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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. AGeneral SAT on formulas in conjunctive normal form
  2. BCircuit satisfiability
  3. C3-SAT, where each clause has exactly three literals
  4. D2-SAT, where each clause has at most two literals
3.Data structuresMedium

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

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

Under simple uniform hashing, what is the expected time of a search in a hash table with chaining, n keys and m buckets?

  1. AO(1 + α), where α = n / m
  2. BO(log n)
  3. CO(n)
  4. DO(m)
5.Graph theoryMedium

A connected graph has 12 vertices and 15 edges. How many edges k must be removed to leave a spanning tree?

|V| = 12|E| = 15
k =
6.Graph theoryHard
1719712182ABCDE

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 3; skipped edges do not count.

k = 3
  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: 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 · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

D · 2-SAT, where each clause has at most two literals
3.Data structures

Answer: A · 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

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

Answer: A · O(1 + α), where α = n / m

Why this is the answer

Computing the hash costs O(1) and the expected chain length is α, so keeping α bounded by resizing gives O(1) expected time.

Answer

A · O(1 + α), where α = n / m
5.Graph theory

Answer: k = 4

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

12 − 1 = 11

Edges to remove

k = 15 − 11 = 4

Answer

k = 4
6.Graph theory

Answer: edge DE

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. CE (1)
2. BE (2)
3. DE (8)
4. AB (17)

edge number 3

DE

Answer

edge DE

ExamProblems · examproblems.com/courses/data-structures-and-algorithms