1. Supplementary: Feasible Labels and Linear Programming  Consider the LP formulation of the shortest s-t path problem: (P) Minimize  (i, j)  A c ij x ij subject to  {j: (i, j)  A} x ij -  {j: (j, i)  A} x ji = +1, for i = s = -1, for i = t = 0, for i  N\{s, t} x ij  0 for all (i, j)  A It is guaranteed that the optimal solution to (P), if exists, is integer-valued (proof later when we study MCF). Hence we can solve the LP instead of IP.  Multiply (-1) on both sides of the constraints, then get  {j: (j, i)  A} x ji -  {j: (i, j)  A} x ij = -1, for i = s = +1, for i = t = 0, for i  N\{s, t} i.e consider (inflow to i) – (outflow from i) Network Theory and Applications 2010 1

2. (P) Minimize  (i, j)  A c ij x ij (S-1) subject to  {j: (j, i)  A} x ji -  {j: (i, j)  A} x ij = -1, for i = s = +1, for i = t = 0, for i  N\{s, t} x ij  0 for all (i, j)  A  The dual of (P) is (D) maximize y t – y s (S-2) subject to y j - y i  c ij for all (i, j)  A  Note that, for a dual feasible solution y, we may subtract any constant  from each dual variable and we still have a dual feasible solution with the same objective value.  we may set y s = 0. Network Theory and Applications 2010 2

3.  We assumed there exists a s-t path in G, hence (P) is feasible. If (P) has a finite optimal value (with directed s-t walk), then by strong duality, (D) is also feasible and its optimal value is the same as (P) (shortest s-t distance = y t ) Conversely, if (D) has a feasible solution, then (P) and (D) have the same optimal value (since (P) is feasible too). (Compare with Thm 5.1)  If the network contains a negative cycle, (P) is unbounded, hence (D) is infeasible, i.e. no y (d(  )) satisfies (5.2) Network Theory and Applications 2010 3

4.  The constraint matrix in (P) has special structure; each column has exactly one 1 and one –1 (except nonnegativity constraints). Adding the rows of the matrix A, we get 0 vector, which implies the rows are linearly dependent (rank is  n-1).  Prop : Let G be a connected directed graph and D = {a e : e  A} be its node- arc incidence matrix ( a e is the column vector for arc e in the constraint matrix). A set {a e : e  T} is a column basis of D iff T is the arc-set of a spanning tree of G (not considering the directions of the arcs). ( G has n nodes, m arcs. D is n  m matrix. Spanning tree has n-1 arcs.) Network Theory and Applications 2010 4

5.  So the rank of D is n-1. To solve the problem by the simplex method, we need nonsingular basis B. Two approaches:  Drop one of the constraints and solve the full row rank problem  Add an artificial variable to any one constraint (objective coefficient is 0). It makes D full row rank (the artificial variable always remains in the basis and has value 0 in any b.f.s).  b.f.s. and dual vector satisfies x e = 0 for all e  T y j – y i = c ij for all (i, j)  T ( From y’B = c B ’ for current basis B) This y vector gives the path length to each node j in the current tree T (If T is directed out-tree). Note that if we add artificial variable for node s (root), then we have y s = 0 in any b.f.s. ( from y’B = c B ’) Network Theory and Applications 2010 5

6.  A b.f.s. for (P) does not necessarily correspond to an directed out-tree. It may be just a spanning tree. However, the simplex iteration can be modified so that we always have directed out-tree, once we started the algorithm with a directed out-tree.  Each step of the simplex method may correspond to multiple Ford’s step; (1) Change of directed out-tree (2) update the node labels so that it gives the actual path length in the directed out-tree. Network Theory and Applications 2010 6

7.  Suppose we solve s-t shortest path problem by LP (with simplex method) and have a feasible directed out-tree (bfs);  entering nonbasic arc selection: choose arc (i, j) with c ij – (y j – y i ) < 0, i.e. c ij + y i < y j. ( from c j – y’A j < 0 since we solve min problem )  leaving basic variable: select the entering arc to node j in the current directed out- tree as leaving basic arc. (updating tree in Ford algorithm)  Update flow (x-value) so that we have value 1 for each arc on the s-t path.  Note that we may solve the shortest s-t path problem by LP. But, in an optimal basis (directed out-tree), we have shortest s-j path, for all j  N. ( If you use the above modified simplex method) ( We have feasible potential (label) y and the s-j paths in the directed out-tree for all j  N gives actual paths with length y j for all j  V.) Network Theory and Applications 2010 7

