Seven-sided dice from five-sided dice

From Rosetta Code
Task
Seven-sided dice from five-sided dice
You are encouraged to solve this task according to the task description, using any language you may know.
Task

(Given an equal-probability generator of one of the integers 1 to 5 as dice5),   create dice7 that generates a pseudo-random integer from 1 to 7 in equal probability using only dice5 as a source of random numbers,   and check the distribution for at least one million calls using the function created in   Simple Random Distribution Checker.


Implementation suggestion: dice7 might call dice5 twice, re-call if four of the 25 combinations are given, otherwise split the other 21 combinations into 7 groups of three, and return the group index from the rolls.

(Task adapted from an answer here)

Ada[edit]

The specification of a package Random_57:

package Random_57 is
 
type Mod_7 is mod 7;
 
function Random7 return Mod_7;
-- a "fast" implementation, minimazing the calls to the Random5 generator
function Simple_Random7 return Mod_7;
-- a simple implementation
 
end Random_57;

Implementation of Random_57:

 with Ada.Numerics.Discrete_Random;
 
package body Random_57 is
type M5 is mod 5;
 
package Rand_5 is new Ada.Numerics.Discrete_Random(M5);
Gen: Rand_5.Generator;
function Random7 return Mod_7 is
N: Natural;
 
begin
loop
N := Integer(Rand_5.Random(Gen))* 5 + Integer(Rand_5.Random(Gen));
-- N is uniformly distributed in 0 .. 24
if N < 21 then
return Mod_7(N/3);
else -- (N-21) is in 0 .. 3
N := (N-21) * 5 + Integer(Rand_5.Random(Gen)); -- N is in 0 .. 19
if N < 14 then
return Mod_7(N / 2);
else -- (N-14) is in 0 .. 5
N := (N-14) * 5 + Integer(Rand_5.Random(Gen)); -- N is in 0 .. 29
if N < 28 then
return Mod_7(N/4);
else -- (N-28) is in 0 .. 1
N := (N-28) * 5 + Integer(Rand_5.Random(Gen)); -- 0 .. 9
if N < 7 then
return Mod_7(N);
else -- (N-7) is in 0, 1, 2
N := (N-7)* 5 + Integer(Rand_5.Random(Gen)); -- 0 .. 14
if N < 14 then
return Mod_7(N/2);
else -- (N-14) is 0. This is not useful for us!
null;
end if;
end if;
end if;
end if;
end if;
end loop;
 
end Random7;
 
function Simple_Random7 return Mod_7 is
N: Natural :=
Integer(Rand_5.Random(Gen))* 5 + Integer(Rand_5.Random(Gen));
-- N is uniformly distributed in 0 .. 24
begin
while N > 20 loop
N := Integer(Rand_5.Random(Gen))* 5 + Integer(Rand_5.Random(Gen));
end loop; -- Now I <= 20
return Mod_7(N / 3);
end Simple_Random7;
 
begin
Rand_5.Reset(Gen);
end Random_57;

A main program, using the Random_57 package:

with Ada.Text_IO, Random_57;
 
procedure R57 is
 
use Random_57;
 
type Fun is access function return Mod_7;
 
function Rand return Mod_7 renames Random_57.Random7;
-- change this to "... renames Random_57.Simple_Random;" if you like
 
procedure Test(Sample_Size: Positive; Rand: Fun; Precision: Float := 0.3) is
 
Counter: array(Mod_7) of Natural := (others => 0);
Expected: Natural := Sample_Size/7;
Small: Mod_7 := Mod_7'First;
Large: Mod_7 := Mod_7'First;
 
Result: Mod_7;
begin
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line("Sample Size: " & Integer'Image(Sample_Size));
Ada.Text_IO.Put( " Bins:");
for I in 1 .. Sample_Size loop
Result := Rand.all;
Counter(Result) := Counter(Result) + 1;
end loop;
for J in Mod_7 loop
Ada.Text_IO.Put(Integer'Image(Counter(J)));
if Counter(J) < Counter(Small) then Small := J; end if;
if Counter(J) > Counter(Large) then Large := J; end if;
end loop;
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line(" Small Bin:" & Integer'Image(Counter(Small)));
Ada.Text_IO.Put_Line(" Large Bin: " & Integer'Image(Counter(Large)));
 
if Float(Counter(Small)*7) * (1.0+Precision) < Float(Sample_Size) then
Ada.Text_IO.Put_Line("Failed! Small too small!");
elsif Float(Counter(Large)*7) * (1.0-Precision) > Float(Sample_Size) then
Ada.Text_IO.Put_Line("Failed! Large too large!");
else
Ada.Text_IO.Put_Line("Passed");
end if;
end Test;
 
begin
Test( 10_000, Rand'Access, 0.08);
Test( 100_000, Rand'Access, 0.04);
Test( 1_000_000, Rand'Access, 0.02);
Test(10_000_000, Rand'Access, 0.01);
end R57;
Output:
Sample Size:  10000
 Bins: 1368 1404 1435 1491 1483 1440 1379
 Small Bin: 1368
 Large Bin:  1491
Passed

Sample Size:  100000
 Bins: 14385 14110 14362 14404 14362 14206 14171
 Small Bin: 14110
 Large Bin:  14404
Passed

Sample Size:  1000000
 Bins: 143765 142384 142958 142684 142799 142956 142454
 Small Bin: 142384
 Large Bin:  143765
Passed

Sample Size:  10000000
 Bins: 1429266 1428214 1428753 1427032 1428418 1428699 1429618
 Small Bin: 1427032
 Large Bin:  1429618
Passed

ALGOL 68[edit]

Translation of: C
- note: This specimen retains the original C coding style.
Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d

C's version using no multiplications, divisions, or mod operators:

PROC dice5 = INT:
1 + ENTIER (5*random);
 
PROC mulby5 = (INT n)INT:
ABS (BIN n SHL 2) + n;
 
PROC dice7 = INT: (
INT d55 := 0;
INT m := 1;
WHILE
m := ABS ((2r1 AND BIN m) SHL 2) + ABS (BIN m SHR 1); # repeats 4 - 2 - 1 #
d55 := mulby5(mulby5(d55)) + mulby5(dice5) + dice5 - 6;
# WHILE # d55 < m DO SKIP OD;
 
m := 1;
WHILE d55>0 DO
d55 +:= m;
m := ABS (BIN d55 AND 2r111); # modulas by 8 #
d55 := ABS (BIN d55 SHR 3) # divide by 8 #
OD;
m
);
 
