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 time complexity of computing the n-th Fibonacci number with a bottom-up table?

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

What did the Cook–Levin theorem establish?

  1. AThe halting problem is undecidable
  2. BThe Boolean satisfiability problem (SAT) is NP-complete
  3. CEvery NP problem can be solved in exponential time
  4. DP is not equal to NP
3.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(n), because later elements must shift
  2. BO(log n), because of binary search
  3. CO(n²), because each element moves n times
  4. DO(1), because the index is known
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. BTombstones speed up hashing
  3. CEmptying the slot could break the probe sequence for other keys
  4. DDeleted keys must be kept forever for security, so slots can never be reused
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
6.Graph theoryHard
201121791314ABCDE

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: B · O(n)

Why this is the answer

Filling the table from 0 to n takes one addition per entry, so the work is linear instead of exponential.

Answer

B · O(n)
2.Algorithms & complexity

Answer: B · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

B · The Boolean satisfiability problem (SAT) is NP-complete
3.Data structures

Answer: A · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

A · O(n), because later elements must shift
4.Data structures

Answer: C · 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

C · Emptying the slot could break the probe sequence for other keys
5.Graph theory

Answer: H = 3

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 = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
6.Graph theory

Answer: edge CE

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. BC (2)
2. CE (9)
3. AC (11)
4. DE (13)

edge number 2

CE

Answer

edge CE

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