# Rand5() to Rand7() [closed]

You are provided with a function Rand5(). This function returns perfectly random (equal distribution) integers between 1 and 5.

Provide the function Rand7(), which uses Rand5() to produce perfectly random integers between 1 and 7.

• Duplicate question Commented Jun 17, 2011 at 5:23
• Obligatory xkcd: xkcd.com/221 Commented Sep 28, 2011 at 14:39
• 1 and 5 inclusive? i.e. from the set {1,2,3,4,5} ? Commented Dec 8, 2011 at 22:54
• What criteria determines a single winner? Commented Jan 26, 2013 at 20:51
• That moment when you realize this is actually an old question.
– user344
Commented Dec 20, 2014 at 12:33

## 31 Answers

Java - 61 chars

int rand7(){int s=0,c=7;while(c-->0)s+=rand5();return s%7+1;}


Test driver for validation:

class Rand {

public static void main(String[] args) {
int[] nums = new int[7];
// get a lot of numbers
for(int i = 0; i < 10000000; i++) nums[rand7()-1]++;
// print the results
for(int i = 0; i < 7; i++) System.out.println((i+1) + ": " + nums[i]);
}

// just for rand5()
static java.util.Random r = new java.util.Random();

static int rand5() {
return r.nextInt(5)+1; // Random.nextInt(n) returns 0..n-1, so add 1
}

static int rand7(){int s=0,c=7;while(c-->0)s+=rand5();return s%7+1;}

}


Results

C:\Documents and Settings\glowcoder\My Documents>java Rand
1: 1429828
2: 1429347
3: 1428328
4: 1426486
5: 1426784
6: 1429853
7: 1429374

C:\Documents and Settings\glowcoder\My Documents>

• Extra points for the "goes to operator" Commented Jun 25, 2011 at 15:46
• shave a char? int rand7(){for(int s=0,c=7;c-->0;s+=rand5());return s%7+1;}
– Ron
Commented Sep 30, 2011 at 15:46
• This answer is not correct: the probabilities of this function returning the values from 1 to 7 are 0.1430656, 0.1430016, 0.1428224, 0.1426432, 0.1426432, 0.1428224 and 0.1430016 respectively. Yes, the difference between the minimum and maximum probabilities is less that 0.0005, but still, the question specified "perfectly random integers". Commented Dec 1, 2011 at 23:12
• @ilmari You're right - I just ran a test and it seems the distribution isn't even... let me think on that... Commented Dec 1, 2011 at 23:22
• @userunknown: Yes, the probabilities I posted are actually not approximations, they exact (0.1430656 = 11177 / 78125, etc.) assuming a perfectly random rand5. I computed them in Maple using simple matrix algebra, but you can do it with pencil and paper in a few minutes if you want. Anyway, it turns out Omar already posted the same figures (sans normalizing factor) in a comment to another answer a couple of days earlier. (Also ps., you can only @notify one user per comment, although the post's author is notified always in any case.) Commented Dec 8, 2011 at 0:27

## Perl - 47 (was 52) chars

sub rand7{($x=5*&rand5+&rand5-3)<24?int($x/3):&rand7}


Plus I get to use the ternary operator AND recursion. Best... day... ever!

OK, 47 chars if you use mod instead of div:

sub rand7{($x=5*&rand5+&rand5)<27?$x%7+1:&rand7}

• So close... replace the 30 with 27 (= 6 + 21) and you'll get a perfectly uniform distribution. Oh, and you can drop the last two & signs to get it down to 46 chars (including the space, which puts your current version at 48). Commented Dec 1, 2011 at 23:25

In Python:

def Rand7():
while True:
x = (Rand5() - 1) * 5 + (Rand5() - 1)
if x < 21: return x/3 + 1

• in Python 3.X, use return x//3 + 1 Commented Mar 9, 2020 at 8:58

Ruby - 54 chars (based on Dan McGrath solution, using loop)

def rand7;x=8;while x>7 do x=rand5+5*rand5-5 end;x;end


Ruby - 45 chars (same solution, using recursion)

def rand7;x=rand5+5*rand5-5;x>7 ?rand7: x;end

• Can be shortened by 1 char by using (x=rand5+5*rand5-5)>7?. Commented May 9, 2011 at 13:17

## JavaScript, 42

Rand7=f=_=>(x=Rand5()+Rand5()*5-5)>7?f():x


Bonus ES5 thing:

Rand7=eval.bind(0,'for(;x=Rand5()+Rand5()*5-5,x>7;);x')


In Common Lisp 70 characters:

(defun rand7()(let((n(-(+(rand5)(* 5(rand5)))5)))(if(> n 7)(rand7)n)))


The parenthesis take up more space than I would like.

