Sequence Comparison.pdf

30 2 Basic Algorithmic Techniques
and
〈P,R,E,S,I,D,E,N,T〉
〈P,R,O,V,I,D,E,N,C,E〉,
〈P,R,D,N〉 is a common subsequence of them, whereas 〈P,R,V 〉 is not. Their LCS
is 〈P,R,I,D,E,N〉.
Now let us formulate the recurrence for computing the length of an LCS of two
sequences. We are given two sequences A = 〈a 1 ,a 2 ,...,a m 〉, and B = 〈b 1 ,b 2 ,...,b n 〉.
Let len[i, j] denote the length of an LCS between 〈a 1 ,a 2 ,...,a i 〉 (a prefix of A) and
〈b 1 ,b 2 ,...,b j 〉 (a prefix of B). They can be computed by the following recurrence:
⎧
⎨ 0 if i = 0or j = 0,
len[i, j]= len[i − 1, j − 1]+1 ifi, j > 0 and a i = b j ,
⎩
max{len[i, j − 1],len[i − 1, j]} otherwise.
In other words, if one of the sequences is empty, the length of their LCS is just
zero. If a i and b j are the same, an LCS between 〈a 1 ,a 2 ,...,a i 〉, and 〈b 1 ,b 2 ,...,b j 〉
is the concatenation of an LCS of 〈a 1 ,a 2 ,...,a i−1 〉 and 〈b 1 ,b 2 ,...,b j−1 〉 and a i .
Therefore, len[i, j] = len[i − 1, j − 1]+1 in this case. If a i and b j are different, their
LCS is equal to either an LCS of 〈a 1 ,a 2 ,...,a i 〉, and 〈b 1 ,b 2 ,...,b j−1 〉, or that of
〈a 1 ,a 2 ,...,a i−1 〉, and 〈b 1 ,b 2 ,...,b j 〉. Its length is thus the maximum of len[i, j − 1]
and len[i − 1, j].
Figure 2.11 gives the pseudo-code for computing len[i, j]. For each entry (i, j),
we retain the backtracking information in prev[i, j]. Iflen[i − 1, j − 1] contributes
the maximum value to len[i, j], then we set prev[i, j]=“↖.” Otherwise prev[i, j] is
set to be “↑” or“←” depending on which one of len[i − 1, j] and len[i, j − 1] contributes
the maximum value to len[i, j]. Whenever there is a tie, any one of them will
Algorithm LCS LENGTH(A = 〈a 1 ,a 2 ,...,a m 〉, B = 〈b 1 ,b 2 ,...,b n 〉)
begin
for i ← 0 to m do len[i,0] ← 0
for j ← 1 to n do len[0, j] ← 0
for i ← 1 to m do
for j ← 1 to n do
if a i = b j then
len[i, j] ← len[i − 1, j − 1]+1
prev[i, j] ←“↖”
else if len[i − 1, j] ≥ len[i, j − 1] then
len[i, j] ← len[i − 1, j]
prev[i, j] ←“↑”
else
len[i, j] ← len[i, j − 1]
prev[i, j] ←“←”
return len and prev
end
Fig. 2.11 Computation of the length of an LCS of two sequences.