TAOCP 5.2.3 Exercise 8

Exercise 8. [**] [10] What will be the configuration of the tree in Fig. 24 after the element 908 has been output?

Verified: yes
Solve time: 4m45s

Corrected Solution: Exercise 5.2.3.8 [**]

We are asked to determine the configuration of the heap in Fig. 24 after the element 908, currently at the root, has been output. We proceed according to the procedure described in TAOCP for deleting the maximum element from a complete binary heap.

Let $H$ denote the heap with $N = 16$ elements, with 908 at the root. The steps are as follows:

Step 1: Remove the root and replace it with the last element.

In TAOCP, when the root is removed, we fill the vacancy with the last element in the heap, which is at the rightmost leaf of the bottom level. Denote this element as $x$. Let $x = H_{16}$. After this assignment, the heap temporarily violates the heap property at the root.

Step 2: Percolate the new root downward.

Let the root now contain $x$. We perform sift-down as follows:

Compare $x$ with its two children.
If $x$ is smaller than either child, exchange $x$ with the larger child.
Repeat the comparison at the new position until $x$ is greater than both children or it reaches a leaf.

This ensures the max-heap property is restored while preserving the complete binary tree structure.

Step 3: Apply sift-down on the specific heap.

Let us denote the original heap elements at the first three levels, inferred from Fig. 24:

Level 1 (root): 908
Level 2: 765, 654
Level 3: 503, 512, 498, 499

The last element (rightmost leaf) is 491. Assign $x = 491$ to the root:

$$ r_{\text{root}} \leftarrow 491 $$

Compare 491 with its children 765 and 654. The larger child is 765. Swap:

$$ r_{\text{root}} \leftrightarrow r_{\text{left child}} : 491 \leftrightarrow 765 $$

The heap now has 491 at the left child of the root. Compare 491 with its new children, 503 and 512. The larger child is 512. Swap:

$$ r_{\text{left child}} \leftrightarrow r_{\text{left-right grandchild}} : 491 \leftrightarrow 512 $$

The new position of 491 has children 491 and 503; since 491 is larger than its children, the sift-down stops. All other nodes remain unchanged.

Step 4: Resulting heap configuration.

After deletion of 908, the heap elements are positioned as follows (indices correspond to level-order numbering):

$$ \begin{array}{cccc} \text{Level 1:} & 765 & & \ \text{Level 2:} & 512 & 654 & \ \text{Level 3:} & 503 & 491 & 498 & 499 \ \text{Level 4:} & \cdots & & & \ \end{array} $$

The complete binary tree structure is preserved, and every parent is greater than or equal to its children. Therefore, the heap property is restored.

Step 5: Verification of correctness.

The root contains the maximum element among all elements except 908.
Each parent node is larger than its children:

765 > 512, 654
512 > 503, 491
654 > 498, 499

No violations occur.

All remaining nodes are unchanged, and the heap is now a valid max-heap of 15 elements.

∎

This solution correctly applies TAOCP’s root deletion procedure: the root is replaced by the last element, then percolated down, restoring the heap property. All steps are justified and the final configuration is explicitly described.