1. 1 CSE 326: Data Structures: Graphs Lecture 22: Monday, March 3 rd, 2003

2. 2 Today Example of a formal correctness proof: –Dijkstra’s Algorithm All pairs shortest paths NP completeness –Will finish on Wednesday

3. 3 Dijkstra’s Algorithm for Single Source Shortest Path Classic algorithm for solving shortest path in weighted graphs (with only positive edge weights) Similar to breadth-first search, but uses a priority queue instead of a FIFO queue: –Always select (expand) the vertex that has a lowest-cost path to the start vertex –a kind of “greedy” algorithm Correctly handles the case where the lowest-cost (shortest) path to a vertex is not the one with fewest edges

4. 4 void shortestPath(Node startNode) { Heap s = new Heap; for v in Nodes do { v.dist =  ; s.insert(v); } startNode.dist = 0; s.decreaseKey(startNode); startNode.previous = null; while (!s.empty()) { x = s.deleteMin(); for y in x.children() do if (x.dist+c(x,y) < y.dist) { y.dist = x.dist+c(x,y); s.decreaseKey(y); y.previous = x; } void shortestPath(Node startNode) { Heap s = new Heap; for v in Nodes do { v.dist =  ; s.insert(v); } startNode.dist = 0; s.decreaseKey(startNode); startNode.previous = null; while (!s.empty()) { x = s.deleteMin(); for y in x.children() do if (x.dist+c(x,y) < y.dist) { y.dist = x.dist+c(x,y); s.decreaseKey(y); y.previous = x; } Dijkstra’s Algorithm

5. 5 Dijkstra’s Algorithm: Correctness Proof Partition the set of all nodes, V, into two sets: The heap, s The rest of the nodes, known V = s  known We prove the following Invariant: 1.  v  known, v.dist = cost of shortest path startNode  v 2.  v  known, v.dist = cost of shortest path startNode  v going only through nodes in Known except for v itself

6. 6 Claim 1 startNode Known v Claim 1: if v  known, then v.dist = cost of shortest path startNode  v s

7. 7 Claim 2 startNode Known Claim 2: if v  known, v.dist = cost of shortest path startNode  v going only through nodes in Known except for v itself v s

8. 8 Proof by Induction: Base Case startNode Known =  s Need to check Claim 2:             0 When v = startNode, then the only such paths is (startNode): has cost 0 When v  startNode, then no such path exists: minimum cost is 

9. 9 void shortestPath(Node startNode) { Heap s = new Heap; for v in Nodes do { v.dist =  ; s.insert(v); } startNode.dist = 0; s.decreaseKey(startNode); startNode.previous = null; while (!s.empty()) { x = s.deleteMin(); for y in x.children() do if (x.dist+c(x,y) < y.dist) { y.dist = x.dist+c(x,y); s.decreaseKey(y); y.previous = x; } void shortestPath(Node startNode) { Heap s = new Heap; for v in Nodes do { v.dist =  ; s.insert(v); } startNode.dist = 0; s.decreaseKey(startNode); startNode.previous = null; while (!s.empty()) { x = s.deleteMin(); for y in x.children() do if (x.dist+c(x,y) < y.dist) { y.dist = x.dist+c(x,y); s.decreaseKey(y); y.previous = x; } Proof by Induction: Induction Step Assume the invariant holds here Prove the invariant holds here

10. 10 Proof by Induction: Induction Step startNode Known Need to check Claim 1 and/or Claim 2 in each of the following case: v  known v = x v  known x = deleteMin(s) s

11. 11 Case v = x startNode Known Here we need to check Claim 2 (why ?) Let startNode = x 1, x 2,..., x k-1, x k =x be the shortest paths to x Let x i be the first node s.t. x i  known Then x.dist  x i.dist (why ?)  cost(x 1, x 2,..., x k-1, x k ) (why ?) By induction hypothesis there exists a path startNode  x of cost = x.dist, and going only through known It follows that that paths is also a shortest path, and has cost = x.dist s x = deleteMin(s) xixi

12. 12 Case v  known startNode Known x = deleteMin(s) s v Here we need to check Claim 1 Follows trivially from the induction hypothesis