PROC distcheck = (PROC INT dice, INT count, upb)VOID: (
[upb]INT sum; FOR i TO UPB sum DO sum[i] := 0 OD;
FOR i TO count DO sum[dice]+:=1 OD;
FOR i TO UPB sum WHILE print(whole(sum[i],0)); i /= UPB sum DO print(", ") OD;
print(new line)
);
 
main:
(
distcheck(dice5, 1000000, 5);
distcheck(dice7, 1000000, 7)
)
Output:
200598, 199852, 199939, 200602, 199009
143529, 142688, 142816, 142747, 142958, 142802, 142460

AutoHotkey[edit]

dice5()
{ Random, v, 1, 5
Return, v
}
 
dice7()
{ Loop
{ v := 5 * dice5() + dice5() - 6
IfLess v, 21, Return, (v // 3) + 1
}
}
Distribution check:

Total elements = 10000

Margin = 3% --> Lbound = 1386, Ubound = 1471

Bucket 1 contains 1450 elements.
Bucket 2 contains 1374 elements. Skewed.
Bucket 3 contains 1412 elements.
Bucket 4 contains 1465 elements.
Bucket 5 contains 1370 elements. Skewed.
Bucket 6 contains 1485 elements. Skewed.
Bucket 7 contains 1444 elements.

BBC BASIC[edit]

      MAXRND = 7
FOR r% = 2 TO 5
check% = FNdistcheck(FNdice7, 10^r%, 0.1)
PRINT "Over "; 10^r% " runs dice7 ";
IF check% THEN
PRINT "failed distribution check with "; check% " bin(s) out of range"
ELSE
PRINT "passed distribution check"
ENDIF
NEXT
END
 
DEF FNdice7
LOCAL x% : x% = FNdice5 + 5*FNdice5
IF x%>26 THEN = FNdice7 ELSE = (x%+1) MOD 7 + 1
 
DEF FNdice5 = RND(5)
 
DEF FNdistcheck(RETURN func%, repet%, delta)
LOCAL i%, m%, r%, s%, bins%()
DIM bins%(MAXRND)
FOR i% = 1 TO repet%
r% = FN(^func%)
bins%(r%) += 1
IF r%>m% m% = r%
NEXT
FOR i% = 1 TO m%
IF bins%(i%)/(repet%/m%) > 1+delta s% += 1
IF bins%(i%)/(repet%/m%) < 1-delta s% += 1
NEXT
= s%
Output:
Over 100 runs dice7 failed distribution check with 4 bin(s) out of range
Over 1000 runs dice7 failed distribution check with 2 bin(s) out of range
Over 10000 runs dice7 passed distribution check
Over 100000 runs dice7 passed distribution check

C[edit]

int rand5()
{
int r, rand_max = RAND_MAX - (RAND_MAX % 5);
while ((r = rand()) >= rand_max);
return r / (rand_max / 5) + 1;
}
 
int rand5_7()
{
int r;
while ((r = rand5() * 5 + rand5()) >= 27);
return r / 3 - 1;
}
 
int main()
{
printf(check(rand5, 5, 1000000, .05) ? "flat\n" : "not flat\n");
printf(check(rand7, 7, 1000000, .05) ? "flat\n" : "not flat\n");
return 0;
}
Output:
flat
flat

C++[edit]

This solution tries to minimize calls to the underlying d5 by reusing information from earlier calls.

template<typename F> class fivetoseven
{
public:
fivetoseven(F f): d5(f), rem(0), max(1) {}
int operator()();
private:
F d5;
int rem, max;
};
 
template<typename F>
int fivetoseven<F>::operator()()
{
while (rem/7 == max/7)
{
while (max < 7)
{
int rand5 = d5()-1;
max *= 5;
rem = 5*rem + rand5;
}
 
int groups = max / 7;
if (rem >= 7*groups)
{
rem -= 7*groups;
max -= 7*groups;
}
}
 
int result = rem % 7;
rem /= 7;
max /= 7;
return result+1;
}
 
int d5()
{
return 5.0*std::rand()/(RAND_MAX + 1.0) + 1;
}
 
fivetoseven<int(*)()> d7(d5);
 
int main()
{
srand(time(0));
test_distribution(d5, 1000000, 0.001);
test_distribution(d7, 1000000, 0.001);
}

C#[edit]

Translation of: Java
 
using System;
 
public class SevenSidedDice
{
Random random = new Random();
 
static void Main(string[] args)
{
SevenSidedDice sevenDice = new SevenSidedDice();
Console.WriteLine("Random number from 1 to 7: "+ sevenDice.seven());
Console.Read();
}
 
int seven()
{
int v=21;
while(v>20)
v=five()+five()*5-6;
return 1+v%7;
}
 
int five()
{
return 1 + random.Next(5);
}
}

Clojure[edit]

Uses the verify function defined in Verify distribution uniformity/Naive#Clojure

(def dice5 #(rand-int 5))
 
(defn dice7 []
(quot (->> dice5 ; do the following to dice5
(repeatedly 2) ; call it twice
(apply #(+ %1 (* 5 %2))) ; d1 + 5*d2 => 0..24
#() ; wrap that up in a function
repeatedly ; make infinite sequence of the above
(drop-while #(> % 20)) ; throw away anything > 20
first) ; grab first acceptable element
3)) ; divide by three rounding down
 
(doseq [n [100 1000 10000] [num count okay?] (verify dice7 n)]
(println "Saw" num count "times:"
(if okay? "that's" " not") "acceptable"))
Saw 0 10 times:    not acceptable
Saw 1 19 times:    not acceptable
Saw 2 12 times:    not acceptable
Saw 3 15 times: that's acceptable
Saw 4 11 times:    not acceptable
Saw 5 11 times:    not acceptable
Saw 6 22 times:    not acceptable
Saw 0 142 times: that's acceptable
Saw 1 158 times:    not acceptable
Saw 2 151 times: that's acceptable
Saw 3 153 times: that's acceptable
Saw 4 118 times:    not acceptable
Saw 5 139 times: that's acceptable
Saw 6 139 times: that's acceptable
Saw 0 1498 times: that's acceptable
Saw 1 1411 times: that's acceptable
Saw 2 1436 times: that's acceptable
Saw 3 1434 times: that's acceptable
Saw 4 1414 times: that's acceptable
Saw 5 1408 times: that's acceptable
Saw 6 1399 times: that's acceptable

