Optimal Binary Search Trees Knuth
I am thinking if jk = ik, let's denote jk=ik=m, then we think of subsequences of [0,m-1] and [m+1,n], treat [m+1,n] as a standalone tree. And according to the paper, T and T' will be of same weighted path length if αn = α. So let's make a tree of subsequence [m+1,n] with last item being α, with weighted path length sum being Tk. And then replace the right-subtree of jk/ik with Tk for both T and T'.
Let's denote the weighted path length sum of [0, m-1] of T and T' being T0 and T0'. Because jk = ik, the levels of T and T' are both k. SoC(T) = T0 + k*Ak + Tk(starting from level k+1); C(T') = T1 + k*Ak + Tk(starting from level k+1); And C(T) should be equal to C(T'). So T1 = T1'
Replace right subtree of ik of T with right subtree of jk of T' will yield T'' with same weighted path length as T'.
PS: actually I don't quite know how it proves j2&lt;=i2, j3&lt;=i3 using induction and left-right symmetry, anyone can help answer that?