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There are three useful distinctions between dynamic programming and a conventional divide-and-conquer algorithm.

A divide-and-conquer algorithm typically splits one instance into mostly separate smaller instances; dynamic programming deliberately recognizes overlapping subproblems that recur along different decision histories.
A dynamic programming state is solved once and its result reused through a table or memoization. A direct divide-and-conquer algorithm normally combines the results of its recursive children without retaining a table indexed by all repeated states.
The dependency structure of a direct recursive decomposition is a recursion tree; dynamic programming evaluates a Directed acyclic graph of states in dependency order, either bottom-up or by memoization. Its cost is the number of distinct states times the work per state, rather than the number of occurrences in the expanded recursion tree.

These are paradigm-level distinctions, not mutually exclusive definitions: a recursive algorithm can use memoization, and dynamic programming need not be implemented iteratively. For optimization, the dynamic programming state and optimal substructure justify combining stored optimum values; storing repeated work alone does not prove the state is sufficient.

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Functional programming, design and analysis of algorithms / 7 / a / Solution 2026-10-25

Use unit-cost insertion, deletion and substitution of one symbol. Their minimum total cost is the edit distance. Let

D [i, j]

be the edit distance between the first

i

symbols of the first sequence and the first

j

of the second. Initialize

D [i, 0] = i

and

D [0, j] = j

. For

i, j > 0

compute

D [i, j] = min {D [i - 1, j] + 1, D [i, j - 1] + 1, D [i - 1, j - 1] + [x_{i} \neq = y_{j}]} .

The cases delete the final source symbol, insert the final target symbol, or align the two final symbols, paying one for substitution and zero for a match. Every optimal alignment ends in one of these cases, and each case extends a valid shorter alignment. Induction on

i + j

proves the dynamic programming recurrence correct.

Fill rows or columns in increasing order so every dependency is already available. There are

(m + 1) (n + 1)

entries with constant work per interior entry: $O (mn)$ running time for nonempty sequences, or

O ((m + 1) (n + 1))

including the boundary cases. The full table has space complexity

O (mn)

; two rows give

O (n)

auxiliary space when only the distance is wanted. Predecessor choices in the full table recover an optimal edit sequence.

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Functional programming, design and analysis of algorithms / 7 / b / ii / Solution 2026-10-25

During the computation in part i, record an index

last [t]

attaining the minimum at every positive amount. Starting with

t = v

, print

x_{last [t]}

and replace

t

t - x_{last [t]}

, continuing until

t = 0

Every step decreases the amount and reduces the table entry by exactly one. The printed denominations therefore sum to

v

and their number is exactly

M [v]

, proving optimality. The total running time is $O (n v + M [v])$ , including the output;

M [v] \leq v

because denomination

1

is available. The predecessor array adds

O (v)

to the space complexity. This is reconstruction from the dynamic programming certificate, not the largest-denomination greedy algorithm, which need not be optimal for arbitrary denominations.

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Functional programming, design and analysis of algorithms / 7 / c / i / Solution 2026-10-25

The printed formula for the edge label has mismatched variable names. The type-correct reading is

x_{i} = σ (v_{i - 1}, v_{i})

. Interpret the requested route as a directed walk of consecutive labelled edges, allowing repeated vertices; the distinction from a vertex-simple directed path is discussed below.

Let

R_{t}

be the set of vertices reachable from the specified start after reading exactly the first

t

labels. Initialize

R_{0} = {v_{0}}

and update

R_{t} = {v : there is (u, v) \in E with u \in R_{t - 1} and σ (u, v) = x_{t}} .

