Discrete Mathematics with Graph Theory (3rd Edition) 346

# Discrete Mathematics with Graph Theory (3rd Edition) 346 -...

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344 Solutions to Exercises od: for j to n do indeg[i] .- indeg[i] + A[j,i]: > t := 1: > found := 0: #found = 1 when we have found #the next vertex > while(t <= n) do i := 0: found := 0: while((found =0) and (i <n)) do i := i+1: if (member (i,V)=true) then fi: if (indeg[i]=O) v[t] := i; t := t+1: found .- 1: V := V minus {i}; in V do #remove vertex i #find indegrees for if (A[i,j]=l) indeg[j] indeg [ j ] - 1: S := seq(v[i],i=l .. n): lprint(/A canonical labeling is , S); new V 5. (a) [BB] If g has a cycle, then there is no canonical labeling so (with notation as in Theorem 12.4.3) some S = V " {vo, VI, V2, .•• , Vk} has the property that Sv #- 0 for all V E S; that is, for every v E S, there exists an arc of the form xv with xES. Let Uo E S be arbitrary and choose UI E S such that UI is an arc. Choose U2 E S such that
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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