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
<lang c>#include <stdlib.h>
- include <stdio.h>
- include <math.h>
inline int rand5() { int r, rand_max = RAND_MAX - (RAND_MAX % 5); while ((r = rand()) >= rand_max); return r / (rand_max / 5) + 1; }
inline int rand5_7() { int r; while ((r = rand5() * 5 + rand5()) >= 27); return r / 3 - 1; }
/* assumes gen() returns a value from 1 to n */ int check(int (*gen)(), int n, int cnt, double delta) /* delta is relative */ { int i = cnt, *bins = calloc(sizeof(int), n); double ratio; while (i--) bins[gen() - 1]++; for (i = 0; i < n; i++) { ratio = bins[i] * n / (double)cnt - 1; if (ratio > -delta && ratio < delta) continue;
printf("bin %d out of range: %d (%g%% vs %g%%), ", i + 1, bins[i], ratio * 100, delta * 100); break; } free(bins); return i == n; }
int main() { int cnt = 1; while ((cnt *= 10) <= 1000000) { printf("Count = %d: ", cnt); printf(check(rand5_7, 7, cnt, 0.03) ? "flat\n" : "NOT flat\n"); }
return 0;
}</lang>output
Count = 10: bin 1 out of range: 1 (-30% vs 3%), NOT flat Count = 100: bin 1 out of range: 11 (-23% vs 3%), NOT flat Count = 1000: bin 1 out of range: 150 (5% vs 3%), NOT flat Count = 10000: bin 3 out of range: 1477 (3.39% vs 3%), NOT flat Count = 100000: flat Count = 1000000: flat
C++
<lang cpp>#include <map>
- include <iostream>
- 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>
Clojure
The code could be shortened if the verify function did the output itself, but the "Clojure way" is to have functions, whenever possible, avoid side effects (like printing) and just return a result. Then the "application-level" code uses doseq and println to display the output to the user. The built-in (rand-int) function is used for all three runs of the verify function, but first with small N to simulate experimental error in a small sample size, then with larger N to show it working properly on large N. <lang clojure> (defn verify [rand n & [delta]]
(let [rands (frequencies (repeatedly n rand)) avg (/ (reduce + (map val rands)) (count rands)) max-delta (* avg (or delta 1/10)) acceptable? #(<= (- avg max-delta) % (+ avg max-delta))] (for [[num count] (sort rands)] [num count (acceptable? count)])))
(doseq [n [100 1000 10000]
[num count okay?] (verify #(rand-int 7) n)] (println "Saw" num count "times:" (if okay? "that's" " not") "acceptable"))</lang>
Saw 1 13 times: that's acceptable Saw 2 15 times: that's acceptable Saw 3 11 times: not acceptable Saw 4 10 times: not acceptable Saw 5 19 times: not acceptable Saw 6 17 times: not acceptable Saw 0 121 times: not acceptable Saw 1 128 times: not acceptable Saw 2 161 times: not acceptable Saw 3 146 times: that's acceptable Saw 4 134 times: that's acceptable Saw 5 170 times: not acceptable Saw 6 140 times: that's acceptable Saw 0 1480 times: that's acceptable Saw 1 1372 times: that's acceptable Saw 2 1411 times: that's acceptable Saw 3 1442 times: that's acceptable Saw 4 1395 times: that's acceptable Saw 5 1432 times: that's acceptable Saw 6 1468 times: that's acceptable
Common Lisp
<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>
D
<lang d>import std.math: abs; import std.string: format; import std.stdio: writefln;
/** Bin the answers to fn() and check bin counts are within +/- delta % of repeats/bincount.