Common Lisp[edit]

Translation of: C
(defun d5 ()
(1+ (random 5)))
 
(defun d7 ()
(loop for d55 = (+ (* 5 (d5)) (d5) -6)
until (< d55 21)
finally (return (1+ (mod d55 7)))))
> (check-distribution 'd7 1000)
Distribution potentially skewed for 1: expected around 1000/7 got 153.
Distribution potentially skewed for 2: expected around 1000/7 got 119.
Distribution potentially skewed for 3: expected around 1000/7 got 125.
Distribution potentially skewed for 7: expected around 1000/7 got 156.
T
#<EQL Hash Table{7} 200B5A53>

> (check-distribution 'd7 10000)
NIL
#<EQL Hash Table{7} 200CB5BB>

D[edit]

Translation of: C++
import std.random;
import verify_distribution_uniformity_naive: distCheck;
 
/// Generates a random number in [1, 5].
int dice5() /*pure nothrow*/ @safe {
return uniform(1, 6);
}
 
/// Naive, generates a random number in [1, 7] using dice5.
int fiveToSevenNaive() /*pure nothrow*/ @safe {
immutable int r = dice5() + dice5() * 5 - 6;
return (r < 21) ? (r % 7) + 1 : fiveToSevenNaive();
}
 
/**
Generates a random number in [1, 7] using dice5,
minimizing calls to dice5.
*/

int fiveToSevenSmart() @safe {
static int rem = 0, max = 1;
 
while (rem / 7 == max / 7) {
while (max < 7) {
immutable int rand5 = dice5() - 1;
max *= 5;
rem = 5 * rem + rand5;
}
 
immutable int groups = max / 7;
if (rem >= 7 * groups) {
rem -= 7 * groups;
max -= 7 * groups;
}
}
 
immutable int result = rem % 7;
rem /= 7;
max /= 7;
return result + 1;
}
 
void main() [email protected]*/ {
enum int N = 400_000;
distCheck(&dice5, N, 1);
distCheck(&fiveToSevenNaive, N, 1);
distCheck(&fiveToSevenSmart, N, 1);
}
Output:
1 80365
2 79941
3 80065
4 79784
5 79845

1 57186
2 57201
3 57180
4 57231
5 57124
6 56832
7 57246

1 57367
2 56869
3 57644
4 57111
5 57157
6 56809
7 57043

E[edit]

Translation of: Common Lisp
This example is in need of improvement:
Write dice7 in a prettier fashion and use the distribution checker once it's been written.
def dice5() {
return entropy.nextInt(5) + 1
}
 
def dice7() {
var d55 := null
while ((d55 := 5 * dice5() + dice5() - 6) >= 21) {}
return d55 %% 7 + 1
}
def bins := ([0] * 7).diverge()
for x in 1..1000 {
bins[dice7() - 1] += 1
}
println(bins.snapshot())

Elixir[edit]

defmodule Dice do
def dice5, do: :rand.uniform( 5 )
 
def dice7 do
dice7_from_dice5
end
 
defp dice7_from_dice5 do
d55 = 5*dice5 + dice5 - 6 # 0..24
if d55 < 21, do: rem( d55, 7 ) + 1,
else: dice7_from_dice5
end
end
 
fun5 = fn -> Dice.dice5 end
IO.inspect VerifyDistribution.naive( fun5, 1000000, 3 )
fun7 = fn -> Dice.dice7 end
IO.inspect VerifyDistribution.naive( fun7, 1000000, 3 )
Output:
:ok
:ok

Erlang[edit]

 
-module( dice ).
 
-export( [dice5/0, dice7/0, task/0] ).
 
dice5() -> random:uniform( 5 ).
 
dice7() ->
dice7_small_enough( dice5() * 5 + dice5() - 6 ). % 0 - 24
 
task() ->
verify_distribution_uniformity:naive( fun dice7/0, 1000000, 1 ).
 
 
 
dice7_small_enough( N ) when N < 21 -> N div 3 + 1;
dice7_small_enough( _N ) -> dice7().
 
Output:
76> dice:task().
ok

Factor[edit]

USING: kernel random sequences assocs locals sorting prettyprint
math math.functions math.statistics math.vectors math.ranges ;
IN: rosetta-code.dice7
 
! Output a random integer 1..5.
: dice5 ( -- x )
5 [1,b] random
;
 
! Output a random integer 1..7 using dice5 as randomness source.
: dice7 ( -- x )
0 [ dup 21 < ] [ drop dice5 5 * dice5 + 6 - ] do until
7 rem 1 +
;
 
! Roll the die by calling the quotation the given number of times and return
! an array with roll results.
! Sample call: 1000 [ dice7 ] roll
: roll ( times quot: ( -- x ) -- array )
[ call( -- x ) ] curry replicate
;
 
! Input array contains outcomes of a number of die throws. Each die result is
! an integer in the range 1..X. Calculate and return the number of each
! of the results in the array so that in the first position of the result
! there is the number of ones in the input array, in the second position
! of the result there is the number of twos in the input array, etc.
: count-dice-outcomes ( X array -- array )
histogram
swap [1,b] [ over [ 0 or ] change-at ] each
sort-keys values
;
 
! Verify distribution uniformity/Naive. Delta is the acceptable deviation
! from the ideal number of items in each bucket, expressed as a fraction of
! the total count. Sides is the number of die sides. Die-func is a word that
! produces a random number on stack in the range [1..sides], times is the
! number of times to call it.
! Sample call: 0.02 7 [ dice7 ] 100000 verify
:: verify ( delta sides die-func: ( -- random ) times -- )
sides
times die-func roll
count-dice-outcomes
dup .
times sides / :> ideal-count
ideal-count v-n vabs
times v/n
delta [ < ] curry all?
[ "Random enough" . ] [ "Not random enough" . ] if
;
 
