TAOCP 5.3.4 Exercise 21
**False.
Section 5.3.4: Networks for Sorting
Exercise 21. [21] True or false: Inserting a new standard comparator into any standard sorting network yields another standard sorting network.
Verified: no
Solve time: 8m23s
Solution
False. Inserting a comparator into a sorting network can destroy the sorting property. A counterexample with (n=4) is given below.
Base sorting network
Consider the following 6‑comparator network (N) for four inputs: [ (1,3),\quad (1,2),\quad (3,4),\quad (1,3),\quad (2,4),\quad (2,3). ] By the zero‑one principle it suffices to test the (16) zero‑one sequences. A direct verification (or the trace below) shows that (N) correctly sorts every input; hence (N) is a standard sorting network.
Modified network
Insert a new standard comparator ((2,3)) after the third comparator of (N). The resulting network (N') is [ (1,3),\quad (1,2),\quad (3,4),\quad \mathbf{(2,3)},\quad (1,3),\quad (2,4),\quad (2,3). ]
Failure on a specific input
Take the zero‑one input (0100) (i.e. (x_1=0,;x_2=1,;x_3=0,;x_4=0)). Trace the execution of (N'):
| Step | Comparator | State before | State after |
|---|---|---|---|
| 1 | ((1,3)) | (0;1;0;0) | (0;1;0;0) |
| 2 | ((1,2)) | (0;1;0;0) | (0;1;0;0) |
| 3 | ((3,4)) | (0;1;0;0) | (0;1;0;0) |
| 4 | ((2,3)) | (0;1;0;0) | (0;0;1;0) |
| 5 | ((1,3)) | (0;0;1;0) | (0;0;1;0) |
| 6 | ((2,4)) | (0;0;1;0) | (0;0;1;0) |
| 7 | ((2,3)) | (0;0;1;0) | (0;1;0;0) |
The final output is (0100), which is not sorted (the sorted order would be (0001)). Therefore (N') is not a sorting network.
Since we have exhibited a standard sorting network (N) and a standard comparator whose insertion yields a network that does not sort, the statement is false.
[ \boxed{\text{False}} ]