8.  The identification of the basis of the LP as a spanning tree (not necessarily directed out-tree) is used in the network simplex algorithms for the minimum cost network flow problem. (Chapter 11) Network Theory and Applications 2010 8

9. Other Interpretation of Ford-Bellman Algorithm  Reference: Combinatorial Optimization: Networks and Matroids, E. Lawler  Bellman’s equations: u 1 = 0 u j = min k  j { u k +a kj }, j=2, 3, …, n (necessary and sufficient conditions for shortest paths when G does not contain a nonpositive cycle.)  Solve Bellman’s eq. iteratively. Still effective if no negative cycle in G.  Initial values u 1 (1) = 0 u j (1) = a 1j, j  1 Iteration u j (m+1) = min { u j (m), min k  j { u k (m) + a kj }}  Interpretation u j (m) = the length of a shortest path from the origin to j, subject to the condition that the path contains no more than m arcs. (compare it with the presentation given in the text) Network Theory and Applications 2010 9

10. 5.6 All-Pairs Shortest Path Probelm  Assume G is strongly connected. (add (s, i) and (i, s) for i  N\{s} with large costs if not exist)  Two approaches:  Repeated shortest path algorithm (suitable for sparse graph)  All-pairs label-correcting algorithm (suitable for dense graph)  Repeated Shortest Path Algorithm: Let S(n, m, C) denote the time needed to solve a shortest path problem with nonnegative arc lengths. Select a node s as a source node and use FIFO label-correcting algorithm. Then, have c ij d = c ij + d(i) – d(j)  0 for all arcs (i, j). Using c ij d  0, solve single-source shortest path problem (n-1) times. From Property 5.2.(b) (  (i, j)  P c ij d =  (i, j)  P c ij + d(k) – d(l)), actual shortest path length from k to l is (  (i, j)  P c ij d + (d(l) – d(k) ) Running time: O(nm + nS(n, m, C)) = O(nS(n, m, C)). Network Theory and Applications 2010 10

11.  All-Pairs Shortest Path Optimality Conditions:  Thm 5.5. Let d[i, j] represent the length of some directed path from node i to node j. Then d[i, j] optimal  d[i, j]  d[i, k] + d[k, j] for all i, j, and k. (5.3) Pf)  ) If d[i, k] + d[k, j] < d[i, j], use path from i → k → j walk. Contradiction to the optimality of d[i, j].  ) d[i, j] length of some path  upper bound on the shortest length Let i = i 1 – i 2 – i 3 - … - i k = j be any i-j path P  d[i, j] = d[i 1, i k ]  d[i 1, i 2 ] + d[i 2, i k ]  c i(1)i(2) + d[i 2, i k ], d[i 2, i k ]  c i(2)i(3) + d[i 3, i k ], … d[i k-1, i k ]  c i(k-1)i(k).  d[i, j]  c i(1)i(2) + c i(2)i(3) + … + c i(k-1)i(k) =  (i, j)  P c ij. Hence d[i, j] is a lower bound on any i-j path. d[i, j] also upper bound, so optimal. Network Theory and Applications 2010 11

12.  All-Pairs Generic Label Correcting Algorithm: algorithm all-pairs label-correcting; begin set d[i, j] :=  for all [i, j]  N  N; set d[i, i] := 0 for all i  N; for each (i, j)  A do d[i, j] := c ij ; while the network contains three nodes i, j, and k satisfying d[i, j] > d[i, k] + d[k, j] do d[i, j] := d[i, k] + d[k, j]; end; Network Theory and Applications 2010 12

13.  Correctness: Whenever d[i, j] < ,  walk of length d[i, j] from i to j. The walk may decompose into a path P and some directed cycles. At termination of the algorithm, all of the directed cycles must have length 0, hence the path length is d[i, j]. Also labels d[i, j] satisfy the optimality condition (5.3).  Finiteness: Distance label is bounded from above by nC, from below by –nC. Each iteration decreases some d[i, j]. Hence, total O(n 3 C) iterations.  O(n 3 ) implementation: Floyd-Warshall Algorithm Network Theory and Applications 2010 13

14.  Floyd-Warshall Algorithm: d k [i, j]: i-j shortest path length using only the nodes 1, 2, …, k-1 as internal nodes. Then, d n+1 [i, j] represents shortest path length from i to j. Starting from d 1 [i, j], update the labels recursively.  Property 5.6. d k+1 [i, j] = min {d k [i, j], d k [i,k] + d k [k, i]}. Two cases: 1.Do not use node k. Then d k+1 [i, j] = d k [i, j] 2.Use node k. Then d k+1 [i, j] = d k [i,k] + d k [k, i] Network Theory and Applications 2010 14

