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

What is the difference between an undecidable problem and an intractable one?

  1. AIntractable problems cannot be solved by any algorithm, while undecidable ones only need exponential time
  2. BBoth mean the problem has no algorithm at all
  3. CUndecidable problems are in NP, while intractable problems are in P
  4. DAn undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
3.Data structuresMedium

How is a new element inserted into a binary heap?

  1. ARebuild the whole heap from scratch
  2. BPlace it at the end of the array and sift it up while it violates the heap property
  3. CInsert it at its sorted position by shifting all of the larger elements one place right
  4. DPlace it at the root and sift it down
4.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(1)
  2. BO(n log n)
  3. CO(n)
  4. DO(log n)
5.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 6
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
1454610992ABCDEF

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =

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 · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow

Why this is the answer

The halting problem has no algorithm at all; an NP-hard problem such as TSP can be solved by brute force, just not in known polynomial time.

Answer

D · An undecidable problem has no algorithm for all inputs; an intractable one has one, but it is too slow
3.Data structures

Answer: B · Place it at the end of the array and sift it up while it violates the heap property

Why this is the answer

Adding at the end keeps the tree complete; swapping with the parent restores the order in O(log n) steps.

Answer

B · Place it at the end of the array and sift it up while it violates the heap property
4.Data structures

Answer: D · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

D · O(log n)
5.Graph theory

Answer: Euler path, no circuit

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) = 2, deg(B) = 3, deg(C) = 3, deg(D) = 2, deg(E) = 2

Odd vertices

B, C → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

6.Graph theory

Answer: d(F) = 20

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 14, C: 5
C (5): B: 9, E: 15
B (9): D: 15, E: 11
E (11): F: 20
D (15)
F (20)

Shortest path

A → C → B → E → F
d(F) = 20

Answer

d(F) = 20

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