your routine to show here the output for at least the following (inclusive) ranges of integer inputs:
0..25, 250..265, 1000..1025
Note: apostrophes are now optional to allow correct apostrophe-less English.
| #Yabasic | Yabasic |
sub ordinal$ (n)
NMod10 = mod(n, 10)
NMod100 = mod(n, 100)
if (NMod10 = 1) and (NMod100 <> 11) then
return "st"
else
if (NMod10 = 2) and (NMod100 <> 12) then
return "nd"
else
if (NMod10 = 3) and (NMod100 <> 13) then
return "rd"
else
return "th"
end if
end if
end if
end sub
sub imprimeOrdinal(a, b)
for i = a to b
print i, ordinal$(i), " ";
next i
print
end sub
imprimeOrdinal (0, 25)
imprimeOrdinal (250, 265)
imprimeOrdinal (1000, 1025)
end
|
http://rosettacode.org/wiki/Mutual_recursion | Mutual recursion | Two functions are said to be mutually recursive if the first calls the second,
and in turn the second calls the first.
Write two mutually recursive functions that compute members of the Hofstadter Female and Male sequences defined as:
F
(
0
)
=
1
;
M
(
0
)
=
0
F
(
n
)
=
n
−
M
(
F
(
n
−
1
)
)
,
n
>
0
M
(
n
)
=
n
−
F
(
M
(
n
−
1
)
)
,
n
>
0.
{\displaystyle {\begin{aligned}F(0)&=1\ ;\ M(0)=0\\F(n)&=n-M(F(n-1)),\quad n>0\\M(n)&=n-F(M(n-1)),\quad n>0.\end{aligned}}}
(If a language does not allow for a solution using mutually recursive functions
then state this rather than give a solution by other means).
| #PHP | PHP | <?php
function F($n)
{
if ( $n == 0 ) return 1;
return $n - M(F($n-1));
}
function M($n)
{
if ( $n == 0) return 0;
return $n - F(M($n-1));
}
$ra = array();
$rb = array();
for($i=0; $i < 20; $i++)
{
array_push($ra, F($i));
array_push($rb, M($i));
}
echo implode(" ", $ra) . "\n";
echo implode(" ", $rb) . "\n";
?> |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Quackery | Quackery | [ $ "bigrat.qky" loadfile ] now!
[ [ 64 bit ] constant
dup random dup *
over random dup * +
swap dup * < ] is hit ( --> b )
[ 0 swap times
[ hit if 1+ ] ] is sims ( n --> n )
[ dup echo say " trials "
dup sims 4 *
swap 20 point$ echo$ cr ] is trials ( n --> )
' [ 10 100 1000 10000 100000 1000000 ] witheach trials |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #R | R | # nice but not suitable for big samples!
monteCarloPi <- function(samples) {
x <- runif(samples, -1, 1) # for big samples, you need a lot of memory!
y <- runif(samples, -1, 1)
l <- sqrt(x*x + y*y)
return(4*sum(l<=1)/samples)
}
# this second function changes the samples number to be
# multiple of group parameter (default 100).
monteCarlo2Pi <- function(samples, group=100) {
lim <- ceiling(samples/group)
olim <- lim
c <- 0
while(lim > 0) {
x <- runif(group, -1, 1)
y <- runif(group, -1, 1)
l <- sqrt(x*x + y*y)
c <- c + sum(l <= 1)
lim <- lim - 1
}
return(4*c/(olim*group))
}
print(monteCarloPi(1e4))
print(monteCarloPi(1e5))
print(monteCarlo2Pi(1e7)) |
http://rosettacode.org/wiki/Morse_code | Morse code | Morse code
It has been in use for more than 175 years — longer than any other electronic encoding system.
Task
Send a string as audible Morse code to an audio device (e.g., the PC speaker).
As the standard Morse code does not contain all possible characters,
you may either ignore unknown characters in the file,
or indicate them somehow (e.g. with a different pitch).
| #PARI.2FGP | PARI/GP | sleep(ms)={
while((ms-=gettime()) > 0,);
};
dot()=print1(Strchr([7]));sleep(250);
dash()=print1(Strchr([7]));sleep(10);print1(Strchr([7]));sleep(10);print1(Strchr([7]));sleep(250);
Morse(s)={
s=Vec(s);
for(i=1,#s,
if(s[i] == ".", dot(),
if(s[i] == "-", dash(), sleep(250))
)
)
};
Morse("...---...") |
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #M2000_Interpreter | M2000 Interpreter |
Module CheckIt {
Enum Strat {Stay, Random, Switch}
total=10000
Print $("0.00")
player_win_stay=0
player_win_switch=0
player_win_random=0
For i=1 to total {
Dim doors(1 to 3)=False
doors(Random(1,3))=True
guess=Random(1,3)
Inventory other
for k=1 to 3 {
If k <> guess Then Append other, k
}
If doors(guess) Then {
Mont_Hall_show=other(Random(0,1)!)
} Else {
If doors(other(0!)) Then {
Mont_Hall_show=other(1!)
