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| • |
Finding
the best alignment
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has
become finding the best
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path
through a table
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Position
(i,j) in the table can
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be
reached from
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– |
Position
(i,j-1)
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At
cost -1
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– |
Position
(i-1,j)
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At
cost -1
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– |
Position
(i-1,j-1)
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At
cost <table entry>
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BestPathCost
(table T[1..n,1..n]):
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// C[i,j] represents best cost
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// to reach position (i,j) in T
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C[0,0] = 0
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for i =
1 to n do C[i,0] = -i;
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for j =
1 to n do C[0,j] = -j;
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for i =
1 to n do
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for j = 1 to n do
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C[i,j] = max{ C[i,j-1] - 1,
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C[i-1,j] - 1,
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C[i-1,j-1] + T[i,j] };
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return
C[n,n];
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