Algorithms at Scale (Week 2) Puzzle of the Day: A bag contains a - PowerPoint PPT Presentation

Approximate Connected Components Key Idea 3: Approximate Cost cost(w) = 1/6 Ignore low cost cost(x) = 1/6 w x components: A If cost(u) < 𝜁 /2, round up. Total added cost ≤ 𝜁 n /2 . cost(y) = 1/3 B y c cost(z) = 1 z

Approximate Connected Components Key Idea 3: Approximate Cost cost(w) = 1/6 Define: per-node cost cost(x) = 1/6 w x Let n(u) = number of nodes in the A connected component containing node u. Let ñ(u) = min(n(u), 2/ 𝜁 ). cost(y) = 1/3 B y Let cost(u) = max(1/n(u), 𝜁 /2). c = 1/ñ(u). cost(z) = 1 z

Approximate Connected Components Key Idea 3: Approximate Cost Define: per-node cost Define: Let n(u) = number of nodes in the connected component containing node u. Note: Let ñ(u) = min(n(u), 2/ 𝜁 ). Let cost(u) = max(1/n(u), 𝜁 /2). = 1/ñ(u).

Approximate Connected Components Key Idea 3: Approximate Cost Define: per-node cost Define: Let n(u) = number of nodes in the connected component containing node u. Note: Let ñ(u) = min(n(u), 2/ 𝜁 ). Let cost(u) = max(1/n(u), 𝜁 /2). = 1/ñ(u).

Approximate Connected Components Close enough approximation: Intuition: By rounding cost(u) up to 𝜁 /2, we increase the error at most 𝜁 n/2.

Approximate Connected Components Close enough approximation: Intuition: By rounding cost(u) up to 𝜁 /2, we increase the error at most 𝜁 n/2.

Approximate Connected Components Close enough approximation: Intuition: By rounding cost(u) up to 𝜁 /2, we increase the error at most 𝜁 n/2.

Approximate Connected Components Close enough approximation: Intuition: By rounding cost(u) up to 𝜁 /2, we increase the error at most 𝜁 n/2.

Approximate Connected Components Close enough approximation: Intuition: By rounding cost(u) up to 𝜁 /2, we increase the error at most 𝜁 n/2.

Approximate Connected Components Close enough approximation: Intuition: By rounding cost(u) up to 𝜁 /2, we increase the error at most 𝜁 n/2.

Approximate Connected Components Algorithm 3 sum = 0 cost(w) = 1/6 cost(x) = 1/6 w for j = 1 to s: x Choose u uniformly at random. A sum = sum + cost(u) return n∙(sum/s) We have shown: cost(y) = 1/3 B y Sufficient to approximate c cost(u) by rounding up. cost(z) = 1 z

Approximate Connected Components Algorithm 3 Define: per-node cost How to efficiently compute cost(u)? Let n(u) = number of nodes in the connected component containing node u. Let ñ(u) = min(n(u), 2/ 𝜁 ). Let cost(u) = max(1/n(u), 𝜁 /2). = 1/ñ(u).

Approximate Connected Components Algorithm 3 Define: per-node cost How to efficiently compute cost(u)? Let n(u) = number of nodes in the connected component containing node u. Let ñ(u) = min(n(u), 2/ 𝜁 ). Let cost(u) = max(1/n(u), 𝜁 /2). = 1/ñ(u).

Approximate Connected Components Algorithm 3 sum = 0 for j = 1 to s: Choose u uniformly at random. Perform a BFS from u; stop after seeing 2/ 𝜁 nodes. if BFS found > 2/ 𝜁 nodes: sum = sum + 𝜁 /2 else if BFS found n(u) nodes: sum = sum + 1/n(u) return n∙(sum/s)

Approximate Connected Components Analysis Goal:

Approximate Connected Components Analysis Goal: Implies:

Approximate Connected Components Algorithm 3 Analysis Define random variables: Y 1 , Y 2 , … , Y s Rounded up cost

