TAOCP 5.3.3: Minimum-Comparison Selection
Section 5.3.3 exercises: 27/27 solved.
Section 5.3.3. Minimum-Comparison Selection
Exercises from TAOCP Volume 3 Section 5.3.3: 27/27 solved.
| # | Rating | Category | Status | Time |
|---|---|---|---|---|
| 1 | [15] | simple | verified | 3m39s |
| 2 | ▶ [M25] | math-medium | verified | 4m03s |
| 3 | [20] | medium | solved | 4m01s |
| 4 | ▶ [M25] | math-medium | solved | 3m41s |
| 5 | [10] | simple | solved | 4m08s |
| 6 | ▶ [M26] | math-hard | solved | 8m24s |
| 7 | [M20] | math-medium | solved | 4m59s |
| 8 | [M21] | math-medium | verified | 2m22s |
| 9 | ▶ [20] | medium | solved | 4m48s |
| 10 | [35] | hard | solved | 5m50s |
| 11 | [38] | project | solved | 4m07s |
| 12 | [21] | medium | solved | 3m19s |
| 13 | ▶ [HM28] | hm-hard | verified | 3m25s |
| 14 | ▶ [20] | medium | solved | 3m49s |
| 15 | [22] | medium | solved | 4m30s |
| 16 | ▶ [25] | medium | solved | 4m36s |
| 17 | [20] | medium | solved | 15m12s |
| 18 | [M20] | math-medium | solved | 6m20s |
| 19 | [M42] | math-project | solved | 4m38s |
| 20 | [M42] | math-project | solved | 20m17s |
| 21 | [32] | hard | solved | 17m31s |
| 22 | [24] | medium | solved | 18m53s |
| 23 | [M42] | math-project | solved | 12m31s |
| 24 | [HM40] | hm-project | solved | 20m25s |
| 25 | ▶ [M35] | math-hard | solved | 16m09s |
| 26 | [M32] | math-hard | solved | 16m19s |
| 27 | ▶ [M34] | math-hard | solved | 16m09s |
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...
TAOCP 5.3.3 Exercise 16
A corrected solution is given below.
TAOCP 5.3.3 Exercise 17
We combine the algorithm of Exercise 16 (finding the maximum and minimum simultaneously) with the tree‑selection method of Kislitsyn’s theorem (formula (6) in the text).
TAOCP 5.3.3 Exercise 18
The median-of-medians algorithm (Blum, Floyd, Pratt, Rivest, Tarjan) finds the \(t\)th largest element by recursively using a pivot obtained as the median of medians of small groups.
TAOCP 5.3.3 Exercise 19
Let \(V_t(n)\) denote the minimum number of comparisons required to select the \(t\)‑th largest element from a set of \(n\) elements, as defined in Section 5.
TAOCP 5.3.3 Exercise 20
We consider the problem of finding the second largest element among \(n\) distinct elements using comparisons.
TAOCP 5.3.3 Exercise 21
We are given integers \(k \ge t \ge 2\) and set \(n = 2^k + 2^{k+1-t}\).
TAOCP 5.3.3 Exercise 22
We need to solve Exercise 22 from Section 5.
TAOCP 5.3.3 Exercise 23
We are asked for the asymptotic value of \(V_{\lceil n/2 \rceil}(n)\) as \(n\to\infty\), where \(V_t(n)\) is the minimum number of comparisons required in the worst case to determine the \(t\)‑th larg...
TAOCP 5.3.3 Exercise 24
We need to solve Exercise 24 from TAOCP Volume 3, Section 5.
TAOCP 5.3.3 Exercise 25
We need to prove that the minimum number of comparisons needed to find the t-th largest element (selection) satisfies V_t(n) >= n + t - 2 for t <= ceil(n/2).