TAOCP 5.3.1 Exercise 18

Section 5.3.1: Minimum-Comparison Sorting

Exercise 18. [20] If the procedure whose first steps are shown in Fig. 36 had produced the linear graph •, •, •, •, •, •, •, •, •, •, •, • with efficiency $12!/2^{29}$, would this have proved that $S(12) = 29$?

Verified: no
Solve time: 4m51s

Solution

The efficiency of a comparison tree that sorts (n) elements (with no redundant comparisons) is defined as the number of leaves divided by (2^{h}), where (h) is the height of the tree (the maximum number of comparisons on any path from the root to a leaf). Since such a tree has exactly (n!) leaves (one for each permutation), its efficiency is (n!/2^{h}).

The information‑theoretic lower bound states that any comparison‑based sorting algorithm must satisfy (S(n) \ge \lceil \lg n! \rceil). For (n=12) we have [ \lceil \lg 12! \rceil = 29, ] because (2^{28} < 12! < 2^{29}). Hence (S(12) \ge 29).

If the procedure shown in Fig. 36 (when completed to a full comparison tree) produced the linear graph , i.e., a completely sorted order , and the resulting tree had efficiency (12!/2^{29}), then [ \frac{12!}{2^{h}} = \frac{12!}{2^{29}} \quad\Longrightarrow\quad h = 29. ] Thus the tree would have height (29), meaning the algorithm uses at most (29) comparisons in the worst case. This would give (S(12) \le 29). Together with the lower bound (S(12) \ge 29) we would obtain (S(12) = 29).

Therefore, yes, such a tree would prove that (S(12) = 29). ∎

Notes

The linear graph (\bullet!-!\bullet!-!\cdots!-!\bullet) (12 nodes) represents one of the (12!) permutations; the hypothesis concerns the efficiency of the entire comparison tree, not of a single path. The argument uses only the definition of efficiency and the standard lower bound (\lceil \lg n! \rceil).