Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Functional programming, design and analysis of algorithms / 6 / a

Store a complete binary tree in a one-indexed array $A[\mathrm{1..}n]$, level by level. Index $i$ has parent in a rooted tree $\lfloor i/2\rfloor$ for $i>1$ and possible children in a rooted tree $2i$ and $2i+1$, present only when their indices are at most $n$. The max-heap condition is $A[\lfloor i/2\rfloor ]\ge A[i]$ for every $i>1$.

Following these inequalities from the root of a tree to any entry proves that $A[1]$ is a maximum. The heap order does not sort siblings or unrelated branches. The complete binary tree has height $\lfloor {\log }_{2}n\rfloor$, which bounds how far any single repair can travel. Store the current heap size separately from the array's capacity.

Assume the subtrees rooted at the children in a rooted tree of $i$ already obey the max-heap condition. Compare $A[i]$ with each present child and choose an index of the largest of these at most three keys. If that index is $i$, stop. Otherwise exchange the two keys, continue at the chosen child, and repeat. This is sift down, with the inequality reversed from a minimum-oriented binary heap.

The exchange repairs the old root position because its new key dominates both child roots; only the subtree receiving the displaced key can still be defective. This recursion invariant proves correctness. Each exchange descends one level and costs constant work, so the worst-case running time is $O(h)$ for subtree height $h$, hence $O(\log n)$. A small key at the root of a suitably ordered heap must descend to a leaf, so the logarithmic worst-case bound is attained.

Leaves already obey the heap order. Process indices $\lfloor n/2\rfloor ,\lfloor n/2\rfloor -1,\dots ,1$, calling the preceding repair at each. When an index is processed, its descendants have already been processed, so the repair's precondition holds. The resulting array is a max-heap.

The bound obtained by charging $O(\log n)$ to every call is unnecessarily large. The number of nodes with height at least $h$ is at most $\lfloor n/2^h\rfloor$. ThereforeThe total repair work is thus $O(n)$, and the loop itself has a linear number of iterations, giving $\Theta (n)$ for this bottom-up heap construction. Most nodes are near the leaves and require little repair.