Spliced alignment algorithm

To solve the problem, we use dynamic programming. We start with a few notations:

• A j-prefix of block $B_k=g_i\ldots g_j\ldots g_l$ is $B_k(j)=g_i\ldots g_j$.
• If B_k contains the position i, the i-prefix of $\Gamma ^{*} = B_1\ast...B_k$ is $\Gamma ^{*}(i) = B_1\ast...B_k(i)$.
• For block $B_k=g_i\ldots g_j$ First(B_k) = i and ${\rm Last}(B_k) = j$.
• A Chain $\Gamma ^{*} = B_1\ast\ldots\ast B_k$ ends at ${\rm Last}(B_k)$, and it ends before position i, if ${\rm Last}(B_k) < i$.
• $\mathcal{B}[i]$ is a set of blocks from $\mathcal{B}$ ending before i, i.e. if $B_k \in \mathcal{B}[i]$ than ${\rm Last}(B_k) < i$.
• $S(i,j,k) = \max \limits _{\Gamma ~ \vert ~ B_k \in \Gamma} S(\Gamma ^{*}(i),T(j))$.

The recursive relation of S(i,j,k) is :

$S(i,j,k) = \max \cases{ S(i,j-1,k) + (-,t_j) & if $i \geq {\rm First}(B_k)$... ... S({\rm Last}(B_l),j,l) + (g_i, -) & if $i = {\rm First}(B_k)$\space \cr }$

The Final score $\max \limits _{B_k \in \mathcal{B}} S({\rm Last}(B_k),m,k)$ can therefore be computed in time complexity of O(nmL²) and in space complexity of O(nmL).