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

What is the worst-case time complexity of quicksort?

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

Which technique gives an expected O(n log n) running time for quicksort on any input?

  1. AAlways choosing the first element
  2. BUsing recursion only on the left side
  3. CSorting the input before running quicksort
  4. DChoosing the pivot uniformly at random
3.Data structuresMedium

In a hash table using open addressing, what is the maximum possible load factor?

  1. AThere is no maximum
  2. B1
  3. C0.5
  4. D2
4.Data structuresHard

In a hash table using open addressing, why is deletion often done with a "tombstone" marker?

  1. AIt makes the table sorted
  2. BEmptying the slot could break the probe sequence for other keys
  3. CDeleted keys must be kept forever for security, so slots can never be reused
  4. DTombstones speed up hashing
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| = 10
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
12471214569ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F

3Solutions

1.Algorithms & complexity

Answer: B · O(n²)

Why this is the answer

With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).

Answer

B · O(n²)
2.Algorithms & complexity

Answer: D · Choosing the pivot uniformly at random

Why this is the answer

With a random pivot, the expected running time is O(n log n) for every input; bad cases still exist but are very unlikely.

Answer

D · Choosing the pivot uniformly at random
3.Data structures

Answer: B · 1

Why this is the answer

With open addressing all keys are stored in the table itself, so n cannot exceed m; separate chaining can have a load factor above 1.

Answer

B · 1
4.Data structures

Answer: B · Emptying the slot could break the probe sequence for other keys

Why this is the answer

If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.

Answer

B · Emptying the slot could break the probe sequence for other keys
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) = 4, deg(E) = 3, deg(F) = 2, deg(G) = 3

Odd vertices

A, C, E, G → 4
neither

Answer

neither
6.Graph theory

Answer: B

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.

Dijkstra round by round

1. A (0): C: 12
2. C (12): B: 16, E: 24
3. B (16): D: 23
4. D (23): F: 28
5. E (24)
6. F (28)

round 3

B

Answer

B

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