perl6 Permutation match - match

Question

I am still trying to work on permutation match, and I wonder if anyone has better way to do it. I want to match all patterns in an array in any order, i.e., match permutations of items (string or other objects) in an array. E.g., if array is (1,2,3), then it is true if a string contains 1 and 2 and 3 in any order; i.e, true if a string contains permutation of (1,2,3).
What I have now is this:
my #x = < one eins uno yi two zwei dos er one one one two two two >;
my #z = < one eins uno yi two zwei dos er one one one two two two foo >;
my $y = "xxx one eins uno yi two zwei dos er xxx";
sub matchAllWords($aString, #anArray) {
my $arraySize = #anArray.elems;
if $arraySize == 0 { False; }
elsif $arraySize == 1 {
($aString.match(/:i "#anArray[0]" /)).Bool;
} else {
my $firstCheck = ($aString.match(/:i "#anArray[0]"/)).Bool;
if $firstCheck {
$firstCheck
and
(matchAllWords($aString, #anArray[1..*]));
} else {
return False;
}
}
}
say matchAllWords($y, #x);
# result is True, but it should NOT be True because $y should not
# match permutations of #x which contains multiple identical elements
# of "one" and "two"
say matchAllWords($y, #z); # False as expected;
The problems is that my function matches all unique words in the array, but is unable to differentiate permutations of duplicate words. I can add more and more codes to tell if a word has been matched, but more codes to accomplish a simple idea, "permutation match", is un-perl-ly. Any suggestions? Thanks

Answer 1

New answer
Based on everyone's comments, here's a restatement of the problem as I now understand it, followed by a new solution:
Test that Y, a string, contains all of the strings in Z, a Bag (multiset) of strings, with correct copy count / multiplicity.
my \Z = < one eins uno yi two zwei dos er two > .Bag ;
my \Y = "xxx one eins uno yi two zwei dos er two xxx" ;
sub string-matches-bag ($string, $bag) {
for $bag.kv -> $sub-string, $copy-count {
fail unless ($string ~~ m:g/ $sub-string /).elems == $copy-count
}
True
}
say string-matches-bag Y, Z
Old answer
say so $y.words.all eq #z.any
An explanation for this line of code is in the last part of this answer.
I found your question pretty confusing. But I'm hopeful this answer is either what you want or at least enough to move things in the right direction.
I found your data confusing. There are two 'xxx' words in your $y but none in either array. So that bit can't match. There's a 'foo' in your #z. Was that supposed to be 'xxx'? There's a 'one' in your $y but both arrays have at least two 'one's. Is that an issue?
I found your narrative confusing too.
For this answer I've assumed that #z has a xxx at the end, and that the key comment is:
a simple idea, "permutation match"
say so $y.words.all eq #z.any
so returns the boolean evaluation (True or False) of the expression on its right.
The expression on so's right uses Junctions. An English prose summary of it is 'all of the "words" in $y, taken one at a time, are string equal to at least one element of #z'.
Is this the simple solution you're asking for?