After

k

stages, answer yes exactly when $R_{k} \neq = \emptyset$ . A directed walk realizing a prefix decomposes into a walk realizing the shorter prefix and one correctly labelled final edge; conversely every such extension is valid. This proves the dynamic programming invariant by induction on

t

. Store reachable vertices in Boolean arrays and scan the edges per stage. The running time is

O (k (∣ V ∣ + ∣ E ∣))

with

O (∣ V ∣)

rolling-table space complexity, in addition to graph storage. The empty label sequence is realized by staying at

v_{0}

If “path” is instead required to mean vertex-simple, augment the state to

(S, v)

, where

S

is the set of visited vertices. Start with

R_{0} ({v_{0}}, v_{0})

and extend a matching edge to

w

only when

w \in / S

, replacing

S

S \cup {w}

. This gives an exact dynamic programming algorithm for that interpretation too; reject immediately when

k \geq ∣ V ∣

. A straightforward bound is

O (k 2^{∣ V ∣} (∣ V ∣ + ∣ E ∣))

running time with

O (2^{∣ V ∣} ∣ V ∣)

rolling-table storage. The exponential distinction is substantive: all labels equal and

k = ∣ V ∣ - 1

include the Hamiltonian path problem from a fixed start. The efficient layered recurrence answers the consecutive-edge interpretation without pretending to enforce distinct vertices.

Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Functional programming, design and analysis of algorithms (resit) / 8 / b / Solution 2026-10-25

Write

c_{j} = 2^{j} + 1

for

0 \leq j \leq k

, with a separate unit denomination. Choose, among all minimum-coin representations of the target, one whose vector of counts is lexicographically largest when larger denominations come first. Such a choice exists because a fixed positive target has only finitely many representations.

There cannot be two unit coins: replacing them by one

c_{0} = 2

decreases the count. For

j < k

, two copies of

c_{j}

can be exchanged using

2 c_{j} = c_{j + 1} + 1.

This keeps the coin count unchanged but increases the first differing larger-denomination count, contradicting the chosen lexicographic maximum. Thus there exists an optimal representation using at most one of each denomination below the largest. The largest denomination may repeat.

It remains to show why that normalization implies greedy optimality; uniqueness alone is not enough for an arbitrary coin system. Fix a denomination

c_{j}

. Let

L

be the total value of all coins smaller than

c_{j}

in the chosen optimal representation, and let

r

be their number. These smaller coins are distinct. If

I \subseteq {0, \dots, j - 1}

records the selected non-unit denominations, then

L = \sum_{i \in I} 2^{i} + r, \sum_{i \in I} 2^{i} \leq 2^{j} - 1.

(2)

L \geq c_{j}

, replace all

r

smaller coins by one

c_{j}

and

L - c_{j}

unit coins. The new number of coins is

1 + L - c_{j} = L - 2^{j} \leq r - 1,

(3)

contradicting optimality. Therefore the total value below every denomination $c_{j}$ is strictly less than $c_{j}$ .

At the largest denomination, this forces its count to be

⌊ n / c_{k} ⌋

: its residual value is nonnegative and smaller than

c_{k}

. Remove those coins. The same smaller-denomination property then forces the next quotient, and so on down to the unit coin. These are exactly greedy's choices, proving greedy is optimal for every positive target. The argument also covers

k = 0

, where the denominations are two and one. Independent comparison with coin-change dynamic programming confirmed the formula for all

0 \leq k \leq 8

and targets from one to 2,000; that finite check supplements the proof.

Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Functional programming, design and analysis of algorithms (resit) / 6 / a / Solution 2026-10-25

Use positive integral item weights and a nonnegative integral capacity. For the unbounded knapsack problem, define

K (c)

to be the best value with total weight at most

c

. Then

K (c) = max ({0} \cup {v_{i} + K (c - w_{i}) : 1 \leq i \leq n, w_{i} \leq c}), K (0) = 0.

The zero candidate is the empty selection, so unused capacity is allowed and an empty set of fitting items causes no problem. Every nonempty feasible selection has some last item

i

; removing one copy leaves a feasible solution at capacity

c - w_{i}

. Its value is bounded by the corresponding candidate. Conversely, taking an optimal smaller selection and adding that item attains the candidate, proving the recurrence relation by optimal substructure. Positive weights make every dependency strictly smaller.

