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Re: Random 1-1 mapping

by BerntB (Deacon)
on May 28, 2006 at 18:44 UTC ( [id://552202] : note . print w/replies, xml ) Need Help??


in reply to Random 1-1 mapping

What you want is a perfect hash function, that is inexpensive in computing time?

Why not ask for something simple, like a perpetum mobile? :-)

The $seed $input is the place in the original list? Maybe permutation encryption algorithms? But to keep them below $max...?

Update: If you could keep $max to a power of 2, you could maybe XOR with a constant? (Update2: The $seed (constant) would also have to be as many bits as $max.)

Code example for Update (n.b. $max must be 2^X):

my $max = 8; my $list = [10, 20, 30, 40, 50, 60, 70, 80]; for(my $seed = 0; $seed < $max; $seed++) { print "With seed $seed: "; for(my $i = 0; $i < @$list; $i++) { my $place = get_rand($i, $seed); print "$place:", $list->[$place], " "; } print "\n"; } sub get_rand { my($item, $seed) = @_; return $item ^ $seed; } __END__ # ------------------------------------------- # output: With seed 0: 0:10 1:20 2:30 3:40 4:50 5:60 6:70 7:80 With seed 1: 1:20 0:10 3:40 2:30 5:60 4:50 7:80 6:70 With seed 2: 2:30 3:40 0:10 1:20 6:70 7:80 4:50 5:60 With seed 3: 3:40 2:30 1:20 0:10 7:80 6:70 5:60 4:50 With seed 4: 4:50 5:60 6:70 7:80 0:10 1:20 2:30 3:40 With seed 5: 5:60 4:50 7:80 6:70 1:20 0:10 3:40 2:30 With seed 6: 6:70 7:80 4:50 5:60 2:30 3:40 0:10 1:20 With seed 7: 7:80 6:70 5:60 4:50 3:40 2:30 1:20 0:10

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Re^2: Random 1-1 mapping
by tomazos (Deacon) on May 28, 2006 at 18:54 UTC
    Update: Your solution is cool but $max is not guaranteed to be a power of 2. In fact, it almost certainly isn't.

    I just scribbled this on a piece of paper, I think I heard something about it to do with the RSA encryption algorithm once...

    $result = ($prime_number * $input + $seed) % $max;

    Example for $max = 10, $prime_number = 7 and $seed = 5..

    shuffle(5,10,0) == (7 * 0 + 5) % 10 == 5 % 10 == 5 shuffle(5,10,1) == (7 * 1 + 5) % 10 == 12 % 10 == 2 shuffle(5,10,2) == (7 * 2 + 5) % 10 == 19 % 10 == 9 shuffle(5,10,3) == (7 * 3 + 5) % 10 == 26 % 10 == 6 shuffle(5,10,4) == (7 * 4 + 5) % 10 == 33 % 10 == 3 shuffle(5,10,5) == (7 * 5 + 5) % 10 == 40 % 10 == 0 shuffle(5,10,6) == (7 * 6 + 5) % 10 == 47 % 10 == 7 shuffle(5,10,7) == (7 * 7 + 5) % 10 == 54 % 10 == 4 shuffle(5,10,8) == (7 * 8 + 5) % 10 == 61 % 10 == 1 shuffle(5,10,9) == (7 * 9 + 5) % 10 == 68 % 10 == 8

    So that works for 7 and 10. Does it work for any $max and any $prime_number? If not what conditions will it work for? I am no good with proofs.

    -Andrew.


    Andrew Tomazos  |  andrew@tomazos.com  |  www.tomazos.com
      A mapping of the form
      f(x) = (a*x + b) % m
      is 1-to-1 (restricted to 0 <= x < m) as long as gcd(a,m)==1, since that's the only time you can find an inverse to a mod m, and invert the function.

      I would suggest splitting your "seed" value into the a & b coefficients in the following way:

      • a = largest number less than seed satisfying gcd(a,m)=1
      • b = seed - a
      sub gcd { $_[1] ? gcd($_[1], $_[0] % $_[1]) : $_[0] } sub shuffle { my ($seed, $max, $i) = @_; my $ca = $seed; $ca-- until gcd($ca, $max) == 1; my $cb = $seed - $ca; ($ca * $i + $cb) % $max; } for my $seed (1 .. 10) { my @result = map shuffle($seed, 15, $_), 0 .. 14; print "seed=$seed ==> @result\n"; }
      But take this approach for what it's worth -- The only kinds of mapping you'll get by this process are simple linear (affine) mappings, which may not "look random enough":
      seed=1 ==> 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 seed=2 ==> 0 2 4 6 8 10 12 14 1 3 5 7 9 11 13 seed=3 ==> 1 3 5 7 9 11 13 0 2 4 6 8 10 12 14 seed=4 ==> 0 4 8 12 1 5 9 13 2 6 10 14 3 7 11 seed=5 ==> 1 5 9 13 2 6 10 14 3 7 11 0 4 8 12 seed=6 ==> 2 6 10 14 3 7 11 0 4 8 12 1 5 9 13 seed=7 ==> 0 7 14 6 13 5 12 4 11 3 10 2 9 1 8 seed=8 ==> 0 8 1 9 2 10 3 11 4 12 5 13 6 14 7 seed=9 ==> 1 9 2 10 3 11 4 12 5 13 6 14 7 0 8 seed=10 ==> 2 10 3 11 4 12 5 13 6 14 7 0 8 1 9
      For more "unpredictable" orders, you could have more tools at your disposal if $max is always a prime. Then you take one of the linear sequences above and use it as a sequence of powers of a generator element for the field mod $max.

      blokhead