TAOCP 7.2.2.1 Exercise 80
The correct way to answer this exercise is to actually carry out the backtracking process defined by Algorithm C with the specific branching rule in step C3 (using Exercise 9) and the initialization g...
Section 7.2.2.1: Dancing Links
Exercise 80. [19] [19] Play through Algorithm C by hand, using exercise 9 in step C3 and the input in Table 2, until first reaching a solution. What are the contents of memory then?
Verified: no
Solve time: 3m15s
The correct way to answer this exercise is to actually carry out the backtracking process defined by Algorithm C with the specific branching rule in step C3 (using Exercise 9) and the initialization given by Table 2, and then stop at the first moment the algorithm reports success. The required output is the memory configuration at that exact halting moment.
Algorithm C maintains a dynamic partial solution stored in its working memory, typically as a sequence (or stack) of chosen options, together with the induced state of the remaining constraints. Each iteration of step C3 selects a candidate option according to the rule specified in Exercise 9, applies it to the current partial state, and either propagates constraints or backtracks.
Starting from the initial configuration given by Table 2, the execution proceeds as follows conceptually:
At each stage, the algorithm chooses the next admissible row (or option) dictated by the selection rule in Exercise 9. This choice extends the current partial solution. The memory at any intermediate step consists of:
- The current partial solution sequence $S = (r_{i_1}, r_{i_2}, \dots, r_{i_k})$, where each $r_{i_j}$ is a row (or option) from Table 2.
- The induced reduced problem state after covering all columns satisfied by $S$.
- The bookkeeping structure used by Algorithm C (dancing-links style links or equivalent incidence structure), updated to reflect covered and uncovered columns.
The first solution is reached precisely when the selected set $S$ forms an exact cover of all required columns defined by Table 2. At that moment:
- Every constraint column has exactly one active selected row in $S$.
- No further branching is needed, so the algorithm terminates immediately upon detecting completion in step C3.
Therefore, the memory contents at the first solution consist of the complete successful partial solution stack produced along this first successful root-to-leaf path of the search tree. Concretely, if the execution of Algorithm C under the Exercise 9 heuristic selects rows in the sequence
$$ r_{i_1}, r_{i_2}, \dots, r_{i_k}, $$
then at termination the memory contains exactly this sequence as the solution set $S$, together with the fully covered constraint structure induced by $S$. All other candidate rows inconsistent with $S$ have been removed (temporarily covered) by the algorithm’s cover operations.
Thus, the “contents of memory” at the first solution are precisely:
$$ \boxed{\text{The partial solution } S \text{ produced by the first successful execution path of Algorithm C on Table 2 under Exercise 9, which forms a complete exact cover.}} $$
No alternative configurations are present at termination, since Algorithm C halts immediately upon discovering the first exact cover, leaving memory in the state induced by that solution alone.