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 Incoming links: List-based set representation

Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Functional programming, data structures and algorithms, procedural programming, and digital hardware / 2 / a / Solution 2026-10-31

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Let

m, n

be the two input lengths and suppose numeric comparisons cost constant time. The list-based set representation determines which representation invariant each answer must preserve.

For unsorted lists allowing repetition, retain every left-hand occurrence whose value appears on the right for set intersection, and use list concatenation for set union:

interR xs ys = [x | x <- xs, x `elem` ys]
unionR xs ys = xs ++ ys

The resulting lists represent exactly the desired sets; repeated entries are permitted. Membership by sequential search makes interR take

O (mn)

time complexity in the worst case. The list concatenation in unionR costs

O (m)

to build its copied left spine and shares ys; fully enumerating the resulting list costs

O (m + n)

For unsorted, duplicate-free inputs, the same intersection keeps uniqueness, while the union retains xs and appends only right-hand elements absent from xs:

interU xs ys = [x | x <- xs, x `elem` ys]
unionU xs ys = xs ++ [y | y <- ys, not (y `elem` xs)]

Neither filtered list has internal repetitions, and the filter in unionU also excludes overlap with xs. Both operations have worst-case

O (mn + m + n)

time complexity, customarily written

O (mn)

when both lengths are positive. Using the original xs for each membership test avoids searching an ever-growing output.

For sorted, duplicate-free lists, compare only the two current heads:

interS [] ys = []
interS xs [] = []
interS (x:xs) (y:ys)
  | x < y = interS xs (y:ys)
  | x > y = interS (x:xs) ys
  | otherwise = x : interS xs ys

unionS [] ys = ys
unionS xs [] = xs
unionS (x:xs) (y:ys)
  | x < y = x : unionS xs (y:ys)
  | x > y = y : unionS (x:xs) ys
  | otherwise = x : unionS xs ys

When

x < y

, sortedness proves that

x

cannot occur anywhere in the right list, so it can be discarded from the set intersection or immediately emitted to the set union. The symmetric argument handles

y < x

; equal heads are consumed together. Each step consumes at least one input head, so both have

O (m + n)

worst-case time complexity. These bounds assume the sorted representation invariant is already established; sorting arbitrary inputs has its own cost.

Representation	Set intersection	Set union, including output traversal
Unsorted, repetitions allowed	$O (mn + m + n)$	$O (m + n)$
Unsorted, unique	$O (mn + m + n)$	$O (mn + m + n)$
Sorted, unique	$O (m + n)$	$O (m + n)$

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