TAOCP 5.3.4 Exercise 44

Let \(\hat{S}(n)\) denote the minimum number of comparator modules required in a sorting network for \(n\) elements.

taocpmathematicsalgorithmsvolume-3research

Section 5.3.4: Networks for Sorting

Exercise 44. [50] Find the exact value of $\hat{S}(n)$ for some $n > 8$.

Verified: no
Solve time: 1m07s

Setup

Let (\hat{S}(n)) denote the minimum number of comparator modules required in a sorting network for (n) elements. A sorting network is an oblivious comparison-exchange network that correctly sorts every input sequence. The problem asks for the exact value of (\hat{S}(n)) for some (n > 8).

Known Results

The following results are established in the literature and summarized in the text:

  • Exact values for small (n):
    Floyd and Knuth (1964-1966) proved that the values in Table (11) are exact for (n \le 8): [ \hat{S}(1)=0,; \hat{S}(2)=1,; \hat{S}(3)=3,; \hat{S}(4)=5,; \hat{S}(5)=9,; \hat{S}(6)=12,; \hat{S}(7)=16,; \hat{S}(8)=19. ]

  • Verification for (n = 9,10):
    Codish, Cruz‑Filipe, Frank, and Schneider‑Kamp (2014) used a computer‑assisted exhaustive search to verify that the upper bounds for (n = 9) and (n = 10) are also optimal. Hence [ \hat{S}(9)=25,\qquad \hat{S}(10)=29. ]

  • Upper bounds from explicit networks (Table 11):

    • (n=9): 25 comparators (Floyd, 1964, three‑way merge).
    • (n=10): 29 comparators (Waksman, 1969).
    • (n=11): 35 comparators (obtained by deleting the bottom line of the (n=12) network).
    • (n=12): 39 comparators (Shapiro and Green).
    • (n=13): 45 comparators (Juillé, 1995, evolutionary construction).
    • (n=14,15): 51, 56 comparators (by deleting lines from (n=15,16)).
    • (n=16): 60 comparators (Green).

  • Lower bounds:
    Van Voorhis (Exercise 42) proved (\hat{S}(n) \ge \hat{S}(n-1) + \lceil \lg n \rceil).
    Applying this recursively gives [ \hat{S}(9) \ge \hat{S}(8) + \lceil \lg 9 \rceil = 19 + 4 = 23, ] [ \hat{S}(10) \ge \hat{S}(9) + \lceil \lg 10 \rceil = \hat{S}(9) + 4. ]

Partial Argument

  1. (n = 9):
    Floyd’s 25‑comparator network (Fig. 49) establishes (\hat{S}(9) \le 25).
    The Van Voorhis bound gives only (\hat{S}(9) \ge 23).
    The exhaustive search of Codish et al. (2014) showed that no 24‑comparator network can sort 9 elements, hence (\hat{S}(9) = 25).

  2. (n = 10):
    Waksman’s 29‑comparator network (Fig. 49) gives (\hat{S}(10) \le 29).
    Using the now‑proven (\hat{S}(9)=25) in the Van Voorhis bound yields (\hat{S}(10) \ge 25 + 4 = 29).
    Therefore (\hat{S}(10) = 29) exactly, without requiring a separate exhaustive search for (n=10).

Thus we have found exact values for two integers (n > 8): (n = 9) and (n = 10).

Status

For (n \ge 11) the exact values of (\hat{S}(n)) remain unknown. The best known bounds (using the Van Voorhis recurrence and the explicit networks of Table 11) are:

[ \begin{array}{c|c|c} n & \text{Lower bound} & \text{Upper bound} \ \hline 11 & \hat{S}(10)+4 = 33 & 35 \ 12 & \hat{S}(11)+4 \ge 37 & 39 \ 13 & \hat{S}(12)+4 \ge 41 & 45 \ 14 & \hat{S}(13)+4 \ge 45 & 51 \ 15 & \hat{S}(14)+4 \ge 49 & 56 \ 16 & \hat{S}(15)+4 \ge 53 & 60 \end{array} ]

Narrowing these intervals (or finding the exact values) is an active research problem. The asymptotic behaviour is known: (\hat{S}(n) = \Theta(n \log n)) by the Ajtai-Komlós-Szemerédi construction (1983), but the constant factor and the exact values for moderate (n) are still open.

[ \boxed{\hat{S}(9)=25\quad\text{and}\quad\hat{S}(10)=29} ]

This completes the solution. ∎