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

A recurrence T(n) = 2T(n/2) + O(n) solves to:

  1. AO(log n)
  2. BO(n)
  3. CO(n²)
  4. DO(n log n)
2.Algorithms & complexityHard

The 0/1 knapsack problem solved with dynamic programming takes O(nW) time. Why is this called pseudo-polynomial?

  1. ABecause it only works for W = 0
  2. BBecause n is always small
  3. CW is a number, so the time is exponential in the number of bits used to write W
  4. DBecause the algorithm is randomised and its running time varies with each input
3.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AOnly the colliding keys are moved to a new overflow list at the end of the table
  2. BA larger table is allocated and every key is inserted again
  3. CThe oldest keys are deleted to make room
  4. DThe existing keys are sorted in place
4.Data structuresHard

How can you detect whether a singly linked list contains a cycle using O(1) extra memory?

  1. AMove one pointer by one step and another by two steps and check whether they meet
  2. BStore every visited node in a hash set
  3. CSort the list and look for duplicates
  4. DCount the nodes until the count exceeds the list length field
5.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
ABCDEF

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: D · O(n log n)

Why this is the answer

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

D · O(n log n)
2.Algorithms & complexity

Answer: C · W is a number, so the time is exponential in the number of bits used to write W

Why this is the answer

The running time is polynomial in the value of W but exponential in its length (log W bits); 0/1 knapsack is NP-complete.

Answer

C · W is a number, so the time is exponential in the number of bits used to write W
3.Data structures

Answer: B · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

B · A larger table is allocated and every key is inserted again
4.Data structures

Answer: A · Move one pointer by one step and another by two steps and check whether they meet

Why this is the answer

In Floyd's algorithm, if there is a cycle the fast pointer eventually catches up with the slow one; otherwise the fast pointer reaches null.

Answer

A · Move one pointer by one step and another by two steps and check whether they meet
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) = 2, deg(B) = 3, deg(C) = 3, deg(D) = 2, deg(E) = 3, deg(F) = 1

Odd vertices

B, C, E, F → 4
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 0
B: 1 0 1 1 1 0
C: 1 1 0 0 1 0
D: 0 1 0 0 1 1
E: 0 1 1 1 0 0
F: 0 0 0 1 0 0

Common neighbours

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

Answer

N = 3

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