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

A greedy algorithm:

  1. AUses only recursion
  2. BTries every possible solution
  3. CMakes the locally best choice at each step
  4. DAlways gives the optimal solution
2.Algorithms & complexityHard

Why does binary search not give O(log n) time on a singly linked list?

  1. AReaching the middle element takes O(n) steps, since there is no direct index access
  2. BComparisons on linked list nodes are slower
  3. CLinked lists cannot be sorted
  4. DBinary search needs at least twice the memory
3.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AThe oldest keys are deleted to make room
  2. BA larger table is allocated and every key is inserted again
  3. CThe existing keys are sorted in place
  4. DOnly the colliding keys are moved to a new overflow list at the end of the table
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 keep the array sorted
  3. CTo avoid repeated resizing when pushes and pops alternate near the boundary
  4. DBecause the memory allocator can only release memory in quarters of a block
5.Graph theoryMedium
ABCDEFEuler 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| = 6|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
20413109141931815ABCDEFG

Prim’s algorithm builds a minimum spanning tree starting from vertex A. Click the edge it adds as edge number 5.

start: Ak = 5
  1. 1edge AC
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge EF
  7. 7edge GB
  8. 8edge GC
  9. 9edge GD
  10. 10edge GE

3Solutions

1.Algorithms & complexity

Answer: C · Makes the locally best choice at each step

Why this is the answer

Greedy algorithms are fast and optimal for some problems (e.g. Dijkstra, Huffman coding) but not in general (e.g. coin change with some coin systems).

Answer

C · Makes the locally best choice at each step
2.Algorithms & complexity

Answer: A · Reaching the middle element takes O(n) steps, since there is no direct index access

Why this is the answer

Binary search needs O(1) access to the middle; on a linked list walking there costs linear time, so the total is O(n).

Answer

A · Reaching the middle element takes O(n) steps, since there is no direct index access
3.Data structures

Answer: B · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

B · A larger table is allocated and every key is inserted again
4.Data structures

Answer: C · 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

C · 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) = 1, deg(B) = 4, deg(C) = 2, deg(D) = 2, deg(E) = 3, deg(F) = 2

Odd vertices

A, 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: edge DE

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.

Prim’s algorithm

1. AC (20)
2. GC (3)
3. BC (4)
4. CE (10)
5. DE (9)
6. EF (14)

edge number 5

DE

Answer

edge DE

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