Lecture 20 10/15/10

(Guest post by Jeffrey Chu)

Lemma2: If G is a graph, then G has a topological sorting.

Recall:

Topological Sorting: ordering of the vertices

v1, v2,….,vn

for all(vi,vj) in E i < j

Q: Given topological sorting

v1,….,vn

# of incoming edges to v1? answer: 0 because all edges go forward and v1 is the first node.

Lemma3: If G is a DAG then there always exist a v in V such that v has 0 incoming edges.

Algo:

Topo-Sort(G)

1) Find a v such that v has 0 incoming edges

2) Compute G’ = G\{v} <- remove v + all edges incident to v

3) Output: v, Topo-Sort(G)

Correctness of the algo

1) Can find v by Lemma3

3) Claim: G’ is algo a DAG

Prove by induction on # vertices in G.

-> n = 1 (checked)

-> n > 1

|v(G’)| = |v(G)| -1

Topo-Sort(G’) is a topological sorting for G’

Prove by Contradiction

prove by picture

-there is a cycle

-if there is a cycle in G’ then there is a cycle in G

-v(G’) subset v(G)

-E(G’) subset E(G)

Proof idea for Lemma3:

G is a DAG

For contradiction,

assume for all v in V

v has at least one incoming edge.

prove by picture:

-every v has incoming edge

w1 <- w2 <- w3 <- w4 ….. wn <- wn+1

as |v| = n

=> two wi & wj repeat            i.e. wi = wj

gives you a cycle, contradicts that G is a DAG

Running Time:

# calls to Topo-sort? answer: n times

Claim: step 1 + step 2 is O(n)

over all is O(n^2)

G -> adjacency list

check if incoming link list is empty

O(1) to check

so for n nodes its O(n) time (step 1)

(Step 2)maintain array Present[v] = T if u in v(G)