- /
void distCheck(TF)(TF func, int nrepeats, double delta) {
int[int] freqs; for (int i; i < nrepeats; i++) freqs[func()]++; double target = nrepeats / cast(double)freqs.length; int deltaCount = cast(int)(delta / 100.0 * target);
foreach (k, count; freqs) if (abs(target - count) >= deltaCount) throw new Exception(format("distribution potentially skewed for '%s': '%d'\n", k, count));
writefln(freqs);
}</lang>
Fortran
<lang fortran>subroutine distcheck(randgen, n, delta)
interface function randgen integer :: randgen end function randgen end interface real, intent(in) :: delta integer, intent(in) :: n integer :: i, mval, lolim, hilim integer, allocatable :: buckets(:) integer, allocatable :: rnums(:) logical :: skewed = .false. allocate(rnums(n)) do i = 1, n rnums(i) = randgen() end do
mval = maxval(rnums) allocate(buckets(mval)) buckets = 0 do i = 1, n buckets(rnums(i)) = buckets(rnums(i)) + 1 end do
lolim = n/mval - n/mval*delta hilim = n/mval + n/mval*delta do i = 1, mval if(buckets(i) < lolim .or. buckets(i) > hilim) then write(*,"(a,i0,a,i0,a,i0)") "Distribution potentially skewed for bucket ", i, " Expected: ", & n/mval, " Actual: ", buckets(i) skewed = .true. end if end do
if (.not. skewed) write(*,"(a)") "Distribution uniform" deallocate(rnums) deallocate(buckets)
end subroutine</lang>
Go
<lang go>package main
import (
"fmt" "math" "rand"
)
// "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.Fmax(max, math.Fabs(float64(c)-expected)) } return max, max < delta
}
// Driver, produces output satisfying both tasks. // // Go global random number (used by dice5) is always initialized with the // same seed, to allow repeatability. By happy coincidence, a threshold // of 500 nicely shows one run where the distribution is "flat enough" // and one where it isn't. func main() {
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)
}</lang> Output:
Max delta: 677.8571428571304 Flat enough: false Max delta: 446.1428571428696 Flat enough: true
Haskell
<lang haskell>import System.Random import Data.List import Control.Monad import Control.Arrow
distribCheck :: IO Int -> Int -> Int -> IO [(Int,(Int,Bool))] distribCheck f n d = do
nrs <- replicateM n f let group = takeWhile (not.null) $ unfoldr (Just. (partition =<< (==). head)) nrs avg = (fromIntegral n) / fromIntegral (length group) ul = round $ (100 + fromIntegral d)/100 * avg ll = round $ (100 - fromIntegral d)/100 * avg return $ map (head &&& (id &&& liftM2 (&&) (>ll)(<ul)).length) group</lang>
Example: <lang haskell>*Main> mapM_ print .sort =<< distribCheck (randomRIO(1,6)) 100000 3 (1,(16911,True)) (2,(16599,True)) (3,(16670,True)) (4,(16624,True)) (5,(16526,True)) (6,(16670,True))</lang>
Icon and Unicon
This example assumes the random number generator is passed to verify_uniform
as a co-expression. The co-expression rnd
is prompted for its next value using @rnd
. The co-expression is created using create (|?10)
where the vertical bar means generate an infinite sequence of what is to its right (in this case, ?10
generates a random integer in the range [1,10]).
