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Call each eyeball visit a "stage" We prove by induction that, after any stage, the values of Dl for vertices visited by the eyeball form the shortest path and that the values of Dl for the other verticesform the shortest path using only vertices visited by the eyeball as intermediates Because the first vertex visited is the starting vertex, this statement is correct through the first stage Assume that it is correctfor the first k stages Let v be the vertex chosen by the eyeball in stage k + 1 Suppose, for the purpose of showing a contradiction, that there is a path from S to v of length less than D, This path must go through an intermediate vertex that has not yet been visited by the eyeball Call the first intermediate vertex on the path not visited by the eyeball u This situation is shown in Figure 1525 The path to u uses only vertices visited by the eyeball as intermediates, so by induction D, represents the optimal distance to u Moreover; D, < D, because u is on the supposed shorter path to v This inequaliry is a contradiction because then we would have moved the eyeball to u instead of v The proof is completed by showing that all the D, values remain correct for nonvisited nodes, which is clear by the update rule

Figure 1525 If D, is minimal among all unseen vertices and if all edge costs are nonnegative, DL, represents the shortest path

L(Yi -

where we have used (12.29). Also we are working in the relativistic limit me = 0 and have made use of

Figure 1526 shows the stages of Dijkstra's algorithm The remaining issue is the selection of an appropriate data structure For dense graphs, we can scan down the graph table looking for the appropriate vertex As with the unweighted shortest-path algorithm, this scan will take O(I V12, time, which is optimal for a dense graph For a sparse graph, we want to do better

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Certainly, a queue does not work The fact that we need to find the vertex v with minimum D, suggests that a priority queue is the method of choice There are two ways to use the priority queue One is to store each vertex in the priority queue and use the distance (obtained by consulting the graph table) as the ordering function When we alter any D,, we must update the priority queue by reestablishing the ordering property This action amounts to a decreaseKey operation To take it we need to be able to find the location of w in the priority queue Many implementations of the priority queue, including the STL priority queue, do not support decreaseKeyOne that does is the pairing heap; we discuss use of the pairing heap for this application in 23 Rather than use a fancy priority queue, we use a method that works with the STL priority queue Our method involves inserting an object consisting of w and D , in the priority queue whenever we lower D, To select a new vertex v for visitation, we repeatedly remove the minimum item (based on distance) from the priority queue until an unvisited vertex emerges Because the size of the priority queue could be as large as I El and there are at most (El priority queue insertions and d e l e t i o n s , the running time i s O(IEl1oglEl) Because IEl I /VI2 implies loglEl 5 2log/VI, we have the same O(IElloglVI) algorithm that we would have if we used the first method (in which the priority queue size is at most IV/)

The priority queue is an appropriate data structureThe easiest method is to add a new entry, consisting of a vertex and a distance, to the priority queue every time a vertex has its distance lowered We can find the new vertex to move to by repeatedly removing the minimum distance vertex from the priority queue until an unvisited vertex emerges

s = (k + p)2 = 2k P = 2k ' p'.

A unary binder adapter is used to generate a unary function from a binary function by making constant one of the parameters to the binary function

3.11a. To show that R2 can be expressed as a function of F, let Sf = yy, the sum of squared residuals in the full model, and let Sr = L(Yi - y)2, the sum of squared residuals in the reduced model Y = f30 + e with

Our discussion of unary binder adapters suggested an application in which we perform a search, basing matches on a predicate This is precisely what ind-i does

(12.58)

Iterator find-if( Iterator begin, Iterator end, Predicate gred )

returns an iterator representing the first object that satisfies predThe search range is from begin to end,not including end If no match is found, end is returned To illustrate how f ind-if is used, let us suppose that v is a vector<int> and that we want to find the first occurrence of an item that is less than 50 We can easily define a function object class that expresses this condition:

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