 
! Call verify with 1, 10, 100, ... 1000000 rolls of 7-sided die.
: verify-all ( -- )
{ 1 10 100 1000 10000 100000 1000000 }
[| times | 0.02 7 [ dice7 ] times verify ] each
;
Output:
USE: rosetta-code.dice7 verify-all
{ 0 0 0 1 0 0 0 }
"Not random enough"
{ 0 2 3 1 1 1 2 }
"Not random enough"
{ 17 12 15 11 13 13 19 }
"Not random enough"
{ 140 130 141 148 143 155 143 }
"Random enough"
{ 1457 1373 1427 1433 1443 1382 1485 }
"Random enough"
{ 14225 14320 14216 14326 14415 14084 14414 }
"Random enough"
{ 142599 141910 142524 143029 143353 142696 143889 }
"Random enough"

Forth[edit]

Works with: GNU Forth
require random.fs
 
: d5 5 random 1+ ;
: discard? 5 = swap 1 > and ;
: d7
begin d5 d5 2dup discard? while 2drop repeat
1- 5 * + 1- 7 mod 1+ ;
Output:
cr ' d7 1000000 7 1 check-distribution .
lower bound = 141429  upper bound = 144285 
1 : 143241 ok
2 : 142397 ok
3 : 143522 ok
4 : 142909 ok
5 : 142001 ok
6 : 142844 ok
7 : 143086 ok
-1

Fortran[edit]

Works with: Fortran version 95 and later
module rand_mod
implicit none
 
contains
 
function rand5()
integer :: rand5
real :: r
 
call random_number(r)
rand5 = 5*r + 1
end function
 
function rand7()
integer :: rand7
 
do
rand7 = 5*rand5() + rand5() - 6
if (rand7 < 21) then
rand7 = rand7 / 3 + 1
return
end if
end do
end function
end module
 
program Randtest
use rand_mod
implicit none
 
integer, parameter :: samples = 1000000
 
call distcheck(rand7, samples, 0.005)
write(*,*)
call distcheck(rand7, samples, 0.001)
 
end program
Output:
Distribution Uniform

Distribution potentially skewed for bucket 1  Expected: 142857  Actual: 143142
Distribution potentially skewed for bucket 2  Expected: 142857  Actual: 143454
Distribution potentially skewed for bucket 3  Expected: 142857  Actual: 143540
Distribution potentially skewed for bucket 4  Expected: 142857  Actual: 142677
Distribution potentially skewed for bucket 5  Expected: 142857  Actual: 142511
Distribution potentially skewed for bucket 6  Expected: 142857  Actual: 142163
Distribution potentially skewed for bucket 7  Expected: 142857  Actual: 142513

Go[edit]

package main
 
import (
"fmt"
"math"
"math/rand"
"time"
)
 
// "given"
func dice5() int {
return rand.Intn(5) + 1
}
 
// function specified by task "Seven-sided dice from five-sided dice"
func dice7() (i int) {
for {
i = 5*dice5() + dice5()
if i < 27 {
break
}
}
return (i / 3) - 1
}
 
// function specified by task "Verify distribution uniformity/Naive"
//
// Parameter "f" is expected to return a random integer in the range 1..n.
// (Values out of range will cause an unceremonious crash.)
// "Max" is returned as an "indication of distribution achieved."
// It is the maximum delta observed from the count representing a perfectly
// uniform distribution.
// Also returned is a boolean, true if "max" is less than threshold
// parameter "delta."
func distCheck(f func() int, n int,
repeats int, delta float64) (max float64, flatEnough bool) {
count := make([]int, n)
for i := 0; i < repeats; i++ {
count[f()-1]++
}
expected := float64(repeats) / float64(n)
for _, c := range count {
max = math.Max(max, math.Abs(float64(c)-expected))
}
return max, max < delta
}
 
// Driver, produces output satisfying both tasks.
func main() {
rand.Seed(time.Now().UnixNano())
const calls = 1000000
max, flatEnough := distCheck(dice7, 7, calls, 500)
fmt.Println("Max delta:", max, "Flat enough:", flatEnough)
max, flatEnough = distCheck(dice7, 7, calls, 500)
fmt.Println("Max delta:", max, "Flat enough:", flatEnough)
}
Output:
Max delta: 356.1428571428696 Flat enough: true
Max delta: 787.8571428571304 Flat enough: false

Groovy[edit]

random = new Random()
 
int rand5() {
random.nextInt(5) + 1
}
 
int rand7From5() {
def raw = 25
while (raw > 21) {
raw = 5*(rand5() - 1) + rand5()
}
(raw % 7) + 1
}

Test:

def test = {
(1..6). each {
def counts = [0g, 0g, 0g, 0g, 0g, 0g, 0g]
def target = 10g**it
def popSize = 7*target
(0..<(popSize)).each {
def i = rand7From5() - 1
counts[i] = counts[i] + 1g
}
BigDecimal stdDev = (counts.collect { it - target}.collect { it * it }.sum() / popSize) ** 0.5g
def countMap = (0..<counts.size()).inject([:]) { map, index -> map + [(index+1):counts[index]] }
 
println """\
counts: ${countMap}
population size: ${popSize}
std dev: ${stdDev.round(new java.math.MathContext(3))}
"""

}
}
 
4.times {
println """
TRIAL #${it+1}
=============="""

test(it)
}
Output:
TRIAL #1
==============
         counts: [1:16, 2:10, 3:9, 4:7, 5:12, 6:8, 7:8]
population size: 70
        std dev: 0.910

         counts: [1:85, 2:97, 3:108, 4:110, 5:95, 6:105, 7:100]
population size: 700
        std dev: 0.800

         counts: [1:990, 2:1008, 3:992, 4:1060, 5:1008, 6:997, 7:945]
population size: 7000
        std dev: 0.995

         counts: [1:9976, 2:10007, 3:10009, 4:9858, 5:10109, 6:9988, 7:10053]
population size: 70000
        std dev: 0.714

         counts: [1:100310, 2:99783, 3:99843, 4:100353, 5:99804, 6:99553, 7:100354]
population size: 700000
        std dev: 0.968

         counts: [1:999320, 2:1000942, 3:1000201, 4:1000878, 5:999181, 6:999632, 7:999846]
population size: 7000000
        std dev: 0.654


