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A divide-and-conquer algorithm splits an instance into smaller instances of the same kind, solves them recursively, and combines their answers. The smaller instances should be sufficiently smaller to ensure algorithm termination; the combination cost determines whether splitting saves work. The two halves of merge sort illustrate this separation.

Dynamic programming identifies a collection of subproblems and dependencies among them, solves each distinct subproblem once, and stores its answer. A table in dependency order or memoization prevents repeated evaluation of overlapping subproblems. For optimization, an optimal substructure argument justifies constructing an optimal answer from optimal smaller answers.

A greedy algorithm repeatedly makes a locally best feasible choice and commits to it. It avoids exploring alternative continuations. Algorithm correctness therefore needs a greedy-choice property, often proved by replacing the first choice of an optimal solution with the greedy choice without worsening it.

divide and combine; reuse subproblem answers; commit to safe local choices .

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On two finite total lists, the equations agree. If either list is empty they return the other; if both have heads, expanding twiddle twice gives x : y : twiddle xs ys, exactly the third equation for twiddle'. The same finite-prefix reasoning proves agreement on infinite inputs with defined spines.

They differ on partial lists. Take the first argument [1] and the second argument

⊥

. The original definition can emit its first head immediately:

twiddle [1] ⊥ = 1 : twiddle ⊥ [] = 1 : ⊥.

After recognizing the nonempty first argument, twiddle' must inspect the second argument to distinguish its second and third equations. That inspection diverges, so

twiddle^{'} [1] ⊥ = ⊥, twiddle^{'} \neq = twiddle .

For example, demanding only the head distinguishes them: the original returns 1, while the primed version diverges. Equality of functions under lazy evaluation requires agreement on these partial inputs too.

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For two proper lazy lists, twiddle alternates heads, swapping which list is served after each output. Thus both input positions have finite waiting time. A tempting extension is remalldups . foldr twiddle [], but arbitrary empty rows make that insufficient: for [ [1,2] ] ++ repeat [], it emits 1 and then demands the combination of infinitely many empty rows before it can return to 2.

Use diagonal merging of lazy lists to avoid that obstruction. Maintain a finite active collection, add at most one new row per round, and take one head from every nonempty active row. The supplied twiddle combines those singleton contributions into a finite batch:

union' = remalldups . rounds []

rounds [] [] = []
rounds active [] = emit active []
rounds active (row:rows) = emit (active ++ [row]) rows

emit active rows = batch ++ rounds tails rows
  where (batch, tails) = sweep active

sweep [] = ([], [])
sweep ([]:rows) = sweep rows
sweep ((x:xs):rows) = (twiddle [x] heads, xs:tails)
  where (heads, tails) = sweep rows

Here rounds takes the finite active collection first and the not-yet-introduced outer list second. Because heads is a finite list, twiddle [x] heads emits x followed by those heads; the finite batch is then completely consumed before the next round. Empty exhausted rows are removed by sweep. No step tries to combine infinitely many not-yet-introduced rows.

Index rows and positions from zero. Row

i

is introduced in round

i

and contributes its element

j

in round

i + j

, if that element exists. Every earlier round inspects only finitely many rows. Therefore every occurrence has a finite emission time, proving fair enumeration. Applying first-occurrence duplicate removal gives

union^{'} = remalldups \circ finite-round diagonal enumeration .

All outer and inner spines must be proper lazy lists and equality tests must terminate. Infinitely many empty rows after the final contribution still prevent certifying a terminating empty tail, just as part (c) predicts. Undefined input spines have the separate partial-list interleaving obstruction; the construction makes no promise to bypass them.

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List concatenation exhausts one inner list before reaching the next. For example, with rows [0,2..] and [1], concat outputs the infinite even row and never reaches 1. First-occurrence duplicate removal cannot recover a value that its input stream never supplies. Hence

ordinary concatenation does not give a fair enumeration of infinite rows .

The obstruction is separate from the undefined final tail in part (c): here a required distinct value is never emitted at all. A fair enumeration must give every row and every element within it a finite opportunity to contribute.

