Disclaimer: trees and rings are concepts found in Mathematics and Computer Science, but in this post they refer to the biology concepts ;)

### Biosphere 2

Earlier this year I went to Arizona and visited the Biosphere 2. Its initial goal was to simulate a close environment where a group of 6 humans should survive for 2 years without any exchange with the exterior (except sunlight, of course). They had different biomes, including a rain forest, an ocean to a desert.

The plan didn’t go well because they had missed interaction of the plants in one of the seasons, which causes the level of oxygen to fall under the accepted level, so they had to get extern supply. The mission did last 2 years, but was not a perfect closed environment.

Later, another mission was attempted, but had to be abandoned after a few months. The project lost its funding and no more attempts were performed. Nowadays, the place is used by biologists to study environments under controlled conditions and also open to the public for tours (the Nautilus magazine recently wrote more about this).

The tour includes a small museum explaining some of the research, one of them is analyzing trees to understand the ecosystem from the past. According to one of the exhibits, tree rings can be used to determine the age of the tree and also some climate from that period, a process called dendrochronology. The more rings a tree has, the old it is and the width of the tree varies with the climate during the period of its growth.

A lot of the trees alive today are not too old. On the other hand they have fossil of older trees. Since they possibly co-existed for a period of time, it’s possible to combine the data from the rings of both trees by matching the outer rings of the fossil tree with the inner rings of the recent tree.

### The Longest Prefix-Suffix String Overlap problem

I thought about the tree ring matching for a while and realized this can be reduced to the problem of finding the longest overlap between the suffix and prefix of two strings. More formally, given strings `A`

and `B`

, find the longest suffix of `A`

that is also a prefix of `B`

.

This problem can be easily solved using a simple quadratic algorithm, which is is probably fast enough for real world instances, possibly in the order of thousands of rings. In Python, we can use the array access to do:

def longest_suffix_prefix_naive(suffix, prefix): minlen = min(len(prefix), len(suffix)) max_match = 0 for match_len in range(1, minlen + 1): if prefix[:match_len] == suffix[-match_len:]: max_match = match_len return max_match

In the code above, `A`

is named `suffix`

and `B`

`prefix`

.

But can we do better? This problem resembles a string search problem, where we are given a text `T`

and a query string `Q`

and want to find whether `Q`

appears in `T`

as substring. One famous algorithm to solve this problem is the KMP, named after their authors Knuth Morris and Pratt (Morris came up with the algorithm independently from Knuth and Pratt) which is linear on the size of `T`

and `Q`

.

We’ll see how to use the ideas behind the KMP to solve our problem. So let’s start by reviewing the KMP algorithm.

### The KMP Algorithm: A refresher

The most straightforward way to search for a query `Q`

in a text is for every character in `T`

, check if `Q`

occurs in that position:

def substr(Q, T): tlen = len(T) qlen = len(Q) for i, _ in enumerate(T): match = 0 # Searches for Q starting from position i in T while match < qlen and \ i + match < tlen and \ Q[match] == T[i + match]: match += 1 if match == qlen: print 'found substring in position', i

#### Avoiding redundant searches

Imagine that our query string had no repeated characters. How could we improve the search?

Q = abcde T = abcdfabcde ^ mismatch in position 4

If we were using our naive idea, after failing to find `Q`

in position 4, we’d need to start the search over from `T(1) = 'b'`

. But since all characters are different in `Q`

, we know that there’s no `'a'`

between positions 1 and 3 because they matched the rest of strings of `Q`

, so we can safely ignore positions 1 to 3 and continue from position 4.

The only case we would need to look back would be if `'a'`

appeared between 1 and 3, which would also mean `'a'`

appeared more than once in `Q`

. For example, we could have

Q = ababac T = ababadac... ^ mismatch in position 5

When there is a mismatch in position 5, we need to go back to position 2, because `'abababac'`

could potentially be there. Let `MM`

be the current *mismatched string*, `'ababad'`

above and `M`

the *string we had right before the mismatch*, i.e., `MM`

without the last letter, in our case, `'ababa'`

. Note that `M`

is necessarily a prefix of `Q`

(because `MM`

is the first mismatch).

The string `M = 'ababa'`

, has a potential of containing `Q`

because it has a longest **proper** suffix that is also a prefix of `Q`

, `'aba'`

. Why is it important to be a **proper** prefix? We saw that `M`

is a prefix of `Q`

, hence the longest suffix of `M`

that is a prefix of `Q`

would be `M`

itself. We are already searched for `M`

and know it will be a mismatch, so we’re interested in something besides it.

Let’s define `LSP(A, B)`

as being the length of the longest proper suffix of `A`

that is a prefix of `B`

. For example, `LSP(M, Q) = len('aba') = 3`

. If we find a mismatch `MM`

, we know we need to go back `LSP(M, Q)`

positions in `M`

to restart the search, but we also know that the first `LSP(M, Q)`

positions will be a match in Q, so we can skip the first `LSP(M, Q)`

letters from `Q`

. In our previous example we have:

Q = ababac T = ababadac... ^ continue search from here (not from position 2)

We’ll have another mismatch but now `MM = 'abad'`

and `LSP(M, Q) = len('a') = 1`

. Using the same logic the next step will be:

Q = ababac T = ababadac... ^ continue search from here (not from position 2)

Now `MM = 'ad'`

and `LSP(M, Q) = 0`

. Now we shouldn’t go back any positions, but also won’t skip any characters from `Q`

:

Q = ababac T = ababadac... ^ continue search from here

If we know `LSP(M, Q)`

it’s easy to know how many characters to skip. We can simplify `LSP(M, Q)`

by noting that `M`

is a prefix of `Q`

and can be unambiguously represented by its length, that is `M = prefix(Q, len(M))`

. Let `lsp(i) = LSP(prefix(Q, i), Q)`

. Then `LSP(M, Q) = lsp(len(M))`

. Since `lsp()`

only depends on `Q`

, so we can pre-compute `lsp(i)`

for all `i = 0`

to `len(Q) - 1`

.

