CPSC 490: Problem Solving in Computer Science Considering SCCs is - PowerPoint PPT Presentation

Lecture 4: 2-SAT, Trees, Minimum spanning tree Henry Xia, Brandon Zhang based on CPSC 490 slides from 2014-2018 2019-01-15 University of British Columbia CPSC 490: Problem Solving in Computer Science

• Considering SCCs is useful when answering questions about reachability. • We can partition directed graphs into strongly connected components. • We can compute strongly connected components in linear time with DFS! 1 Last time...

city, and pass through cities and roads multiple times. 2 Problem 1 – Tourist Input : A list of cities, connected by one-way roads. Output : whether it is possible to visit all the cities. It is allowed to start and end at any

• Consider the component graph (where we compress each SCC into a node). • Find the strongly connected components of the graph. • Check if there is one path that passes through all the nodes (with topological sort!) 3 Problem 1 – Solution

true, or determine it is impossible. We can solve this with SCC! 4 2-SAT Input : An expression of the form ( a ∨ b ) ∧ ( c ∨ d ) ∧ ( e ∨ f ) ∧ . . . , where a , b , c , . . . are of the form x i or ¬ x i . Output : An assignment to the boolean variables x 1 , x 2 . . . which makes the expression

Build a graph. • Convert each clause into these two implications. component graph. 5 2-SAT – Solution • There is a node representing each variable x i and a node for the negation ¬ x i . • For each clause ( a ∨ b ) : • ( a ∨ b ) is logically equivalent to ( ¬ a ⇒ b ) ∧ ( ¬ b ⇒ a ) . • For each implication u ⇒ v , make an edge u → v . Find the SCCs of this graph. There is a solution if and only if there is no x i such that x i and ¬ x i lie in the same SCC. If there is a solution, we can find a satisfying assignment to the x i by looking at the

6 Planks may be placed vertically or horizontally. Problem 2 – Wooden Beams Input : A r × c rectangular grid with n pegs, where wooden planks may be placed. Each plank has the same length 2 k + 1. The center of the plank is placed on the peg. Output : whether it is possible to place a plank on each peg without overlaps. Constraints : n , r , c ≤ 1000.

Reduce the problem to 2-SAT. • For each pair of pegs and each orientation the planks can be in, determine if they overlap. Each of these turns into a clause. • For example, if a vertical plank placed on the i th peg overlaps with a horizontal plank • We can place a plank on each peg with no overlaps if and only if there is a solution to this 2-SAT problem. 7 Problem 2 – Solution • Variable x i is true if the plank placed on the i th peg is vertical. placed on the j th peg, we should add the clause ¬ x i ∨ x j .

to work with. We wish to assign these people to K teams, each of which corresponds to a specialization. Each person must be assigned to a team corresponding to their specialization, and their team should only consist of people with whom they like to work with. 8 Problem 3 – Team Selection Input : N ≤ 100 people, each with two specializations and a list of people they would like Output : whether it’s possible to assign everyone to a team while keeping everyone happy.

Use 2-SAT! • If person i doesn’t like person j and they share a specialization, add a clause ensuring they won’t end up on the same team. 9 Problem 3 – Solution • The variables x i represent the team that the i th person is on.

A tree is: • A connected graph with no cycles. (These definitions are equivalent.) Some nice properties: • Unique simple path between any two nodes. • Adding any edge forms exactly one cycle. • Removing any edge splits the tree into two trees. Oħten, we consider rooted trees, where we assign one node to be the root. We can then talk about a node’s parent and children. 10 Trees • A connected graph with | V | − 1 edges. A leaf is a node in a tree with degree 1.

The center of a tree is the node which minimizes the distance to the farthest node from it. (There can be one or two centers.) 11 Problem 4 – Center of a tree Input : A tree with N ≤ 10 5 vertices. Output : the center of the tree.

Keep removing leaves until we only have one or two nodes leħt. How do we implement this? 12 Problem 4 – Solution Given the tree T , consider T ′ = T with all its leaves removed. Claim: the center of T ′ is the same as the center of T !

Keep removing leaves until we only have one or two nodes leħt. How do we implement this? 12 Problem 4 – Solution Given the tree T , consider T ′ = T with all its leaves removed. Claim: the center of T ′ is the same as the center of T !

13 6 The center is the last node (or last two nodes) processed. 12 11 10 9 8 Looks like topological sort! 7 5 3 1 4 2 Problem 4 – Solution degree = array of length N, storing the degree of each node q = new queue for u = 1 .. N: if degree[u] == 1: push(q, u) while q is not empty: u = pop(q) for v in neighbours(u): decrement degree[v] if degree[v] == 1: push(q, v)

contained in G where the sum of the edge weights is minimized/maximized. Given a (connected) graph G , a minimum/maximum spanning tree is a spanning tree How to find MSTs? It turns out we can be greedy! 14 Minimum spanning tree

15 4 9 8 7 6 Recall Dijkstra: 5 2 3 1 Dijkstra’s algorithm q = new priority queue, sorted by distance, containing (s, 0) while q is not empty: (u, cost) = pop(q) if u is visited: continue mark u as visited dist[u] = cost for each edge u → v with cost c: enqueue(q, (v, cost + c))

16 4 9 8 7 6 Uses the same greedy approach as Dijkstra. 5 2 3 1 Prim’s algorithm q = new priority queue, sorted by weight, containing (s, 0) while q is not empty: (u, cost) = pop(q) if u is visited: continue mark u as visited add this edge to the MST for each edge u → v with cost c: enqueue(q, (v, c))

Alternative greedy: add smallest (resp. largest) edges that don’t make a cycle. 1 2 3 4 5 (A union-find data structure lets us quickly check whether two nodes are in difgerent sets, and merge two sets together.) 17 Kruskal’s algorithm dsu = new union-find data structure for each edge (u, v), in order of weight: if u, v are in different sets: add this edge to the MST merge(u, v) Complexity: O ( | E | log | E | ) (assuming an effjcient implementation of union-find).

The bandwidth of a path is the minimum edge weight on the path. 18 Problem 4 – Bandwidth Input : A weighted graph with N ≤ 1000 nodes, and non-negative edge weights. Output : For every pair of nodes, the maximum-bandwidth path between them.

19 Build a maximum spanning tree! Problem 4 – Solution

Why does this work? Because of the greedy selection of edges! • Instead of picking this edge, we previously picked edges that form a u - v path. • These edges must have had greater weight, so there’s already a u - v path in the MST 20 Problem 4 – Solution • Suppose there’s an edge ( u , v ) with weight w uv that we don’t pick for the MST. with bandwidth ≥ w uv .

The cities must be connected by roads. Each contract proposes to build one road for some cost. You were bribed to give exactly K contracts to company A. 21 Problem 5 – Building Roads Input : N cities, and a set of contract proposals for roads from two companies. Output : The minimum cost to build roads between the cities.

Connect the graph with minimum total cost – MST! But, we can’t enforce that we use K of company A’s roads. • Suppose we choose more that K roads from company A. • Apply a constant penalty P to each of company A’s proposals. • Run MST on the new set of proposals. • Similarly, if we choose fewer than K roads then apply an incentive P . • Still don’t use K roads? Change P and try again! • Binary search on P . 22 Problem 5 – Solution

Some edge cases to watch for: • Tree with zero or one node • Tree which is a stick • Tree with very large breadth • “Broomstick” tree: large number of sticks glued together 23 Watch out!