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    <p>First, we read in the second file and put the values into an <em>array</em>. I further assume that this <code>chr1</code> is constant and can be discarded safely:</p> <pre><code>#!/usr/bin/perl use strict; use warnings; my @file2; open my $fh2, "&lt;", "file2" or die $!; while (&lt;$fh2&gt;) { my (undef, $num) = split; die "the number contains illegal characters" if $num =~ /\D/; push @file2, $num; } @file2 = sort {$a &lt;=&gt; $b} @file2; # sort ascending # remove previous line if sorting is already guaranteed. </code></pre> <p>Then, we define a sub to find the two values in our array. It is just a variation of a basic algorithm to find a certain value in a sorted list (in <em>O(log n)</em>), and should perform better than iterating over each value, at least on large sets. Also, it doesn't require reversing the whole list for each value.</p> <pre><code>sub find { my ($num, $arrayref) = @_; # exit if array is too small return unless @$arrayref &gt;= 2; # exit if $num is outside the values of this array (-1 is last element) return if $num &lt;= $arrayref-&gt;[0] or $arrayref-&gt;[-1] &lt; $num; my ($lo, $hi) = (1, $#$arrayref); my $i = int(($lo+$hi)/2); # start in the middle # iterate until # a) the previous index contains a number that is smaller than $num and # b) the current index contains a number that is greater or equal to $num. until($arrayref-&gt;[$i-1] &lt; $num and $num &lt;= $arrayref-&gt;[$i]) { # make $i the next lower or upper bound. # instead of going into an infinite loop (which would happen if we # assign $i to a variable that already holds the same value), we discard # the value and move on towards the middle. # $i is too small if ($num &gt; $arrayref-&gt;[$i] ) { $lo = ($lo == $i ? $i+1 : $i) } # $i is too large elsif ($num &lt;= $arrayref-&gt;[$i-1]) { $hi = ($hi == $i ? $i-1 : $i) } # in case I made an error: else { die "illegal state" } # calculate the next index $i = int(($lo+$hi)/2); } return @{$arrayref}[$i-1, $i]; } </code></pre> <p>The rest is trivial:</p> <pre><code>open my $fh1, "&lt;", "file1" or die $!; while (&lt;$fh1&gt;) { my ($chr, $num) = split; die "the number contains illegal characters" if $num =~ /\D/; if (my ($lo, $hi) = find($num, \@file2)) { if ($hi == $num) { print join("\t", $chr, $num, "Yes"), "\n"; } else { print join("\t", $chr, $num, "No", $hi-$num, $lo-$num), "\n"; } } else { # no matching numbers were found in file 2 print join("\t", $chr, $num, "No-match"), "\n"; } } </code></pre> <p>Output:</p> <pre><code>chr1 10227 No 790 -977 chr1 447989 No 6100 -8406 chr1 535362 No 9349 -75345 chr1 856788 Yes </code></pre>
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    1. POPerl: Compare keys of 2 hashes & print the difference between the closest keys
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    1. COHello again sir! Thanks for your answer. Could you please elaborate the subroutine step by step? I have been trying to understand it since last night(hence the lack of comments till now) and while I was able to figure out what `->` means and even make it work with the alternate `$$` and can also see the implementation is a `binary search` algorithm, the exact sequence of flow is eluding me..
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    2. CO@Neal I improved the algorithm; it now coveres more cases and won't go into an infinite loop any longer. Also, I added some comments that should make it easier to understand. `->[$index]` ist just used to dereference values from the array reference. The output code now distinguishes between a found match and a missed match.
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    3. CO**Example:** Array from `0` to `8`. We want to find `5`. **Init:** `$lo=1`, `$hi=8` and `$i=int(4.5)=4`. **First iteration:** The value at `$i-1` is `3`, and the value at `$i` is `4`. `$i` is too small, so now `$lo=4`. New `$i` is `6`. **Next iteration:** The value at `$i-1` is `5` and the value at `$i` is `6`. `$i` is too large, so now `$hi=6`. New `$i` is `5`. **Next iteration:** The value at `$i-1` is `4` and the value at `$i` is `5`. The termination condition matches. The two values are returned.
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