<lang Icon>
- rnd : a co-expression, which will generate the random numbers
- n : the number of numbers to test
- delta: tolerance in non-uniformity
- This procedure fails if after the sampling the difference
- in uniformity exceeds delta, a proportion of n / number-of-alternatives
procedure verify_uniform (rnd, n, delta)
# generate a table counting the outcome of the generator results := table (0) every (1 to n) do results[@rnd] +:= 1 # retrieve the statistics smallest := n largest := 0 every num := key(results) do { # record result and limits write (num || " " || results[num]) if results[num] < smallest then smallest := results[num] if results[num] > largest then largest := results[num] } # decide if difference is within bounds defined by delta return largest-smallest < delta * n / *results
end
procedure main ()
gen_1 := create (|?10) # uniform integers, 1 to 10 if verify_uniform (gen_1, 1000000, 0.01) then write ("uniform") else write ("skewed") gen_2 := create (|(if ?2 = 1 then 6 else ?10)) # skewed integers, 1 to 10 if verify_uniform (gen_2, 1000000, 0.01) then write ("uniform") else write ("skewed")
end </lang>
Output:
5 99988 2 99998 10 99894 7 99948 4 100271 1 99917 9 99846 6 99943 3 99824 8 100371 uniform 5 49940 2 50324 10 50243 7 49982 4 50295 1 50162 9 49800 6 549190 3 50137 8 49927 skewed
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 to left of adverb freqtable=. /:~ (~. sample) ,. #/.~ sample expected=. n % # freqtable errmsg=. 'Distribution is potentially skewed' errmsg assert (delta * expected) > | expected - {:"1 freqtable freqtable
)</lang>
It is possible to use tacit expressions within an explicit definition enabling a more functional and concise style: <lang j>checkUniformT=: adverb define
0.05 u checkUniformT y : freqtable=. /:~ (~. ,. #/.~) u n=. */y errmsg=. 'Distribution is potentially skewed' errmsg assert ((n % #) (x&*@[ > |@:-) {:"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(delta*expected)>|expected-{:"1 freqtable</lang>
JavaScript
<lang javascript>function distcheck(random_func, times, opts) {
if (opts === undefined) opts = {} opts['delta'] = opts['delta'] || 2;
var count = {}, vals = []; for (var i = 0; i < times; i++) { var val = random_func(); if (! has_property(count, val)) { count[val] = 1; vals.push(val); } else count[val] ++; } vals.sort(function(a,b) {return a-b});
var target = times / vals.length; var tolerance = target * opts['delta'] / 100;
for (var i = 0; i < vals.length; i++) { var val = vals[i]; if (Math.abs(count[val] - target) > tolerance) throw "distribution potentially skewed for " + val + ": expected result around " + target + ", got " +count[val]; else print(val + "\t" + count[val]); }
}
function has_property(obj, propname) {
return typeof(obj[propname]) == "undefined" ? false : true;
}
try {
distcheck(function() {return Math.floor(10 * Math.random())}, 100000); print(); distcheck(function() {return (Math.random() > 0.95 ? 1 : 0)}, 100000);
} catch (e) {
print(e);
}</lang>
output
0 9945 1 9862 2 9954 3 10104 4 9861 5 10140 6 10066 7 10001 8 10101 9 9966 distribution potentially skewed for 0: expected result around 50000, got 95040
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>
PARI/GP
This tests the purportedly random 7-sided die with a slightly biased 1000-sided die. <lang parigp>dice5()=random(5)+1;
dice7()={
my(t); while((t=dice5()*5+dice5()) > 21,); (t+2)\3
};
cumChi2(chi2,dof)={ my(g=gamma(dof/2)); incgam(dof/2,chi2/2,g)/g };
test(f,n,alpha=.05)={ v=vector(n,i,f()); my(s,ave,dof,chi2,p); s=sum(i=1,n,v[i],0.); ave=s/n; dof=n-1; chi2=sum(i=1,n,(v[i]-ave)^2)/ave; p=cumChi2(chi2,dof); if(p<alpha, error("Not flat enough, significance only "p) , print("Flat with significance "p); ) };
