1 / 28

15.053

15.053. Tuesday, April 2. The Shortest Path Problem Dijkstra’s Algorithm for Solving the Shortest Path Problem Handouts: Lecture Notes. The Minimum Cost Flow Problem. A network with costs, capacities, supplies, demands. Directed Graph G = ( N, A ). Node set N , arc set A ;

lang
Download Presentation

15.053

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 15.053 Tuesday, April 2 • The Shortest Path Problem • Dijkstra’s Algorithm for Solving the Shortest Path Problem Handouts: Lecture Notes

  2. The Minimum Cost Flow Problem A network with costs, capacities, supplies, demands Directed Graph G = (N, A). Node set N, arc set A; Capacities uijon arc (i,j) Lower bound of 0 on arc (i,j) Cost cij on arc (i,j) Supply/demand bi for node i. (Positive indicates supply) Minimize the cost of sending flow s.t. Flow out of i - Flow into i= bi 0 ≤xij ≤uij

  3. Formulation In general the LP formulation is given as Minimize subject to

  4. The Shortest Path Problem What is the shortest path from a source node (often denoted as s) to a sink node, (often denoted as t)? What is the shortest path from node 1 to node 6? Assumptions for this lecture: • There is a path from the source to all other nodes. • All arc lengths are non-negative

  5. Formulation as a linear program In general the LP formulation for the shortest path from a source, s, to a sink, t, is given as Minimize subject to

  6. Another Formulation The LP formulation for the shortest path from a source, s, to all other nodes is given as Minimize subject to

  7. Some Questions Concerning theShortest Path Problem • Where does it arise in practice? • Direct applications • Indirect (and often subtle) applications • How does one solve the shortest path problem? • Dijkstra’s algorithm • How does one measure the performance of an algorithm? • CPU time measurements • Performance Guarantees • How does one establish that a solution is really the shortest path? • Connection to LP duality

  8. Possible sports scores • Flumbaya is an unusual water sport in which there are two types of scores possible. One can score a gymbol, which is worth 7 points, or one can score a quasher, which is worth 5 points. An announcer on TV states that a recent game was won by a score of 19 to 18. Is this possible?

  9. More on Flumbaya There is no path from node 0 to node 18. A score of 18 is impossible.

  10. More on Flumbaya Data: Gymbol is worth n1 points Quasher is worth n2 points: determine whether one can score q points The network: G = (N, A), where N = {0, …, q} for each node j = 0 to q – n1 , (j, j+n1) ∈ A for each node j = 0 to q – n2, (j, j+n2) ∈ A Question: Is there a path in G from node 0 to node q? Fact: if n1 and n2 have no common integer divisor (other than 1 and –1), then the number of scores that cannot be obtained is (n-1)(n2-1)/2. Extra credit for proving this fact. (Warning, it is difficult to prove.)

  11. An indirect application: Finding optimalparagraph layouts TeX optimally decomposes paragraphs by selecting the breakpoints for each line optimally. It has a subroutine that computes the attractiveness F(i,j) of a line that begins at word i and ends at word j-1. How can one use F(i,j) to create a shortest path problem whose solution will solve the paragraph problem? TeX optimally decomposes paragraphs by selecting the breakpoints for each line optimally. It has a subroutine that computes the attractiveness F(i,j) of a line that begins at word i and ends at word j-1. How can one use F(i,j) to create a shortest path problem whose solution will solve the paragraph problem?

  12. An indirect application: finding optimalparagraph layouts Each word corresponds to a node and an arc (i,j) indicates that a line begins with word i and ends at word j-1. A path from Tex to “end” corresponds to a paragraph layout. TeX optimally decomposes paragraphs by selecting the breakpoints for each line optimally. It has a subroutine that computes the attractiveness F(i,j) of a line that begins at word i and ends at word j-1. How can one use F(i,j) to create a shortest path problem whose solution will solve the paragraph problem? that line selecting by line j-1 Tex A value of the path is the “ugliness” of the path. shortest end solve

  13. On the paragraph example • n different yes-no decisions • Decision j: Yes means start a line at word j • No: don’t start a line at word j • The cost of each yes decision depends only on the subsequent yes decision • f(i,j) was the cost of starting a line at word i assuming that word j begins the next line. • Create a shortest path problem with nodes 1, 2, … , n+1 where the cost of arc (i,j) is f(i,j). What is the shortest path from 1 to n+1

  14. An Application in Data Compression:Approximating Piecewise Linear Functions • INPUT: A piecewise linear function • n points a1= (x1,y1), a2= (x2,y2),..., an= (xn,yn). • x1≤ x2≤ ... ≤ xn. • Objective: approximate f with fewer points • c* is the “cost” per point included • cij = cost of approximating the function through points i, i+1, . . ., j-1 by a single line joining point I to point j. (sum of errors, or errors squared.)

