Week 11 - Friday

 What did we talk about last time?  Exam 2 post mortem  Cycle detection  Connectivity

Spellchecker

 Connected for an undirected graph: There is a path from every node to every other node  How can we determine if a graph is connected?

 Startup 1. Set the number of all nodes to 0 2. Pick an arbitrary node u and run DFS( u, 1 )  DFS( node v, int i ) 1. Set number(v) = i++ 2. For each node u adjacent to v If number(u) is 0 DFS( u, i )  If any node has a number of 0, the graph is not connected

 Weakly connected directed graph: If the graph is connected when you make all the edges undirected  Strongly connected directed graph: If for every pair of nodes, there is a path between them in both directions

 Components of a directed graph can be strongly connected  A strongly connected component is a subgraph such that all its nodes are strongly connected  To find strongly connected components, we can use a special DFS  It includes the notion of a predecessor node, which is the lowest numbered node in the DFS that can reach a particular node

 StrongDFS( node v, int i ) 1. Set number(v) = i++ 2. Set predecessor(v) to number(v) 3. Push(v) 4. For each node u adjacent to v If number(u) is 0 StrongDFS( u, i ) predecessor(v) = min(predecessor(v), predecessor(u)) Else if number(u) < number(v) and u is 0n the stack predecessor(v) = min(predecessor(v), number(u)) 5. If predecessor(v) is number(v) w = pop() while w is not v output w w = pop() output w

 StronglyConnected( Vertices V )  Set i = 1  For each node v in V Set number(v) = 0  For each node v in V If number(v) is 0 StrongDFS( v, i )

A B I D F H C E G

Student Lecture

 A directed acyclic graph (DAG) is a directed graph without cycles in it  Well, obviously  These can be used to represent dependencies between tasks  An edge flows from the task that must be completed first to a task that must come after  This is a good model for course sequencing  Especially during advising  A cycle in such a graph would mean there was a circular dependency

 A topological sort gives an ordering of the tasks such that all tasks are completed in dependency ordering  In other words, no task is attempted before its prerequisite tasks have been done  There are usually multiple legal topological sorts for a given DAG

 Give a topological sort for the following DAG:  A F I C G K D J E H A A H H E E J J D D K K G G C C F F I I

 Create list L  Add all nodes with no incoming edges into set S  While S is not empty  Remove a node u from S  Add u to L  For each node v with an edge e from u to v ▪ Remove edge e from the graph ▪ If v has no other incoming edges, add v to S  If the graph still has edges  Print "Error! Graph has a cycle"  Otherwise  Return L

 A bipartite graph is one whose nodes can be divided into two disjoint sets X and Y  There can be edges between set X and set Y  There are no edges inside set X or set Y  A graph is bipartite if and only if it contains no odd cycles

A A B B C C D D E E F F A A B B C C D D E E F F X Y

 A perfect matching is when every node in set X and every node in set Y is matched  It is not always possible to have a perfect matching  We can still try to find a maximum matching in which as many nodes are matched up as possible

1. Come up with a legal, maximal matching 2. Take an augmenting path that starts at an unmatched node in X and ends at an unmatched node in Y 3. If there is such a path, switch all the edges along the path from being in the matching to being out and vice versa 4. If there is another augmenting path, go back to Step 2

A A B B C C D D E E F F A A B B C C D D E E F F X Y Anna Becky CaitlinDaisyErinFiona Adam Ben CarlosDanEvanFred

 All 2n people want to get married  Each woman has ranked all n men in order of preference  Each man has ranked all n women in order of preference  We want to match them up so that the marriages are stable

 Consider two marriages:  Anna and Bob  Caitlin and Dan  This pair of marriages is unstable if  Anna likes Dan more than Bob and Dan likes Anna more than Caitlin or  Caitlin likes Bob more than Dan and Bob likes Caitlin more than Anna  We want to arrange all n marriages such that none are unstable

 n rounds  In each round, every unengaged man proposes to the woman he likes best (who he hasn’t proposed to already)  An unengaged woman must accept the proposal from the one of her suitors she likes best  If a woman is already engaged, she accepts the best suitor only if she likes him better than her fiance

 It’s a graph problem where we never look at a graph  The algorithm is guaranteed to terminate  A woman can’t refuse to get engaged  If a man gets dumped, he will eventually propose to everyone  The algorithm is guaranteed to find stable marriages  Unfortunately, it’s male-optimal and female- pessimal  The lesson: Women should propose to men

 Finish matching and stable marriage  Euler paths and tours

 Finish Project 3  Due tonight before midnight  Start Assignment 6  Due next Friday  Keep reading Chapter 8