Raku Challenge, Week 92, Issue 1

This week’s task has an interesting solution in Raku. So, here’s the task:

OK, so if the two strings are isomorphic, their characters are mapped: for each character from the first string, the character at the same position in the second string is always the same.

In the stings abc and def, a always corresponds to d, b to e, and c to f. That’s a trivial case. But then for the string abca, the corresponding string must be defd.

The letters do not need to go sequentially, so the strings aeiou and bcdfg are isomorphic too, as well as aeiou and gxypq. But also aaeeiioouu and bbccddffgg, or the pair aeaieoiuo and gxgyxpyqp.

The definition also means that the number of different characters is equal in both strings. But it also means that if we make the pairs of corresponding letters, the number of unique pairs is also the same, right? If a matches x, there cannot be any other pair with the first letter a.

Let’s exploit these observation:

sub is-isomorphic($a, $b) {
    +(([==] ($a, $b)>>.chars) && 
      ([==] ($a.comb, $b.comb, ($a.comb Z~ $b.comb))>>.unique));
}

First of all, the strings must have the same length.

Then, the strings are split into characters, and the number of unique characters should also be equal. But the collection of the unique pairs from the corresponding letters from both strings should also be of the same size.

Test it:

use Test;

# . . .

is(is-isomorphic('abc', 'def'), 1);
is(is-isomorphic('abb', 'xyy'), 1);
is(is-isomorphic('sum', 'add'), 0);
is(is-isomorphic('ACAB', 'XCXY'), 1);
is(is-isomorphic('AAB', 'XYZ'), 0);
is(is-isomorphic('AAB', 'XXZ'), 1);
is(is-isomorphic('abc', 'abc'), 1);
is(is-isomorphic('abc', 'ab'), 0);

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