Reversal of a graph

digraph_list <- list(c(2,3), 
                    c(3,6), 
                    c(4,5,6), 
                    c(1,5,6), 
                    c(2), 
                    c(1))
M <- graph_from_adj_list(digraph_list, mode="out") 

reversed_list <- list(c(4,6), 
                      c(1,5), 
                      c(1,2), 
                      c(3), 
                      c(3,4), 
                      c(2,3,4))
M <- graph_from_adj_list(reversed_list, mode="out") 

Reversal Algorithm

RevG(A[1..V])
   initialize B[1..V], same vertex names, empty linked lists
   for i <- 1 through V
      for each out-neighbor e in A[i]
         add i to the beginning of the linked-list B[e] 
   return B[1..V]

• Adding to beginning of linked list is constant time.
  • Adding to end of linked list requires traversing the whole list, which would increase the asymptotic running time.
• Processes each out-neighbor of each vertex exactly once.
• Time: \(O(V+E)\)

Begging the question

You are asked to explain why \(X\) implies \(Y\). “Begging the question” is an argument of the following form:

\(X\) implies \(Y\) because if \(X\) is true, then \(Y\) must be true.

• Example of begging the question: All prime numbers greater than 2 are odd, because, in order for \(n\) to be prime, it has to be odd, unless \(n = 2\).
• The English term is a mistranslation of the Latin petitio principii, which means “assuming the premise.”

Semiconnected graphs: False

IsSemiConnected(G):
   compute scc(G)
   let v_1..v_n be a topological sort of scc(G)
   for i in 1..n
      if v_i v_[i+1] is not an edge of scc(G)
         return False
   return True

\(a.\) The graph scc(G) is a DAG, meaning that you can’t loop back to an earlier component in the topological order, so there is not a path from v[i+1] to v[i] in scc(G). So if there also isn’t an edge from v[i] to v[i+1], there is no connection between v[i] and v[i+1]. This means that if u is a vertex of G that lies in the component v[i] and v is a vertex in G that lies in v[i+1], there can be no path from u to v, and no path from v to u. So the graph is not semi-connected.

Semiconnected graphs: True

IsSemiConnected(G):
   compute scc(G)
   let v_1..v_n be a topological sort of scc(G)
   for i in 1..n
      if v_i v_[i+1] is not an edge of scc(G)
         return False
   return True

\(b.\) The function can only return True if we’ve gone through all the components and all of them had an edge to the component after them. Let u and v be any two vertices of G. If u and v are in the same component, then there is a path between them because the component is strongly connected. Suppose u and v are in different components. If u.post > v.post, then u’s component comes before v’s in topological order, so there is a path from u’s component to v’s component in scc(G). Therefore there is a path from u to v in G. Similarly, if v.post > u.post, there is a path from v to u in G.

Semiconnected graphs: Time

IsSemiConnected(G):
   compute scc(G)
   let v_1..v_n be a topological sort of scc(G)
   for i in 1..n
      if v_i v_[i+1] is not an edge of scc(G)
         return False
   return True

• The running time of compute scc(G) is \(O(V+E)\).
• Topological sort costs \(O(V_\text{scc} + E_\text{scc})\).
• \(V_\text{scc} \in O(V)\) and \(E_\text{scc} \in O(E)\).
• The for-loop requires \(O(V_\text{scc} + E_\text{scc} )\)
• Total Time: \(O(V+E)\).

Study Guide on Canvas

• Questions on the Study Guide?
• See also the slide summary.

Table Groups