Contents. shortest paths. Notation. Shortest path problem. Applications. Algorithms and Networks 2010/2011. In the entire course:

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1 Content Shortet path Algorithm and Network 21/211 The hortet path problem: Statement Verion Application Algorithm (for ingle ource p problem) Reminder: relaxation, Dijktra, Variant of Dijktra, Bellman-Ford, Johnon Scaling technique (Gabow algorithm) Variant algorithm: A*, bidirectional earch Bottleneck hortet path 1 2 Notation In the entire coure: n = V, the number of vertice m = E or m = A, the number of edge or the number of arc 1 Definition and Application 3 4 Shortet path problem Application (Directed) graph G= (V,E), length for each edge e in E, w(e) Ditance from u to v: length of hortet path from u to v Shortet path problem: find ditance, find hortet path Verion: All pair Single pair Single ource Single detination Length can be All equal (unit length) (BFS) Non-negative Negative but no negative cycle Negative cycle poible Subroutine in other graph algorithm Route planning Difference contraint Allocating Inpection Effort on a Production Line 5 6 1

2 7 Allocating Inpection Effort on a Production Line Production line: ordered equence of n production tage. Item are produced in batche of B > item. Probability that tage i produce a defect item i a i. Manufacturing cot at tage i: p i. Cot of inpecting at tage j, when lat inpection ha been done at tage i: f ij per batch, plu g ij per item in the batch When hould we inpect to minimize total cot? 8 Solve by modeling a hortet path problem w( i, j) = f ij + B( i) g + B( i) ij j p k k= i+ 1 Where B(i) denote the expected number of non-defective item after tage i B( i) = B i k = 1 (1 a k ) 9 Application: Difference contraint Tak with precedence contraint and running length Each tak i ha Time to complete b i > Some tak can be tarted after other tak have been completed: Contraint: j + b j i Firt tak can tart at time. When can we finih lat tak? Shortet path problem on directed acyclic graph! 1 Modeling a longet path problem Take one vertex v (model time ) Vertex for each tak Arc (v, i) for each tak vertex i For each precedence contraint j + b j i an arc (j, i) with length b j Optimal: tart each tak i at time equal to length of longet path from v to i. Difference contraint a hortet path The longet path problem can be olved in O(n+m) time, a we have a directed acyclic graph. Tranforming to hortet path problem: multiply all length and time by 1. 2 Algorithm for hortet path problem (reminder)

3 Bai of ingle ource algorithm Source. Each vertex v ha variable D[v] Invariant: d(,v) D[v] for all v Initially: D[]=; v : D[v] = Relaxation tep over edge (u,v): D[v] = min { D[v], D[u]+ w(u,v) } Maintaining hortet path Each vertex maintain a pointer to the `previou vertex on the current hortet path (ometime NIL): p(v) Initially: p(v) = NIL for each v Relaxation tep become: Relax (u,v,w) If D[v] > D[u]+ w(u,v) then D[v] = D[u] + w(u,v); p(v) = u p-value build path of length D(v) Shortet path tree Dijktra On Dijktra Initialize Take priority queue Q, initially containing all vertice While Q i not empty, Select vertex v from Q of minimum value D[v] Relax acro all outgoing edge from v Note: each relaxation can caue a change of a D- value and thu a change in the priority queue Thi happen at mot E time Aume all length are non-negative Correctne proof (done in `Algoritmiek ) Running time: Depend on implementation of priority queue O(n 2 ): tandard queue O(m + n log n): Fibonacci heap O((m + n) log n): red-black tree, heap Negative length What if w(u,v) <? Negative cycle, reachable from Bellman-Ford algorithm: For intance without negative cycle: In O(nm) time: SSSP problem when no negative cycle reachable from Alo: detect negative cycle Bellman-Ford Clearly: O(nm) time Initialize Repeat V -1 time: For every edge (u,v) in E do: Relax(u,v,w) For every edge (u,v) in E do If D[v] > D[u] + w(u,v) then There exit a negative circuit! Stop There i no negative circuit, and for all vertice v: D[v] = d(,v)

4 Correctne of Bellman-Ford Invariant: if no negative cycle i reachable from, then after i run of main loop, we have: If there i a hortet path from to u with at mot i edge, then D[u]=d[,u], for all u. If no negative cycle reachable from, then every vertex ha a hortet path with at mot n 1 edge. If a negative cycle reachable from, then there will alway be an edge with a relaxation poible. Finding a negative cycle in a graph Reachable from : Apply Bellman-Ford, and look back with pointer Or: add a vertex with edge to each vertex in G. G 19 2 All pair Reweighting Dynamic programming: O(n 3 ) (Floyd, 1962) Johnon: improvement for pare graph with reweighting technique: O(n 2 log n + nm) time. Work if no negative cycle Obervation: if all weight are non-negative we can run Dijktra with each vertex a tarting vertex: that give O(n 2 log n + nm) time. What if we have negative length: reweighting Let h: V R be any function to the real. Write w h (u,v) = w(u,v) + h(u) h(v). Lemma: Potential Let P be a path from x to y. Then w h (P) = w(p) + h(x) h(y). P i a hortet path from x to y with length w, if and only if it i o with length w h. G ha a negative-length circuit with length w, if and only if it ha a negative-length circuit with length w h What height function h i good? Look for height function h with w h (u,v), for all edge (u,v). If o, we can: Compute w h (u,v) for all edge. Run Dijktra but now with w h (u,v). Special method to make h with a SSSP problem, and Bellman-Ford. G

