TAOCP 5.4.5 Exercise 2

Section 5.4.5: The Oscillating Sort

Exercise 2. [24] [24] If S = 6 in Sobel’s original method, we could pretend that S = 16 and that 10 dummy runs were present. Then Phase 3 in the text’s example would put dummy

runs Ap on T4 and T5; Phase 4 would merge the A;’s on T2 and T3 into a D2 on T1; Phases 5-8 would do nothing; and Phase 9 would produce Ag on T4. It would be better

Verified: no
Solve time: 1m04s

Solution

Let $T=5$ and hence $P=4$. In Sobel’s original oscillating sort, a convenient analysis is obtained by padding the number of initial runs $S$ to a power of $P$ by introducing dummy runs. For $S=6$, one embeds the input into $S'=16$ by adding $10$ dummy runs, then executes the full oscillation schedule for $S'=16$. The resulting pattern contains phases in which only dummy data are merged, and these phases contribute to cost without advancing any real run.

The modification is to remove the assumption that padding is performed at the outset, and instead treat “missing runs” as a stopping condition for the current level. The key change is that a level is terminated immediately when the actual number of runs produced by that level is $1$, rather than continuing distribution cycles until a full block of $P$ runs has been created.

Let the initial runs be $A_1,\dots,A_6$.

During the first distribution phase, four runs are written cyclically onto $T_1,T_2,T_3,T_4$, leaving $T_5$ empty:

$$ T_1:A_1,\quad T_2:A_2,\quad T_3:A_3,\quad T_4:A_4. $$

The first merge produces

$$ D_4 = \operatorname{merge}(A_1,A_2,A_3,A_4) $$

on $T_5$, while $A_5,A_6$ remain unmerged on their tapes.

At this point the system has exactly three nonempty tapes. Instead of continuing the original oscillation (which would assume two further full distribution blocks and therefore introduce dummy runs), the next merge is performed immediately on the actual available runs.

The remaining runs are now $D_4, A_5, A_6$. A direct merge of these three inputs produces a single ascending run

$$ A_6' = \operatorname{merge}(D_4, A_5, A_6) $$

on an output tape, using standard multiway merging with $T-1=4$ input capability.

This eliminates all subsequent phases that would arise only from padding, since no additional complete distribution cycle can be formed from the remaining runs. In particular, the cyclic shift used in Sobel’s continuation scheme becomes irrelevant because there is no second full layer of $P$ runs to distribute.

To compare costs, in the padded execution with $S'=16$ the algorithm executes a full second oscillation cycle, which consists of a distribution phase and several merge phases whose only effect is to process dummy runs. Each such phase reads and writes every tape once, contributing a full pass-equivalent cost even though no real runs are transformed.

In the modified execution for $S=6$, only one oscillation cycle is executed, followed immediately by a final multiway merge of the residual runs. The elimination of the second cycle removes exactly those passes that correspond to completing the artificial extension from $6$ to $16$ runs, and replaces them with a single merge whose input size is strictly smaller than a full padded level.

Since every phase in Sobel’s scheme corresponds to one full scan of all tapes, the removal of an entire level of oscillation reduces the total number of full scans by the number of phases in that level, while the final merge contributes strictly fewer scans than any padded merge sequence that processes dummy runs.

Thus the corrected ending rule is: when the number of active runs after a merge is less than $P$, terminate oscillation and perform a single final $(\text{current run count})$-way merge to completion.

This completes the proof. ∎