Verify distribution uniformity/Naive

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Task
Verify distribution uniformity/Naive
You are encouraged to solve this task according to the task description, using any language you may know.

This task is an adjunct to Seven-dice from Five-dice.

Create a function to check that the random integers returned from a small-integer generator function have uniform distribution.

The function should take as arguments:

  • The function (or object) producing random integers.
  • The number of times to call the integer generator.
  • A 'delta' value of some sort that indicates how close to a flat distribution is close enough.

The function should produce:

  • Some indication of the distribution achieved.
  • An 'error' if the distribution is not flat enough.

Show the distribution checker working when the produced distribution is flat enough and when it is not. (Use a generator from Seven-dice from Five-dice).

See also:

AutoHotkey

<lang AutoHotkey>MsgBox, % DistCheck("dice7",10000,3)


DistCheck(function, repetitions, delta) { Loop, % 7 ; initialize array 

  {  bucket%A_Index% := 0
  }

  Loop, % repetitions ; populate buckets
  {  v := %function%()
     bucket%v% += 1
  }

  lbnd := round((repetitions/7)*(100-delta)/100)
  ubnd := round((repetitions/7)*(100+delta)/100)
  text := "Distribution check:`n`nTotal elements = " repetitions
        . "`n`nMargin = " delta "% --> Lbound = " lbnd ", Ubound = " ubnd "`n"
  Loop, % 7
  {  text := text "`nBucket " A_Index " contains " bucket%A_Index% " elements."
     If bucket%A_Index% not between %lbnd% and %ubnd%
        text := text " Skewed."
  }
  Return, text

}</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

Library: Judy

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <stdbool.h>
  3. include <math.h>
  1. include <Judy.h>

bool distcheck(int (*dist)(), int n, double D) { 

 Pvoid_t h = (Pvoid_t) NULL;
 PWord_t value;
 PWord_t element;

 Word_t i;
 int h_length;

 // populate hashes
 for(i=0; i < n; i++) {
   int rn = dist();
   JLI(value, h, rn);
   ++*value;
 }

 JLC(h_length, h, 0, -1);

 double target = 1.0 * n / (double)h_length;

 i = 0;
 JLF(element, h, i);
 while(element != NULL) {
   if ( abs(*element - target) > 0.01*n*D ) {
     fprintf(stderr, "distribution potentially skewed for '%d': expected '%d', got '%d'\n",

i, (int)target, *element); 

     JudyLFreeArray(&h, PJE0);
     return false; // bad distr.
   }
   JLN(element, h, i);
 }

 JudyLFreeArray(&h, PJE0);
 return true; // distr. ok

}

int main() { 

 distcheck(rand, 1000000, 1);
 return 0;

}</lang>

C++

<lang cpp>

  1. include <map>
  2. include <iostream>
  3. include <cmath>

template<typename F> 

bool test_distribution(F f, int calls, double delta)

{ 

 typedef std::map<int, int> distmap;
 distmap dist;

 for (int i = 0; i < calls; ++i)
   ++dist[f()];

 double mean = 1.0/dist.size();

 bool good = true;

 for (distmap::iterator i = dist.begin(); i != dist.end(); ++i)
 {
   if (std::abs((1.0 * i->second)/calls - mean) > delta)
   {
     std::cout << "Relative frequency " << i->second/(1.0*calls)
               << " of result " << i->first
               << " deviates by more than " << delta
               << " from the expected value " << mean << "\n";
     good = false;
   }
 }

 return good;

} </lang>

Common Lisp

Translation of: OCaml

<lang lisp>(defun check-distribution (function n &optional (delta 1.0)) 

 (let ((bins (make-hash-table)))
   (loop repeat n do (incf (gethash (funcall function) bins 0)))
   (loop with target = (/ n (hash-table-count bins))
         for key being each hash-key of bins using (hash-value value)
         when (> (abs (- value target)) (* 0.01 delta n))
         do (format t "~&Distribution potentially skewed for ~w:~
                        expected around ~w got ~w." key target value)
         finally (return bins))))</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>

J

This solution defines the checker as an adverb. Adverbs combine with the verb immediately to their left to create a new verb. So for a verb generateDistribution and an adverb checkUniform, the new verb might be thought of as checkGeneratedDistributionisUniform.