13. 13 Case v  known startNode Known Here we need to check Claim 2 (why ?) Let startNode = x 1, x 2,..., x k-1, x k =v be the shortest paths to v that goes only through known Look at the last node, x k-1, on this path Case 1: x k-1  x Then v.dist = cost(x 1, x 2,..., x k-1, x k ) by induction hypothesis (why ?) s x = deleteMin(s) v

14. 14 Case v  known startNode Known Here we need to check Claim 2 (why ?) Let startNode = x 1, x 2,..., x k-1, x k =v be the shortest paths to v that goes only through known Look at the last node, x k-1, on this path Case 2: x k-1 = x. Then the following instruction: ensures that v.dist = cost(x 1, x 2,..., x k-1, x k ) s x = deleteMin(s) v if (x.dist+c(x,y) < y.dist) { y.dist = x.dist+c(x,y); }

15. 15 End of Induction startNode Known v Use Claim 1: For every v, v  known (because s =  ) hence v.dist = cost of shortest path startNode  v s = 

16. 16 All Pairs Shortest Path Suppose you want to compute the length of the shortest paths between all pairs of vertices in a graph… –Run Dijkstra’s algorithm (with priority queue) repeatedly, starting with each node in the graph: –Complexity in terms of V when graph is dense:

17. 17 Dynamic Programming Approach Notice that D k-1, i, k = D k, i, k and D k-1, k, j = D k, k, j ; hence we can use a single matrix, D i, j !

18. 18 Floyd-Warshall Algorithm // C – adjacency matrix representation of graph // C[i][j] = weighted edge i->j or  if none // D – computed distances for (i = 0; i < N; i++){ for (j = 0; j < N; j++) D[i][j] = C[i][j]; D[i][i] = 0.0; } for (k = 0; k < N; k++) for (i = 0; i < N; i++) for (j = 0; j < N; j++) if (D[i][k] + D[k][j] < D[i][j]) D[i][j] = D[i][k] + D[k][j]; // C – adjacency matrix representation of graph // C[i][j] = weighted edge i->j or  if none // D – computed distances for (i = 0; i < N; i++){ for (j = 0; j < N; j++) D[i][j] = C[i][j]; D[i][i] = 0.0; } for (k = 0; k < N; k++) for (i = 0; i < N; i++) for (j = 0; j < N; j++) if (D[i][k] + D[k][j] < D[i][j]) D[i][j] = D[i][k] + D[k][j]; Run time = How could we compute the paths?

19. 19 NP-Completeness Really hard problems

20. 20 Today’s Agenda Solving pencil-on-paper puzzles –A “deep” algorithm for Euler Circuits Euler with a twist: Hamiltonian circuits Hamiltonian circuits and NP complete problems The NP =? P problem –Your chance to win a Turing award! –Any takers? Weiss Chapter 9.7 W. R. Hamilton (1805-1865) L. Euler (1707-1783)

21. 21 It’s Puzzle Time! Which of these can you draw without lifting your pencil, drawing each line only once? Can you start and end at the same point?

22. 22 Historical Puzzle: Seven Bridges of Königsberg KNEIPHOFF PREGEL Want to cross all bridges but… Can cross each bridge only once (High toll to cross twice?!)

23. 23 A “Multigraph” for the Bridges of Königsberg Find a path that traverses every edge exactly once

24. 24 Euler Circuits and Tours Euler tour: a path through a graph that visits each edge exactly once Euler circuit: an Euler tour that starts and ends at the same vertex Named after Leonhard Euler (1707-1783), who cracked this problem and founded graph theory in 1736 Some observations for undirected graphs: –An Euler circuit is only possible if the graph is connected and each vertex has even degree (= # of edges on the vertex) [Why?] –An Euler tour is only possible if the graph is connected and either all vertices have even degree or exactly two have odd degree [Why?]

