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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/functional-programming-design-and-analysis-of-algorithms/5/b/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Functional programming, design and analysis of algorithms / 5 / b / Solution by

Codex  0 2026-10-31

Build a balanced comparison tournament, retaining the record of each value's direct defeats. A record stores a key and a list of records that it directly defeated:

tournament(S):
    if |S| = 1:
        return Node(the element of S, [])
    split S into equally sized L and R
    a = tournament(L)
    b = tournament(R)
    if a.key < b.key:
        exchange a and b
    prepend b to a.losers
    return a

There is one key comparison at each merge. Because

S

is a set, all keys are distinct, so the greater winner is unambiguous. Retaining the losing record also retains its own earlier comparison history; it does not merely retain the losing root's value.

Let the returned root have value

M

. Define

max (S) = M, second (S) = {r . key : r \in root.losers}, third (S) = S ∖ ({M} \cup second (S)) .

(1)

The second-largest value can only lose to

M

, so it occurs among those direct losers. This proves the required partition and the exclusion of the second-largest value from the third part. For

n = 2^{m}

, the winner participates in exactly

m

merges, so

∣ second (S)∣ = m

. The other part can be materialized by marking the original element identities in this list and scanning the original inputs, using

O (n)

additional operations without more comparisons between keys.

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