Table 2 The Inside VMT algorithm

   1:    Allocate a matrix B of size ||s|| × ||s||, and initialize all entries in B with ϕ elements.

   2:    Call Compute-Inside-Sub-Matrix ([0, n], [0, n]), where n is the length of s.

   3:    return B

Compute-Inside-Sub-Matrix (I, J)

pre-condition: The values in B_{[I,J], [I,J]}, excluding the values in B_I,J, are computed, and B_I,J= β_I,(I,J)⊗ β_(I,J),J.

post-condition: B_{[I,J], [I,J]}= β_{[I,J], [I,J]}.

   1:    if I = [i, i] and J = [j, j] then

   2:       If i ≤ j, compute β_i,j(in o (||s||) running time) by querying computed values in B and the value μ_i,jwhich is stored in B_i,j. Update B_i,j← β_i,j.

   3:    else

   4:       if |I| ≤ |J| then

   5:          Let J₁ and J₂ be the two intervals such that J₁J₂ = J, and |J₁| = ⌊|J|/2⌋.

   6:          Call Compute-Inside-Sub-Matrix (I, J₁).

   7:          Let L be the interval such that (I, J)L = (I, J₂).

   8:          Update $B_{I,J_2}\leftarrow B_{I,J_2}\oplus \left(B_{I,L}\otimes B_{L,J_2}\right)$.

   9:          Call Compute-Inside-Sub-Matrix (I, J₂).

   10:       else

   11:          Let I₁ and I₂ be the two intervals such that I₂I₁ = I, and |I₂| = ⌊|I|/2⌋.

   12:          Call Compute-Inside-Sub-Matrix (I₁, J).

   13:          Let L be the interval such that L(I, J) = (I₂, J).

   14:          Update $B_{I_2,J}\leftarrow \left(B_{I_2,L}\otimes B_{L,J}\right)\oplus B_{I_2,J}$.

   15:          Call Compute-Inside-Sub-Matrix (I₂, J).

   16:       end if

   17:    end if