Inspired by Random with your hands tied:

The Goal

The goal of this challenge is to write a program that generates a pseudorandom bit stream, which is a string of 1s and 0s that appears to be purely random, but is actually generated in a deterministic way. Your program should output a string of 1 and 0s (with optional whitespace) and should pass the following requirements:

  1. Given unlimited time and memory, your program must continue to output a string of 1s and 0s forever
  2. Your program must output more than 1000 random bits in about one minute, on a reasonable machine. If this requirement is impossible, then I will lessen it.
  3. The string of bits can repeat, but the length of the repeating section must be more than 1000 bits.
  4. The string of bits must pass as many of the randomness tests (described below) as possible.
  5. The program must not take any input from any external sources or use any built-in rand()-like function.
  6. Because of the above requirement, the program must output the same exact string of bits each time that it is run.

Randomness Test #1

The string of pseudorandom bits must not include any obvious pattern upon visual inspection.

Randomness Test #2 (subject to change based on comments)

The string of bits must contain an equal distribution of 1s and 0s. To test this (and other things too), the stream of bits is broken into segments that are 3 bits long, such as 101|111|001.

Out of all of these segments, 1/8 of them should have three 1s and no 0s, 3/8 of them should have two 1s and one 0, 3/8 of them should have one 1 and two 0s, and 1/8 of them should have no 1s and three 0s.

Randomness Test #3

A "run" is defined as a consecutive series of bits that all have the same value. The string 1001001110 has three runs of size 1 (1..1.....0), two runs of size 2 (.00.00....) and one run of size 3 (......111.). Notice that runs do not overlap.

Out of a string of 1000 random bits, there should be about 250 runs of size 1, 125 runs of size 2, 62 runs of size 3, etc. In general, for run size R, there should be about 1000/(2**(R+1)) runs of that size.

Randomness Test #4

The first 840 bits are split into two halves of 420 bits each. Each bit on the first half is compared to the corresponding bit on the second half. The two bits should match about fifty percent of the time.

Here is the source code of a Perl program that performs tests 2 through 4. As of now, it requires that the string of bits not contain any whitespace.

Objective Winning Criterion Time!

The winner is the program that passes all of the 6 requirements and all of the randomness tests to the degree that it is indistinguishable from randomness. If multiple programs accomplish this, then the one that takes the longest amount of time to repeat will win. If multiple programs accomplish this, then I might have to find more randomness tests to act as tie-breakers.

  • \$\begingroup\$ #2 and #3 aren't really very good criteria for randomness. For #2 especially, a random sample will probably not exhibit this characteristic. Maybe you can do a larger sample size? I would suggest something between 100 and 300. \$\endgroup\$ Aug 2, 2012 at 22:00
  • \$\begingroup\$ A better measurement method would be a moving average, as the mean over a large window on the bitstream will not change much (and should be around 0.5) \$\endgroup\$ Aug 2, 2012 at 22:02
  • \$\begingroup\$ @JoelCornett Thanks for the advice. I don't know much about randomness tests. I'll change #2 to something else, and I'm reading about moving averages. \$\endgroup\$
    – PhiNotPi
    Aug 2, 2012 at 22:28
  • 1
    \$\begingroup\$ No problem. Random sequences tend to clump and not be uniformly distributed, this is a fact that is sometimes used in accounting to detect fraud. (Fraudulent numbers will often be too evenly distributed, because the people inventing them mistake uniformity for randomness) \$\endgroup\$ Aug 2, 2012 at 22:45
  • \$\begingroup\$ Can I use built in crypto functions (Such as AES or SHA-2)? \$\endgroup\$ Aug 6, 2012 at 12:05

11 Answers 11


Mathematica 78 53 chars

The digits of the binary representation of Pi seem to behave as if they are chaotically produced although this is unproven.

The following simple routine deterministically returns as a string the binary digits of pi, corresponding to d decimal digits:



If we request the counterpart of 301 decimal digits of Pi, we receive 1000 binary digits.


(* out *)

1000 (* characters *)

Because Pi is an irrational number, there is no period. However, there will be practical constraints due to the hardware one is running.

Test 1 Looks good to me.