test(dice7, 10^5) test(()->if(random(1000),random(1000),1), 10^5)</lang>
Output:
Flat with significance 1.000000000000000000 ### user error: Not flat enough, significance only 5.391077606003910233 E-3500006
Perl 6
Since the tested function is rolls of a 7 sided die, the test numbers are magnitudes of 10x bumped up to the closest multiple of 7. This reduces spurious error from there not being an integer expected value. <lang perl6>my $d7 = 1..7; sub roll7 { $d7.roll };
my $threshold = 3;
for 14, 105, 1001, 10003, 100002, 1000006 -> $n
{ dist( $n, $threshold, &roll7 ) };
sub dist ( $n is copy, $threshold, &producer ) {
my @dist; my $expect = $n / 7; say "Expect\t",$expect.fmt("%.3f"); loop ($_ = $n; $n; --$n) { @dist[&producer()]++; } for @dist.kv -> $i, $v is copy { next unless $i; $v //= 0; my $pct = ($v - $expect)/$expect*100; printf "%d\t%d\t%+.2f%% %s\n", $i, $v, $pct, ($pct.abs > $threshold ?? '- skewed' !! ); } say ;
} </lang> Sample output:
Expect 2.000 1 2 +0.00% 2 3 +50.00% - skewed 3 2 +0.00% 4 2 +0.00% 5 3 +50.00% - skewed 6 0 -100.00% - skewed 7 2 +0.00% Expect 15.000 1 15 +0.00% 2 17 +13.33% - skewed 3 13 -13.33% - skewed 4 16 +6.67% - skewed 5 14 -6.67% - skewed 6 16 +6.67% - skewed 7 14 -6.67% - skewed Expect 143.000 1 134 -6.29% - skewed 2 142 -0.70% 3 141 -1.40% 4 137 -4.20% - skewed 5 142 -0.70% 6 170 +18.88% - skewed 7 135 -5.59% - skewed Expect 1429.000 1 1396 -2.31% 2 1468 +2.73% 3 1405 -1.68% 4 1442 +0.91% 5 1453 +1.68% 6 1417 -0.84% 7 1422 -0.49% Expect 14286.000 1 14222 -0.45% 2 14320 +0.24% 3 14326 +0.28% 4 14425 +0.97% 5 14140 -1.02% 6 14275 -0.08% 7 14294 +0.06% Expect 142858.000 1 142510 -0.24% 2 142436 -0.30% 3 142438 -0.29% 4 143152 +0.21% 5 142905 +0.03% 6 143232 +0.26% 7 143333 +0.33%
PicoLisp
The following function takes a count, and allowed deviation in per mill (one-tenth of a percent), and a 'prg' code body (i.e. an arbitrary number of executable expressions). <lang PicoLisp>(de checkDistribution (Cnt Pm . Prg)
(let Res NIL (do Cnt (accu 'Res (run Prg 1) 1)) (let (N (/ Cnt (length Res)) Min (*/ N (- 1000 Pm) 1000) Max (*/ N (+ 1000 Pm) 1000) ) (for R Res (prinl (cdr R) " " (if (>= Max (cdr R) Min) "Good" "Bad")) ) ) ) )</lang>
Output:
: (checkDistribution 100000 5 (rand 1 7)) 14299 Good 14394 Bad 14147 Bad 14418 Bad 14159 Bad 14271 Good 14312 Good
PureBasic
<lang PureBasic>Prototype RandNum_prt()
Procedure.s distcheck(*function.RandNum_prt, repetitions, delta.d)
Protected text.s, maxIndex = 0 Dim bucket(maxIndex) ;array will be resized as needed For i = 1 To repetitions ;populate buckets v = *function() If v > maxIndex maxIndex = v Redim bucket(maxIndex) EndIf bucket(v) + 1 Next lbnd = Round((repetitions / maxIndex) * (100 - delta) / 100, #PB_Round_Up) ubnd = Round((repetitions / maxIndex) * (100 + delta) / 100, #PB_Round_Down) text = "Distribution check:" + #crlf$ + #crlf$ text + "Total elements = " + Str(repetitions) + #crlf$ + #crlf$ text + "Margin = " + StrF(delta, 2) + "% --> Lbound = " + Str(lbnd) + ", Ubound = " + Str(ubnd) + #crlf$ For i = 1 To maxIndex If bucket(i) < lbnd Or bucket(i) > ubnd text + #crlf$ + "Bucket " + Str(i) + " contains " + Str(bucket(i)) + " elements. Skewed." EndIf Next ProcedureReturn text
EndProcedure
MessageRequester("Results", distcheck(@dice7(), 1000000, 0.20))</lang> A small delta was chosen to increase the chance of a skewed result in the sample output:
Distribution check: Total elements = 1000000 Margin = 0.20% --> Lbound = 142572, Ubound = 143142 Bucket 1 contains 141977 elements. Skewed. Bucket 6 contains 143860 elements. Skewed.
Python
<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
<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]
- 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