15.  Algorithm Floyd-Warshall; begin for all node pairs [i, j]  N  N d[i, j] :=  and pred[i, j] := 0; for all nodes i  N do d[i, i] := 0; for each arc (i, j)  A do d[i, j] := c ij and pred[i, j] := i; for each k := 1 to n do for each [i, j]  N  N do if d[i, j] > d[i, k] + d[k, j] then begin d[i, j] := d[i, k] + d[k, j]; pred[i, j] := pred[k, j]; end; Network Theory and Applications 2010 15

16.  View as operation on matrix: Network Theory and Applications 2010 16 d[i, j] d[k, j] d[i, k] k k  If there exists no negative cycle, d k [i, i] = 0 for all i, k Hence row k, column k do not change.  maintain only one copy of the matrix. If there exists a negative cycle  d[i, i] < 0 for some i.

17.  Retrieving actual paths: pred[i, j] denotes the last node prior to node j in the tentative shortest path from node i to node j. Consider a path P from node k to node l. Starting from node l, let g=pred[k, l]. Then g is the last node prior to node l in P. Similarly, h = pred[k, g] is the node prior to node g in P, and so on. Repeat until we reach node k.  Thm 5.7. The Floyd-Warshall algorithm runs in O(n 3 ). Network Theory and Applications 2010 17

18.  Detection of negative cycles: Two tests for negative cycle:  If i = j, check whether d[i, i] < 0.  If i  j, check whether d[i, j] < -nC.  Use predecessor graph to retrieve negative cycle. Network Theory and Applications 2010 18

19. Other algebraic view  Reference: Combinatorial Optimization: Networks and Matroids, E. Lawler  Let d[i, j] = length of a shortest path from i to j. d k [i, j] = length of a shortest path from i to j using  k arcs. d 0 [i, i] = 0, d 0 [i, j] = +  (i  j) d k+1 [i, j] = min l { d k [i, l] + c lj } Have d (n-1) [i, j] = d[i, j]  Define the new matrix multiplication  as P = A  B, P ij = min k {a ik + b kj } We have D k+1 = D k  C D (0) = is identity matrix D (0)  C = C multiplication  is associative. D n-1 = D n-2  C = … = ((D 0  C)  C) …  C) = C n-1 Also C m = C n-1, m  n-1. Network Theory and Applications 2010 19

20.  Can compute C n-1 as C → C 2 → C 4 → … → C 2(k), 2 k  n-1. log(n) steps  O(n 3 log n) Floyd-Warshall is a clever implementation of matrix multiplcation. Network Theory and Applications 2010 20

21. 5.7 Minimum Cost-To-Time Ratio Cycle Problem  Tramp steamer problem: Identify a directed cycle W of G with the maximum ratio (  (i, j)  W p ij )/(  (i, j)  W  ij ) Let cost c ij of arc (i, j) as c ij = -p ij, and identify W with minimum ratio. Assume data are integral and  ij  0 for (i, j)  A and  (i, j)  W  ij > 0 for every directed cycle W in G. Let  (W) = (  (i, j)  W c ij )/(  (i, j)  W  ij ) and  * is optimal value. Starting from arbitrary value of , define the length of each arc as l ij = c ij -  ij. Then use negative cycle detection algorithm. Network Theory and Applications 2010 21

22.  Three cases: (1) G contains a negative length cycle W. have  (i, j)  W (c ij -  ij ) < 0. Alternatively,  > (  (i, j)  W c ij )/(  (i, j)  W  ij )   *. Therefore,  is a strict upper bound on  *. (2) G contains no negative cycle, but does contain a zero-length cycle W*. (see Exercise 5.19 on how to identify a zero-length cycle) no negative cycle implies  (i, j)  W (c ij -  ij )  0 for every directed cycle W. Alternatively,   (  (i, j)  W c ij )/(  (i, j)  W  ij ) for every directed cycle W. Also, existence of zero-length cycle W* implies  = (  (i, j)  W* c ij )/(  (i, j)  W*  ij ). Hence  =  *, and W* is a minimum cost-to-time ratio cycle. (3) Every directed cycle W in G has a positive length. have  (i, j)  W (c ij -  ij ) > 0 for every directed cycle W. Alternatively,  < (  (i, j)  W c ij )/(  (i, j)  W  ij ) for every directed cycle W. Hence,  is a strict lower bound on  *. Network Theory and Applications 2010 22

