TAOCP 5.3.2 Exercise 20

Section 5.3.2: Minimum-Comparison Merging

Exercise 20. [20] [20] Show that if $M(m,n+1) < M(m+1,n)$ for all $m < n$, then $M(m,n+1) < 1 + M(m,n)$ for all $m < n$.

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Solution

Fix $m<n$. Let $T = M(m,n+1)$ and $S = M(m,n)$. The hypothesis states that

$M(m,n+1) < M(m+1,n) \quad \text{for all } m<n,$

hence in particular

T < M(m+1,n). \tag{1}

Consider an optimal first comparison in a merging algorithm for $(m,n+1)$. Let it compare $A_i : B_j$.

If $i>1$, then before this comparison the algorithm has not yet distinguished $A_1$ from all elements of $B$. Construct an instance for $(m+1,n)$ by inserting a new element $A_0$ with $A_0 < A_1$. The same first comparison $A_i:B_j$ is valid in the $(m+1,n)$ instance and leads to a decision tree whose remaining subproblems correspond exactly to subinstances of total size at most $m+n$. This yields a strategy for $(m+1,n)$ using at most $T$ comparisons, contradicting $(1)$. Hence $i=1$.

A symmetric argument shows that $j=1$; otherwise, inserting a new element $B_0$ with $B_0 < B_1$ produces an instance $(m,n+2)$ in which the same initial decision would again embed the $(m,n+1)$ strategy into a problem of size $(m+1,n)$ after relabeling, contradicting $(1)$. Hence the first comparison must be $A_1 : B_1$.

After comparing $A_1 : B_1$, two cases arise.

If $A_1 < B_1$, the remaining problem is merging $A_2,\dots,A_m$ with $B_1,\dots,B_{n+1}$, which has cost $M(m-1,n+1)$.

If $A_1 > B_1$, the remaining problem is merging $A_1,\dots,A_m$ with $B_2,\dots,B_{n+1}$, which has cost $M(m,n)$.

Therefore

M(m,n+1) = 1 + \max\big(M(m-1,n+1),, M(m,n)\big). \tag{2}

If $M(m,n+1) \ge 1 + M(m,n)$, then from $(2)$ it follows that

M(m-1,n+1) \le M(m,n). \tag{3}

Applying the hypothesis with $(m-1,n)$ in place of $(m,n)$ gives

M(m-1,n+1) < M(m,n). \tag{4}

The inequalities $(3)$ and $(4)$ are incompatible, hence the assumption $M(m,n+1) \ge 1 + M(m,n)$ is impossible. Therefore

$M(m,n+1) < 1 + M(m,n)$

for all $m<n$.

This completes the proof. ∎