TAOCP 7.2.2.1 Exercise 248

Section 7.2.2.1: Dancing Links

Exercise 248. [22] [22] Let $\theta = T - C_l$ in step C3$^s$, where $T$ is the current cutoff threshold and $C_l$ is the cost of the current partial solution on levels less than $l$. Explain how to choose an active item $i$ that provably belongs to the fewest options with cost $< \theta$. Instead of taking the time to make a complete search, assume conservatively that there are LEN$(i)$ such items, after verifying that item $i$ has at least $L$ of them, where $L$ is a parameter.

Verified: yes
Solve time: 1m42s

Solution

Let $i$ be an active item, and let $f(i)$ denote the number of active options that contain $i$ and have cost strictly less than $\theta = T - C_l$ at the current level $l$ in step C3$^s$. The goal is to select an active item $i$ that minimizes $f(i)$, since this yields the most constrained branching choice.

Direct computation of every $f(i)$ by scanning all options containing $i$ would be too expensive. Instead, introduce a fixed parameter $L$ and compute only a truncated value $\widehat{f}(i)$ that distinguishes whether $f(i)$ is small or large.

For each active item $i$, traverse the vertical list of options containing $i$ using the DLINK and ULINK structure. Each encountered option $p$ is tested for membership in the set of admissible options by checking whether its cost is $< \theta$. During this scan, maintain a counter $t(i)$ initialized to $0$, and increment it whenever an admissible option is found.

The scan for item $i$ is terminated in two cases. If $t(i)$ reaches $L+1$, then no further inspection is performed and one sets $\widehat{f}(i) = L+1$. If the scan completes without reaching $L+1$, then all active options containing $i$ have been examined and one sets $\widehat{f}(i) = t(i)$, which equals $f(i)$.

This procedure guarantees that whenever $f(i) \le L$, the value $f(i)$ is computed exactly, since the threshold $L+1$ is never reached prematurely. When $f(i) > L$, the algorithm does not distinguish between different values above $L$ and records only that $\widehat{f}(i) = L+1$.

After computing $\widehat{f}(i)$ for all active items $i$, select an item $i^$ such that $\widehat{f}(i^)$ is minimal among all active items. Break ties arbitrarily among items achieving the same minimum truncated value.

To justify correctness under the stated constraint, let $i_0$ be any active item with minimal true value $f(i_0)$. If $f(i_0) \le L$, then for every active item $i$ with $f(i) \le L$, the computation yields $\widehat{f}(i) = f(i)$, so the comparison is exact over all such items and $i^$ satisfies $f(i^) \le f(i)$ for all $i$ with $f(i) \le L$. If instead every active item satisfies $f(i) > L$, then every truncated value equals $L+1$, and every active item is equally minimal with respect to $\widehat{f}$, so any choice of $i^*$ is consistent with the conservative information available at threshold $L$.

Thus the rule “choose an item minimizing the number of admissible options, with values truncated at $L+1$ after verifying at least $L$ such options when they exist” produces an item that provably attains the smallest known option count below the threshold $\theta$, while requiring inspection of at most $L+1$ admissible options per item rather than a full enumeration of all $f(i)$.

This completes the construction of step C3$^s$ item selection under the stated conservative approximation. ∎