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

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. ANo, because Big O must always state the exact growth rate of the function
  2. BYes, because Big O is an upper bound and n² grows no faster than n³
  3. CNo, because n³ is smaller than n² for large n
  4. DYes, but only if n is smaller than 100
2.Algorithms & complexityHard

Taking both endpoints of every edge in a maximal matching gives a vertex cover. How good is this cover?

  1. AOf unknown quality compared to the optimum
  2. BAlways a minimum vertex cover
  3. CAt most log n times the minimum
  4. DAt most twice the size of a minimum vertex cover
3.Data structuresMedium

Which data structure is used by depth-first search (iterative version)?

  1. AA hash table only
  2. BA heap
  3. CA queue
  4. DA stack
4.Data structuresHard

A Bloom filter can answer "is this element in the set?" with:

  1. AExact answers always
  2. BPossible false positives but no false negatives
  3. CAnswers only for numbers
  4. DPossible false negatives but no false positives
5.Graph theoryMedium

A connected graph has 10 vertices and 17 edges. How many edges k must be removed to leave a spanning tree?

|V| = 10|E| = 17
k =
6.Graph theoryHard
519111539181613ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex C. Click the edge it adds as edge number 4.

start: Ck = 4
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge DF
  8. 8edge EF
  9. 9edge BE

3Solutions

1.Algorithms & complexity

Answer: B · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

B · Yes, because Big O is an upper bound and n² grows no faster than n³
2.Algorithms & complexity

Answer: D · At most twice the size of a minimum vertex cover

Why this is the answer

The matching edges share no vertices, so the optimum needs at least one vertex per edge, while this method takes two: a 2-approximation.

Answer

D · At most twice the size of a minimum vertex cover
3.Data structures

Answer: D · A stack

Why this is the answer

DFS explores as deep as possible, naturally using a stack (or recursion's call stack); BFS uses a queue.

Answer

D · A stack
4.Data structures

Answer: B · Possible false positives but no false negatives

Why this is the answer

If any bit for an item is 0, it is definitely absent; if all are 1 it is probably present. Bloom filters trade accuracy for very little memory.

Answer

B · Possible false positives but no false negatives
5.Graph theory

Answer: k = 8

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.

Edges of a spanning tree

10 − 1 = 9

Edges to remove

k = 17 − 9 = 8

Answer

k = 8
6.Graph theory

Answer: edge AB

Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.

Prim’s algorithm

1. CE (3)
2. DE (9)
3. BC (11)
4. AB (5)
5. EF (16)

edge number 4

AB

Answer

edge AB

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