Approximate Connected Components Algorithm 3 Analysis Define random variables: Y 1 , Y 2 , … , Y s

Approximate Connected Components Algorithm 3 Analysis Unbiased estimator:

Approximate Connected Components Algorithm 3 Analysis Notice: Expected output of algorithm is:

Approximate Connected Components Algorithm 3 Analysis Goal:

Approximate Connected Components Algorithm 3 Analysis Derivation:

Approximate Connected Components Algorithm 3 Analysis Derivation:

Approximate Connected Components Analysis Goal: Implies:

Approximate Connected Components Algorithm 3 sum = 0 for j = 1 to s: Choose u uniformly at random. Perform a BFS from u; stop after seeing 2/ 𝜁 nodes. if BFS found > 2/ 𝜁 nodes: sum = sum + 𝜁 /2 else if BFS found n(u) nodes: sum = sum + 1/n(u) return n∙(sum/s)

Approximate Connected Components Algorithm 3 We have shown: cost(w) = 1/6 cost(x) = 1/6 w x With probability > 2/3, A output is equal to: CC(G) ± 𝜁 n cost(y) = 1/3 B y c cost(z) = 1 z

Approximate Connected Components Algorithm 3 Cost of BFS: O((2 / 𝜁 )∙d) sum = 0 for j = 1 to s: Choose u uniformly at random. Perform a BFS from u; stop after seeing 2/ 𝜁 nodes. if BFS found > 2/ 𝜁 nodes: sum = sum + 𝜁 /2 else if BFS found n(u) nodes: sum = sum + 1/n(u) return n∙(sum/s)

Approximate Connected Components Algorithm 3 Cost of BFS: O((2 / 𝜁 )∙d) sum = 0 for j = 1 to s: Choose u uniformly at random. Perform a BFS from u; stop after seeing 2/ 𝜁 nodes. if BFS found > 2/ 𝜁 nodes: sum = sum + 𝜁 /2 Total cost: else if BFS found n(u) nodes: O(s(2/ 𝜁 )∙d) = sum = sum + 1/n(u) O((1/ 𝜁 2 )(2/ 𝜁 )d) = return n∙(sum/s) O(d/ 𝜁 3 )

Approximate Connected Components Algorithm 3 We have shown: cost(w) = 1/6 cost(x) = 1/6 w x With probability > 2/3, A output is equal to: CC(G) ± 𝜁 n cost(y) = 1/3 B y Running time: c cost(z) = 1 z

Approximate Connected Components Algorithm 3 We have shown: cost(w) = 1/6 cost(x) = 1/6 w x With probability > 1 - 1/ δ , A output is equal to: CC(G) ± 𝜁 n cost(y) = 1/3 B y Running time: c cost(z) = 1 z

Summary Today: Last Week: Number of connected components in Toy example 1 : array all 0’s? a graph. • Gap-style question: • Approximation algorithm. All 0’s or far from all 0’s? Toy example 2: Faction of 1’s? • Weight of MST Additive ± 𝜁 approximation • • Approximation algorithm. Hoeffding Bound Is the graph connected? • Gap-style question. • O(1) time algorithm. 9 dots • Correct with probability 2/3. 4 lines

Today’s Problem: Minimum Spanning Tree Assumptions: 3 3 Graph G = (V,E) 2 • Undirected 2 2 1 • Weighted, max weight W 2 • Connected 1 • n nodes 3 • m edges 2 • maximum degree d 3 Error term: 𝜁 < 1/2 1 Output: Example: output 16 Weight of MST.

Today’s Problem: Minimum Spanning Tree Approximation: 3 3 Output M such that: 2 2 2 1 2 1 Alternate form: 3 2 3 1 Correct output: w.p. > 2/3 Example: 𝜁 = 1/4 Output ∊ [12,20]

Today’s Problem : Minimum Spanning Tree When is this useful? What are trivial values of 𝜁 ? What are hard values of 𝜁 ? What sort of applications is this useful for? Why multiplicative approximation for MST and additive approximation for connected components?