Test 2

(* out *)

More thorough check:


{{0,1,1},139119}, {{1,0,1},138404},{{1,0,0},137926},{{0,0,1},138462}}

Test 3: Runs

res3=SortBy[Tally@Split@RealDigits[N[Pi,d],2][[1]],Last]/.{a_,b_}:> {Length[a],b}
ListPlot[res3 ,AxesLabel-> {"Run Length","Runs"},AxesOrigin->{0,0}]

I ran a large number of cases to systematically check out the distribution of runs. In approximately 3 million binary digits, there were 830k runs of 1, 416k runs of 2, 208k runs of 3, 104k runs of 4, etc.

runs 2 Test 4: Matching of first and second half of data

The matches are the 212 cases of 0 and 2; the mismatches are the 208 cases where the sum of the respective digits is 1.


(* out *)


It takes under two seconds to calculate 3321928 binary digits (corresponding to 10^6 decimal digits).


  • 1
    \$\begingroup\$ I knew someone would do this... \$\endgroup\$ Aug 9, 2012 at 12:01
  • 1
    \$\begingroup\$ Low-hanging fruit, right? \$\endgroup\$
    – DavidC
    Aug 9, 2012 at 12:19
  • \$\begingroup\$ Couldn't you use e instead of pi to save one byte? \$\endgroup\$ Dec 3, 2015 at 2:15
  • \$\begingroup\$ Is e distributed chaotically? \$\endgroup\$
    – DavidC
    Dec 3, 2015 at 2:49

C, 61


Yeah, I know it's not code golf. This is obviously rather an anti-solution... but it sure enough fulfills your criteria.

$ ./a.out | head -c840 000000001000111101100110110011100100011101110001101001101110011101100111010111010010001001110001111110010001010101101000110101000011110010100101001101110100011111000101000110000111100100010101010100110101101010001110101000011011001010111011110110101000001110011000010100001111110110001000000101100011000110000010110011000010101101011100010011011111111100001010111110111010110001000010110111000010101001000011001001110110001001101011011110111011010100000000001001010100111000000100010011110001010111001110101111100110110111010001101001100000101000111000101000110100001111010110100001100110011011110001011011100010111011100001110100110001010011001110001010110100111111111010110101100110000101111111111000111101100011000111010111010100110001001110101110100011101110001100110110000001100101110100101000101011100000110111101101100110000000101110
$ ./a.out | head -c840 | perl tester.pl
Test 2: 1 (1) 2.93333333333333 (3) 3.1 (3) 0.966666666666667 (1)
Test 3: 214 99 71 24 7 5 1 1 2 2
Test 4: 0.495238095238095

Period length is 2²⁹.

  • 7
    \$\begingroup\$ This goes to show how hard it is to tell randomness from something that is widely known to be one of the worst random number generators in existence. +1. \$\endgroup\$
    – PhiNotPi
    Aug 3, 2012 at 3:32

Python, 90

print(''.join("01"[(g.append((11*g[-1]+13)%1024)or g[-1])>512]for i in range(1000)))

g is the seed value. Random sampling exhibits a remarkably normal distribution repeated random sampling of sample means yielded a mean of 0.506 and a a standard deviation of .0473 (sample size of 1000). Unfortunately, randomness is highly sensitive to initial seed. The seed in the above code gave me the best randomness :p


Let's see how this code holds up to the OP's tests:

Test #1

This one's a bit subjective... but it looks pretty irregular to me.

Test #2

Three 1's: 0.141
Two 1's: 0.371
One 1: 0.353
Zero 1's: 0.135

Test #3

Runs by size:

8: 11
7: 3
6: 7
5: 13
4: 32
3: 67
2: 119
1: 216

Test #4

Ratio of equalities: 0.94 This is a typo. Will update with the correct number soon.

