Five Finger Stat Punch

Wherein the middle part of the story is the really interesting part…


We dance round in a ring and suppose, but the secret sits in the middle and knows.

Robert Frost

Five-number Summary

Submitted by: Mohammad S Anwar

You are given an array of integers.

Write a script to compute the five-number summary of the given set of integers.

You can find the definition and example in the Wikipedia page.


In statistics, mathematical analysis is performed on what are often large datasets. In this way of looking at things, the weight of individual data points is deemphasized, as we look beyond the values in an effort to infer more general statements about the collection as a whole. The data itself is usually empirical in origin, gathered from real-world phenomena, and as such are always subject to some degree of collection or recording error. Much of statistical analysis addresses the precision within the dataset, assigning numerical weights to the confidence in the values, which can be mathematically extrapolated as a caveat onto any conclusions made. The analysis, after all, can only be as good as the data it comes from.

Given some collection of data, when seeking a normal, representative, middle value, it might not be immediately obvious that this idea can be approached in more than manner. There are in fact quite a few ways to choose from. We could for instance find an arithmetic average, a mean value, summing and dividing, but even that is only one of several more complicated mean values available to us. We first explored this idea in PWC157, calculating the Pythagorean Means. Ultimately the derivation of the data itself will determine the best fit for the job.

But even the average, some version of average, isn’t necessarily meaningfully representative. Another thing we could look at is which value occurs most often. To take an extreme case, if the data isn’t numerically related at all this can be a better way to summarize, obviously. A box of red balls with a few green ones interspersed can still be meaningfully called a box of red balls. If one value keeps showing up in the data there is presumably some reason to be inferred, and we could look into that.

One somewhat radical approach to select a representative sample is to first order the data and then take the literal middle value. Arithmetically this doesn’t make much sense, but statistically it does, as one unexpected benefit of this method is its remarkable resiliency against spurious, outlier data points.

Given a collection of values of human heights, one might expect a range between extremes, with babies at maybe 30cm and large adults at somewhat over 200cm. Averaging all of these might then tell you something, like how big a human is, that we could then compare to some other animals. Depending on how many babies are involved the number might be around say, 140-150cm? Just guessing here; it doesn’t really matter. However a single data entry error for, say air travel on business for a year: 15000km, would not only make that average useless but also poison any additional, further use of the value downstream. Using a median, though, the single bad outlier data point would only move the central element slightly, with about equal probability up or down. Our monsterous error has been transformed into a slight loss of precision.

The quartitles are an extension of this data-levelling idea into two more subdivisions on either side of the median, to slot in and combine with the median to provide points at 25%, 50%, and 75%. Exactly how these points are arrived at is not universally accepted.

The problem arrives when we wish to take the median of one-half the data. If the whole set itself has a discrete median, do we include that point when calculating one-half the data? Including it will make each computed half slightly larger than a real half, and not including it will make the cut smaller. There are more exact methods to accommodate a more even division but we have to stop and consider the imprecision of the median itself before we make things unnecessarily more complicated. Uncertainty, we need to remember, carries over through the calculations, and the median is not the mean. It is essentially non-numeric, instead positional.

It should be added that should we require a rigorous mathematical certainty there are well-defined approaches to looking at the way data is distributed within a range, such as the standard deviation. So we have that there waiting if we want it. On the other hand if we’re looking for a middle, a representative low number, a representative high number, and a set of outer boundaries that all the data lies inside, then the 5-number summary has proven to be a pretty practical start at getting the feel of a dataset — arithmetically imperfect as that definition may be.


Depending on whether the number of elements under consideration is even or odd, to find the median we either select the center element from the list or average the two elements either side of the center. In this simplest case of an arithmetic mean we add the two values and divide by 2.

Then, depending on whether the left and right — equal — sides around the median have an even or odd count we again need to select either a discrete center element or average the two surrounding to find that median, both above and below.

