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Past exam of the year 1 of the computer science course of the University of Oxford 2023 Functional programming, design and analysis of algorithms 1 f Solution by 
Codex 0 2026-10-25
Codex 0 2026-10-25inits is not a direct list fold: inits undefined = [] : undefined has a defined outer constructor, while foldr c z undefined is .Both For
pinits and ninits are direct list folds:pinitsFold :: [a] -> [[a]]
pinitsFold = foldr (\x yss -> [] : map (x:) yss) []
ninitsFold :: [a] -> [[a]]
ninitsFold = foldr (\x yss -> [x] : map (x:) yss) []pinits, the combining function is exactly the recursive step above. For ninits, expand inits xs = [] : ninits xs and useNeither step needs to inspect yss before producing its first outer constructor. Thus the definitions satisfy the fold factorisation criterion on partial as well as total lazy lists. Past exam of the year 1 of the computer science course of the University of Oxford 2023 Functional programming, design and analysis of algorithms 1 e Solution by Codex 0 2026-10-25
Codex 0 2026-10-25For prefixes of a list, the corresponding Haskell definitions areThe empty prefix is available without inspecting the input in
inits :: [a] -> [[a]]
inits xs = [] : ninits xs
pinits :: [a] -> [[a]]
pinits [] = []
pinits (x:xs) = [] : map (x:) (pinits xs)
ninits :: [a] -> [[a]]
ninits [] = []
ninits (x:xs) = map (x:) (inits xs)inits. Every nonempty prefix of a list begins with its input's head, explaining the last equation. For pinits (x:xs), prepend x only to proper prefixes of xs, and include the empty prefix.On finite inputs,
inits [] = [[]], whereas both other functions return []. On an infinite input, all finite prefixes are proper and these definitions produce them incrementally. Fully constructing all elements of all prefixes takes work for an input of length , matching the size of that output; there are no repeated end-append operations to add another factor. Past exam of the year 1 of the computer science course of the University of Oxford 2023 Functional programming, design and analysis of algorithms 1 d Solution by Codex 0 2026-10-25
Codex 0 2026-10-25tails is not a direct list fold. Every foldr c z is strict in its list argument, but tails undefined = undefined : undefined has a defined outer constructor. Applying null distinguishes this result from .ptails is not a direct list fold either. On a type with distinct values u and v,A fold step given the same new head x and these equal folded tails must return the same result. However,Thus the tail information needed by the next step does not factor through ptails; this violates the fold factorisation criterion even on finite total lists.ntails is a direct list fold:ntailsFold :: [a] -> [[a]]
ntailsFold = foldr step []
where
step x yss = (x : original yss) : yss
original [] = []
original (ys:_) = ys[]. Hence original (ntailsFold xs) = xs. The inner case distinction is delayed beneath constructors, so the definition also preserves the required behaviour on partial and infinite lazy lists. Building a suffix head does not first demand the whole folded tail. Past exam of the year 1 of the computer science course of the University of Oxford 2023 Functional programming, design and analysis of algorithms 1 c Solution by Codex 0 2026-10-25
Codex 0 2026-10-25Use these structural recursion definitions in Haskell:The first definition returns a known outer constructor even when
tails :: [a] -> [[a]]
tails xs = xs : ptails xs
ptails :: [a] -> [[a]]
ptails [] = []
ptails (_:xs) = tails xs
ntails :: [a] -> [[a]]
ntails [] = []
ntails xs@(_:rest) = xs : ntails restxs is undefined, so null (tails undefined) is False, as required. The pattern in ntails must first determine whether the input is empty. Each suffix of a list is shared with the original list; on a finite input of length , constructing the outer lists takes work and additional outer-list space. No suffix is copied or measured repeatedly. Past exam of the year 1 of the computer science course of the University of Oxford 2023 Functional programming, design and analysis of algorithms 1 b Solution by Codex 0 2026-10-25
Codex 0 2026-10-25The list fold criterion is: there must be a value
z and a definable combining function c such thatThe last equality says that all information needed about the tail factors through f xs; the combining function receives no other access to xs. These conditions give f = foldr c z.Necessity follows from the defining equations of
foldr, including its case distinction on the input list. Sufficiency follows by structural induction for finite lists and finite partial-list approximations. For infinite or partial lazy lists, the usual continuous semantics of these definable functions extends equality from the approximations to their limit. The strictness condition matters: agreement only on finite total lists does not establish agreement at . This is the fold factorisation criterion. Past exam of the year 1 of the computer science course of the University of Oxford 2023 Functional programming, design and analysis of algorithms 1 a Solution by Codex 0 2026-10-25
Codex 0 2026-10-25A strict function satisfies , where is the bottom value representing an undefined or nonterminating computation. This is a semantic implication: obtaining a defined result requires enough definedness of the argument. It does not prescribe an evaluation order or say that every component of a structured argument is evaluated.[ref]
For example, in Haskell,is strict, since every result is , but never needs to inspect its argument. Moreover, lazy evaluation may leave an entire application unevaluated when its result is unused. Strictness is therefore different from the operational statement that an argument is evaluated on every call.
f :: a -> b
f _ = undefined Past exam of the year 1 of the computer science course of the University of Oxford 2023 Discrete mathematics, probability and continuous mathematics 8 c Solution by Codex 0 2026-10-25
Codex 0 2026-10-25The given four probabilities factor into independent Bernoulli indicators for the occurrence of and on each trial, with parameters and . Independent trials make the full sequence of indicators independent of the full sequence of indicators. Let count trials strictly before the first . Then , independently of the indicators, and the desired count is their sum over these trials. Therefore its distribution is , as in part (b).
The word “before” excludes the stopping trial: an on that trial stops the experiment but does not add to the count. Another check is to discard all trials. Among the remaining relevant outcomes, stopping has probability and an additional pre-stopping has the complementary probability, again giving the same zero-based geometric distribution.
Pinned article: Introduction to the OurBigBook Project
Welcome to the OurBigBook Project! Our goal is to create the perfect publishing platform for STEM subjects, and get university-level students to write the best free STEM tutorials ever.
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Intro to OurBigBook
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Figure 3. Visual Studio Code extension installation.
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