TRIAL #2
==============
         counts: [1:10, 2:8, 3:9, 4:9, 5:14, 6:7, 7:13]
population size: 70
        std dev: 0.756

         counts: [1:104, 2:101, 3:97, 4:108, 5:100, 6:87, 7:103]
population size: 700
        std dev: 0.619

         counts: [1:995, 2:970, 3:1001, 4:953, 5:1006, 6:1081, 7:994]
population size: 7000
        std dev: 1.18

         counts: [1:10013, 2:10063, 3:9843, 4:9984, 5:9986, 6:10059, 7:10052]
population size: 70000
        std dev: 0.711

         counts: [1:100048, 2:99647, 3:100240, 4:100683, 5:99813, 6:100320, 7:99249]
population size: 700000
        std dev: 1.39

         counts: [1:1000579, 2:1000541, 3:999497, 4:1000805, 5:999708, 6:999161, 7:999709]
population size: 7000000
        std dev: 0.586


TRIAL #3
==============
         counts: [1:9, 2:8, 3:11, 4:14, 5:10, 6:11, 7:7]
population size: 70
        std dev: 0.676

         counts: [1:100, 2:92, 3:105, 4:107, 5:111, 6:91, 7:94]
population size: 700
        std dev: 0.733

         counts: [1:1010, 2:1053, 3:967, 4:981, 5:1027, 6:959, 7:1003]
population size: 7000
        std dev: 0.984

         counts: [1:9857, 2:10037, 3:9992, 4:10231, 5:9828, 6:10140, 7:9915]
population size: 70000
        std dev: 1.37

         counts: [1:99650, 2:99580, 3:99848, 4:100507, 5:99916, 6:100212, 7:100287]
population size: 700000
        std dev: 1.01

         counts: [1:1001710, 2:999667, 3:1000685, 4:1000411, 5:999369, 6:998469, 7:999689]
population size: 7000000
        std dev: 0.965


TRIAL #4
==============
         counts: [1:12, 2:7, 3:11, 4:12, 5:7, 6:9, 7:12]
population size: 70
        std dev: 0.676

         counts: [1:97, 2:96, 3:101, 4:93, 5:96, 6:124, 7:93]
population size: 700
        std dev: 1.01

         counts: [1:985, 2:1023, 3:1018, 4:1023, 5:995, 6:973, 7:983]
population size: 7000
        std dev: 0.615

         counts: [1:9948, 2:9968, 3:10131, 4:10050, 5:9990, 6:10039, 7:9874]
population size: 70000
        std dev: 0.764

         counts: [1:100125, 2:99616, 3:99912, 4:100286, 5:99674, 6:100190, 7:100197]
population size: 700000
        std dev: 0.787

         counts: [1:1001267, 2:999911, 3:1000602, 4:999483, 5:1000549, 6:998725, 7:999463]
population size: 7000000
        std dev: 0.798

Haskell[edit]

import System.Random
import Data.List
 
sevenFrom5Dice = do
d51 <- randomRIO(1,5) :: IO Int
d52 <- randomRIO(1,5) :: IO Int
let d7 = 5*d51+d52-6
if d7 > 20 then sevenFrom5Dice
else return $ 1 + d7 `mod` 7
Output:
*Main> replicateM 10 sevenFrom5Dice
[2,3,1,1,6,2,5,6,5,3]

Test:

*Main> mapM_ print .sort =<< distribCheck sevenFrom5Dice 1000000 3
(1,(142759,True))
(2,(143078,True))
(3,(142706,True))
(4,(142403,True))
(5,(142896,True))
(6,(143028,True))
(7,(143130,True))

Icon and Unicon[edit]

Translation of: Ruby

Uses verify_uniform from here.

 
$include "distribution-checker.icn"
 
# return a uniformly distributed number from 1 to 7,
# but only using a random number in range 1 to 5.
procedure die_7 ()
while rnd := 5*?5 + ?5 - 6 do {
if rnd < 21 then suspend rnd % 7 + 1
}
end
 
procedure main ()
if verify_uniform (create (|die_7()), 1000000, 0.01)
then write ("uniform")
else write ("skewed")
end
 
Output:
5 142870
2 142812
7 142901
4 142960
1 143113
6 142706
3 142638
uniform

J[edit]

The first step is to create 7-sided dice rolls from 5-sided dice rolls (rollD5):

rollD5=: [: >: ] ?@$ 5:      NB. makes a y shape array of 5s, "rolls" the array and increments.
roll2xD5=: [: rollD5 2 ,~ */ NB. rolls D5 twice for each desired D7 roll (y rows, 2 cols)
toBase10=: 5 #. <: NB. decrements and converts rows from base 5 to 10
keepGood=: #~ 21&> NB. compress out values not less than 21
groupin3s=: [: >. >: % 3: NB. increments, divides by 3 and takes ceiling
 
getD7=: [email protected]@[email protected]

Here are a couple of variations on the theme that achieve the same result:

getD7b=: 0 8 -.~ 3 >.@%~ 5 #. [: <:@rollD5 2 ,~ ]
getD7c=: [: (#~ 7&>:) 3 >.@%~ [: 5&#.&.:<:@rollD5 ] , 2:

The trouble is that we probably don't have enough D7 rolls yet because we compressed out any double D5 rolls that evaluated to 21 or more. So we need to accumulate some more D7 rolls until we have enough. J has two types of verb definition - tacit (arguments not referenced) and explicit (more conventional function definitions) illustrated below:

Here's an explicit definition that accumulates rolls from getD7:

rollD7x=: monad define
n=. */y NB. product of vector y is total number of D7 rolls required
rolls=. '' NB. initialize empty noun rolls
while. n > #rolls do. NB. checks if if enough D7 rolls accumulated
rolls=. rolls, getD7 >. 0.75 * n NB. calcs 3/4 of required rolls and accumulates getD7 rolls
end.
y $ rolls NB. shape the result according to the vector y
)

Here's a tacit definition that does the same thing:

getNumRolls=: [: >. 0.75 * */@[       NB. calc approx 3/4 of the required rolls
accumD7Rolls=: ] , [email protected] NB. accumulates getD7 rolls
isNotEnough=: */@[ > #@] NB. checks if enough D7 rolls accumulated
 
rollD7t=: ] $ (accumD7Rolls ^: isNotEnough ^:_)&''

The verb1 ^: verb2 ^:_ construct repeats x verb1 y while x verb2 y is true. It is like saying "Repeat accumD7Rolls while isNotEnough".

