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 does dynamic programming give no real speed-up for merge sort?

  1. AIts subproblems do not overlap, so there is nothing to reuse
  2. BDynamic programming only works on graph problems, not on arrays
  3. CMerge sort has no recursive structure
  4. DMerge sort is already O(n)
2.Algorithms & complexityHard

How is the travelling salesman problem classified?

  1. AIt is in P because Dijkstra's algorithm solves it by visiting every city once
  2. BThe decision version is in P but the optimisation version is NP-complete
  3. CThe decision version is NP-complete and the optimisation version is NP-hard
  4. DIt is undecidable
3.Data structuresMedium

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo simplify code by removing special cases at the head or tail
  2. BTo prevent the list from growing beyond a fixed maximum number of nodes
  3. CTo make the list sorted
  4. DTo store the number of elements
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 the memory allocator can only release memory in quarters of a block
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CTo keep the array sorted
  4. DBecause halving at one half would lose elements
5.Graph theoryMedium

A connected graph has 8 vertices and 14 edges. How many edges k must be removed to leave a spanning tree?

|V| = 8|E| = 14
k =
6.Graph theoryHard
1131012131458ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 6; A itself is round 1.

start: Around 6
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G

3Solutions

1.Algorithms & complexity

Answer: A · Its subproblems do not overlap, so there is nothing to reuse

Why this is the answer

Merge sort's halves are disjoint and each is solved once; DP helps only when subproblems recur, as in Fibonacci.

Answer

A · Its subproblems do not overlap, so there is nothing to reuse
2.Algorithms & complexity

Answer: C · The decision version is NP-complete and the optimisation version is NP-hard

Why this is the answer

"Is there a tour of length at most k?" is NP-complete; finding the shortest tour is NP-hard, since it is at least as hard.

Answer

C · The decision version is NP-complete and the optimisation version is NP-hard
3.Data structures

Answer: A · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

A · To simplify code by removing special cases at the head or tail
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: k = 7

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

8 − 1 = 7

Edges to remove

k = 14 − 7 = 7

Answer

k = 7
6.Graph theory

Answer: F

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

1. A (0): B: 1
2. B (1): C: 2, D: 4, G: 15
3. C (2): E: 12
4. D (4): E: 5, F: 6, G: 9
5. E (5)
6. F (6)
7. G (9)

round 6

F

Answer

F

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