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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/functional-programming-design-and-analysis-of-algorithms/7/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms / 7 / c / Solution by

Codex  0 2026-10-31

A decreased key can violate heap order only against its parent. An inserted entry begins at the next free position and uses the same repair:

DecreaseKey(A,i,key):
    if i < 0 or i >= A.heap_size:
        error "invalid index"
    if key > A[i]:
        error "key increased"
    A[i] = key
    while i > 0:
        p = Parent(i)
        if A[p] <= A[i]:
            return
        swap A[p], A[i]
        i = p

Insert(A,key):
    ensure capacity for A.heap_size+1 entries
    i = A.heap_size
    A.heap_size = A.heap_size+1
    A[i] = +infinity
    DecreaseKey(A,i,key)

Here

+ \infty

is a sentinel at least as large as every permitted key. One can equivalently write key directly into the new entry and run the upward loop, avoiding any sentinel representation.

At every upward swap, the smaller key moves above its larger parent; the displaced parent remains no larger than the entries below it, as follows from the old heap order. Thus only the new parent edge can be invalid. At most one parent comparison is needed per level, with no scan of siblings. Both operations take $O (lo g_{d} n)$ time in the usual $2 \leq d \leq n$ range, or

O (1 + lo g_{d} (n + 1))

uniformly including tiny heaps and large

d

The heap-operation bound assumes space for the inserted entry. A capacity expansion in a dynamic array can cost

O (n)

for that individual insertion; sufficient preallocation avoids it, while geometric capacity growth adds only constant amortized allocation cost per insertion.

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