• Nice. You can squeeze out two more characters by making n a global variable: (defun rand7()(setq n(-(+(rand5)(* 5(rand5)))5))(if(> n 7)(rand7)n)) Commented Apr 29, 2011 at 18:19
• Even better: (defun rand7()(if(>(setq n(-(+(rand5)(* 5(rand5)))5))7)(rand7)n)) Commented Apr 29, 2011 at 18:51

Translation to PHP, from the answer posted ny Dan McGrath.

function Rand7(){$x=8;while($x>7)$x=rand5()+5*rand5()-5;return$x;}


67 characters.

• Shouldn't this be prefixed by the word "function" (and a space)? Commented Jan 28, 2011 at 0:28
• Yeah... and now it's 67 characters... Commented Jan 28, 2011 at 0:35

In c/c++ using rejection sampling

int rand7(){int x=8;while(x>7)x=rand5()+5*rand5()-5;return x;}


62 characters.

• @barrycarter: The condition is while(x>7), so that would only be satisfied by numbers in the valid range. Commented Apr 25, 2011 at 17:14
• My bad. Deleted my dumb comment.
– user503
Commented Apr 27, 2011 at 17:12
• @barry And then you left another one. ;) Commented May 5, 2011 at 4:49
• It took me a few minutes to realize how the math here produces a uniformly random distribution which can be used for rejection sampling. Commented Dec 11, 2012 at 13:27

# R, 34 characters

In R (a language built for statistical computation), a deliberately cheaterish solution:

# Construct a Rand5 function
Rand5 <- function() sample(seq(5),1)
# And the golf
Rand7=function(r=Rand5())sample(1:(r/r+6),1)
# Or (same character count)
Rand7=function(r=Rand5())sample.int(r/r+6,1)
# Or even shorter(thanks to @Spacedman)
Rand7=function()sample(7)[Rand5()]


Thanks to lazy evaluation of arguments, I eliminated the semicolon and braces.

Output over 10^6 replicates:

> test <- replicate(10^6,Rand7())
> table(test)
test
1      2      3      4      5      6      7
142987 142547 143133 142719 142897 142869 142848

library(ggplot2)
qplot(test)


• If you're going to be cheaterish, you may as well be the best cheater you can be: Rand7=function(){r=Rand5();sample(7)[r]} Commented Dec 2, 2011 at 18:02
• If you are going to do that, why bother with the intermediate storage? Rand7=function(){sample(7)[Rand5()]} Commented Dec 2, 2011 at 18:22
• @BrianDiggs Path-dependency in action.... :-) Commented Dec 2, 2011 at 18:25

## С++

int Rand7()
{
int r = Rand5();
int n = 5;
do {
r = (r - 1) * 5 + Rand5();
int m = n * 5 / 7 * 7;
if (r <= m) {
return r % 7 + 1;
}
r -= m;
n = n * 5 - m;
} while (1);
}


Numbers distribution (1000000 integers):

142935 142751 142652 143299 142969 142691 142703


Average number of calls to Rand5() per every generated integer is about 2.2 (2 to 10+).

1 2      3      4     5    6   7 8  9 10
0 840180 112222 44433 2212 886 0 60 6 1


## scala, 47, 40 59 chars:

def rand7:Int={val r=5*(rand5-1)+rand5
if(r<8)r else rand7}


with 2 inputs from rand5:

\ 1 2 3 4 5
1 1 2 3 4 5
2 6 7 8 ..
3 11 ..
4 ..
5


I multiply the first-1 by 5, and add the second. Most results are ignored, and lead to a new calculation. The result should be an equal distribution of values from 1-25, from which I only pick the first 7 ones. I could accept the first 21 with building a modulo, but this would lead to longer code.

historic code which failed, but not very obviously. Thanks to Ilmari Karonen for pointing it out:

def rand7=(1 to 7).map(_=>rand5).sum%7+1


Thanks to Yoshiteru Takeshita, for this scala-2.8.0-approach which made 'sum' so easy. My solution before:

def rand7=((0/:(1 to 7))((a,_)=>a+rand5-1))%7+1


rand5:

val rnd = util.Random
def rand5 = rnd.nextInt (5) + 1


• User Yoshiteru Takeshita proposed a reduction to 40 chars for Scala 2.8.0 or later as def rand7=(1 to 7).map(_=>rand5).sum%7+1 Commented Sep 28, 2011 at 8:01
• This solution is also not correct, see the comments to glowcoder's answer. Commented Dec 2, 2011 at 18:37
• @IlmariKaronen: You're right - I reworked my solution. Commented Dec 8, 2011 at 4:52

