I'm trying to learn how the "failure table" is constructed in the Knuth-Morris-Pratt algorithm since it seemed nontrivial to me that you could do it in $O(k)$ time (where $k$ is the length of the pattern, $W$). To be clear, we let the failure table $T$ be a length $k$ array where $T[i]$ is the length of the longest proper prefix of $W$ which is also a suffix of $W[1]\cdots W[i]$.

I was trying to understand it better by reading the following code:

    vector<int> T(pattern.size());
    T[0] = 0;
    int j;
    for (int i = 1; i < pattern.size(); i++) {
        j = T[i - 1];
        // Find largest j s.t. T[:j+1] == T[i-j:i+1]
        while (j > 0 && pattern[j] != pattern[i]) {
            j = T[j - 1];
        T[i] = (pattern[j] == pattern[i]) ? j + 1 : j;
    return T;

The part I'm confused about is why pattern[j] == pattern[i] is enough to guarantee that $T[i] = j + 1$. I was trying to prove this inductively but have had no luck. I've phrased the problem as follows:

Suppose $W[1]\cdots W[T[i - 1]] = W[i - T[i - 1]]\cdots W[i - 1]$. If $W[i] == W[j]$, then is it true that $W[1]\cdots W[j] == W[i - j + 1]\cdots W[i]$?

I can't make any headway with this, nor can I see why it should be true. It still seems like we have to check $W[\ell] = W[i - j + \ell]$ for all $\ell = 1...j$. I can't figure out how to use the inductive hypothesis to prove this result. Do I have to use strong induction (where we assume all prior steps are true and not just the previous step) to prove this?

  • $\begingroup$ The code snippet doesn’t look like a correct implementation of KMP. $\endgroup$ Nov 20, 2018 at 14:45
  • $\begingroup$ @DmitriUrbanowicz It passes LeetCode problem 28 (leetcode.com/problems/implement-strstr), however in my own messing around I did change the line j-- from it's original j = T[j - 1] to see if it worked, and it still did. $\endgroup$ Nov 20, 2018 at 14:46
  • $\begingroup$ @DmitriUrbanowicz My logic for changing j = T[j - 1] to j-- is that I proved inductively that T[i+1] <= T[i] + 1. $\endgroup$ Nov 20, 2018 at 14:47
  • $\begingroup$ Your modification fails to compute the table correctly for this input: "xaxxyxaxxx". $\endgroup$ Nov 20, 2018 at 15:07
  • $\begingroup$ Wow, great counterexample! I knew my change wasn't "proven", I just thought I had heuristic reason to try it, and since it passed all the tests I assumed it worked, but I guess not. Regardless, my inability to prove the correctness of either j-- or j = T[j - 1] is due to my lack of understanding of the condition W[i] == W[j]. Do you have any insight into that? $\endgroup$ Nov 20, 2018 at 15:11

1 Answer 1


A proper prefix, which is also a suffix of the same string, is usually called border.

The condition you're asking about basically says this:

  1. If $W[1..i-1]$ has a border of length $j$ and $W[j+1] = W[i]$, then $W[1..i]$ has a border of length $j+1$.
  2. If $W[1..i-1]$ has no border of length $j$ such that $W[j+1] = W[i]$, then the length of the maximal border of $W[1..i]$ is $0$.

Both points follow directly from definition of string equality and definition of borders.

What you need to prove is that the while-loop indeed enumerates all possible borders in descending order.


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