TAOCP 5.3.4 Exercise 47

Section 5.3.4: Networks for Sorting

Exercise 47. [47] Is the function $T(n)$ of exercise 6 strictly less than $\hat{T}(n)$ for some $n$?

Verified: no
Solve time: 11m37s

Setup

Let (T(n)) be the minimum number of stages required to sort (n) numbers by an adaptive parallel algorithm that in each stage performs any set of disjoint comparisons (the choice of comparisons in a stage may depend on the outcomes of all previous stages). This is the “parallel sorting” model described in Exercise 6.

Let (\hat{T}(n)) be the minimum number of stages required to sort (n) numbers by an oblivious sorting network (a fixed sequence of stages, each consisting of disjoint comparators). This is the standard sorting‑network depth, denoted (\hat{T}(n)) in Exercise 7.

The question asks whether there exists an (n) for which (T(n) < \hat{T}(n)), i.e. whether adaptivity can ever reduce the number of parallel stages below the best possible oblivious network depth.

Known Results

1. Oblivious sorting‑network depth (\hat{T}(n)) is known exactly for (n \le 16) (see Fig. 49 and the literature on optimal sorting networks): [ \begin{array}{c|cccccccccccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \ \hline \hat{T}(n) & 0 & 1 & 3 & 3 & 5 & 5 & 6 & 6 & 7 & 7 & 8 & 8 & 9 & 9 & 9 & 9 \end{array} ]

2. Information‑theoretic lower bound for (T(n)).
   Each stage can perform at most (\lfloor n/2 \rfloor) disjoint comparisons. To distinguish the (n!) permutations we need at least (\lceil \lg(n!) \rceil) comparisons in the worst case, hence [ T(n) \ge \left\lceil \frac{\lg(n!)}{\lfloor n/2 \rfloor} \right\rceil. ] For small (n) this gives: [ \begin{array}{c|cccccccccccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \ \hline \text{lower bound} & 0 & 1 & 3 & 3 & 4 & 4 & 5 & 4 & 5 & 5 & 6 & 5 & 6 & 6 & 6 & 6 \end{array} ]

3. Equality for (n \le 4).
   The lower bounds match (\hat{T}(n)) for (n = 1,2,3,4); therefore (T(n) = \hat{T}(n)) for these (n).

4. Best known adaptive algorithms.
   No adaptive parallel sorting algorithm has been discovered that uses fewer stages than the best oblivious network for any (n). The best known upper bounds for (T(n)) coincide with (\hat{T}(n)) for all (n \le 16). In particular:

   • (T(5)) is either (4) (if a 4‑stage adaptive algorithm exists) or (5) (if not).
   • (T(6)) is either (4) or (5).
   • (T(7)) is either (5) or (6).
   • (T(8)) is either (4), (5), or (6).

5. Merging vs. sorting.
   Exercise 46 proves that for merging, the adaptive model achieves the same delay as the bitonic merge ((\lceil \lg(m+n) \rceil)). Whether adaptivity helps for full sorting remains open.

Partial Argument

The zero‑one principle holds for adaptive parallel algorithms as well, so one could in principle test all 0‑1 sequences to verify a candidate adaptive algorithm. The difficulty lies in the enormous number of possible decision trees. For (n=5) the lower bound is (4), while (\hat{T}(5)=5). If a 4‑stage adaptive algorithm for (n=5) could be constructed, it would immediately show (T(5) < \hat{T}(5)). Despite extensive search, no such algorithm has been found, nor has it been proved impossible. The same gap between the information bound and the best known construction persists for several larger (n) (e.g., (n=6,7,8,12,16)).

Status

Open problem. It is not known whether (T(n) < \hat{T}(n)) for any (n). The question is equivalent to asking whether there exists an adaptive parallel sorting algorithm that sorts some number of elements in fewer parallel steps than any oblivious sorting network. Current knowledge: (T(n) = \hat{T}(n)) for (n \le 4); for (n \ge 5) the best known upper bounds for (T(n)) equal (\hat{T}(n)), but the information‑theoretic lower bounds are strictly smaller for many (n). Whether those lower bounds are attainable remains unknown. ∎