Fill capacities in increasing order. There are

W + 1

entries, each scanning

n

items, so the dynamic programming takes

O (nW)

time and

O (W)

storage. A winning item pointer at each capacity reconstructs the chosen multiplicities. These are capacity-dependent, pseudopolynomial bounds. For zero-weight items, any positive value makes the repetition problem unbounded; nonpositive-value zero-weight items may simply be discarded before using the recurrence.

Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Functional programming, design and analysis of algorithms (resit) / 6 / c / Solution 2026-10-25

This version of Kadane's algorithm, named for Joseph Born Kadane, permits the empty subsequence. For

j \geq 1

, let

E_{j}

be the maximum of zero and the weight of a contiguous subsequence ending exactly at position

j

. Let

B_{j}

be the best weight anywhere in the first

j

positions. Use

E_{0} = B_{0} = 0, E_{j} = max (0, E_{j - 1} + a_{j}), B_{j} = max (B_{j - 1}, E_{j}) .

A segment ending at

j

either extends one ending at

j - 1

or starts at

j

; the zero candidate permits discarding a negative predecessor and the empty answer. Any best segment of the first

j

positions either ends at

j

or already occurs in the first

j - 1

. These two decompositions prove the dynamic programming equations by induction.

Here is an implementation returning endpoints, using zero-based indexing:

best_segment(a):
    end_weight = best_weight = 0
    start = best_lo = best_hi = 0
    for j in range(len(a)):
        trial = end_weight + a[j]
        if trial <= 0:
            end_weight = 0
            start = j+1
        else:
            end_weight = trial
        if end_weight > best_weight:
            best_weight = end_weight
            best_lo, best_hi = start, j+1
    return (best_lo, best_hi, best_weight)

There is exactly one addition per input element, hence

n

additions and

O (n)

comparisons. Auxiliary storage is

O (1)

. The all-negative case retains interval

[0, 0)

and weight zero. The example returns

[3, 6)

and weight 8, hence the desired contiguous subsequence.

Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Functional programming, design and analysis of algorithms / 5 / c / Solution 2026-10-25

First run breadth-first search, ignoring weights, to obtain the minimum number of edges

h (v)

from the source to each reachable vertex. Then compute weighted distances only along edges that advance one breadth-first layer:

d[source] = 0; all other d[v] = infinity
for u in nondecreasing BFS-layer order:
    for each outgoing edge (u,v):
        if h[v] == h[u]+1:
            if d[u]+w(u,v) < d[v]:
                d[v] = d[u]+w(u,v)
                predecessor[v] = u

Unreachable vertices keep distance

\infty

. For an undirected graph, scan both orientations of each edge.

To justify discarding other routes, a shortest path can be taken simple because every weight is positive. A simple path of

r \leq n - 1

edges has weight in

[r, r (1 + 1/ n)] \subseteq [r, r + 1) .

Every path with at least

r + 1

edges weighs at least

r + 1

. Consequently every minimum-weight path must also have the minimum possible number of edges: this is minimum-hop dominance for nearly unit weights.

A minimum-hop path to

v

uses exactly

h (v)

edges. Every prefix of such a path must itself have minimum hop count, since replacing a longer prefix by a shorter one would shorten the whole route. Hence it advances from layer

j

to layer

j + 1

at every step. The retained edges form a Directed acyclic graph, ordered by

h

. The displayed dynamic programming recurrence therefore finds the cheapest of all minimum-hop paths, which by the preceding bound is a true shortest path.

Figure 1.
Disjoint weight bands for simple paths with different hop counts
. **Details**.

With an adjacency list, both the breadth-first search and the layer-ordered relaxations take $O (n + m)$ time and $O (n + m)$ total storage, where

m = ∣ E ∣

. The queue order from breadth-first search already supplies the layer order; no sorting or priority queue is needed. This improves the usual

O ((n + m) lo g n)

binary-heap Dijkstra bound and is linear in the input size.

Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Functional programming, design and analysis of algorithms / 6 / c / Solution 2026-10-25

Let

m = ∣ s ∣

ℓ_{i} = ∣ w_{i} ∣

, and let

D (i, j)

be the minimum number of words chosen from

w_{1}, \dots, w_{i}

, in their given order, whose concatenation equals the length-

j

prefix of a list of

s

. Use

\infty

for impossibility. The boundary values are

D (0, 0) = 0, D (0, j) = \infty (j > 0) .

For

i \geq 1

D (i, j) = min {D (i - 1, j), 1 + D (i - 1, j - ℓ_{i}) if j \geq ℓ_{i} and s [j - ℓ_{i} : j] = w_{i}} .

(2)

Here

s [a : b]

denotes indices

a, \dots, b - 1

. The first case omits

w_{i}

. In the second, the last selected word is

w_{i}

, so removing it leaves a solution using only earlier words for the shorter prefix. These are exhaustive alternatives. Conversely either alternative constructs a valid solution, proving the dynamic programming recurrence. Even an empty word causes no cycle because both alternatives use row

i - 1

Fill rows in increasing

i

and store the chosen alternative for every finite entry. If

D (n, m) = \infty

, report no solution. Otherwise trace back from

(n, m)

: a skip moves to

(i - 1, j)

; a take records index

i

and moves to

(i - 1, j - ℓ_{i})

. Reverse the recorded indices to obtain the required subsequence. Ties may choose either valid predecessor.

There are

(n + 1) (m + 1)

states. A substring comparison costs

O (ℓ_{i})

, so under the given bounds

m = O (n)

and

ℓ_{i} = O (1)

the running time is $O (n^{2})$ , with

O (n^{2})

storage including reconstruction choices. Two rows give

O (m)

working storage if only the optimum count is needed. For the source example, the optimum is three words, indices $(2, 5, 6)$ , whose concatenation is 010111. This is ordered word concatenation dynamic programming.

Postorder traversal 2026-10-31

Visit all child subtrees before visiting their parent. A depth-first traversal of a tree produces this order by processing a node on return from its recursive calls. It lets dynamic programming use the already computed answers of every child subtree.

Profile dynamic programming 2026-10-20

A dynamic programming method that advances a boundary across a structured object and records only the information needed on that boundary. For tilings by pieces of bounded height, a bit mask recording occupied cells in a few frontier rows can determine every legal continuation. Its state space is typically exponential in the width but independent of the total height.

Pseudo-polynomial time Created 2026-10-14 Updated 2026-10-20

An algorithm has pseudo-polynomial time when its running time is bounded by a polynomial in the numerical magnitudes of input quantities, rather than necessarily in their encoded length. The dynamic programming table for the 0-1 knapsack problem uses

O (nW)

arithmetic operations for

n

items and capacity

W

. A binary numeral system encoding needs only

O (lo g (W + 1))

bits for that capacity, so the table bound can be exponential in the number of capacity bits. This distinction does not claim that every input requires exponential work.

Rounded dynamic programming for two-bin load balancing 2026-10-14

For integer weights, subset-sum dynamic programming finds the greatest achievable total at most

s /2

in time

O (n s)

, where

s

is the total weight. For rational weights and

0 < ε < 1

, scale by

K = ε s / (2 n)

and round down to

q_{i} = ⌊ w_{i} / K ⌋

. The scaled total is at most

2 n / ε

, so exact dynamic programming costs

O (n^{2} / ε)

arithmetic operations. Lifting its partition to the original weights increases the maximum load by at most

n K \leq ε OPT

. Thus it is a fully polynomial-time approximation scheme for this two-bin problem. Bit complexity also accounts for the rational input and epsilon encodings.