In Java (or C/C++ I suppose)

• using generation formula by Alexandru, in 65 characters:

int rand7(){int x=rand5()*5+rand5()-6;return x>20?rand7():x/3+1;}

• using generation formula by Dan McGrath, in 60 characters

int rand7(){int x=rand5()+5*rand5()-5;return x>7?rand7():x;}


## C++

int Rand4()
{
int r = Rand5();
return r > 4 ? Rand4() : r;
}

inline int Rand8()
{
return (Rand4() - 1) << 2 + Rand4();
}

int Rand7()
{
int r = Rand8();
return r > 7 ? Rand7() : r;
}


## C++ (109)

### Golfed

int Rand4(){int r=Rand5();return r>4?Rand4():r;}int Rand7(){int r=Rand4()-1<<2+Rand4();return r>7?Rand7():r;}

• I don't really think you can call that a "one liner" because semicolons define a line of code in C++. Commented May 3, 2011 at 6:46
• @Peter Oh well, it doesn't even require one-liners anymore. Commented May 3, 2011 at 6:48
• It returned a number from 1 to 8. Commented Dec 20, 2014 at 6:11

Translation to Javascript, from the answer posted by Dan McGrath.

function Rand7(){x=8;while(x>7)x=rand5()+5*rand5()-5;return x}


62 chars

• function Rand7(){for(x=8;x>7;x=rand5()+5*rand5()-5);return x} is little shorter :P Commented Jun 23, 2011 at 3:44

# Python, 56 37 chars

Another solution that may be wrong, in Python:

rand7 = lambda: sum(rand5() for i in range(7)) % 7 + 1


This seems to be too simple, but when I try:

counter = [0] * 7
for i in range(100000):
counter[rand7()] += 1


I get a reasonably even distribution (all between 14000 and 14500).

Okay, now as somebody voted for this post: Is this solution indeed correct? I more posted this here to make people criticize it. Well, if it is correct, my golfed version would be:

rand7=lambda:eval("+rand5()"*7)%7+1


which comes out to 37 chars.

• Your solution is not correct: you base your decision on 7 rolls of a fair 5-sided die, which means there are 5^7 (5 to the 7th power) equiprobable outcomes. Since this is not a multiple of 7, you cannot return 7 equiprobable results. I don't think there's an easy formula for what you return; you can brute-force the computation or work it out by hand on smaller numbers (flip 3 coins (H=1, T=2) and sum the results). Commented Nov 13, 2011 at 1:00
• Wow, the distribution you generate, though not uniform, is remarkably close: the exact proportion of the probabilities of each number is {1: 11177, 2: 11172, 3: 11158, 4: 11144, 5: 11144, 6: 11158, 7: 11172}
– Omar
Commented Nov 29, 2011 at 20:01

## JavaScript, 85

function Rand7(){for(x=0,i=1;i<8;x^=i*((k=Rand5())%2),i*=1+(k<5));return x?x:Rand7()}


I know there's shorter answer, but I wanted to show the test of this puzzle. It turns out that only Clyde Lobo's answer using Dan McGrath's rejection sampling is correct (between JS answers).

Java, 65 chars:

int rand7(){int r;do{r=rand5()+5*rand5()-5;}while(r>7);return r;}


## Java - 66 chars

int rand7(){int s;while((s=rand5()*5+rand5())<10);return(s%7+1);}


Longer than previous routine, but I think this one returns uniformly distributed numbers in less time.

# Clojure - 58 chars

(defn rand7[](#(if(<% 8)%(rand7))(+(rand5)(*(rand5)5)-5)))


# Python, 70 chars

def rand7():
while True:
n=5*(rand5()-1)+(rand5()-1)
if n<21:return n%7+1


but completely correct based on the reasoning here.

## Perl, 43 chars, iterative rejection sampling

sub rand7{1while($_=5*&rand5-rand5)>6;$_+1}


This gives a warning about Ambiguous use of -rand5 resolved as -&rand5(), but works correctly. Prepending an & also to the second rand5 call fixes it at the cost of one stroke. (Conversely, the other & can also be removed if rand5 has been defined with a () prototype.)

Ps. The following 46-char version is about three times faster:

sub rand7{1while($_=5*&rand5-rand5)>20;$_%7+1}


PostScript (46)

This uses binary token encoding, therefore, here is a hexdump:

00000000  2f 72 61 6e 64 37 7b 38  7b 92 38 37 92 61 7b 92  |/rand7{8{.87.a{.|
00000010  40 7d 69 66 92 75 32 7b  72 61 6e 64 35 7d 92 83  |@}if.u2{rand5}..|
00000020  35 92 6c 92 01 35 92 a9  7d 92 65 7d 92 33        |5.l..5..}.e}.3|
0000002e


To try it out, you can also download it.

