Sardinas–Patterson algorithm

Idea of the algorithm

Consider the code { a ↦ 1 , b ↦ 011 , c ↦ 01110 , d ↦ 1110 , e ↦ 10011 } . This code, which is based on an example by Berstel, is an example of a code which is not uniquely decodable, since the string

can be interpreted as the sequence of codewords

but also as the sequence of codewords

Two possible decodings of this encoded string are thus given by cdb and babe.

In general, a codeword can be found by the following idea: In the first round, we choose two codewords x 1 and y 1 such that x 1 is a prefix of y 1 , that is, x 1 w = y 1 for some "dangling suffix" w . If one tries first x 1 = 011 and y 1 = 01110 , the dangling suffix is w = 10 . If we manage to find two sequences x 2 , … , x p and y 2 , … , y q of codewords such that x 2 ⋯ x p = w y 2 ⋯ y q , then we are finished: For then the string x = x 1 x 2 ⋯ x p can alternatively be decomposed as y 1 y 2 ⋯ y q , and we have found the desired string having at least two different decompositions into codewords.

In the second round, we try out two different approaches: the first trial is to look for a codeword that has w as prefix. Then we obtain a new dangling suffix w', with which we can continue our search. If we eventually encounter a dangling suffix that is itself a codeword (or the empty word), then the search will terminate, as we know there exists a string with two decompositions. The second trial is to seek for a codeword that is itself a prefix of w. In our example, we have w = 10 , and the sequence 1 is a codeword. We can thus also continue with w'=0 as the new dangling suffix.

Precise description of the algorithm

The algorithm is described most conveniently using quotients of formal languages. In general, for two sets of strings D and N, the (left) quotient N − 1 D is defined as the residual words obtained from D by removing some prefix in N. Formally, N − 1 D = { y ∣ x y ∈ D   and   x ∈ N } . Now let C denote the (finite) set of codewords in the given code.

The algorithm proceeds in rounds, where we maintain in each round not only one dangling suffix as described above, but the (finite) set of all potential dangling suffixes. Starting with round i = 1 , the set of potential dangling suffixes will be denoted by S i . The sets S i are defined inductively as follows:

S 1 = C − 1 C ∖ { ε } . Here, the symbol ε denotes the empty word.

S i + 1 = C − 1 S i ∪ S i − 1 C , for all i ≥ 1 .

The algorithm computes the sets S i in increasing order of i . As soon as one of the S i contains a word from C or the empty word, then the algorithm terminates and answers that the given code is not uniquely decodable. Otherwise, once a set S i equals a previously encountered set S j with j < i , then the algorithm would enter in principle an endless loop. Instead of continuing endlessly, it answers that the given code is uniquely decodable.

Termination and correctness of the algorithm

Since all sets S i are sets of suffixes of a finite set of codewords, there are only finitely many different candidates for S i . Since visiting one of the sets for the second time will cause the algorithm to stop, the algorithm cannot continue endlessly and thus must always terminate. More precisely, the total number of dangling suffixes that the algorithm considers is at most equal to the total of the lengths of the codewords in the input, so the algorithm runs in polynomial time as a function of this input length. By using a suffix tree to speed the comparison between each dangling suffix and the codewords, the time for the algorithm can be bounded by O(nk), where n is the total length of the codewords and k is the number of codewords. The algorithm can be implemented using a pattern matching machine. The algorithm can also be implemented to run on a nondeterministic Turing machine that uses only logarithmic space; the problem of testing unique decipherability is NL-complete, so this space bound is optimal.

A proof that the algorithm is correct, i.e. that it always gives the correct answer, is found in the textbooks by Salomaa and by Berstel et al.