  • 1
    \$\begingroup\$ You can remove the whitespace before 'for'. \$\endgroup\$
    – daniero
    Aug 5, 2012 at 19:01

Haskell 74 58

main=print$iterate(read.take 9.show.(^3))7>>=show.(`mod`2)

Thanks to shiona for the simplification. Results:

$ ~/golf$ ./pseudorandom | head -c 1000 111110110101011110001011110101011110001111011001110101100100111000100100000011000100011101010111000000001110100111100010110000100000001001111011100001101101001111110100111011111110011011101111010110001110011100010000001011011100010111011110011110100110100000110110010011101010001111101011111110000110110110011111011101000011110010000011011010011110110100010011100011011111011010110001100111100100100001110100001110111011111000001101001010100001100111010111011011011001110000100000100000010111011010111011111011011101001100100011111101110101101010111101001111110101110101001011111011111100111110001001001010001011100100101000010101011001101011011001100111101010001110001011110001011111110110100110011000010101100010011011010100000110100001000010110110111101101010000001110001100011010100000011111010111100101110101001000111001000000101111110010110001011011000011111011001100001101000101001111000101011101010001010110010111111110111111101000111010001001010010011101011100101011101000101101000010101100

./pseudorandom | head -c 1000 | perl test.pl

Test 2: 0.966666666666667 (1) 2.4 (3) 3.3 (3) 1.33333333333333 (1)

Test 3: 260 108 66 33 15 11 5 2

Test 4: 0.495238095238095

This is also a terrible pseudo-random generator (similar to one used by von-Neuman). For those that were not aware concatMap == (=<<) == flip . (>>=) (for lists)

  • \$\begingroup\$ You can replace \x->if odd x then"1"else"0" with show.(`mod`2). \$\endgroup\$
    – shiona
    Aug 6, 2012 at 17:48

C, 52 chars


This is a 10 bit LFSR, test results:

$ ./a.out |head -c 1000 | perl randtest.pl
Test 2: 1.13333333333333 (1) 2.86666666666667 (3) 3.16666666666667 (3) 0.833333333333333 (1)
Test 3:  251 122 64 32 16 8 4 2  1
Test 4: 0.466666666666667
  • \$\begingroup\$ a should start out as 1, (assuming it's called with no arguments). Also you could stick the a= in the middle, something like a=a/2^-!putchar(49-a%2)%576 (taking some liberties with the algorithm) \$\endgroup\$
    – walpen
    Aug 9, 2012 at 1:47
  • \$\begingroup\$ @walpen: My initial implementation didn't set a, I changed it because of "The program must not take any input from any external sources" \$\endgroup\$
    – Hasturkun
    Aug 9, 2012 at 9:26

The question is essentially equivalent to "implement a stream cipher". So I implement RC4, since it's relatively simple.

I use no key, and drop the first 100000 bits, because the beginning of RC4 is a bit biased, especially since I skipped the key schedule. But I'd expect it to pass your test even without that (saving 20 chars of code).

Normally one would output a full byte per cycle, but converting to binary is rather ugly in C#, so I simply discard everything except the least significant bit.

var s=Enumerable.Range(0,256).ToArray();
byte i=0,j=0;
for(int k=0;;k++)
    var t=s[i];s[i]=s[j];s[j]=t;

Or without spaces:

var s=Enumerable.Range(0,256).ToArray();byte i=0,j=0;for(int k=0;;k++){i++;j+=(byte)s[i];var t=s[i];s[i]=s[j];s[j]=t;if(k>99999)Console.Write(s[i]+s[j]&1);}

C#, 156 chars, works in LinqPad's statement mode. For a full C# program add the usual boilerplate.

We could also use built in crypto primitives(Cheater solution):

var h=SHA256.Create();for(BigInteger i=0;;i++){Console.Write(h.ComputeHash(i.ToByteArray())[0]%2);}

(C#, 99 chars, works in LinqPad's statement mode. For the normal C# compiler you'll need to add a bit of boilerplate)

The output of cryptographic hash functions is designed to be indistinguishable from random data, so I expect it to pass all randomness tests(die harder,...) you throw at it, but I'm too lazy to test.



This program prints the rightmost binary digits that are common to every sufficiently tall exponentiation tower of form 3333... For all that could ever be feasibly generated, these are the rightmost binary digits of Graham's number. The digit sequence is infinite, and is not periodic.

m = 1; x = 3; last = 0
while True:
    m *= 2; x = pow(3,x,m); l = len(bin(x))
    print '1' if l > last else '0',
    last = l

For 1000 digits, this took less than 2 seconds; however, the time will increase much faster than linearly in the number of digits.