23.  Search procedures: Guess a value  and use shortest path algorithm If negative cycle,    *. Use smaller value of . If zero-length cycle, then optimal. If positive length cycle,  <  *. Use larger value of .  Sequential search algorithm: Let  0 be a known upper bound on  *. Solve the shortest path problem with (c ij -  0  ij ) as arc lengths. Either find a zero-length cycle W or a negative cycle W. In the former case, optimal found, stop. In the latter case, let  1 = (  (i, j)  W c ij )/(  (i, j)  W  ij ) as our next guess. Have  0 >  1   *. Obtain sequence of values  0 >  1 … >  k =  *. Network Theory and Applications 2010 23

24.  Binary search algorithm: Let [  L,  U ] be the interval that contains  *. (e.g. [-C, C]) Use  0 = (  L +  U )/2, and check for negative cycle. If a negative cycle exists,  0 >  *. Reset  U =  0. Otherwise,  0   *. Reset  L =  0. After sufficiently large number of iterations, the search interval has a unique solution.  Let c(W) and  (W) denote the cost and travel time of any directed cycle W, and let  0 = max{  ij : (i, j)  A}. Claim: The interval of size at most 1/  0 2 contains at most one value from the set {c(W)/  (W): W is a directed cycle} Let W 1 and W 2 be two directed cycles with distinct ratios. Then |c(W 1 )/  (W 1 ) - c(W 2 )/  (W 2 )|  0, or | {c(W 1 )/  (W 2 ) - c(W 2 )/  (W 1 )} / {  (W 1 )  (W 2 )} |  0. Smallest value of the left-hand side is 1/  0 2. Hence if the interval becomes smaller than 1/  0 2, using  U as the value of  identifies the optimal cycle W. Network Theory and Applications 2010 24

25. Network Theory and Applications 2010 25  Handout: Combinatorial Optimization: Networks and Matroids, E. Lawler  Enumerate simple paths in nondecreasing order with respect to the weights.  Let P be the set of all 1-n paths. Find shortest path P 1 initially. Modify the lengths using c d ij  0. Partition P – {P 1 } into subsets P 1, P 2, …, P k, where k  n-1, so that we can determine the shortest path in each subset. Let P 2 be the shortest of these shortest paths identified. Suppose P 2  P j. We then partition P j - {P 2 } into subsets in the same way, and so on.  The procedure can be described as a tree. How to partition the subset so that we do not lose any path and we can find a shortest path in each subset easily? M Shortest Paths without Repeated Nodes

26.  Suppose we have shortest paths P 1, P 2, …, P m. Let P be the set of all 1-n paths. We divide the set P – {P 1, P 2, …, P m } into P 1, P 2, …, P k, where k  n- 1. Each P j includes paths with the condition that (a) they include the arcs in a certain specified path from node 1 to another node p, and (b) certain arcs from node p are excluded. Then we can find a shortest path in P j by finding a shortest path from node p to n after deleting nodes in the path from 1 to p, and deleting the arcs excluded from node p.  In general, suppose that P m is the shortest path (1, 2), (2, 3), …, (q-1, q), (q, n) in P j. Paths in P j have restrictions that they should include arcs (1, 2), (2, 3), …, (p-1, p), where p  q, and that certain arcs from node p are excluded. Partition P j – {P m } as follows:  If p = q, apply Dijkstra’s to obtain 1-n shortest path, subject to the conditions that arcs (1, 2), (2, 3), …, (p-1, p) are included, and that (p, n) is excluded, in addition to the arcs from p excluded for P m. Network Theory and Applications 2010 26

27.  If p < q, apply Dijkstra’s to find the shortest path from 1 to n, subject to the following sets of conditions: (1) Arcs (1, 2), (2, 3), …, (p-1, p) are included and arc (p, p+1) is excluded, in addition to the arcs from p excluded for P m. (2) Arcs (1, 2), (2, 3), …, (p, p+1) are included and arc (p+1, p+2) is excluded. … (q-p-2) Arcs (1, 2), (2, 3), …, (q-2, q-1) are included and arc (q-1, q) is excluded. (q-p-1) Arcs (1, 2), (2, 3), …, (q-1, q) are included and arc (q, n) is excluded.  Network Theory and Applications 2010 27