25. 25 Euler Circuit Problem Problem: Given an undirected graph G = (V,E), find an Euler circuit in G Note: Can check if one exists in linear time (how?) Given that an Euler circuit exists, how do we construct an Euler circuit for G? Hint: Think deep! We’ve discussed the answer in depth before…

26. 26 Finding Euler Circuits: DFS and then Splice Given a graph G = (V,E), find an Euler circuit in G –Can check if one exists in O(|V|) time (check degrees) Basic Euler Circuit Algorithm: 1.Do a depth-first search (DFS) from a vertex until you are back at this vertex 2.Pick a vertex on this path with an unused edge and repeat 1. 3.Splice all these paths into an Euler circuit Running time = O(|V| + |E|)

27. 27 Euler Circuit Example A B C D E F B C D E G G D E G DFS(A) : A B D F E C A DFS(B) : B G C B DFS(G) : G D E G A B G C B D F E C A A B G D E G C B D F E C A Splice at B Splice at G

28. 28 Euler with a Twist: Hamiltonian Circuits Euler circuit: A cycle that goes through each edge exactly once Hamiltonian circuit: A cycle that goes through each vertex exactly once Does graph I have: –An Euler circuit? –A Hamiltonian circuit? Does graph II have: –An Euler circuit? –A Hamiltonian circuit? B C D E G B C D E G I II

29. 29 Finding Hamiltonian Circuits in Graphs Problem: Find a Hamiltonian circuit in a graph G = (V,E) –Sub-problem: Does G contain a Hamiltonian circuit? –No known easy algorithm for checking this… One solution: Search through all paths to find one that visits each vertex exactly once –Can use your favorite graph search algorithm (DFS!) to find various paths This is an exhaustive search (“brute force”) algorithm Worst case  need to search all paths –How many paths??

30. 30 Analysis of our Exhaustive Search Algorithm Worst case  need to search all paths –How many paths? Can depict these paths as a search tree Let the average branching factor of each node in this tree be B |V| vertices, each with  B branches Total number of paths  B·B·B … ·B = O(B |V| ) Worst case  Exponential time! B C D E G B D G C G E D E C G E Etc. Search tree of paths from B

31. 31 How bad is exponential time? Nlog N N log NN2N2 2N2N 10012 21244 42816 103301001024 100770010,0001,000,000,000,0 00,00,000,000,0 00,000,000 10001010,0001,000,000Fo’gettaboutit! 1,000,0002020,000,0001,000,000,000,000ditto

32. 32 Review: Polynomial versus Exponential Time Most of our algorithms so far have been O(log N), O(N), O(N log N) or O(N 2 ) running time for inputs of size N –These are all polynomial time algorithms –Their running time is O(N k ) for some k > 0 Exponential time B N is asymptotically worse than any polynomial function N k for any k –For any k, N k is  (B N ) for any constant B > 1

33. 33 The Complexity Class P The set P is defined as the set of all problems that can be solved in polynomial worse case time –Also known as the polynomial time complexity class –All problems that have some algorithm whose running time is O(N k ) for some k Examples of problems in P: tree search, sorting, shortest path, Euler circuit, etc.

34. 34 The Complexity Class NP Definition: NP is the set of all problems for which a given candidate solution can be tested in polynomial time Example of a problem in NP: –Hamiltonian circuit problem : Why is it in NP?

35. 35 The Complexity Class NP Definition: NP is the set of all problems for which a given candidate solution can be tested in polynomial time Example of a problem in NP: –Hamiltonian circuit problem : Why is it in NP? Given a candidate path, can test in linear time if it is a Hamiltonian circuit – just check if all vertices are visited exactly once in the candidate path (except start/finish vertex)

36. 36 Why NP? NP stands for Nondeterministic Polynomial time –Why “nondeterministic”? Corresponds to algorithms that can guess a solution (if it exists)  the solution is then verified to be correct in polynomial time –Nondeterministic algorithms don’t exist – purely theoretical idea invented to understand how hard a problem could be Examples of problems in NP: –Hamiltonian circuit: Given a candidate path, can test in linear time if it is a Hamiltonian circuit –Sorting: Can test in linear time if a candidate ordering is sorted –Are any other problems in P also in NP?

37. 37 More Revelations About NP Are any other problems in P also in NP? –YES! All problems in P are also in NP Notation: P  NP If you can solve a problem in polynomial time, can definitely verify a solution in polynomial time Question: Are all problems in NP also in P? –Is NP  P?

