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evinda
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Hello! (Wave)I am reading about the Method of Four Russians from here.
We divide the dynamic programming table $D$ into $t$-blocks such that the last row of a $t$ − block is shared with the first row of the $t$ − block below it (if any), and the last column of a $t$ − block is shared with the first column of the $t$ − block to its right (if any).
Instead of filling all the cells of the matrix $D$, we just fill the cells that are on the boundary of each $t$-block, so just these that are on the last column and last row of each $t$-block.
We fill firstly the first row and column of the matrix $D$ and then we fill the last row and column of each $t$-block.
We fill these cells with the offset vectors, or not?
These offset vectors consists of the values of the function $\delta$, i.e., the costs of insertion, deletion, match und mismatch, right?
How exactly do we get the optimal alignment-score?
Why can we just fill the cells that are on the boundary of each $t$-block and get in that way the optimal alignment?
We divide the dynamic programming table $D$ into $t$-blocks such that the last row of a $t$ − block is shared with the first row of the $t$ − block below it (if any), and the last column of a $t$ − block is shared with the first column of the $t$ − block to its right (if any).
Instead of filling all the cells of the matrix $D$, we just fill the cells that are on the boundary of each $t$-block, so just these that are on the last column and last row of each $t$-block.
We fill firstly the first row and column of the matrix $D$ and then we fill the last row and column of each $t$-block.
We fill these cells with the offset vectors, or not?
These offset vectors consists of the values of the function $\delta$, i.e., the costs of insertion, deletion, match und mismatch, right?
How exactly do we get the optimal alignment-score?
Why can we just fill the cells that are on the boundary of each $t$-block and get in that way the optimal alignment?