6
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Background

Two strings s and t are called k-Abelian equivalent (shortened to k-equivalent in the following) for a positive integer k if the following conditions hold:

  • The length-k-1 prefixes of s and t are equal.
  • The length-k-1 suffixes of s and t are equal.
  • The strings s and t have the same multisets of length-k contiguous substrings. In other words, every string of length k occurs an equal number of times in s and t. Overlapping occurrences are counted as distinct.

Note that k-equivalent strings are always k-1-equivalent if k > 1, and the third condition implies that s and t have the same length. It is also known, and possibly useful, that the three conditions above are equivalent to the following one:

  • Every non-empty string of length k or k-1 occurs an equal number of times in s and t.

An Example

For example, consider the strings s = "abbaaabb" and t = "aabbaabb". Because the strings have the same number of each character, 4 of a and 4 of b, they are 1-equivalent. Consider then 2-equivalence. Their first and last characters are the same (both strings begin with a and end with b), so the first two conditions are satisfied. The length-2 substrings of s are aa (occurs twice), ab (twice), ba (once), and bb (twice), and those of t are exactly the same. This means that the strings are 2-equivalent. However, since their second letters are different, the strings are not 3-equivalent.

Input

Two alphanumeric strings s and t of the same length n > 0.

Output

The largest integer k between 1 and n (inclusive) such that s and t are k-equivalent. If no such k exists, the output is 0. In the above example, the correct output would be 2.

Rules

You can write a full program or a function. The lowest byte count wins, and standard loopholes are disallowed. Crashing on malformed input is perfectly fine.

Test Cases

"abc" "def" -> 0
"abbaabbb" "aabbbaab" -> 0
"abbaaabb" "aabbbaab" -> 2
"abbaaabb" "aabbaabb" -> 2
"aaabbaabb" "aabbaaabb" -> 3
"aaabbaabb" "aaabbaabb" -> 9
"abbabaabbaababbabaababbaabbabaab" "abbabaabbaababbaabbabaababbabaab" -> 9
"yzxyxx" "yxyzxx" -> 2
"xxxxxyxxxx" "xxxxyxxxxx" -> 5
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  • \$\begingroup\$ Why is the solution for the third case "abbaaabb" "aabbbaab" = 1? The prefixes a, a and the suffixes b, b are the same, and the substrings aa, ab, bb appear each twice, and the substring ba appears once in each string. \$\endgroup\$ – Jakube Feb 6 '15 at 16:11
  • \$\begingroup\$ @Jakube prefix of N length \$\endgroup\$ – Optimizer Feb 6 '15 at 16:17
  • \$\begingroup\$ @Optimizer It says: "The length-k-1 prefixes of s and t are equal." \$\endgroup\$ – Jakube Feb 6 '15 at 16:21
  • \$\begingroup\$ @Jakube Oh right \$\endgroup\$ – Optimizer Feb 6 '15 at 16:25
  • \$\begingroup\$ @Jakube Good find. There may be something wrong with my reference implementation, or it may be my mistake. I'll change it to a 2 for now, and investigate this later. \$\endgroup\$ – Zgarb Feb 6 '15 at 16:40
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Pyth: 26 22 bytes

lfq_TTCmmSm<>dbhkldldQ

Input are two strings like "abc","def". Try it online: Pyth Compiler/Executor

Explanation:

Quite some complicated code. But I'll start easy.

First we want to create all substrings of length k of the string d.

Sm<>dbhkld
 m      ld      apply a function ... to all elements b 
                in the list [0, 1, 2, ..., len(d)-1]
   >db              take the string d starting from the position b
  <   hk            but only hk = k + 1 characters
S               Sort the resulting list. 

For the string "acbde" and k = 3 this results in ['abc', 'bde', 'cbd', 'de', 'e']. Notice, that 'de' and 'e' are part of this list. This will be quite important in a bit.

I want the substrings for each k, one map again.

mSm<>dbhkldld
m          ld   apply a function (create all substrings of lenght k) to 
                all elements k in the list [0, 1, 2, ..., len(d)-1]

For the string "acbd" this results in [['a', 'b', 'c', 'd'], ['ac', 'bd', 'cb', 'd'], ['acb', 'bd', 'cbd', 'd'], ['acbd', 'bd', 'cbd', 'd']].

I want this for both strings, another map.

mmSm<>dbhkldldQ
m             Q   apply the function (see above) for all d in input()

[[['a', 'b', 'c'], ['ab', 'bc', 'c'], ['abc', 'bc', 'c']], [['d', 'e', 'f'], ['de', 'ef', 'f'], ['def', 'ef', 'f']]] for the input "abc","def", and after a zip C, this looks [(['a', 'b', 'c'], ['d', 'e', 'f']), (['ab', 'bc', 'c'], ['de', 'ef', 'f']), (['abc', 'bc', 'c'], ['def', 'ef', 'f'])].

Now comes the exciting part: We are of course only interest in the ks, where the substrings of both words are the same.

lfq_TTCmmSm<>dbhkldldQ
 f                         I filter the list for elements T, where
  q_TT                         reversed(T) == T
                           (basically T[0] == T[1])
l                          print the lenght of the resulting list.
                           (= the number of times, where the substrings of length k 
                              of string 1 are equal to the substrings of 
                              length k of string 2)

Since we also have the suffixes of length < k in Ts, this also checks, if the suffixes are equal.

The only thing, which I don't check is, if the prefixes are equal. But as it turns out, this isn't necessary. Lemma 2.3 (4<=>5) of the linked paper says: If the substrings of length k are the same, than pref_{k-1}(string 1) = pref_{k-1}(string 2) is equivalent to suff_{k-1}(string 1) = suff_{k-1}(string 2). Since the suffixes are equal, the prefixes are also equal.

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  • \$\begingroup\$ the 'ab' in g"abbab"3 is from the last two characters, not the prefix. So no, you are not correctly following the spec as you do not check for k-1 length prefix at all \$\endgroup\$ – Optimizer Feb 6 '15 at 17:46
  • \$\begingroup\$ @Optimizer Lemma 2.3 in the linked paper does the job. I'll add some more explanation. \$\endgroup\$ – Jakube Feb 6 '15 at 18:00
  • \$\begingroup\$ Are you saying that checking for k-1 length prefix is not at all required ? \$\endgroup\$ – Optimizer Feb 6 '15 at 18:04
  • \$\begingroup\$ @Optimizer Exactly \$\endgroup\$ – Jakube Feb 6 '15 at 18:09
  • \$\begingroup\$ @Zgarb Thanks for accepting my answer. Too sad, this question didn't get more recognition. I found this was one of the best challenges in a while, and am quite proud of my solution. \$\endgroup\$ – Jakube Feb 23 '15 at 9:46
1
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CJam, 51 49 47 41 34 bytes

WqN/:Q0=,,{Qf{_,,\f>\)f<$}~=},+W=)

Just to get something started. This can surely be golfed a lot. Looking at a different algorithm too.

Input goes as the two strings without quotes on separate lines.

Example:

aaabbaabb
aabbaaabb

Output:

3

Try it online here

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