in reply to Re: Answer: How can I print all the numbers from 1 to n in random order?
in thread How can I print all the numbers from 1 to n in random order?

The above comment would be true if it did $c = int(rand @_), but it doesn't. It just uses the floating point value returned from rand as a hash key. There are unlikely to be any collisions, even shuffling 1000000 numbers. You can test this with: 
perl -wle'undef $x{rand()} while keys(%x) == $x++; print $x'

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which adds keys until it gets a collision. Be prepared to wait a very long time.

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Re: Re: Re: Answer: How can I print all the numbers from 1 to n in random order?
by halley (Prior) on Apr 23, 2004 at 14:50 UTC
    Though the books always say that the hash is traversed in a "random" order, you should not take that to mean it is traversed in a TRULY RANDOM order. They just meant you should not write your program such that it will depend on any PARTICULAR order.

    A hashtable implementation might decide to visit the values %foo in a key order which looks suspiciously artificial: 0.05, 0.10, 0.15, 0.20, 0.06, 0.11, 0.16, 0.21, etc. You are hoping that the hashtable implementation doesn't actually bear any relationship to the pseudo-random number generator. Well, as seen in a recent security bug-fix, hash functions CAN be tied to random functions. You might end up aligning the two functions and defeating any appearance of unpredictability.

    This hashtable technique is interesting, but I wouldn't trust it to return things in a random order.

    --
    [ e d @ h a l l e y . c c ]

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      That was my initial thought on looking at this, but it doesn't bear out. It's the values you want in random order; random keys are selected for each value. The fact that values() doesn't return a truly random order based on the keys doesn't matter.

      Update:
      You could bypass all issues of ordering based on hash storage by replacing the values %foo with @foo{sort keys %foo}. Then (assuming a hypothetical keysof() function that returns the key for a value), values would be returned in order of sorted keysof(x), while plain values(%foo) (assuming a hypothetical order() function that returns the index into the keys %foo list of a particular key) is in order of sorted order(keysof(x)). Since keysof(x) is a random number between [0 and 1), order(keysof(x)) is equally random for each x, notwithstanding order()'s lack of randomness wrt its argument.

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