TAOCP 5.2.2 Exercise 40
Algorithm R (radix exchange sort) sorts a file of \(N\) records \(R_1,\dots,R_N\) with keys \(K_1,\dots,K_N\) by examining the bits of the keys from the most significant to the least significant.
Section 5.2.2: Sorting by Exchanging
Exercise 40. [21] Explain how to modify Algorithm R so that it works with reasonable efficiency when sorting files containing numerous equal keys.
Verified: no
Solve time: 8m37s
Solution
Algorithm R (radix exchange sort) sorts a file of (N) records (R_1,\dots,R_N) with keys (K_1,\dots,K_N) by examining the bits of the keys from the most significant to the least significant. It uses a stack to manage subfiles; each stack entry is a triple ((l,r,d)) indicating that the subfile (R_l,\dots,R_r) should be sorted using bits (d,d-1,\dots,1). The partitioning step for a subfile ((l,r,d)) works as follows:
- Set (i \leftarrow l), (j \leftarrow r).
- While (i \le j):
- If the (d)-th bit of (K_i) is 0, increase (i) by 1.
- Else if the (d)-th bit of (K_j) is 1, decrease (j) by 1.
- Else exchange (R_i \leftrightarrow R_j), increase (i) by 1, decrease (j) by 1.
- After the loop, push ((l,j,d-1)) and ((i,r,d-1)) onto the stack (if they contain at least two records).
When a file contains numerous equal keys, a subfile may consist entirely of identical keys. For such a subfile, every key has the same bit at position (d); consequently the partitioning step makes no exchanges and produces one subfile of the same size (either ((l,r,d-1)) or ((i,r,d-1))) and one empty subfile. The algorithm then repeats the same partitioning for the next bit position, and so on, until the least significant bit is processed. This results in (w) partitioning passes over the same (N) records, where (w) is the word length (e.g., 32 or 64). Although the asymptotic complexity remains (O(Nw)), the constant factor (w) can be large, and the work is entirely unnecessary because equal keys are already in sorted order.
To eliminate this overhead, we modify the partitioning step so that it detects when all keys in the current subfile are equal. The detection can be done during the partitioning scan with negligible extra cost:
- Before the partitioning loop, initialize (\mathit{min} \leftarrow K_l) and (\mathit{max} \leftarrow K_l).
- Whenever a key (K_i) or (K_j) is examined in the loop, update [ \mathit{min} \leftarrow \min(\mathit{min}, K_i, K_j),\qquad \mathit{max} \leftarrow \max(\mathit{max}, K_i, K_j). ]
- After the loop terminates, if (\mathit{min} = \mathit{max}), then every key in the subfile is identical. In this case the subfile is already sorted, so we do not push the resulting subfiles onto the stack for further processing.
This modification adds only a few integer comparisons per element during the partitioning pass. For a subfile where all keys are equal, the algorithm now performs exactly one partitioning pass ((O(N)) operations) instead of (w) passes. For subfiles that are not homogeneous, the extra comparisons are a tiny fraction of the total work. Thus the modified Algorithm R achieves reasonable efficiency on files with many equal keys. ∎