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Each of the finitely many distinct values has a first occurrence at a finite position, so every one is eventually emitted. After the last new value, however, the list filters keep searching the remaining infinite input for another new value. They never observe a terminating []. Therefore the result has a finite visible prefix followed by the bottom value:

x_{1} : x_{2} : \dots : x_{k} : ⊥, rather than [x_{1}, \dots, x_{k}] .

For example, remalldups (repeat 1) exposes 1 : ..., but demanding its tail diverges. This is the absence-detection obstruction for infinite lists: no finite inspection can certify that an unseen new value will never occur. The emitted values are correct, but computing the output length or proving the output empty after its last value does not terminate.

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Assume a proper lazy list with defined elements and terminating equality tests. If infinitely many distinct values occur, first-occurrence duplicate removal produces an infinite list of distinct values, in first-occurrence order. To produce the next new value after

k

have appeared, the nested list filters need skip only a finite input prefix: an infinite collection of distinct values cannot all belong to the

k

already seen. The next new value therefore occurs at a finite position and is reached in finite work. Consequently

infinitely many distinct values ⟹ productive infinite output .

This is a stream productivity claim, not a bounded-delay claim: arbitrarily long runs of repetitions can make successive delays arbitrarily large. Undefined elements or diverging equality tests invalidate the stated assumptions.

Codex  0 2026-10-31

To agree with the example, preserve the first occurrence of each value. This first-occurrence duplicate removal is a right fold:

remalldups = foldr step []
  where step x ys = x : filter (/= x) ys

The folded tail has no repetitions by the induction hypothesis. The list filter removes any occurrence of the new head from that tail; prepending the head then gives exactly one occurrence of it, in its first input position. Other values retain their relative order. For the supplied example the intermediate folded results, from right to left, are [1], [3,1], [2,3,1], and [1,2,3]. Thus

remalldups [1, 2, 3, 1] = [1, 2, 3] .

Under lazy evaluation, constructing x : ... does not first demand the folded tail. That observation is essential for the infinite cases below. The alternative step that first asks whether x belongs to the folded tail would lose this property and would also keep last occurrences instead of matching the stated example.

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Use permutation generation by insertion: recursively arrange the tail, then insert the removed head in every possible position.

inserts x [] = [[x]]
inserts x (y:ys) = (x:y:ys) : map (y:) (inserts x ys)

perms [] = [[]]
perms (x:xs) = concat (map (inserts x) (perms xs))

The map function in inserts prefixes y to each placement after the first position. For a list of length

k

, the number of results obeys

I (0) = 1

and

I (k + 1) = 1 + I (k)

, so mathematical induction gives

I (k) = k + 1

. Every result still contains the same

k

old elements and the one inserted element.

Let

P (n)

be the length of perms xs for an input of length

n

. There is exactly one permutation of the empty list, namely the empty list itself: hence perms [] is [[]], with length one. For a nonempty input, each recursively obtained permutation has length

n - 1

and gives

n

placements. Thus

P (0) = 1 = 0!, P (n) = n P (n - 1), P (n) = n! .

Length alone would not exclude repeated answers. Since the input elements are distinct, removing the distinguished element

x

from any result uniquely recovers both its placement and the shorter permutation. Hence different construction choices give different outputs, and every desired permutation is obtained by inserting

x

into its restriction to xs. For [a,b,c], this code yields [a,b,c], [b,a,c], [b,c,a], [a,c,b], [c,a,b], [c,b,a], each exactly once.

Codex  0 2026-10-31

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	$O (m + n)$	$O (m + n)$

Codex  0 2026-10-31

A particularly transparent design uses six D flip-flops in a one-hot encoding. Let

Q_{i} = 1

mean that face

i

is displayed. Initialize

Q_{1} = 1

and

Q_{2} = \dots = Q_{6} = 0

, and wire the next-state inputs as a ring:

D_{1} = Q_{6}, D_{2} = Q_{1}, D_{3} = Q_{2}, D_{4} = Q_{3}, D_{5} = Q_{4}, D_{6} = Q_{5} .

All six D flip-flops share the same active clock edge. Each pulse moves the single stored

1

to the next position, giving the state-transition diagram cycle

1 \to 2 \to 3 \to 4 \to 5 \to 6 \to 1

. The representation invariant “exactly one

Q_{i}

is high” holds after initialization and is preserved by rotation. Six D flip-flops are more than the three-bit theoretical minimum, but their simple wiring gives a complete implementation without requiring that minimum.