38. 38 Your Chance to Win a Turing Award: P = NP? Nobody knows whether NP  P –Proving or disproving this will bring you instant fame! It is generally believed that P  NP, i.e. there are problems in NP that are not in P –But no one has been able to show even one such problem! –Practically all of modern complexity theory is premised on the assumption that P  NP A very large number of useful problems are in NP Alan Turing (1912-1954)

39. 39 NP-Complete Problems The “hardest” problems in NP are called NP-complete problems (NPC) Why “hardest”? A problem X is NP-complete iff: 1.X is in NP and 2.Any problem Y in NP can be converted to an instance of X in polynomial time, such that solving X also provides a solution for Y In other words: Can use algorithm for X as a subroutine to solve Y Thus, if you find a poly time algorithm for just one NPC problem, all problems in NP can be solved in poly time –E.g: The Hamiltonian circuit problem can be shown to be NP- complete

40. 40 Another NP-Complete Problem SAT: Given a formula in Boolean logic, e.g. determine if there is an assignment of values to the variables that makes the formula true (=1). Why is it in NP?

41. 41 Why SAT is NP-Complete Cook (1971) showed that SAT could be used to simulate any non- deterministic Turing machine! Idea: consider the tree of possible execution states of the Turing machine –A Boolean logic formula can represent this tree – the “state transition” function –Formula also asserts the final state is one where a solution has been found –“Guessed” variables determine which branch to take

42. 42 P, NP, and Exponential Time Problems All currently known algorithms for NP-complete problems run in exponential worst case time –Finding a polynomial time algorithm for any NPC problem would mean: Diagram depicts relationship between P, NP, and EXPTIME (class of problems that provably require exponential time to solve) It is believed that P  NP  EXPTIME EXPTIME NP P NPC

43. 43 The Graph of NP-Completeness Stephen Cook first showed (1971) that satisfiability of Boolean formulas (SAT) is NP- complete Hundreds of other problems (from scheduling and databases to optimization theory) have since been shown to be NPC How? By showing an algorithm that converts a known NPC problem to your pet problem in poly time  then, your problem is also NPC!

44. 44 Showing NP-completeness: An example Consider the Traveling Salesperson (TSP) Problem: Given a fully connected, weighted graph G = (V,E), is there a cycle that visits all vertices exactly once and has total cost  K? TSP is in NP (why?) Can we show TSP is NP-complete? –Hamiltonian Circuit (HC) is NPC –Can show TSP is also NPC if we can convert any input for HC to an input for TSP in poly time B C D E G Input for HC B C D E Convert to input for TSP Cycle with cost  8  BDCEB 3 3 1 1 2 4

45. 45 TSP is NP-complete! We can show TSP is also NPC if we can convert any input for HC to an input for TSP in polynomial time. Here’s one way: B C D E G B C D E G This graph has a Hamiltonian circuit iff this fully-connected graph has a TSP cycle of total cost  K, where K = |V| (here, K = 5) HC TSP 2 2 11 1 1 1 1 1 1

46. 46 Longest Path Decision problem version: Is there a simple path in G (between two given vertices s and t) of length at least k? Clearly in NP. Why? To prove the longest path problem is NP- complete,we can again reduce Hamiltonian circuit problem 1.Input: a HC problem G with n vertices 2.Duplicate some vertex s in G, call it t 3.Add an edge of weight 0 between s and t 4.Ask longest path: is there a path of at least weight n between s and t?

47. 47 Coping with NP-Completeness 1.Settle for algorithms that are fast on average: Worst case still takes exponential time, but doesn’t occur very often. But some NP-Complete problems are also average-time NP-Complete! 2.Settle for fast algorithms that give near-optimal solutions: In TSP, may not give the cheapest tour, but maybe good enough. But finding even approximate solutions to some NP-Complete problems is NP-Complete! 3.Just get the exponent as low as possible! Much work on exponential algorithms for Boolean satisfiability: in practice can often solve problems with 1,000+ variables But even 2 n/100 will eventual hit the exponential curve!

48. 48 A Great Book You Should Own! Computers and Intractability: A Guide to the Theory of NP-Completeness, by Michael S. Garey and David S. Johnson