} Else Mont_Hall_show=other(0!)
Delete Other, Mont_Hall_show
}
Strategy=Each(Strat)
While Strategy {
Select Case Eval(strategy)
Case Random
{
If Random(1,2)=1 Then {
If doors(guess) Then player_win_Random++
} else If doors(other(0!)) Then player_win_Random++
}
Case Switch
If doors(other(0!)) Then player_win_switch++
Else
If doors(guess) Then player_win_stay++
End Select
}
}
Print "Stay: ";player_win_stay/total*100;"%"
Print "Random: ";player_win_Random/total*100;"%"
Print "Switch: ";player_win_switch/total*100;"%"
}
CheckIt
|
http://rosettacode.org/wiki/Monty_Hall_problem | Monty Hall problem |
Suppose you're on a game show and you're given the choice of three doors.
Behind one door is a car; behind the others, goats.
The car and the goats were placed randomly behind the doors before the show.
Rules of the game
After you have chosen a door, the door remains closed for the time being.
The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
If both remaining doors have goats behind them, he chooses one randomly.
After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door.
Imagine that you chose Door 1 and the host opens Door 3, which has a goat.
He then asks you "Do you want to switch to Door Number 2?"
The question
Is it to your advantage to change your choice?
Note
The player may initially choose any of the three doors (not just Door 1), that the host opens a different door revealing a goat (not necessarily Door 3), and that he gives the player a second choice between the two remaining unopened doors.
Task
Run random simulations of the Monty Hall game. Show the effects of a strategy of the contestant always keeping his first guess so it can be contrasted with the strategy of the contestant always switching his guess.
Simulate at least a thousand games using three doors for each strategy and show the results in such a way as to make it easy to compare the effects of each strategy.
References
Stefan Krauss, X. T. Wang, "The psychology of the Monty Hall problem: Discovering psychological mechanisms for solving a tenacious brain teaser.", Journal of Experimental Psychology: General, Vol 132(1), Mar 2003, 3-22 DOI: 10.1037/0096-3445.132.1.3
A YouTube video: Monty Hall Problem - Numberphile.
| #Mathematica.2FWolfram_Language | Mathematica/Wolfram Language | montyHall[nGames_] :=
Module[{r, winningDoors, firstChoices, nStayWins, nSwitchWins, s},
r := RandomInteger[{1, 3}, nGames];
winningDoors = r;
firstChoices = r;
nStayWins = Count[Transpose[{winningDoors, firstChoices}], {d_, d_}];
nSwitchWins = nGames - nStayWins;
Grid[{{"Strategy", "Wins", "Win %"}, {"Stay", Row[{nStayWins, "/", nGames}], s=N[100 nStayWins/nGames]},
{"Switch", Row[{nSwitchWins, "/", nGames}], 100 - s}}, Frame -> All]] |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Seed7 | Seed7 | const func bigInteger: modInverse (in var bigInteger: a,
in var bigInteger: b) is func
result
var bigInteger: modularInverse is 0_;
local
var bigInteger: b_bak is 0_;
var bigInteger: x is 0_;
var bigInteger: y is 1_;
var bigInteger: lastx is 1_;
var bigInteger: lasty is 0_;
var bigInteger: temp is 0_;
var bigInteger: quotient is 0_;
begin
if b < 0_ then
raise RANGE_ERROR;
end if;
if a < 0_ and b <> 0_ then
a := a mod b;
end if;
b_bak := b;
while b <> 0_ do
temp := b;
quotient := a div b;
b := a rem b;
a := temp;
temp := x;
x := lastx - quotient * x;
lastx := temp;
temp := y;
y := lasty - quotient * y;
lasty := temp;
end while;
if a = 1_ then
modularInverse := lastx;
if modularInverse < 0_ then
modularInverse +:= b_bak;
end if;
else
raise RANGE_ERROR;
end if;
end func; |
http://rosettacode.org/wiki/Modular_inverse | Modular inverse | From Wikipedia:
In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that
a
x
≡
1
(
mod
m
)
.
{\displaystyle a\,x\equiv 1{\pmod {m}}.}
Or in other words, such that:
∃
k
∈
Z
,
a
x
=
1
+
k
m
{\displaystyle \exists k\in \mathbb {Z} ,\qquad a\,x=1+k\,m}
It can be shown that such an inverse exists if and only if a and m are coprime, but we will ignore this for this task.