Example usage:

   rollD7t 10         NB. 10 rolls of D7
6 4 5 1 4 2 4 5 2 5
rollD7t 2 5 NB. 2 by 5 array of D7 rolls
5 1 5 1 3
3 4 3 5 6
rollD7t 2 3 5 NB. 2 by 3 by 5 array of D7 rolls
4 7 7 5 7
3 7 1 4 5
5 4 5 7 6
 
1 1 7 6 3
4 4 1 4 4
1 1 1 6 5
 
NB. check results from rollD7x and rollD7t have same shape
([email protected] -: [email protected]) 10
1
([email protected] -: [email protected]) 2 3 5
1

Java[edit]

Translation of: Python
import java.util.Random;
public class SevenSidedDice
{
private static final Random rnd = new Random();
public static void main(String[] args)
{
SevenSidedDice now=new SevenSidedDice();
System.out.println("Random number from 1 to 7: "+now.seven());
}
int seven()
{
int v=21;
while(v>20)
v=five()+five()*5-6;
return 1+v%7;
}
int five()
{
return 1+rnd.nextInt(5);
}
}

JavaScript[edit]

Translation of: Ruby
function dice5()
{
return 1 + Math.floor(5 * Math.random());
}
 
function dice7()
{
while (true)
{
var dice55 = 5 * dice5() + dice5() - 6;
if (dice55 < 21)
return dice55 % 7 + 1;
}
}
 
distcheck(dice5, 1000000);
print();
distcheck(dice7, 1000000);
Output:
1       199792
2       200425
3       199243
4       200407
5       200133

1       143617
2       142209
3       143023
4       142990
5       142894
6       142648
7       142619 

Julia[edit]

dice5() = rand(1:5)
 
function dice7()
r = 5*dice5() + dice5() - 6
r < 21 ? (r%7 + 1) : dice7()
end

Distribution check:

julia> hist([dice5() for i=1:10^6])
(0:1:5,[199932,200431,199969,199925,199743])

julia> hist([dice7() for i=1:10^6])
(0:1:7,[142390,143032,142837,142999,142800,142642,143300])

Liberty BASIC[edit]

 
n=1000000 '1000000 would take several minutes
print "Testing ";n;" times"
if not(check(n, 0.05)) then print "Test failed" else print "Test passed"
end
 
'function check(n, delta) is defined at
'http://rosettacode.org/wiki/Verify_distribution_uniformity/Naive#Liberty_BASIC
 
function GENERATOR()
'GENERATOR = int(rnd(0)*10) '0..9
'GENERATOR = 1+int(rnd(0)*5) '1..5: dice5
 
'dice7()
do
temp =dice5() *5 +dice5() -6
loop until temp <21
GENERATOR =( temp mod 7) +1
 
end function
 
function dice5()
dice5=1+int(rnd(0)*5) '1..5: dice5
end function
 
Output:
Testing 1000000 times
minVal        Expected      maxVal
135714        142857        150000
Bucket        Counter       pass/fail
1             143310
2             143500
3             143040
4             145185
5             140998
6             142610
7             141357
Test passed

Lua[edit]

dice5 = function() return math.random(5) end
 
function dice7()
x = dice5() * 5 + dice5() - 6
if x > 20 then return dice7() end
return x%7 + 1
end

Mathematica[edit]

sevenFrom5Dice := (tmp$ = 5*RandomInteger[{1, 5}] + RandomInteger[{1, 5}] - 6; 
If [tmp$ < 21, 1 + Mod[tmp$ , 7], sevenFrom5Dice])
CheckDistribution[sevenFrom5Dice, 1000000, 5]
->Expected: 142857., Generated :{142206,142590,142650,142693,142730,143475,143656}
->"Flat"

OCaml[edit]

let dice5() = 1 + Random.int 5 ;;
 
let dice7 =
let rolls2answer = Hashtbl.create 25 in
let n = ref 0 in
for roll1 = 1 to 5 do
for roll2 = 1 to 5 do
Hashtbl.add rolls2answer (roll1,roll2) (!n / 3 +1);
incr n
done;
done;
let rec aux() =
let trial = Hashtbl.find rolls2answer (dice5(),dice5()) in
if trial <= 7 then trial else aux()
in
aux
;;

PARI/GP[edit]

dice5()=random(5)+1;
 
dice7()={
my(t);
while((t=dice5()*5+dice5()) > 21,);
(t+2)\3
};

Perl[edit]

Using dice5 twice to generate numbers in the range 0 to 24. If we consider these modulo 8 and re-call if we get zero, we have eliminated 4 cases and created the necessary number in the range from 1 to 7.

sub dice5 { 1+int rand(5) }
 
sub dice7 {
while(1) {
my $d7 = (5*dice5()+dice5()-6) % 8;
return $d7 if $d7;
}
}
 
my %count7;
my $n = 1000000;
$count7{dice7()}++ for 1..$n;
printf "%s: %5.2f%%\n", $_, 100*($count7{$_}/$n*7-1) for sort keys %count7;
 
Output:
1:  0.05%
2:  0.16%
3: -0.43%
4:  0.11%
5:  0.01%
6: -0.15%
7:  0.24%

Perl 6[edit]

Works with: Rakudo Star version 2010.09

Since rakudo is still pretty slow, we've done some interesting bits of optimization. We factor out the range object construction so that it doesn't have to be recreated each time, and we sneakily subtract the 1's from the 5's, which takes us back to 0 based without having to subtract 6.

my $d5 = 1..5;
sub d5() { $d5.roll; } # 1d5
 
sub d7() {
my $flat = 21;
$flat = 5 * d5() - d5() until $flat < 21;
$flat % 7 + 1;
}

Here's the test. We use a C-style for loop, except it's named loop, because it's currently faster than the other loops--and, er, doesn't segfault the GC on a million iterations...

my @dist;
my $n = 1_000_000;
my $expect = $n / 7;
 
loop ($_ = $n; $n; --$n) { @dist[d7()]++; }
 
say "Expect\t",$expect.fmt("%.3f");
for @dist.kv -> $i, $v {
say "$i\t$v\t" ~ (($v - $expect)/$expect*100).fmt("%+.2f%%") if $v;
}
Output:
Expect	142857.143
1	142835	-0.02%
2	143021	+0.11%
3	142771	-0.06%
4	142706	-0.11%
5	143258	+0.28%
6	142485	-0.26%
7	142924	+0.05%