Because again there oddly is no rigorous consistency in defining the calculation of the 1st and 3rd quartiles, and these differing approaches result in differing outcomes, we will need to just pick one. I don’t really have a strong opinion on this, so we will use the common method of halving the array if the median point lies between two values, and if the median is a discreet element not including it in the half-array length.

We are left with four cases to evaluate, for each of the four combinations of even and odd numbers of data points in both the full dataset and our chosen method of determining the half-size.

I still find it exceedingly odd that the method for determining the 25% and 75% medians, if you will, are not well-defined. However in the statistical context of the summary, a closer look suggests an essentially parallel ambiguity in what we want from our middle value. So in a weird way the ambiguity is consistent. It’s positional, and treating it as purely numeric is misleading.

After all, the statistics are based on the median, not the mean.


use feature ":5.26";
use feature qw(signatures);
no warnings 'experimental::signatures';

my @arr = @ARGV;

## default data array
@arr == 0 and @arr = (1,2,3,4,5,6,7,8,9);

## main
@arr = sort { $a <=> $b } @arr;
say " Data Values: [ @arr ]\n";
my @five = ( $arr[0], quartiles(\@arr), $arr[-1] );
my @names = ( "Minimum", "1st Quartile", "Median", "3rd Quartile", "Maximum" );
say sprintf "%12s: %-s", $names[$_], $five[$_] for (0..4);

## subs
sub quartiles ( $arr ) {
    my $count = scalar $arr->@*;
    my $half  = int($count/2);          ## note integer truncation
    my ($q1, $q2, $q3);
    if ( $count & 1 ) {                     ## odd num elements
        $q2 = $arr->[$half];
        if ( $half & 1 ) {                          ## odd halfway index
            $q1 = $arr->[$half/2];
            $q3 = $arr->[$half + $half/2 + 1];
        else {                                      ## odd halfway index
            $q1 = avg($arr->[$half/2-1], $arr->[$half/2]);
            $q3 = avg($arr->[$half+$half/2], $arr->[$half+$half/2+1]);
    else {                                  ## even num elements
        $q2 = avg($arr->[$half-1], $arr->[$half]);
        if ( $half & 1 ) {                          ## odd halfway index
            $q1 = $arr->[$half/2];
            $q3 = $arr->[$half + $half/2];
        else {                                      ## even halfway index 
            $q1 = avg($arr->[$half/2-1], $arr->[$half/2]);
            $q3 = avg($arr->[$half+$half/2-1], $arr->[$half+$half/2]);
    return ( $q1, $q2, $q3 );

sub avg( $n1, $n2 ) {
    return ($n1 + $n2)/2;

The code is broken down into four cases, with subtle differences between them as to which elements are included in the lower and higher subsets. To keep track of it all I made a chart so summarize them:

the four cases, broken down: 
(array length, computed half-array length)

[0, 1, 2, 3,    4,    5, 6, 7, 8]           
count = 9
half  = 4
q2 = half
q1 = (1+2)/2 = avg(half/2-1, half/2)
q3 = (6+7)/2 = avg(half+half/2, half+half/2+1)

[0, 1, 2, 3,    4, 5, 6, 7]                 
count = 8
half  = 4
q2 = avg(half-1,half)
q1=(1+2)/2 = avg(half/2-1, half/2)
q3=(5+6)/2 = avg(half+half/2-1, half+half/2)
[0, 1, 2,    3,    4, 5, 6]                 
count = 7
half  = 3
q2 = 3 = half
q1 = 1 = half/2
q3 = 5 = half + half/2 + 1

[0, 1, 2,     3, 4, 5]                      
count = 6
half  = 3
q2 = avg(half-1,half) 
q1 = 1 = half/2
q3 = 4 = half + half/2

The Perl Weekly Challenge, that idyllic glade wherein we stumble upon the holes for these sweet descents, is now known as

The Weekly Challenge – Perl and Raku

It is the creation of the lovely Mohammad Sajid Anwar and a veritable swarm of contributors from all over the world, who gather, as might be expected, weekly online to solve puzzles. Everyone is encouraged to visit, learn and contribute at

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