The delta is given as an optional left argument (x), defaulting to 5%. The right argument (y) is any valid argument to the distribution generating verb. <lang j> checkUniform=: adverb define 

 0.05 u checkUniform y
 :
 n=. */y
 delta=. x
 sample=. u n              NB. the u refers to the verb given to the left of the adverb
 freqtable=. /:~ (~. ,. #/.~) sample
 expected=. n % # freqtable
 errmsg=. 'Distribution is potentially skewed'
 errmsg assert expected (delta&*@[ > |@:-) {:"1 freqtable
 freqtable

) </lang>

Show usage using rollD7t given in Seven-dice from Five-dice:

<lang j> 

  0.05 rollD7t checkUniform 1e5

1 14082 2 14337 3 14242 4 14470 5 14067 6 14428 7 14374 

  0.05 rollD7t checkUniform 1e2

|Distribution is potentially skewed: assert | errmsg assert(lbound&<*.ubound&>){:"1 freqtable </lang>

OCaml

<lang ocaml>let distcheck fn n ?(delta=1.0) () = 

 let h = Hashtbl.create 5 in
 for i = 1 to n do
   let v = fn() in
   let n = 
     try Hashtbl.find h v 
     with Not_found -> 0 
   in
   Hashtbl.replace h v (n+1)
 done;
 Hashtbl.iter (fun v n -> Printf.printf "%d => %d\n%!" v n) h;
 let target = (float n) /. float (Hashtbl.length h) in
 Hashtbl.iter (fun key value ->
   if abs_float(float value -. target) > 0.01 *. delta *. (float n)
   then (Printf.eprintf
     "distribution potentially skewed for '%d': expected around %f, got %d\n%!"
      key target value)
 ) h;
</lang>

Python

Works with: Python version 3.1

<lang python>from collections import Counter from pprint import pprint as pp

def distcheck(fn, repeats, delta): 

   \
   Bin the answers to fn() and check bin counts are within +/- delta %
   of repeats/bincount
   bin = Counter(fn() for i in range(repeats))
   target = repeats // len(bin)
   deltacount = int(delta / 100. * target)
   assert all( abs(target - count) < deltacount
               for count in bin.values() ), "Bin distribution skewed from %i +/- %i: %s" % (
                   target, deltacount, [ (key, target - count)
                                         for key, count in sorted(bin.items()) ]
                   )
   pp(dict(bin))</lang>

Sample output: 

>>> distcheck(dice5, 1000000, 1)
{1: 200244, 2: 199831, 3: 199548, 4: 199853, 5: 200524}
>>> distcheck(dice5, 1000, 1)
Traceback (most recent call last):
  File "<pyshell#30>", line 1, in <module>
    distcheck(dice5, 1000, 1)
  File "C://Paddys/rand7fromrand5.py", line 54, in distcheck
    for key, count in sorted(bin.items()) ]
AssertionError: Bin distribution skewed from 200 +/- 2: [(1, 4), (2, -33), (3, 6), (4, 11), (5, 12)]

R

<lang r> distcheck <- function(fn, repetitions=1e4, delta=3) { 

  if(is.character(fn))
  {
     fn <- get(fn)
  }
  if(!is.function(fn))
  {
     stop("fn is not a function")
  }
  samp <- fn(n=repetitions)   
  counts <- table(samp)
  expected <- repetitions/length(counts)
  lbound <- expected * (1 - 0.01*delta)
  ubound <- expected * (1 + 0.01*delta)
  status <- ifelse(counts < lbound, "under", 
     ifelse(counts > ubound, "over", "okay"))
  data.frame(value=names(counts), counts=as.vector(counts), status=status)

} distcheck(dice7.vec) </lang>

Ruby

Translation of: Tcl

<lang ruby>def distcheck(n, delta=1) 

 unless block_given?
   raise ArgumentError, "pass a block to this method"
 end
 
 h = Hash.new(0)
 n.times {h[ yield ] += 1}
 
 target = 1.0 * n / h.length
 h.each do |key, value| 
   if (value - target).abs > 0.01 * delta * n
     raise StandardError,
       "distribution potentially skewed for '#{key}': expected around #{target}, got #{value}"
   end
 end
 
 h.keys.sort.each {|k| print "#{k} #{h[k]} "}
 puts

end

if __FILE__ == $0 

 begin
   distcheck(100_000) {rand(10)}
   distcheck(100_000) {rand > 0.95} 
 rescue StandardError => e
   p e
 end

end</lang>

output: 

0 9986 1 9826 2 9861 3 10034 4 9876 5 10114 6 10329 7 9924 8 10123 9 9927 
#<StandardError: distribution potentially skewed for 'false': expected around 50000.0, got 94841>

Tcl

<lang tcl>proc distcheck {random times {delta 1}} { 

   for {set i 0} {$i<$times} {incr i} {incr vals([uplevel 1 $random])}
   set target [expr {$times / [array size vals]}]
   foreach {k v} [array get vals] {
       if {abs($v - $target) > $times  * $delta / 100.0} {
          error "distribution potentially skewed for $k: expected around $target, got $v"
       }
   }
   foreach k [lsort -integer [array names vals]] {lappend result $k $vals($k)}
   return $result

}</lang> Demonstration: <lang tcl># First, a uniformly distributed random variable puts [distcheck {expr {int(10*rand())}} 100000]

  1. Now, one that definitely isn't!

puts [distcheck {expr {rand()>0.95}} 100000]</lang> Which produces this output (error in red): 

0 10003 1 9851 2 10058 3 10193 4 10126 5 10002 6 9852 7 9964 8 9957 9 9994
distribution potentially skewed for 0: expected around 50000, got 94873
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