Figure 1.
One-hot ring state register and the six lamp patterns, including the original opposite diagonals on faces 2 and 3
. **Details**.

Decode the seven lamps by their positions, with

T, M, B

denoting top, middle and bottom, and

L, R

denoting left and right. Let

L_{C}

be the center lamp. In the original figure, face 2 uses top-left and bottom-right, while face 3 uses top-right and bottom-left plus center; the two diagonals must therefore have different controls.^[ref] The required OR gate decoder is

L_{T L} = L_{BR} L_{TR} = L_{B L} L_{M L} = L_{MR} L_{C} = Q_{2} \lor Q_{4} \lor Q_{5} \lor Q_{6}, = Q_{3} \lor Q_{4} \lor Q_{5} \lor Q_{6}, = Q_{6}, = Q_{1} \lor Q_{3} \lor Q_{5} .

For example, only the center is high in state

Q_{1}

, the first diagonal is high in state

Q_{2}

, and in state

Q_{6}

both outer columns are high while the center is low. This verifies all six layouts directly.

To implement the switch behavior, use a rapid common clock signal and a sampled enable

S

, high while the switch is pressed. Each next-state input is then selected between its advancing value and its own stored value:

D_{i} = (S \land Q_{prev (i)}) \lor (\overline{S} \land Q_{i}), prev (1) = 6, prev (i) = i - 1 (i > 1) .

(3)

One multiplexer per D flip-flop implements that selection. While enabled the electronic dice advances; otherwise every D flip-flop reloads its own value and the display holds. The enable supplied to this synchronous design must be stable at sampling edges. A reset or preset facility must establish the specified one-hot initial state: an all-zero ring does not start itself.

With a rapid clock signal, uncertainty in human release timing makes the final face practically unpredictable. The sequential circuit itself is deterministic; this observation alone is not a proof that its six outcomes have equal probabilities.

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A D flip-flop stores one bit. At its specified active clock edge, it samples

D

and, after its clock-to-Q delay, makes

Q

equal to that sampled value. Between active clock edges it retains the stored value. The input must meet the setup and hold time requirements around the sampling edge; a level-sensitive latch would have different behavior.

For a finite-state sequential circuit, encode the present state in a bank of D flip-flops. Feed their outputs

q

and the external inputs

u

to combinational logic computing the next-state function

F (q, u)

, and connect those outputs to the flip-flop

D

inputs. A common clock signal then performs the simultaneous update

q^{+} = F (q, u) .

Output combinational logic computes

H (q)

for a Moore machine or

H (q, u)

for a Mealy machine. Since any finite next-state and output table is a Boolean function, it has a Boolean circuit implementation. Initialization specifies the starting state, and unused state encodings require either a recovery rule or an initialization assumption. The clock period must accommodate propagation through the D flip-flops and intervening combinational logic before the next sampling edge.

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Use function extensionality: show

f (x) = g (x)

for every value

x

of the argument type. In lazy evaluation semantics, “every value” includes the bottom value and partial values. For list arguments, structural induction on lazy lists supplies the empty, bottom and cons cases, with continuity justifying passage to infinite lists.

The original printed equation has the same right fold on both sides, including the same operator and seed; this is visible in the scan.^[ref] Thus its literal answer is

foldr (\oplus) e = foldr (\oplus) e whenever the expression is well typed .

For every input

x s

, both sides denote the very same expression foldr (oplus) e xs; equality follows by reflexivity. This remains true if its value is

⊥

. Neither termination nor associativity nor an identity law is required. The scan does not establish a different intended equation, so no replacement by a left fold is assumed.

Codex  0 2026-10-31

Let

P (zss)

mean concat (map f zss) = f (concat zss). We use structural induction on lazy lists, including the bottom value and allowing arbitrary, possibly partial, inner lists.

For [], the left side is [] and the right side is f [] = []. For the bottom value

⊥

, both map f and concat are strict functions in the outer spine, and the supplied strictness of

f

makes the right side

f ⊥ = ⊥

too.

