By K. Mehlhorn

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1. that Min, Deletemin take time O(a log n/log a) and Insert, Demote* take time O(log n/log a) if PQ is realized as an unordered (a,2a)-tree for integer a choice of a? Note that loop (3) - ~ 2. What is a good (12) is executed at most n times since no vertex is added to U twice by our choice of line (5"). Hence line (5") is executed at most n times. Moreover, lines (7) to (9) are executed at most once for each edge (u,v) E E. Hence the total running time of the algorithm is O(an log n/log a + e log n/log a) Theorem 2: The single source least cost path problem on networks with non-negative edge costs can be solved in time a) 0(n 2 ) b) O(e log n) c) O(e log n/max(1, log e/n» Proof: a) Choose a = n.

Then each edge of E' is used at most once in this cycle. It cannot contain an edge of E' - E because then the length of the cycle were at least large ~ O. Hence N contains a cycle of negative cost. 3. to find out whether N' hence N) has a cycle of negative cost and if not to determine Ic(u,v) I (and ~(s,v) for all v E V. In the first case the algorithm stops, in the second case we use a(v) = ~(s,v) to transform the all pairs problem on a gen- eral network into a set of n single source problems on a non-negative network.

Lemma 1: a) ~(s,v) > -co for all v E V iff the algorithm terminates. b) If the algorithm terminates then ~(s,v) = COST [v] for all v E V after termination. Proof: a) II~II: The following claim is easily proved by induction on the number of iterations of the loop. Claim: Before any execution of line (6) the following is true: PATH[v] is a path of cost COST[v] from s to v and if PATH[v] = (vO, .... • ,v i ) was the content of PATH[v i ] previously. Proof: Immediate from the formulation of the algorithm.