algorithm - Graph Connectivity Proof -


suppose have adjacency matrix g representing n-node graph. g[i, j] 1 if there's edge j , 0 otherwise. can think of each entry turned on slip of paper 1 or 0 on it.

my text claims lower bound on number of times need query matrix determine if graph connected n * (n-1) / 2. i'm bit confused proof of this.

proof: here strategy adversary: when algorithm asks flip on slip of paper, return 0 unless force graph disconnected.

1) seems imply in cases, if return 0, it's impossible add edges later lead path u v. surely graph can still connected if return 0 above...right?

claim: maintain invariant y un-asked pair (u, v) graph revealed far has no path u v

proof: suppose there path u v. can remove last edge (u', v') revealed on path. have answered 0 , kept same connectivity in graph having edge (u, v). contradicts definition of our adversary strategy.

end of proof: suppose there algorithm finished without examining every slip of paper. consider unasked pair (u, v). if algorithm claims graph connected, show all-zeroes remaining unasked edges and means there no path u v, algorithm wrong.

2. explain italics? i'm not seeing connection...

on other hand, if algorithm says disconnected, show all-ones remaining edges , algorithm wrong definition of our adversary strategy.

3. couldn't graph still disconnected if show ones?

thanks!

  1. suppose have four-node graph. if algorithm has queried , gotten "0" answers 13 , 14, , asks 12, if answer "0", means graph cannot connected, because 1 not connected 2 (there's no direct path, , 1 has no other possible connection).

  2. at given time, can think of edge being "yes", "maybe", or "no". edges "maybes". whenever algorithm asks edge, becomes "yes" or "no". graph yes , maybe edges connected input consistent observations. graph yes edges least connected input consistent observations. algorithm can return when these graphs have same connectivity status.


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