Task
Either by implementing the algorithm, by using a dedicated library or by using a built-in function in
your language, compute the modular inverse of 42 modulo 2017.
| #Sidef | Sidef | say 42.modinv(2017) |
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.
| #Fantom | Fantom |
class Main
{
static Void multiplicationTable (Int n)
{
// print column headings
echo (" |" + (1..n).map |Int a -> Str| { a.toStr.padl(4)}.join("") )
echo ("-----" + (1..n).map { "----" }.join("") )
// work through each row
(1..n).each |i|
{
echo ( i.toStr.padl(4) + "|" +
Str.spaces(4*(i-1)) +
(i..n).map |Int j -> Str| { (i*j).toStr.padl(4)}.join("") )
}
}
public static Void main ()
{
multiplicationTable (12)
}
}
|
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
| #Processing | Processing |
int n = 8;
int[] b = new int[n];
int s = 0;
int y = 0;
void setup() {
size(400, 400);
textAlign(CENTER, CENTER);
textFont(createFont("DejaVu Sans", 44));
b[0] = -1;
}
void draw() {
if (y >= 0) {
do {
b[y]++;
} while ((b[y] < n) && unsafe(y));
if (b[y] < n) {
if (y < (n-1)) {
b[++y] = -1;
} else {
drawBoard();
}
} else {
y--;
}
} else {
textSize(18);
text("Press any key to restart", width / 2, height - 20);
}
}
boolean unsafe(int y) {
int x = b[y];
for (int i = 1; i <= y; i++) {
int t = b[y - i];
if (t == x ||
t == x - i ||
t == x + i) {
return true;
}
}
return false;
}
void drawBoard() {
float w = width / n;
for (int y = 0; y < n; y++) {
for (int x = 0; x < n; x++) {
fill(255 * ((x + y) % 2));
square(x * w, y * w, w);
if (b[y] == x) {
fill(255 - 255 * ((x + y) % 2));
textSize(42);
text("♕", w / 2 + x *w, w /2 + y * w);
}
}
}
fill(255, 0, 0);
textSize(18);
text("Solution " + (++s), width / 2, height / 90);
}
void keyPressed() {
b = new int[n];
s = 0;
y = 0;
b[0] = -1;
}
|
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.
| #zkl | zkl | #if 0
fcn addSuffix(n){
z:=n.abs()%100;
if(11<=z<=13) return(String(n,"th"));
z=z%10;
String(n,(z==1 and "st") or (z==2 and "nd") or (z==3 and "rd") or "th");
}
#else
fcn addSuffix(n){
var suffixes=T("th","st","nd","rd","th","th","th","th","th","th"); //0..10
z:=n.abs()%100;
String(n,(z<=10 or z>20) and suffixes[z%10] or "th");
}
#endif |
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).
| #Picat | Picat | table
f(0) = 1.
f(N) = N - m(f(N-1)), N > 0 => true.
table
m(0) = 0.
m(N) = N - f(m(N-1)), N > 0 => true. |
http://rosettacode.org/wiki/Monte_Carlo_methods | Monte Carlo methods | A Monte Carlo Simulation is a way of approximating the value of a function
where calculating the actual value is difficult or impossible.
It uses random sampling to define constraints on the value
and then makes a sort of "best guess."
A simple Monte Carlo Simulation can be used to calculate the value for
π
{\displaystyle \pi }
.
If you had a circle and a square where the length of a side of the square
was the same as the diameter of the circle, the ratio of the area of the circle
to the area of the square would be
π
/
4
{\displaystyle \pi /4}
.
So, if you put this circle inside the square and select many random points
inside the square, the number of points inside the circle
divided by the number of points inside the square and the circle
would be approximately
π
/
4
{\displaystyle \pi /4}
.
Task
Write a function to run a simulation like this, with a variable number of random points to select.
Also, show the results of a few different sample sizes.
For software where the number
π
{\displaystyle \pi }
is not built-in,
we give
π
{\displaystyle \pi }
as a number of digits:
3.141592653589793238462643383280
| #Racket | Racket | #lang racket
(define (in-unit-circle? x y) (<= (sqrt (+ (sqr x) (sqr y))) 1))
;; point in ([-1,1], [-1,1])
(define (random-point-in-2x2-square) (values (* 2 (- (random) 1/2)) (* 2 (- (random) 1/2))))
;; Area of circle is (pi r^2). r is 1, area of circle is pi
;; Area of square is 2^2 = 4
;; There is a pi/4 chance of landing in circle
;; .: pi = 4*(proportion passed) = 4*(passed/samples)
(define (passed:samples->pi passed samples) (* 4 (/ passed samples)))
;; generic kind of monte-carlo simulation
(define (monte-carlo run-length report-frequency
sample-generator pass?
interpret-result)
(let inner ((samples 0) (passed 0) (cnt report-frequency))
(cond
[(= samples run-length) (interpret-result passed samples)]
[(zero? cnt) ; intermediate report
(printf "~a samples of ~a: ~a passed -> ~a~%"
samples run-length passed (interpret-result passed samples))
(inner samples passed report-frequency)]
[else
(inner (add1 samples)
(if (call-with-values sample-generator pass?)
(add1 passed) passed) (sub1 cnt))])))
;; (monte-carlo ...) gives an "exact" result... which will be a fraction.
;; to see how it looks as a decimal we can exact->inexact it
(let ((mc (monte-carlo 10000000 1000000 random-point-in-2x2-square in-unit-circle? passed:samples->pi)))
(printf "exact = ~a~%inexact = ~a~%(pi - guess) = ~a~%" mc (exact->inexact mc) (- pi mc))) |
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