TAOCP 5.3.3: Minimum-Comparison Selection
Section 5.3.3 exercises: 16/16 solved.
Section 5.3.3. Minimum-Comparison Selection
Exercises from TAOCP Volume 3 Section 5.3.3: 16/16 solved.
| # | Rating | Category | Status | Time |
|---|---|---|---|---|
| 1 | [**] | verified | 3m39s | |
| 2 | [**] | verified | 4m03s | |
| 3 | [**] | solved | 4m01s | |
| 4 | [**] | solved | 3m41s | |
| 5 | [**] | solved | 4m08s | |
| 6 | [25] | medium | solved | 8m24s |
| 7 | [20] | medium | solved | 4m59s |
| 8 | [M20] | math-medium | verified | 2m22s |
| 9 | [M42] | math-project | solved | 4m48s |
| 10 | [M47] | math-research | solved | 5m50s |
| 11 | [32] | hard | solved | 4m07s |
| 12 | [24] | medium | solved | 3m19s |
| 13 | [M49] | math-research | verified | 3m25s |
| 14 | [HM40] | hm-project | solved | 3m49s |
| 15 | [M35] | math-hard | solved | 4m30s |
| 16 | [M32] | math-hard | solved | 4m36s |
TAOCP 5.3.3 Exercise 1
The previous solution fails because it replaces Carroll’s actual tournament mechanism with an invented dominance-based rule and then reasons about that artificial system.
TAOCP 5.3.3 Exercise 2
We give a clean decision-tree argument that avoids the gap in the previous solution and directly relates worst-case depths.
TAOCP 5.3.3 Exercise 3
Let $W_t(n)$ denote the worst-case number of comparisons required by any comparison-based algorithm for the structure defined in the exercise.
TAOCP 5.3.3 Exercise 4
Let $A$ be an optimal comparison-based algorithm that finds the third largest element, and let its worst-case number of comparisons be $V_3(n)$.
TAOCP 5.3.3 Exercise 5
Let $n$ be the total number of distinct elements.
TAOCP 5.3.3 Exercise 6
We give a complete proof consisting of a tight upper bound (algorithm) and a correct adversary lower bound.
TAOCP 5.3.3 Exercise 7
The previous argument fails because it replaces Floyd’s comparison accounting with informal “reuse” claims and an invalid decomposition into independent subproblems.
TAOCP 5.3.3 Exercise 8
Let the selection algorithm be modified so that the input is partitioned into groups of 5 instead of groups of 7, and the median-of-medians is used as the pivot exactly as in the proof of Theorem L.
TAOCP 5.3.3 Exercise 9
We construct all values for $V_t(8)$ using a single consistent method: an optimal 8-element tournament followed by explicit optimal selection in the induced comparison structure.
TAOCP 5.3.3 Exercise 10
Solution to TAOCP 5.3.3 Exercise 10.
TAOCP 5.3.3 Exercise 11
Let $t > 2$ and $k > t$ be integers.
TAOCP 5.3.3 Exercise 12
Let $o$ be the integer satisfying $2^o < \frac{n-1}{4} < 2^{o+1}$, equivalently $4\cdot 2^o < n-1 < 5\cdot 2^o.$ Write $n-1 = 4\cdot 2^o + r,\qquad 0 < r < 2^o.$ Form four disjoint knockout trees $T_1...
TAOCP 5.3.3 Exercise 13
We restart the argument from the definition used in this section of TAOCP, where $V_r(n)$ denotes the generalized power sum V_r(n) = \sum_{k=1}^n k^r, extended to complex $r$ by analytic continuation.
TAOCP 5.3.3 Exercise 14
The previous argument fails because it treats comparison sharing and adversary accounting heuristically, and it never establishes a valid cost model for selecting the two boundary order statistics or...
TAOCP 5.3.3 Exercise 15
The previous proof failed because it incorrectly treated reachability in a general DAG as if it could be incremented only locally per comparison, and then incorrectly decomposed two dependent costs as...