The test results using the OP's program are

Test 2: 1.26666666666667 (1) 3.16666666666667 (3) 2.8 (3) 0.766666666666667 (1)
Test 3:  268 126 61 30 20 7 2  1 1
Test 4: 0.466666666666667

(See Are the rightmost digits of G random? for more than 32000 digits and additional statistical tests.)


Java, 371 317

Based on a 128 bit LFSR (bit taps are from xilinx app note 52)

EDIT: I was not satisfied with using BigInteger so this version does not. Saved some characters. Output might be a little less random as I could not think of a good 'seeding' method.

New Code: Arguments: BITS_TO_PRINT

class R{public static void main(String[]a){int L=65536;int[]v={0,128,126,101,99};int[]b=new int[L];for(int x=0;x<L;x++)b[x]=(x*x)&1;for(int i=0;i<Integer.parseInt(a[0])+L;i++){if(1!=(b[v[1]]^b[v[2]]^b[v[3]]^b[v[4]]))b[v[0]]=1;else b[v[0]]=0;if(i>L)System.out.print(b[v[0]]);for(int j=0;j<5;j++)v[j]=(v[j]-1)&(L-1);}}}

Old Version: Arguments: SEED, BITS_TO_PRINT

import java.math.BigInteger;class R{public static void main(String[]a){BigInteger v=new BigInteger(a[0]);BigInteger m=new BigInteger("ffffffffffffffffffffffffffffffff",16);for(int i=Integer.parseInt(a[1]);i>0;i--){v=v.shiftLeft(1);if(!(v.testBit(128)^v.testBit(126)^v.testBit(101)^v.testBit(99))){v=v.setBit(0);}v=v.and(m);java.lang.System.out.print(v.testBit(0)?1:0);}}}

New Version: Example output, bits=100:

  • 2
    \$\begingroup\$ BTW, I assume that both Noah accounts from this post are the same person. If that's so, you can ask a moderator to merge them at meta.codegolf.stackexchange.com \$\endgroup\$ Sep 20, 2012 at 7:05

JavaScript - 1ms to 2ms for 1000 pseudo-random bits (139ms to 153ms for 100000 bits)

This solution uses the fact that square roots are irrational, and thus pretty much random. Basically, it takes the square root of 2 to start, converts it to binary, throws out the leading part that matched the previous root, appends that to the random string, repeats with the next higher number (or back to 2 if the number repeated and was at least 30 bits long), and returns the random string once it is long enough.

var getDeterministicPseudoRandString = function(length){
    var randString = '';

    var i = 2;
    var prevRand = '';

    while(randString.length < length){
        var nextRand, nextFullRand = Math.sqrt(i++).toString(2).substring(1).replace('.', '');
        nextRand = nextFullRand;
        for(var j = prevRand.length; j > 0; j--){
            var replaceString = prevRand.substring(0, j);

            nextRand = nextFullRand;

            if(nextFullRand.indexOf(replaceString) == 0){
                if(j == prevRand.length && j > 30){
                    //start i over at 2
                    console.log('max i reached: ' + i);

                    i = 2;
                    continue outerLoop;
                } else {
                    nextRand = nextFullRand.replace(replaceString, '');

        prevRand = nextFullRand;

        randString += nextRand;

    return randString.substring(0, length);//Return the substring with the appropriate length

I haven't ran it through the tests yet, but I imagine it will do well on them. Here's a fiddle so you can see it in action. For my times, I just ran the program several times and took the fastest and slowest values as the ranges.



import hashlib
while 1:
    print ord(x[0])%2

Should have a period of about 2^512.


perl, 44 bytes

I know this isn't code golf, but I've always been a fan of taking the low order bits of a simple quadratic function, eg:

$x=1/7;print substr($x*=4-4*$x,9,1)%2while 1

Period is longer than 3 billion, but I've run out of disk space to calculate more.

  • 1
    \$\begingroup\$ you can save 3 chars by juxtaposing numeric constants and keywords and also distributing that 4: $x=1/7;print substr($x*=4-4*$x,9,1)%2while 1 \$\endgroup\$
    – ardnew
    Apr 21, 2014 at 15:49

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