Here is the ungolfed and commented code, together with testing code.

% This is the actual rand7 procedure.
/rand7{
8{                      % potentialResult
% only if the random number is less than or equal to 7, we're done
dup 7 le{             % result
exit                % result
}if                   % potentialResult
pop                   % -/-
2{rand5}repeat        % randomNumber1 randomNumber2
5 mul add 5 sub       % randomNumber1 + 5*randomNumber2 - 5 = potentialResult
}loop
}def

%Now, some testing code.

% For testing, we use the built-in rand operator;
% Doesn't really give a 100% even distribution as it returns numbers
% from 0 to 2^31-1, which is of course not divisible by 5.
/rand5 {
rand 5 mod 1 add
}def

% For testing, we initialize a dict that counts the number of times any number
% has been returned. Of course, we start the count at 0 for every number.
<<1 1 7{0}for>>begin

% Now we're calling the function quite a number of times
% and increment the counters accordingly.
1000000 {
rand7 dup load 1 add def
}repeat

% Print the results
currentdict{
2 array astore ==
}forall

int result = 0;

for (int i = 0; i++; i<7)
if (((rand(5) + rand(5)) % 2) //check if odd
result += 1;

return result + 1;

• This won't give a uniform distribution. Look at the distribution of rand(5)+rand(5) over 10000 iterations to see why Commented Nov 20, 2011 at 22:25
• result can be any number from 1 to 8 in your code...
– Omar
Commented Nov 29, 2011 at 20:30
• Plus, as gnibbler said, the distribution is not uniform: (rand(5)+rand(5))%2 is biased towards 0, it produces 0 13 times for every 12 times it produces 1; i.e., the probabilities are proportional to {0: 13, 1: 12}. With that notation, the probalities for your function are proportional to {1: 62748517, 2 : 405451956, 3: 1122790032, 4: 1727369280, 5: 1594494720, 6: 883104768, 7: 271724544, 8: 35831808} (quite heavily skewed towards larger numbers). Or, fixing the loop to run 6 times, {1: 4826809, 2: 26733096, 3: 61691760, 4: 75928320, 5: 52565760, 6: 19408896, 7: 2985984}
– Omar
Commented Nov 29, 2011 at 20:35

# R (30 characters)

Define rand7:

rand7=function(n)sample(7,n,T)


Because R was written with statistical analysis in mind, this task is trivial, and I use the built-in function sample with replacement set to TRUE.

Sample output:

> rand7(20)
[1] 4 3 6 1 2 4 3 2 3 2 5 1 4 6 4 2 4 6 6 1
> rand7(20)
[1] 1 2 5 2 6 4 6 1 7 1 1 3 7 6 4 7 4 2 1 2
> rand7(20)
[1] 6 7 1 3 3 1 5 4 3 4 2 1 5 4 4 4 7 7 1 5

• It does say you have to use Rand5. Doesn't say how, but you have to use it... Commented Dec 2, 2011 at 18:03
• @Spacedman Yes, I explicitly ignored it. That is use by non-reference. Commented Dec 2, 2011 at 18:10

# Groovy

rand7={if(b==null)b=rand5();(b=(rand5()+b)%7+1)}


example distribution over 35,000 iterations:

[1:5030, 2:4909, 3:5017, 4:4942, 5:5118, 6:4956, 7:5028]


Is it bad that it's stateful?

## Mathematica, 30

Rand7=Rand5[]~Sum~{7}~Mod~7+1&


How about this?

int Rand7()
{
return Rand5()+ Rand5()/2;
}

• Whatever language that is, does its / operator do integer math? What happens to your results if it does decimal, floating-point, or integer math? Commented Jan 26, 2013 at 20:50
• Assuming integer division, this function has the following distribution: [2/25, 4/25, 5/25, 5/25, 5/25, 3/25, 1/25]. Not exactly uniform. Commented Jan 28, 2013 at 8:52
• primo is right. adding random numbers is generally going to skew the probabilities toward the middle values. Commented Feb 1, 2013 at 2:35

# Ruby (43 bytes)

def rand7;(0..7).reduce{|i|i+rand5}%7+1;end


cemper93's solution ported to Ruby is three bytes shorter ;) (34 bytes)

def rand7;eval("+rand5"*7)%7+1;end


## Java - 54

int m=0;int rand7(){return(m=m*5&-1>>>1|rand5())%7+1;}


Distribution test: [1000915, 999689, 999169, 998227, 1001653, 1000419, 999928]

Algorithm:

• Keep a global variable
• multiply by 5, so there get 5 places free at the least significant end
• Truncate the sign bit to make it positive (not necessary if unsigned numbers were supported)
• Modulo 7 is the answer

> The numbers are not mutually uncorrelated anymore, but individually perfectly random.

