The Algorithm:

1.

Compute V(A, B) while saving the values of the ${{n}\over{2}}$-th row. Denote D(A,B) as the Forward Matrix F.

2.

Compute V(A^r, B^r) while saving the ${{n}\over{2}}$-th row. Denote D(A^r, B^r) as the Backward Matrix B.

3.

Find the column k^* so that the crossing point ( ${{n}\over{2}}$, k^*) satisfies:

$F({{n}\over{2}}, k^{*}) + B({{n}\over{2}}, m - k^{*}) = F(n,m)$

4.

Now that k^* is found, recursively partition the problem to two sub problems:

(i) Find the path from (0,0) to ( ${{n}\over{2}}$, k^*).
(ii) Find the path from (n, m) to ( ${{n}\over{2}}$, m - k^*).

Lemma 2.11   Time complexity of Hirschberg's algorithm is O(nm).

Proof:Time complexity
Let T^*(n,m) = Time to find the value of a $n\times{m}$ problem. Let T(n,m) = Time to find the path (solution) of a $n\times{m}$ problem.

$$T^{*}(n,m) = 2T(n,m) + T^{*}({{n}\over{2}}, k^{*}) + T^{*}({{n}\over{2}}, m - k^{*})$$ (2.1)

T(n,m)=cnm, therefore T^*(n,m) = 4T(n,m) = 4cnm. according to 2.1: $4cnm = 2cnm + 4c({{n}\over{2}} \times k^{*})+4c({{n}\over{2}} \times (m-k^{*}))$. Therefore, the time complexity will remain O(nm).

Lemma 2.12   Space complexity of Hirschberg's algorithm is O(m)

Proof:Space Complexity
Each Dynamic Programming computation requires storing one additional row (middle one) which can be discarded once the middle point is found. Therefore the space complexity will be O(m).