TAOCP 5.2.3 Exercise 10
Let the standard heapsort “sift-down” step be denoted by the variables of Algorithm H, where a key at position $k$ is moved downward by repeatedly comparing it with its children at $2k$ and $2k+1$, an...
Section 5.2.3: Sorting by Selection
Exercise 10. [20] [20] (J. W. J. Williams.) Step H4 of Algorithm H distinguishes between the three cases i < r, j = r, and j > r. Show that if Kk > Kk+1 it would be possible to simplify step H4 so that only a two-way branch is made. How could the condition K > K+1 be ensured throughout the heapsort process, by modifying step H2?
Verified: yes
Solve time: 2m10s
Solution
Let the standard heapsort “sift-down” step be denoted by the variables of Algorithm H, where a key at position $k$ is moved downward by repeatedly comparing it with its children at $2k$ and $2k+1$, and where step H4 distinguishes three cases depending on whether $i < r$, $j = r$, or $j > r$.
The distinction in H4 arises solely from the fact that the right child $K_{2k+1}$ may or may not exist in the current heap segment $1 \le j \le r$. The comparison structure must therefore decide whether both children are present before selecting the larger child for comparison with $K_k$.
Assume that throughout the heapsort process the additional structural condition
$K_{2k} > K_{2k+1} \quad \text{whenever } 2k+1 \le r$
is maintained for every internal node $k$. Under this condition, whenever both children exist, the left child is guaranteed to be strictly larger than the right child. Hence, the candidate to be exchanged with the parent during sift-down is always $K_{2k}$, and the value of $K_{2k+1}$ is irrelevant to the decision at node $k$.
With this ordering invariant in place, step H4 no longer needs to distinguish whether the right child exists. The only relevant question becomes whether a left child exists, since the left child alone determines the comparison.
The original three-way branching in H4 can therefore be replaced by a two-way branch of the form: test whether $2i \le r$, and if so compare $K_i$ with $K_{2i}$; otherwise terminate the sift-down at $i$.
To ensure the invariant $K_{2k} > K_{2k+1}$ throughout heapsort, step H2 (and every operation that modifies the heap structure) must be strengthened to enforce sibling ordering locally. Whenever a new heap is constructed or a key is moved into position $k$, and both positions $2k$ and $2k+1$ lie within $1 \le j \le r$, an additional comparison is performed between $K_{2k}$ and $K_{2k+1}$, followed by an exchange if $K_{2k} < K_{2k+1}$.
This maintenance ensures that after every structural change at node $k$, the subtree rooted at $k$ satisfies both the heap condition $K_k \ge K_{2k}, K_k \ge K_{2k+1}$ and the strengthened sibling condition $K_{2k} > K_{2k+1}$. Since every modification of a subtree root is followed by restoration of this local property, induction on the height of affected subtrees shows that the condition holds globally throughout the execution of heapsort.
With this invariant maintained, step H4 reduces to a single comparison with the left child when it exists, eliminating the need for the three-way case distinction.
This completes the solution. ∎