C/C++ code the core code has one line only!

static unsigned int gi = 0;

int rand7()
{
return (((rand() % 5 + 1) + (gi++ % 7)) % 7) + 1;
}

//call this seed before rand7
//maybe it's not best seed, if yo have any good idea tell me please
//and thanks JiminP again, he remind me to do this
void srand7()
{
int i, n = time(0);
for (i = 0; i < n % 7; i++)
rand7();
}


The srand7() is the seed of rand7, must call this function before rand7, just like call srand before rand in C.

This is a very good one, because it call rand() only one time, and no loop thing, no expends extra memories.

Let me explain it: consider a integer array with size of 5:

1st get one number from 1 2 3 4 5 by rand5
2nd get one number from 2 3 4 5 6
3rd get one number from 3 4 5 6 7
4th get one number from 4 5 6 7 1
5th get one number from 5 6 7 1 2
5th get one number from 6 7 1 2 3
7th get one number from 7 1 2 3 4


So we got the TABLE, each one of 1-7 appears 5 times in it, and has all 35 numbers, so the probability of each number is 5/35=1/7. And next time,

8th get one number from 1 2 3 4 5
9th get one number from 2 3 4 5 6
......


After enough times, we can get the uniform distribution of 1-7.

So, we can allocate a array to restore the five elements of 1-7 by loop-left-shift, and get one number from array each time by rand5. Instead, we can generate the all seven arrays before, and using them circularly. The code is simple also, has many short codes can do this.

But, we can using the properties of % operation, so the table 1-7 rows is equivalent with (rand5 + i) % 7, that is : a = rand() % 5 + 1 is rand5 in C language, b = gi++ % 7 generates all permutations in table above, and 0 - 6 replace 1 - 7 c = (a + b) % 7 + 1, generates 1 - 7 uniformly. Finally, we got this code:

(((rand() % 5 + 1) + (gi++ % 7)) % 7) + 1


But, we can not get 6 and 7 at first call, so we need a seed, some like srand for rand in C/C++, to disarrange the permutation for first formal call.

Here is the full code to testing:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

static unsigned int gi = 0;

//a = rand() % 5 + 1 is rand5 in C language,
//b = gi++ % 7 generates all permutations,
//c = (a + b) % 7 + 1, generates 1 - 7 uniformly.
//Dont forget call srand7 before rand7
int rand7()
{
return (((rand() % 5 + 1) + (gi++ % 7)) % 7) + 1;
}

//call this seed before rand7
//maybe it's not best seed, if yo have any good idea tell me please
//and thanks JiminP again, he remind me to do this
void srand7()
{
int i, n = time(0);
for (i = 0; i < n % 7; i++)
rand7();
}

void main(void)
{
unsigned int result[10] = {0};
int k;

srand((unsigned int)time(0)); //initialize the seed for rand
srand7() //initialize the rand7

for (k = 0; k < 100000; k++)
result[rand7() - 1]++;

for (k = 0; k < 7; k++)
printf("%d : %.05f\n", k + 1, (float)result[k]/100000);
}

• It 'passes' the 'test', but that doesn't mean this is a good random function. Can I get 6 or 7 by calling it once? Commented Oct 5, 2011 at 13:57
• But there's good kinds and bad kinds of approximation. And this code is bad - because it does not gives uniform distribution when called only once. If one wrote something like int main(){if(rand7()==6) printf("Hello, world!");}, approximation using loop will print 'Hello, world!' 1 in 7 times, but your code doesn't. Commented Oct 5, 2011 at 14:06
• thank you @JiminP! you are right for 6,7 at first time. i need a seed to disarrange before call rand7, the seed just like the srand in C/C++. i fixed my code, and thank you again!!! Commented Oct 5, 2011 at 15:46
• hm....the srand10 does not work, the last 3 numbers can not get at 10, 20, 30 ...positions. sorry @JiminP, but how to modify it? i think this is hopeful way. Commented Oct 5, 2011 at 18:37
• Different calls to this function are not independent of each other. The spec here doesn't require this, but that is typically expecetd of random number generators. Otherwise, you could say, return a random uniform number the first time and in future calls just return (previous+1)%7...
– Omar
Commented Nov 29, 2011 at 20:38