PicoLisp[edit]

(de dice5 ()
(rand 1 5) )
 
(de dice7 ()
(use R
(until (> 21 (setq R (+ (* 5 (dice5)) (dice5) -6))))
(inc (% R 7)) ) )
Output:
: (let R NIL
   (do 1000000 (accu 'R (dice7) 1))
   (sort R) )
-> ((1 . 142295) (2 . 142491) (3 . 143448) (4 . 143129) (5 . 142701) (6 . 143142) (7 . 142794))

PureBasic[edit]

Translation of: Lua
Procedure dice5()
ProcedureReturn Random(4) + 1
EndProcedure
 
Procedure dice7()
Protected x
 
x = dice5() * 5 + dice5() - 6
If x > 20
ProcedureReturn dice7()
EndIf
 
ProcedureReturn x % 7 + 1
EndProcedure

Python[edit]

from random import randint
 
def dice5():
return randint(1, 5)
 
def dice7():
r = dice5() + dice5() * 5 - 6
return (r % 7) + 1 if r < 21 else dice7()

Distribution check using Simple Random Distribution Checker:

>>> distcheck(dice5, 1000000, 1)
{1: 200244, 2: 199831, 3: 199548, 4: 199853, 5: 200524}
>>> distcheck(dice7, 1000000, 1)
{1: 142853, 2: 142576, 3: 143067, 4: 142149, 5: 143189, 6: 143285, 7: 142881}

Racket[edit]

 
#lang racket
(define (dice5) (add1 (random 5)))
 
(define (dice7)
(define res (+ (* 5 (dice5)) (dice5) -6))
(if (< res 21) (+ 1 (modulo res 7)) (dice7)))
 

Checking the uniformity using math library:

 
-> (require math/statistics)
-> (samples->hash (for/list ([i 700000]) (dice7)))
'#hash((7 . 100392)
(6 . 100285)
(5 . 99774)
(4 . 100000)
(3 . 100000)
(2 . 99927)
(1 . 99622))
 

R[edit]

5-sided die.

dice5 <- function(n=1) sample(5, n, replace=TRUE)

Simple but slow 7-sided die, using a while loop.

dice7.while <- function(n=1) 
{
score <- numeric()
while(length(score) < n)
{
total <- sum(c(5,1) * dice5(2)) - 3
if(total < 24) score <- c(score, total %/% 3)
}
score
}
system.time(dice7.while(1e6)) # longer than 4 minutes

More complex, but much faster vectorised version.

dice7.vec <- function(n=1, checkLength=TRUE) 
{
morethan2n <- 3 * n + 10 + (n %% 2) #need more than 2*n samples, because some are discarded
twoDfive <- matrix(dice5(morethan2n), nrow=2)
total <- colSums(c(5, 1) * twoDfive) - 3
score <- ifelse(total < 24, total %/% 3, NA)
score <- score[!is.na(score)]
#If length is less than n (very unlikely), add some more samples
if(checkLength)
{
while(length(score) < n)
{
score <- c(score, dice7(n, FALSE))
}
score[1:n]
} else score
}
system.time(dice7.vec(1e6)) # ~1 sec

REXX[edit]

/*REXX program simulate a  7─sided  die    based  on a    5─sided throw.      */
parse arg trials sample . /*obtain optional arguments from the CL*/
if trials=='' then trials=1 /*Not specified? Then use the default.*/
if sample=='' then sample=1000000 /* " " " " " " */
 
do t=1 for trials /*performs the number of desired trials*/
die.=0; k=0
do until k==sample; r=5*random(1,5) + random(1,5) - 6
if r>20 then iterate
k=k+1; r=r//7 + 1; die.r=die.r + 1
end /*until*/
expect=sample%7
say
say center('trial:'right(t,4) ' ' sample 'samples, expect='expect,79,'─')
say
do j=1 for 7
say ' side' j "had " die.j ' occurrences',
' difference from expected:'right(die.j-expect, length(sample))
end /*j*/
end /*t*/
/*stick a fork in it, we're all done. */

output   when using the following input is used:   11

─────────────────trial:   1     1000000 samples, expect=142857─────────────────

      side 1 had  142879  occurrences       difference from expected:     22
      side 2 had  142980  occurrences       difference from expected:    123
      side 3 had  142366  occurrences       difference from expected:   -491
      side 4 had  143129  occurrences       difference from expected:    272
      side 5 had  143316  occurrences       difference from expected:    459
      side 6 had  142742  occurrences       difference from expected:   -115
      side 7 had  142588  occurrences       difference from expected:   -269

─────────────────trial:   2     1000000 samples, expect=142857─────────────────

      side 1 had  142583  occurrences       difference from expected:   -274
      side 2 had  142797  occurrences       difference from expected:    -60
      side 3 had  143093  occurrences       difference from expected:    236
      side 4 had  142168  occurrences       difference from expected:   -689
      side 5 had  143203  occurrences       difference from expected:    346
      side 6 had  143260  occurrences       difference from expected:    403
      side 7 had  142896  occurrences       difference from expected:     39

─────────────────trial:   3     1000000 samples, expect=142857─────────────────

      side 1 had  143089  occurrences       difference from expected:    232
      side 2 had  142815  occurrences       difference from expected:    -42
      side 3 had  142645  occurrences       difference from expected:   -212
      side 4 had  142959  occurrences       difference from expected:    102
      side 5 had  142764  occurrences       difference from expected:    -93
      side 6 had  143205  occurrences       difference from expected:    348
      side 7 had  142523  occurrences       difference from expected:   -334

─────────────────trial:   4     1000000 samples, expect=142857─────────────────

      side 1 had  142535  occurrences       difference from expected:   -322
      side 2 had  142316  occurrences       difference from expected:   -541
      side 3 had  143101  occurrences       difference from expected:    244
      side 4 had  143159  occurrences       difference from expected:    302
      side 5 had  143130  occurrences       difference from expected:    273
      side 6 had  142551  occurrences       difference from expected:   -306
      side 7 had  143208  occurrences       difference from expected:    351

