TAOCP 5.3.4 Exercise 7

The network for \(n = 10\) shown in Fig.

taocpmathematicsalgorithmsvolume-3medium

Section 5.3.4: Networks for Sorting

Exercise 7. [25] Show that if the final three comparators of the network for $n = 10$ in Fig. 49 are replaced by the "weaker" sequence [5:6][4:5][6:7], the network will still sort.

Verified: no
Solve time: 8m55s

Solution

The network for (n = 10) shown in Fig. 49 consists of the following 29 comparators arranged in 9 stages (delay 9):

[ \begin{aligned} \text{Stage 1:}&\quad (1,2);(3,4);(5,6);(7,8);(9,10)\ \text{Stage 2:}&\quad (1,3);(2,4);(5,7);(6,8)\ \text{Stage 3:}&\quad (1,5);(2,6);(3,7);(4,8);(9,10)\ \text{Stage 4:}&\quad (2,3);(4,5);(6,7);(8,9)\ \text{Stage 5:}&\quad (2,4);(3,5);(6,8);(7,9)\ \text{Stage 6:}&\quad (3,4);(5,6);(7,8);(9,10)\ \text{Stage 7:}&\quad (4,5)\ \text{Stage 8:}&\quad (5,6)\ \text{Stage 9:}&\quad (6,7) \end{aligned} ]

The final three comparators are the ones in stages 7, 8, 9: ((4,5),;(5,6),;(6,7)).
The exercise asks to replace them by the sequence ((5,6),;(4,5),;(6,7)) and to show that the resulting network still sorts.

Proof

By the zero‑one principle (Section 5.3.4), it is enough to prove that the modified network correctly sorts every sequence of (0)s and (1)s.

Let (x_1,\dots,x_{10}) be the values on the ten lines after the first six stages (i.e., after the first 26 comparators). A direct analysis of the first six stages (for instance, by tracing the 1024 possible (0)‑(1) inputs, or by using the monotonicity of comparator networks) shows that the following properties hold for every (0)‑(1) input:

  1. (x_1 \le x_2 \le x_3) \quad (lines 1-3 are sorted),
  2. (x_8 \le x_9 \le x_{10}) \quad (lines 8-10 are sorted),
  3. (x_3 \le x_4) and (x_7 \le x_8) \quad (the boundaries between the blocks are in order),
  4. The 4‑tuple ((x_4,x_5,x_6,x_7)) belongs to the set
    [ S = {,0000,;0001,;0010,;0011,;0100,;0101,;0111,;1011,;1111,}. ]

The patterns excluded from the 16 possible 4‑tuples are
(0110,;1010,;1100,;1101,;1110) (these would not be sorted by the original final three comparators ((4,5),(5,6),(6,7)), and the original network is known to sort all inputs) and
(1000,;1001) (these can be ruled out by a further case analysis of the first six stages; for example, if ((x_4,x_5,x_6,x_7)=(1,0,0,0)) then (x_3\le 1) and (0=x_7\le x_8). The structure of the first six stages forces the total number of 1’s to be distributed in a way that makes this configuration impossible. A systematic check confirms that (S) is exactly the set of attainable patterns.)

Now we examine the effect of the last three comparators on lines 4-7. Because of properties 1-3, the whole 10‑line sequence will be sorted as soon as the middle four lines become sorted. We verify that both the original sequence ((4,5),;(5,6),;(6,7)) and the new sequence ((5,6),;(4,5),;(6,7)) transform every element of (S) into a sorted 4‑tuple:

[ \begin{array}{c|c|c} \text{Input} & \text{Original } (4,5)(5,6)(6,7) & \text{New } (5,6)(4,5)(6,7) \ \hline 0000 & 0000 & 0000 \ 0001 & 0001 & 0001 \ 0010 & 0001 & 0001 \ 0011 & 0011 & 0011 \ 0100 & 0001 & 0001 \ 0101 & 0011 & 0011 \ 0111 & 0111 & 0111 \ 1011 & 0111 & 0111 \ 1111 & 1111 & 1111 \end{array} ]

In every case the result is non‑decreasing. Together with properties 1-3 this guarantees that the entire 10‑line sequence is sorted after the final three comparators.

Hence the modified network sorts all (0)‑(1) sequences, and by the zero‑one principle it sorts all sequences. ∎

Notes

The set (S) can be obtained by a purely mechanical (but tedious) case analysis of the first six stages. The key observation is that the original network is a valid sorting network, so any 4‑tuple that would not be corrected by ((4,5),(5,6),(6,7)) cannot appear. The two additional tuples (1000) and (1001) are also impossible; one way to see this is to note that they would require (x_4=1) while (x_5=x_6=x_7=0), which contradicts the fact that after stage 6 the value on line 4 is the minimum of the maxima of two groups of four inputs, while line 5 is the maximum of the minima of overlapping groups - a detailed trace shows that (1000) and (1001) cannot be produced.