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

Why do many library sorting routines switch to insertion sort for very small subarrays?

  1. ARecursive algorithms cannot handle arrays below 10 elements
  2. BInsertion sort is O(n log n) in the worst case
  3. CInsertion sort has little overhead and is fast on small or nearly sorted inputs
  4. DInsertion sort uses less memory than any other sorting algorithm on large inputs
2.Algorithms & complexityHard

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. ANo, the lower bound applies to the worst case over all inputs, not to every input
  2. BYes, but only for arrays shorter than 100
  3. CNo, because insertion sort is not a comparison sort
  4. DYes, the lower bound is therefore wrong
3.Data structuresMedium

An in-order traversal of a binary search tree visits the keys in:

  1. ADescending order always
  2. BLevel order
  3. CAscending sorted order
  4. DInsertion order
4.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause halving at one half would lose elements
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CBecause the memory allocator can only release memory in quarters of a block
  4. DTo keep the array sorted
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| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
414814510142ABCDEF

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: C · Insertion sort has little overhead and is fast on small or nearly sorted inputs

Why this is the answer

For small n the simple loop of insertion sort beats the recursion overhead of quicksort or merge sort, even though it is O(n²).

Answer

C · Insertion sort has little overhead and is fast on small or nearly sorted inputs
2.Algorithms & complexity

Answer: A · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

A · No, the lower bound applies to the worst case over all inputs, not to every input
3.Data structures

Answer: C · Ascending sorted order

Why this is the answer

Visiting the left subtree, then the node, then the right subtree outputs BST keys in sorted order.

Answer

C · Ascending sorted order
4.Data structures

Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
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) = 2, deg(D) = 2, deg(E) = 1

Odd vertices

B, E → 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) = 32

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): C: 4
C (4): B: 18, E: 18
B (18): D: 26
E (18): D: 23, F: 32
D (23)
F (32)

Shortest path

A → C → E → F
d(F) = 32

Answer

d(F) = 32

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