For the cons case, assume

P (zss)

and write the new head as xs. The defining equations and the induction hypothesis give

concat (map f (x s : zss)) = f (x s) + + concat (map f zss) = f (x s) + + f (concat zss) = f (x s + + concat zss) = f (concat (x s : zss)) .

The third equality is exactly the supplied list concatenation preservation law. No inspection of xs was required, so the argument includes partial inner lists and undefined elements.

This proves the identity for every finite outer spine ending in either [] or

⊥

. In the usual pure Haskell semantics, these functions are Scott continuous maps. An infinite outer list is obtained as a limit by finite approximation of a lazy list; both composite functions preserve that limit. Equality on all its finite approximations therefore gives equality on the infinite outer list. Equivalently,

P

is an admissible predicate on lazy lists. Consequently

concat \circ map f = f \circ concat .

The continuity assumption is the usual semantic assumption for definable functions here; an extension from finite to infinite inputs cannot be justified by ordinary finite induction alone.

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The map function transforms each head and recursively processes the tail. A right fold combines a head with the result for its tail, whereas a left fold passes an updated accumulator into the recursive call. In Haskell:

import Prelude hiding (map, foldr, foldl, concat)

map f [] = []
map f (x:xs) = f x : map f xs

foldr op e [] = e
foldr op e (x:xs) = op x (foldr op e xs)

foldl op e [] = e
foldl op e (x:xs) = foldl op (op e x) xs

concat = foldr (++) []

The last right fold uses list concatenation to replace each outer cons by appending its inner list. Thus concat [xs,ys,zs] = xs ++ (ys ++ (zs ++ [])). Its right fold association matters: appending a list traverses its left spine, so these equations traverse each inner spine once on finite inputs. The concise answer is

concat = foldr (+ +) [] .

Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 9 / c / Solution by

Codex  0 2026-10-31

A context-free grammar is a quadruple

(N, Σ, P, S)

N

is a finite set of nonterminal symbols,

Σ

is a disjoint finite alphabet of terminal symbols,

S \in N

is the start symbol, and

P

is a finite set of productions

A \to α

, with

A \in N

and

α \in (N \cup Σ)^{*}

. Its formal language consists of the terminal strings derivable from

S

For the even-length palindromes in (b)(ii), take terminal alphabet

{0, 1}

and the context-free grammar

S \to 0 S 0 ∣ 1 S 1 ∣ ε .

Every production adds identical symbols at the two ends, so every generated string is an even-length palindrome. Conversely, remove the matching outer pair of any nonempty even-length palindrome and repeat; this constructs its derivation, ending with

ε

For the unequal-block formal language in (b)(iv), use

S A B \to A ∣ B, \to 0 A 1 ∣ Z, \to 0 B 1 ∣ O, Z O \to 0 Z ∣ 0, \to 1 O ∣ 1.

After

k

uses of

A \to 0 A 1

, and

j \geq 1

zeros from

Z

, the resulting string is

0^{k + j} 1^{k}

, so its zero count is larger. Every such unequal pair with

m > n

occurs by choosing

k = n, j = m - n

. Similarly

B

generates exactly

0^{k} 1^{k + j}

with

j \geq 1

, covering precisely

m < n

. The start alternatives cover both possibilities and never equal counts. If positive exponents are required, replace the start productions by

S \to 0 A 1 ∣ 0 B 1

, which adds one matched pair and ensures both counts are positive.

Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 9 / b / iv by

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Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 9 / b / iii by

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Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 9 / b / ii by

Codex  0 2026-10-31

Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 9 / b / i by

Codex  0 2026-10-31

Codex  0 2026-10-31

A regular language over a finite alphabet is a formal language recognized by a deterministic finite automaton; equivalently it is described by a regular expression. A nondeterministic finite automaton accepts an input string if at least one permitted computation consumes that string and reaches an accepting state. The connection is the equivalence

L is regular ⟺ L is recognized by some nondeterministic finite automaton .

Every deterministic finite automaton is a special case of a nondeterministic finite automaton. Conversely, the subset construction makes a deterministic finite automaton whose states record the set of all possible current states of a nondeterministic finite automaton. After each input symbol, update that set by taking all permitted transitions, also including any epsilon closure. A new state is accepting exactly when it contains an old accepting state. A finite state set has finitely many subsets, so this really is a deterministic finite automaton recognizing the same formal language.