Suppose we have `lsp`

ready. The KMP algorithm would be:

def kmp(Q, T): lsp = precompute_lsp(Q) i = 0 j = 0 qlen = len(Q) tlen = len(T) while i < tlen and j < qlen: if j < qlen and Q[j] == T[i]: j += 1 i += 1 elif j > 0: j = lsp[j - 1] else: i += 1 # string was found if j == qlen: print i - j

We have to handle 3 cases inside the loop:

1: A match, in which case we keep searching both in T, Q.

2: A mismatch, but there’s a chance Q could appear inside the mismatched part, so we move Q the necessary amount, but leave T.

3: A mismatch, but there’s no chance Q could appear in the mismatched part, in which case we can advance T.

This algorithm is simple but it’s not easy to see it’s linear. The argument behind it is very clever. First, observe that `lsp[j] < j`

, because we only account for proper suffixes. It’s clear that i is bounded by `len(T)`

, but it might not increase in all iterations of the while loop. The key observation is that `(i - j)`

is also bounded by `len(T)`

. Now in the first condition, i increases while `(i - j)`

remains constant. In the second, i might remain constant but j decreases and hence `(i - j)`

increases. So every iteration either `i`

or `(i - j)`

increases, so at most `2T`

iterations can occur.

#### The pre-computed lsp

How can we pre-compute the lsp array? One simple idea is to, for every suffix of `Q`

, find the maximum prefix it forms with `Q`

. In the following code, `i`

denotes the start of the suffix and `j`

the start of the prefix.

def precompute_lsp_naive(Q): qlen = len(Q) lsp = [0]*qlen i = 1 while i < qlen : j = 0 while (i + j < qlen and Q[j] == Q[i + j]): lsp[i + j] = max(lsp[i + j], j + 1) j += 1 i += 1 return lsp

The problem is that this code is `O(len(Q)^2)`

. In practice `len(Q)`

should be much less than `len(T)`

, but we can do in `O(len(Q))`

, using a similar idea from the main idea of the KMP, but this time trying to find `Q`

in itself.

The main difference now is that we’ll compute `lsp`

on the fly. While searching for `Q`

in `Q`

, when we have a matching prefix, we update the `lsp`

. When we have a mismatch, instead of re-setting the search to the beginning of Q we skip some characters given by the `lsp`

. This works because `j < i`

and we had `lsp(k)`

for `k <= i`

defined. In the following code, the only changes we performed were adding the lsp attribution and having `T = Q`

.

def precompute_lsp(Q): qlen = len(Q) lsp = [0]*qlen i = 1 j = 0 while i < qlen: if Q[j] == Q[i]: j += 1 i += 1 # We only added this line lsp[i - 1] = j elif j > 0: j = lsp[j - 1] else: i += 1 return lsp

We can use exactly the same arguments from the main code of KMP to prove this routine is `O(len(Q))`

. The total complexity of KMP is `O(len(Q) + len(T))`

.

### Solving our problem

At this point we can observe that the problem we were initially trying to solve, that is, find the `lsp(A, B)`

, is the core of the KMP algorithm. We can search `B`

, the prefix, in `A`

, the suffix. If we can keep track of the indices i and j, we just need to find when i equals to the length of `T`

, which means there was a (partial) match of B in the suffix of `A`

.

We can generalize our kmp function to accept a callback that yields the indices of the strings at any step:

def kmp_generic(Q, T, yield_indices): lsp = precompute_lsp(Q) i = 0 j = 0 qlen = len(Q) tlen = len(T) while i < tlen: if j < qlen and Q[j] == T[i]: j += 1 i += 1 elif j > 0: j = lsp[j - 1] else: i += 1 yield_indices(i, j)

Now, if we want an array with the indices of all occurrences of Q in T, we can do:

def kmp_all_matches(Q, T): matches = [] qlen = len(Q) def match_accumulator(i, j): if j == qlen: matches.append(i - j) kmp_generic(Q, T, match_accumulator) return matches

Here we use a nested function as our callback. We can also use `kmp_generic`

to solve our original problem:

def longest_suffix_prefix(suffix, prefix): slen = len(suffix) max_match = [None] def max_matcher(i, j): if max_match[0] is None and i == slen: max_match[0] = j # Search prefix in suffix kmp(prefix, suffix, max_matcher) return 0 if max_match[0] is None else max_match[0]

Note that the nested function has *read-only* access to the variables in the scope of the outer function, like `max_match`

and `slen`

. To be able to share data with the nested function, we have to work with references, so we define `max_match`

as a single-element array.

### Conclusion

I used to participate in programming competitions. One of the most interesting parts of the problems are their statements. It’s usually an interesting random story from which you have to extract the actual computer science problem. A lot of the times though, the problem writer thinks about a problem and then how to create a story around that problem. It’s nice when we can find real-world problems that can be modelled as classic computer science problems.

I’ve used the KMP several times in the past but never took the time to study it carefully until now, which only happened because I wrote it down as a post.

### References

[1] Wikipedia – Knuth–Morris–Pratt algorithm

[2] Annual Growth Rings