  15. Approximating Piecewise Linear Functions • Objective: approximate f with fewer points • c* is the “cost” per point included • c36 = |a4 - b4| + |a5 - b5| = sum of errors. (other metrics would also be OK.)

  16. On approximating functions • n different yes-no decisions • Decision j: yes means select point j • No: don’t select point j • The cost of each yes decision depends only on the subsequent yes decision • cij is the cost of selecting point i followed by point j, and takes into account the cost of selecting i, and the costs of approximating points i+1, …, j-1. • Create a shortest path problem with nodes 1, … , n where the cost of arc (i,j) is cij. What is the shortest path from 1 to n?

  17. Dijkstra’s Algorithm for the ShortestPath Problem Exercise with your partner. Find the shortest paths by inspection. Exercise: find the shortest path from node 1 to all other nodes. Keep track of distances using labels, d(i) and each node’s immediate predecessor, pred(i). d(1)= 0, pred(1)=0; d(2) = 2, pred(2)=1 Find the other distances, in order of increasing distance from node 1.

  18. A Key Step in Shortest Path Algorithms • Let d( ) denote a vector of temporary distance labels. • d(j) is the length of some path from the origin node 1 to node j. • Procedure Update(i)for each (i,j) ∈ A(i) doif d(j) > d(i) + cijthen d(j) : = d(i) + cij and pred(j) : = i; Up to this point, the best path from 1 to j has length 78

  19. A Key Step in Shortest Path Algorithms • Let d( ) denote a vector of temporary distance labels. • d(j) is the length of some path from the origin node 1 to node j. • Procedure Update(i)for each (i,j) ∈ A(i) doif d(j) > d(i) + cijthen d(j) : = d(i) + cij and pred(j) : = i; P(1,j) is a “path” from 1 to j of length 72.

  20. Dijkstra’s Algorithm Initialize distances. LIST = set of temporary nodes begin d(s) : = 0 and pred(s) : = 0; d(j) : = ∞ for each j ∈ N - {s}; LIST : = {s}; while LIST ≠ φ do begin let d(i) : = min {d(j) : j ∈ LIST}; remove node i from LIST; update(i) if d(j) decreases, place j in LIST end end Select the node i on LIST with minimum distance label, and then update(i)

  21. Scan the arcs out of i, and update d( ), pred( ), and LIST An Example The End Find the node i on LIST with minimum distance.

  22. The Output from Dijkstra’sAlgorithm To find the shortest path from node j, trace back from the node to the source. Dijkstra provides a shortest path from node 1 to all other nodes. It provides a shortest path tree.

  23. Comments on Running time • Dijkstra’s algorithm is efficient in its current form. The running time grows as n2. • It can be made much more efficient • In practice it runs in time linear in the number of arcs (or almost so).

  24. The string solution and LP duality Let d(j) denote the distance to node j from the source. d(1) = 0 d(2) <= d(1) + 2; d(5) <= d(2) + 2; d(2) <= d(5) + 2 d(5) <= d(3) + 3; d(3) <= d(5) + 3 etc. Dual: Max d(t)-d(s) s.t. d(s) = 0 d(j) <= d(i) + cij

  25. The string solution Imagine replacing each arc by a string of the same length. Thus arc (1,3) would be replaced by a string of length 4 inches joining node 1 to node 3. Now hold node 1 in one hand and node 6 in the other, and pull until the string is tight.

  26. The string solution Does one get the shortest path from node 1 to node 6? If so, why? Note: In some sense we are maximizing the physical distance from node 1 to node 6.

  27. Summary • Direct and indirect applications for the shortest path problem • Dijkstra’s algorithm finds the shortest path from node 1 to all other nodes in increasing order of distance from the source node. • The bottleneck operation is identifying the minimum distance label. One can speed this up, and get an incredibly efficient algorithm • The string solution optimizes the dual LP as well as the shortest path problem.

  28. Some final comments • The shortest path problem shows up again and again in network optimization • There is an interesting connection with dynamic programming • There are other solution techniques as well. We’ll see one in a later lecture.

More Related