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

Which statement about a recursive function without a reachable base case is correct?

  1. AThe compiler always turns it into a loop that ends safely
  2. BIt returns 0 automatically
  3. CIt keeps calling itself until the call stack overflows
  4. DIt runs in O(1) time
2.Algorithms & complexityHard

The 0/1 knapsack problem solved with dynamic programming takes O(nW) time. Why is this called pseudo-polynomial?

  1. ABecause the algorithm is randomised and its running time varies with each input
  2. BW is a number, so the time is exponential in the number of bits used to write W
  3. CBecause it only works for W = 0
  4. DBecause n is always small
3.Data structuresMedium

How is a new element inserted into a binary heap?

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

An AVL node is unbalanced because its left child's right subtree became too tall (left-right case). Which fix is applied?

  1. AA single right rotation at the node
  2. BA right rotation at the left child, then a left rotation at the node
  3. CA left rotation at the left child, then a right rotation at the node
  4. DA single left rotation at the node
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
1113681181198ABCDEF

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

start: Aend: E
d(E) =

3Solutions

1.Algorithms & complexity

Answer: C · It keeps calling itself until the call stack overflows

Why this is the answer

Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.

Answer

C · It keeps calling itself until the call stack overflows
2.Algorithms & complexity

Answer: B · W is a number, so the time is exponential in the number of bits used to write W

Why this is the answer

The running time is polynomial in the value of W but exponential in its length (log W bits); 0/1 knapsack is NP-complete.

Answer

B · W is a number, so the time is exponential in the number of bits used to write W
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: C · A left rotation at the left child, then a right rotation at the node

Why this is the answer

The first rotation converts the left-right case into a left-left case, which a right rotation at the unbalanced node then fixes.

Answer

C · A left rotation at the left child, then a right rotation at the node
5.Graph theory

Answer: Euler 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) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
6.Graph theory

Answer: d(E) = 19

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: 11, C: 13
B (11): D: 19, E: 19
C (13)
D (19): F: 30
E (19): F: 28
F (28)

Shortest path

A → B → E
d(E) = 19

Answer

d(E) = 19

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