Here's a new solution for question 3a. This is not strict in the sense that you have to turn off strict for the set-up code to work. However, the actual test \$foo->[1] == \$foo->{1} runs under strict, unlike in my original symbolic ref solution.

I think that this solution differs from all the solutions given so far, so I submit it even though this puzzle is now quite old.

<Reveal this spoiler or all in this thread>

In reply to Re: How's your Perl? by ambrus
in thread How's your Perl? by xmath

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