Stochastic control Created 2026-10-07 Updated 2026-10-14

A stochastic control problem chooses an admissible control input using information available in a filtration to optimize an expected objective for a random dynamical system. Dynamic programming often characterizes its value function through a Hamilton-Jacobi-Bellman equation.

Strictly increasing subsequence 2026-10-14

A strictly increasing subsequence selects indices

i_{1} < \dots < i_{m}

of a sequence with

x_{i_{1}} < \dots < x_{i_{m}}

. Equal values cannot both occur in it. Its maximum possible length is

L_{<} (x)

, with

L_{<} (()) = 0

; in general

L_{<} (x) \leq L (x)

for the weak longest increasing subsequence. Use strict comparisons in the dynamic programming recurrence relation and replace the first tail at least as large as the next value in the binary search algorithm.

Synchronous Bellman-Ford recurrence 2026-10-18

For the single-source problem, let

D_{k} (v)

be the minimum cost of a directed walk from

s

v

using at most

k

edges. Set

D_{0} (s) = 0

and

D_{0} (v) = \infty

for

v \neq = s

, use

\infty + c = \infty

for finite

c

, and let a minimum over an empty set be

\infty

. Then

D_{k} (v) = min {D_{k - 1} (v), min_{(u, v) \in E} (D_{k - 1} (u) + w_{uv})} .

Every entry on the right comes from the previous round, requiring separate old and new arrays. The proof is the last-edge decomposition: a permitted directed walk either already uses at most

k - 1

edges, or ends with some

u \to v

after a prefix using at most

k - 1

. Conversely, each candidate describes a permitted directed walk. This dynamic programming viewpoint gives an exact meaning to every intermediate label, even when negative-cost directed cycles exist. Apply the formula to the source too: its initial zero is not a permanent constraint.

If no negative-cost directed cycle is reachable from

s

, removing repeated-vertex portions of a directed walk cannot increase its cost. Whenever a destination is reachable, some shortest directed path therefore uses at most

∣ V ∣ - 1

edges, and

D_{∣ V ∣ - 1} (v) = δ (s, v)

. Unreachable vertices keep distance

\infty

. For in-place passes, every finite label still comes from a discovered directed walk; moreover, pass

k

has offered every prefix of length

k

of a shortest directed path, regardless of scan order. Its labels can improve faster than the synchronous labels, while the same

∣ V ∣ - 1

upper bound on passes remains valid. This is why repeated local corrections eventually settle a global shortest path problem.

Totient chain 2026-10-20

The sequence

n, φ (n), φ (φ (n)), \dots, 1

obtained by iterating Euler's totient function on a positive integer. Since

1 \leq φ (n) < n

for

n > 1

, the sequence reaches 1. Counting both endpoints gives the length recurrence relation

L (1) = 1

and

L (n) = 1 + L (φ (n))

; increasing order of

n

therefore suffices for dynamic programming.

Treewidth 2026-10-25

The minimum width of a tree decomposition. A tree has width at most one, while a clique of

k

vertices forces width at least

k - 1

. Every pair of vertices in the clique shares a bag; their connected bag connected subgraphs of a tree have a common intersection by the Helly property of connected connected subgraphs of a tree, placing the whole clique in one bag. Small treewidth is useful for dynamic programming on finite-domain constraint satisfaction problems.

Weighted interval scheduling 2026-10-18

Weighted interval scheduling assigns each job a reward

w_{i}

and maximizes their sum over compatible jobs. Earliest finish can fail for this different objective function; dynamic programming can compare alternatives.^[ref]

For a separate two-job example, let

X = [0, 2)

have reward one and

Y = [1, 3)

reward ten. Earliest finish accepts

X

and rejects

Y

, earning one; the optimum accepts

Y

, earning ten. Replacing

Y

X

preserves job count but loses reward. Thus the interval scheduling exchange argument proves the unit-value theorem only: its preserved quantity must match the intended objective function.