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 does the recurrence T(n) = T(n/2) + O(1) solve to?

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

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AKeys must be sorted first
  2. BArrays have O(n) access
  3. CHash functions are slow
  4. DMany keys may collide into the same bucket
3.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AHeaps never contain more than 100 elements
  2. BAll keys in a heap are equal
  3. CIt is a complete binary tree, so positions follow simple index formulas
  4. DArrays automatically keep the keys in sorted order, so no pointers are needed
4.Data structuresHard

Without union by rank or path compression, what is the worst-case cost of a single find in a union-find structure with n elements?

  1. AO(1)
  2. BO(α(n))
  3. CO(log n)
  4. DO(n)
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
6.Graph theoryHard
71012916202ABCDE

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 2; skipped edges do not count.

k = 2
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge BE

3Solutions

1.Algorithms & complexity

Answer: D · O(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

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

Answer: D · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

D · Many keys may collide into the same bucket
3.Data structures

Answer: C · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

C · It is a complete binary tree, so positions follow simple index formulas
4.Data structures

Answer: D · O(n)

Why this is the answer

Naive unions can link trees into a long chain, so find may follow n − 1 parent pointers; the two heuristics bring it close to constant.

Answer

D · O(n)
5.Graph theory

Answer: H = 360

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.

Count

H = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
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.

Kruskal’s algorithm

1. BE (2)
2. AB (7)
3. BD (9)
4. AC (10)

edge number 2

AB

Answer

edge AB

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