─────────────────trial:   5     1000000 samples, expect=142857─────────────────

      side 1 had  143335  occurrences       difference from expected:    478
      side 2 had  142296  occurrences       difference from expected:   -561
      side 3 had  142500  occurrences       difference from expected:   -357
      side 4 had  142775  occurrences       difference from expected:    -82
      side 5 had  142581  occurrences       difference from expected:   -276
      side 6 had  143363  occurrences       difference from expected:    506
      side 7 had  143150  occurrences       difference from expected:    293

─────────────────trial:   6     1000000 samples, expect=142857─────────────────

      side 1 had  142593  occurrences       difference from expected:   -264
      side 2 had  143012  occurrences       difference from expected:    155
      side 3 had  142703  occurrences       difference from expected:   -154
      side 4 had  143702  occurrences       difference from expected:    845
      side 5 had  142867  occurrences       difference from expected:     10
      side 6 had  142158  occurrences       difference from expected:   -699
      side 7 had  142965  occurrences       difference from expected:    108

─────────────────trial:   7     1000000 samples, expect=142857─────────────────

      side 1 had  143065  occurrences       difference from expected:    208
      side 2 had  142892  occurrences       difference from expected:     35
      side 3 had  142681  occurrences       difference from expected:   -176
      side 4 had  142637  occurrences       difference from expected:   -220
      side 5 had  142563  occurrences       difference from expected:   -294
      side 6 had  142934  occurrences       difference from expected:     77
      side 7 had  143228  occurrences       difference from expected:    371

─────────────────trial:   8     1000000 samples, expect=142857─────────────────

      side 1 had  142745  occurrences       difference from expected:   -112
      side 2 had  142906  occurrences       difference from expected:     49
      side 3 had  142894  occurrences       difference from expected:     37
      side 4 had  142976  occurrences       difference from expected:    119
      side 5 had  142265  occurrences       difference from expected:   -592
      side 6 had  142909  occurrences       difference from expected:     52
      side 7 had  143305  occurrences       difference from expected:    448

─────────────────trial:   9     1000000 samples, expect=142857─────────────────

      side 1 had  142320  occurrences       difference from expected:   -537
      side 2 had  142510  occurrences       difference from expected:   -347
      side 3 had  142867  occurrences       difference from expected:     10
      side 4 had  143711  occurrences       difference from expected:    854
      side 5 had  142732  occurrences       difference from expected:   -125
      side 6 had  143115  occurrences       difference from expected:    258
      side 7 had  142745  occurrences       difference from expected:   -112

─────────────────trial:  10     1000000 samples, expect=142857─────────────────

      side 1 had  142725  occurrences       difference from expected:   -132
      side 2 had  142865  occurrences       difference from expected:      8
      side 3 had  143076  occurrences       difference from expected:    219
      side 4 had  142759  occurrences       difference from expected:    -98
      side 5 had  142593  occurrences       difference from expected:   -264
      side 6 had  142750  occurrences       difference from expected:   -107
      side 7 had  143232  occurrences       difference from expected:    375

─────────────────trial:  11     1000000 samples, expect=142857─────────────────

      side 1 had  142825  occurrences       difference from expected:    -32
      side 2 had  142870  occurrences       difference from expected:     13
      side 3 had  142919  occurrences       difference from expected:     62
      side 4 had  143098  occurrences       difference from expected:    241
      side 5 had  142627  occurrences       difference from expected:   -230
      side 6 had  143163  occurrences       difference from expected:    306
      side 7 had  142498  occurrences       difference from expected:   -359

Ruby[edit]

Translation of: Tcl

Uses distcheck from here.

require './distcheck.rb'
 
def d5
1 + rand(5)
end
 
def d7
loop do
d55 = 5*d5 + d5 - 6
return (d55 % 7 + 1) if d55 < 21
end
end
 
distcheck(1_000_000) {d5}
distcheck(1_000_000) {d7}
Output:
1 200227
2 200264
3 199777
4 199387
5 200345
1 143175
2 143031
3 142731
4 142716
5 142931
6 142605
7 142811

Sidef[edit]

Translation of: Perl
func dice5 { 1 + 5.rand.int }
 
func dice7 {
loop {
var d7 = ((5*dice5() + dice5() - 6) % 8);
d7 && return d7;
}
}
 
var count7 = Hash.new;
 
var n = 1e6;
n.times { count7{dice7()} := 0 ++ }
count7.keys.sort.each { |k|
printf("%s: %5.2f%%\n", k, 100*(count7{k}/n * 7 - 1));
}
Output:
1: -0.00%
2:  0.02%
3:  0.23%
4:  0.42%
5: -0.23%
6: -0.54%
7:  0.10%

Tcl[edit]

Any old D&D hand will know these as a D5 and a D7...

proc D5 {} {expr {1 + int(5 * rand())}}
 
proc D7 {} {
while 1 {
set d55 [expr {5 * [D5] + [D5] - 6}]
if {$d55 < 21} {
return [expr {$d55 % 7 + 1}]
}
}
}

Checking:

% distcheck D5 1000000
1 199893 2 200162 3 200075 4 199630 5 200240
% distcheck D7 1000000
1 143121 2 142383 3 143353 4 142811 5 142172 6 143291 7 142869

VBScript[edit]

Option Explicit
 
function dice5
dice5 = int(rnd*5) + 1
end function
 
function dice7
dim j
do
j = 5 * dice5 + dice5 - 6
loop until j < 21
dice7 = j mod 7 + 1
end function

zkl[edit]

var die5=(1).random.fp(6); // [1..5]
fcn die7{ while((r:=5*die5() + die5())>=27){} r/3-1 }
 
fcn rtest(N){ //test spread over [0..9]
dist:=L(0,0,0,0,0,0,0,0,0,0);
do(N){ dist[die7()]+=1 }
sum:=dist.sum();
dist=dist.apply('wrap(n){ "%.2f%%".fmt(n.toFloat()/sum*100) }).println();
}
 
println("Looking for ",100.0/7,"%");
rtest(0d1_000_000);
Output:
Looking for 14.2857%
L("0.00%","14.28%","14.36%","14.22%","14.26%","14.34%","14.33%","14.21%","0.00%","0.00%")