Data Structures and Algorithms: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

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1Formulas and methods

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

2Practice problems

1.Algorithms & complexityMedium

Why does dynamic programming give no real speed-up for merge sort?

  1. ADynamic programming only works on graph problems, not on arrays
  2. BMerge sort has no recursive structure
  3. CIts subproblems do not overlap, so there is nothing to reuse
  4. DMerge sort is already O(n)
2.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AA decision tree with n possible outcomes must have depth at least log₂ n
  2. BBinary search is the only possible algorithm, and it needs log₂ n comparisons
  3. CComparisons take logarithmic time each
  4. DEvery element must be compared at least once
3.Data structuresMedium

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AThe leftmost node of its right subtree
  2. BIts parent, always
  3. CThe rightmost node of its left subtree
  4. DIts right child, always
4.Data structuresHard

What is the main advantage of a skip list over a sorted linked list?

  1. AIt uses no extra memory
  2. BExpected O(log n) search by using several levels of forward pointers
  3. CIt stores keys without order
  4. DGuaranteed O(1) search because every node stores a pointer to every other node
5.Graph theoryMedium
ABCDEFGEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 7|E| = 9
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
ABCDE

How many closed walks of length 2 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · Its subproblems do not overlap, so there is nothing to reuse

Why this is the answer

Merge sort's halves are disjoint and each is solved once; DP helps only when subproblems recur, as in Fibonacci.

Answer

C · Its subproblems do not overlap, so there is nothing to reuse
2.Algorithms & complexity

Answer: A · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

A · A decision tree with n possible outcomes must have depth at least log₂ n
3.Data structures

Answer: A · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

A · The leftmost node of its right subtree
4.Data structures

Answer: B · Expected O(log n) search by using several levels of forward pointers

Why this is the answer

Randomised levels allow searches to skip many nodes, giving expected O(log n) search, insert and delete.

Answer

B · Expected O(log n) search by using several levels of forward pointers
5.Graph theory

Answer: neither

Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.

Degrees

deg(A) = 1, deg(B) = 4, deg(C) = 3, deg(D) = 3, deg(E) = 3, deg(F) = 1, deg(G) = 3

Odd vertices

A, C, D, E, F, G → 6
neither

Answer

neither
6.Graph theory

Answer: N = 3

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 0 1
C: 1 1 0 0 1
D: 0 0 0 0 1
E: 0 1 1 1 0

Common neighbours

(M²)CC = 1·1 + 1·1 + 0·0 + 0·0 + 1·1 = 3

Answer

N = 3

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