Apparently you're right on both accounts. That of the nested range is a stupid mistake of mine, thinking that the end of the second range would always be the larger one because its start is. In fact, what I had in my head before I wrote anything down, was using max(): 
use List::Util qw(max);
$e = $range[$i][1] = max($e, $range[$j][1]);

[download]
But in fact, your code is simpler, thus, better.

But that of the infinite loop surprised me, a lot. I had assumed that the condition, the second expression in the for(;;) line, was being tested before any attempt at running the loop body, for each loop. 

print "before\n";
for(my $i = 5; $i < 2; $i++) {
    print " # $i\n";
}
print "after\n";

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Result: 
before
after

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But apparently, something goes wrong if you do a redo: 
print "before\n";
for(my $i = 5; $i < 10; $i++) {
    print " # $i\n";
    if($i==7) {
        $i = 12;
        redo;
    }
}
print "after\n";

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Result: 
before
 # 5
 # 6
 # 7
 # 12
after

[download]
 Ouch!

Hmm... perldoc -f redo apparently says:

The "redo" command restarts the loop block without evaluating the conditional again. The "continue" block, if any, is not executed.
I recalled the latter, but forgot about the former. I guess it applies here as well. But what seems odd: the variable is initialized again, but the condition is not tested? Heh? While for the first loop, it is initialized and then tested? That is awkward.

I haste to say this... but a goto almost seems better: 

for (my $i = 0; $i < @range; $i++) {
REDO:
    my $e = $range[$i][1];
    for(my $j = $i+1; $j < @range; $j++) {
        unless($range[$j][0] > $e) {
            $range[$i][1] = $range[$j][1] if $range[$j][1] > $e;
            splice @range, $j, 1;
            goto REDO;
        }
    }
}

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Much as I despise goto, it seems like using redo as in your code, with the extra condition, is actually worse. Well, I can work around it: 
@range = sort { $a->[0] <=> $b->[0] } @range;
for (my $i = 0; $i < @range; $i++) {
CURRENT:
    {
        my $e = $range[$i][1];
        for(my $j = $i+1; $j < @range; $j++) {
            unless($range[$j][0] > $e) {
                $range[$i][1] = $range[$j][1] if ($range[$j][1] > $e);
                splice @range, $j, 1;
                redo CURRENT;
            }
        }
    }
}

[download]
which seems OK.

In reply to Re: Re: Re: Overlapping portions of sub strings by bart
in thread Overlapping portions of sub strings by Anonymous Monk

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