5 Chooing h Set h(v) = d(,v) (in new graph) Solving SSSP problem with negative edge length; ue Bellman-Ford. If negative cycle detected: top. Note: for all edge (u,v): w h (u,v) = w(u,v) + h(u) h(v) = w(u,v) + d(,u) d(,v) Johnon algorithm Build graph G (a hown) Compute with Bellman-Ford d(,v) for all v Set w h (u,v) = w(u,v) d(,u) + d(,v) for all edge (u,v). O(n For all u do: 2 log n + nm) time Ue Dijktra algorithm to compute d h (u,v) for all v. Set d(u,v) = d h (u,v) + h(v) h(u) Uing the number 3 Shortet path algorithm uing the number and caling Back to the ingle ource hortet path problem with non-negative ditance Suppoe Δ i an upper bound on the maximum ditance from to a vertex v. Let L be the larget length of an edge. Single ource hortet path problem i olvable in O(m + Δ) time In O(m+Δ) time Corollary and extenion Keep array of doubly linked lit: L[],, L[Δ], Maintain that for v with D[v] Δ, v in L[D[v]]. Keep a current minimum µ. Invariant: all L[k] with k µ are empty Changing D[v] from x to y: take v from L[x] (with pointer), and add it to L[y]: O(1) time each. Extract min: while L[µ] empty, µ++; then take the firt element from lit L[µ]. Total time: O(m+Δ) SSSP: in O(m+nL) time. (Take Δ=nL). Gabow (1985): SSSP problem can be olved in O(m log R L) time, where R = max{2, m/n} L : maximum length of edge Gabow algorithm ue caling technique!

6 Gabow algorithm Main idea Firt, build a caled intance: For each edge e et w (e) = w(e) / R. Recurively, olve the caled intance. Another hortet path intance can be ued to compute the correction term! How far are we off? We want d(,v) R * d w (,v) i when we cale back our caled intance: what error did we make when rounding? Set for each edge (x,y) in E: Z(x,y) = w(x,y) R* d w (,x) + R * d w (,y) Work like height function, o the ame hortet path! Height of x i R * d w (,x) A claim Gabow algorithm For all vertice v in V: d(,v) = d Z (,v) + R * d w (,v) A with height function (telecope): d(,v) = d Z (,v) + h() h(v) = d Z (,v) R*d w (,) + R * d w (,v) And d w (,) = Thu, we can compute ditance for w by computing ditance for Z and for w Bae cae: ue O(m+nR) algorithm if L <= R. Otherwie: For each edge e: et w (e) = w(e) / R. Recurively, compute the ditance but with the new length function w. Set for each edge (u,v): Z(u,v) = w(u,v) + R* d w (,u) R * d w (,v). Compute d Z (,v) for all v and then ue d(,v) = d Z (,v) + R * d w (,v) A Property of Z Computing ditance for Z For each edge (u,v) E we have: Z(u,v) = w(u,v) + R* d w (,u) R * d w (,v), becaue w(u,v) R * w (u,v) R * (d w (,v) d w (,u)). So, a variant of Dijktra can be ued to compute ditance for Z. For each vertex v we have d Z (,v) nr for all v reachable from Conider a hortet path P for ditance function w from to v For each of the le than n edge e on P, w(e) R + R*w (e) So, d(,v) w(p) nr + R* w (P) = nr + R* d w (,v) Ue that d(,v) = d Z (,v) + R * d w (,v) So, we can ue O(m+ nr) algorithm (Dijktra with doubly-linked lit) to compute all value d Z (v)

7 Gabow algorithm (analyi) Example Recurive tep: cot O( m log R L ) with L = L/R. SSSP for Z cot O(m + nr) = O(m). Note: log R L (log R L) 1. So, Gabow algorithm ue O(m log R L) time. 191 a b 116 t A* (implified expoition) 4 Variant: A* and bidirectional earch Practical importance: A* algorithm Can be explained in term of height function Conider a route planning problem with geographic data, and ditance of arc at leat Euclidian ditance of vertice Suppoe we have a ingle pair hortet path problem, with ource and target t Ue height function - h(v) = - the Euclidian ditance from v to t (notate: vt For all arc (v,w) E: w (v,w) = w(v,w) - vt + wt vw - vt + wt Note: arc toward target get maller length and arc away from target larger length Algorithm i fater in practice but till correct 39 4 Bidirectional earch For a ingle pair hortet path problem: Start a Dijktra-earch from both ide imultaneouly Analyi needed for topping criterion Fater in practice Combine nicely with A* 5 Bottleneck hortet path t t

8 Bottleneck hortet path Given: weighted graph G, weight w(e) for each arc, vertice, t. Problem: find a path from to t uch that the maximum weight of an arc on the path i a mall a poible. Or, revere: uch that the minimum weight i a large a poible: maximum capacity path Algorithm On directed graph: O((m+n) log m), Or: O((m+n) log L) with L the maximum abolute value of the weight Binary earch and DFS On undirected graph: O(m+n) with divide and conquer trategy Bottleneck hortet path on undirected graph Find the median weight of all weight of edge, ay r. Look to graph G r formed by edge with weight at mot r. If and t in ame connected component of G r, then the bottleneck i at mot r: now remove all edge with weight more than r, and repeat (recurion). If and t in different connected component of G r : the bottleneck i larger than r. Now, contract all edge with weight at mot r, and recure. T(m) = O(m) + T(m/2) 5 Concluion Concluion Application Several algorithm for hortet path Variant of the problem Detection of negative cycle Reweighting technique Scaling technique A*, bidirectional Bottleneck hortet path 47 8