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.
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 1000000 calls using the function created in Simple Random Distribution Checker.
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)
AutoHotkey
<lang AutoHotkey>dice5() { Random, v, 1, 5
Return, v
}
dice7() { Loop
{ v := 5 * dice5() + dice5() - 6
IfLess v, 21, Return, (v // 3) + 1
}
}</lang>
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.
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <math.h>
void distcheck(int (*)(), int, double);
int dice5() {
return 1 + (int)(5.0*rand() / (RAND_MAX + 1.0));
}
int dice7() {
int d55;
do {
d55 = 5*dice5() + dice5() - 6;
} while(d55 >= 21);
return d55 % 7 + 1;
}
int main() {
distcheck(dice5, 1000000, 1); distcheck(dice7, 1000000, 1); return 0;
}</lang>
C++
This solution tries to minimize calls to the underlying d5 by reusing information from earlier calls.
<lang cpp>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);
}</lang>
Common Lisp
<lang lisp>(defun d5 ()
(1+ (random 5)))
(defun d7 ()
(loop for d55 = (+ (* 5 (d5)) (d5) -6)
until (< d55 21)
finally (return (1+ (mod d55 7)))))</lang>
> (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
<lang d>import std.random: rand; import distcheck: distCheck;
/// generates a random number in [1, 5] int dice5() {
return 1 + cast(int)((rand() / cast(double)uint.max) * 5.0);
}
/// Naive version, generates a random number in [1, 7] using dice5
int dice7() {
int r = dice5() + dice5() * 5 - 6; return (r < 21) ? (r % 7) + 1 : dice7();
}
/**
Generates a random number in [1, 7] using dice5,
minimizing calls to dice 5.
- /
struct FiveToSeven(alias d5) {
int opCall() {
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;
}
private:
int rem = 0, max = 1;
}
void main() {
const int N = 1_000_000; distCheck(&dice5, N, 1); distCheck(&dice7, N, 1); // naive version FiveToSeven!(dice5) dice7b; distCheck(&dice7b.opCall, N, 1);
}</lang>
Output: [1:200416,2:199418,3:199471,4:200016,5:200679] [1:142443,2:143605,3:142878,4:143038,5:142595,6:142205,7:143236] [1:143354,2:143240,3:142882,4:142641,5:142715,6:142779,7:142389]
E
<lang e>def dice5() {
return entropy.nextInt(5) + 1
}
def dice7() {
var d55 := null
while ((d55 := 5 * dice5() + dice5() - 6) >= 21) {}
return d55 %% 7 + 1
}</lang>
<lang e>def bins := ([0] * 7).diverge() for x in 1..1000 {
bins[dice7() - 1] += 1
} println(bins.snapshot())</lang>
Haskell
<lang haskell>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</lang>
Output: <lang haskell>*Main> replicateM 10 sevenFrom5Dice [2,3,1,1,6,2,5,6,5,3]</lang> Test: <lang haskell>*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))</lang>
J
The first step is to create 7-sided dice rolls from 5-sided dice rolls (rollD5):
<lang j>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=: groupin3s@keepGood@toBase10@roll2xD5</lang>
Here are a couple of variations on the theme that achieve the same result: <lang j>getD7b=: 0 8 -.~ 3 >.@%~ 5 #. [: <:@rollD5 2 ,~ ] getD7c=: [: (#~ 7&>:) 3 >.@%~ [: 5&#.&.:<:@rollD5 ] , 2:</lang>
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:
<lang j>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 > #res 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
)</lang>
Here's a tacit definition that does the same thing: <lang j>getNumRolls=: [: >. 0.75 * */@[ NB. calc approx 3/4 of the required rolls accumD7Rolls=: ] , getD7@getNumRolls NB. accumulates getD7 rolls isNotEnough=: */@[ > #@] NB. checks if enough D7 rolls accumulated
rollD7t=: ] $ (accumD7Rolls ^: isNotEnough ^:_)&</lang>
The verb1 ^: verb2 ^:_ construct repeats x verb1 y while x verb2 y is true. It is like saying "Repeat accumD7Rolls while isNotEnough".
Example usage: <lang j> 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
($@rollD7x -: $@rollD7t) 10
1
($@rollD7x -: $@rollD7t) 2 3 5
1</lang>
JavaScript
<lang javascript>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);</lang> 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
OCaml
<lang ocaml>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
- </lang>
Python
<lang python>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()</lang>
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}
R
5-sided die. <lang r>dice5 <- function(n=1) sample(5, n, replace=TRUE)</lang> Simple but slow 7-sided die, using a while loop. <lang r>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</lang> More complex, but much faster vectorised version. <lang r>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</lang>
Ruby
Uses distcheck from here.
<lang ruby>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}</lang>
output
1 200478 2 199986 3 199582 4 199560 5 200394 1 142371 2 142577 3 143328 4 143630 5 142553 6 142692 7 142849
Tcl
Any old D&D hand will know these as a D5 and a D7... <lang tcl>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}]
}
}
}</lang> 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