Simple Minimum Spanning Tree Assume all weights 1 or 2 Which edges must be in MST? 2 2 2 2 2 How many weight-2 edges in 1 2 MST? 1 1 2 Best (exact) algorithm? 2 2 1

Simple Minimum Spanning Tree Assume all weights 1 or 2 Let G 1 = graph containing only 2 2 edges of weight 1. 2 2 2 1 2 1 1 2 2 2 1

Simple Minimum Spanning Tree Assume all weights 1 or 2 Let G 1 = graph containing only 2 2 edges of weight 1. 2 2 2 1 2 Let C 1 = number of connected 1 components in G 1 . 1 2 2 2 1 Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Let G 1 = graph containing only 2 2 edges of weight 1. 2 2 2 1 2 Let C 1 = number of connected 1 components in G 1 . 1 2 2 Claim: MST contains example 2 C 1 -1 edges of weight 2. 1 Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Claim: MST contains example 2 2 C 1 -1 edges of weight 2. 2 2 2 1 2 Basic MST Property: 1 For any cut, minimum weight 1 2 edge across cut is in MST. 2 2 1 Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Claim: MST contains example 2 2 C 1 -1 edges of weight 2. 2 2 2 1 2 Algorithm: 1 For any connected component, 1 2 add minimum weight outgoing edge. 2 2 Here all the edges have weight 2, 1 so add C 1 -1 edges of weight 2. Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Claim: MST contains example 2 2 C 1 -1 edges of weight 2. 2 2 2 1 2 Weight of MST? 1 1 2 2 2 1 Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Claim: MST contains example 2 2 C 1 -1 edges of weight 2. 2 2 2 1 2 Weight of MST? 1 1 2 2 2 1 Ex: 10 + 6 – 2 = 14 Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Weight of MST: n + C 1 - 2 2 2 2 Algorithm idea? 2 2 1 2 1 1 2 2 2 1 Ex: C 1 = 6

Simple Minimum Spanning Tree Assume all weights 1 or 2 Weight of MST: n + C 1 - 2 2 2 2 Algorithm idea: 2 2 1 2 Approximate connected components of G 1 . 1 1 2 2 2 1 Ex: C 1 = 6

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Let G 1 = graph containing only 3 edges of weight 1. 2 2 2 1 2 Let G 2 = graph containing only 1 edges of weight {1, 2}. 3 … 2 3 Let G j = graph containing only 1 edges of weights {1, 2, … , j}. Ex: G 2

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Let C 1 = number CC in G 1 . 3 2 2 2 Let C 2 = number CC in G 2 . 1 2 … 1 3 Let C j = number CC in G j . 2 3 1 Ex: G 2

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Claim: 3 MST(G) contains C j – 1 edges 2 2 2 of weight > j. 1 2 1 3 2 3 1 Ex: G 2

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Claim: 3 MST(G) contains C j – 1 edges 2 2 2 of weight > j. 1 2 1 Why? 3 There are C j connected 2 components in G j . There 3 much be C j – 1 edges 1 connecting them, and each must have weight > j. Ex: G 2

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Lemma: 3 2 2 2 1 2 1 3 2 3 1 Ex: G 2

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Edges of weight 1: 3 2 2 2 n – 1 edges total in MST 1 2 C 1 – 1 edges of weight > 1 1  3 (n – 1) – (C 1 – 1) edges of 2 weight 1. 3  1 (n – C 1 ) edges of weight 1. Ex: G 2

Approximate Minimum Spanning Tree Weights {1, 2, … , W} 3 Edges of weight j+1: 3 2 2 2 C j – 1 edges of weight > j 1 2 C j+1 – 1 edges of weight > j+1 1  3 (C j – 1) – (C j+1 – 1) edges of 2 weight j+1. 3  1 (C j – C j+1 ) edges of weight j+1. Ex: G 2 Note: C j ≥ C j+1