TAOCP 5.2.3 Exercise 6
We analyze the algorithm of Exercise 5.
Section 5.2.3: Sorting by Selection
Exercise 6. [**] [M25] What is the average number of comparisons performed by the algorithm of exercise 8, for random input?
Verified: yes
Solve time: 3m39s
We analyze the algorithm of Exercise 5.2.3.5 (left-to-right maximum search used repeatedly in a selection-type procedure).
Let $C(N)$ be the average number of comparisons performed when selecting maxima repeatedly on a random permutation of $N$ keys using the modified search rule:
In a pass over a segment $K_i,\dots,K_j$, the maximum is found by scanning left-to-right, and in the next pass the scan starts again from the left boundary of the remaining segment.
We compute the expected number of comparisons directly from first principles.
1. Structure of one pass
Consider a segment of length $k$. The algorithm finds its maximum by scanning left to right:
- The first element is taken as current maximum.
- Each subsequent element is compared once with the current maximum.
- If it is larger, it becomes the new maximum.
Hence a pass over $k$ elements always performs exactly $k-1$ comparisons.
This part is deterministic.
Thus the total cost is determined entirely by how the segment sizes evolve across passes.
2. Evolution of segment sizes
Each pass removes the maximum element of the current segment and places it in its final position (as in standard selection sort behavior).
Therefore, the algorithm performs exactly $N$ passes over segments of sizes
$$ N,; N-1,; N-2,; \dots,; 1. $$
This is the same sequence of segment lengths as ordinary selection sort; only the comparison pattern inside each pass is different.
3. Key observation: modified scanning changes reuse structure, not pass lengths
The flaw in the previous solution is assuming a uniform “halving per comparison” effect. The correct approach is instead to compute how many comparisons are actually performed by the left-to-right maximum scan relative to the standard right-to-left scan.
We analyze a single pass carefully.
4. Expected comparisons in left-to-right maximum search
Fix a segment of size $k$, and consider the process of scanning from left to right.
Let the elements be a uniformly random permutation of distinct keys.
Define indicator variables:
For $2 \le t \le k$, let
$$ X_t = \begin{cases} 1 & \text{if the } t\text{-th element is compared with the current maximum},\ 0 & \text{otherwise.} \end{cases} $$
Then the number of comparisons in the pass is
$$ X = \sum_{t=2}^k X_t. $$
We compute $\mathbb{E}[X_t]$.
The $t$-th element is compared unless it is smaller than the current maximum. A comparison is skipped only in the degenerate situation that the current maximum is already greater than all remaining elements and the scan structure allows no further update. In the left-to-right standard maximum scan, however, every element from position $2$ onward is always compared exactly once with the current maximum.
Thus for every $t \ge 2$,
$$ X_t = 1 \quad \text{deterministically}. $$
Hence
$$ \mathbb{E}[X] = k-1. $$
So the modified rule does not change the cost per pass.
5. Total cost over all passes
Since each pass over a segment of size $k$ costs exactly $k-1$ comparisons, the total number of comparisons is
$$ C(N) = \sum_{k=1}^{N} (k-1). $$
Compute:
$$ C(N) = \sum_{k=1}^{N} k - \sum_{k=1}^{N} 1 = \frac{N(N+1)}{2} - N = \frac{N(N-1)}{2}. $$
6. Correction of the flawed “factor of $1/2$” argument
The previous solution incorrectly treated comparisons as if each were independently retained with probability $1/2$. This ignores the strong dependence introduced by the evolving maximum and the deterministic nature of comparisons in a sequential scan.
No comparison is “canceled” by randomness; every element after the first is still explicitly compared once per pass. Therefore no halving occurs.
Final result
$$ \boxed{C(N) = \frac{N(N-1)}{2}} $$
This is the same as ordinary selection sort, since the left-to-right maximum search performs exactly one comparison per element after the first in each pass.