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- 1. Algorithm Design and Complexity Course 6
- 2. Overview Classes of Problems P vs NP Polynomial Reduction NP-hard and NP-complete Backtracking n-Queens Problem Graph Coloring Problem
- 3. Classes of Problems Complexity of an algorithm notation Used to compare algorithms in order to determine which one is better Complexity of a problem? We would like to know how difficult a problem is in order to know what solution to look for Used to compare problems in order to determine which one is more difficult, regardless of the algorithm used to solve it! In other words, by taking into consideration the best algorithms we could devise to solve the problems
- 4. Classes of Problems (2) We are able to define classes of problems, but not as many than the classes of algorithms P = any problem for which we can find a correct solution in polynomial time NP = any problem for which we can verify that a solution is correct in polynomial time We can solve the problem in polynomial time There exists at least such a solution/an algorithm The problem is called tractable We may not solve the problem in polynomial time Polynomial time = O(nk) k - constant
- 5. P vs NP Therefore, P and NP are classes of problems Any problem in P is also in NP Not any problem in NP is in P? If we can find a solution in polynomial time, we surely can verify that a solution is correct in polynomial time P NP It is easier to verify that a “guessed” solution is correct or not than to compute this solution This has not been proved until now! NP P ??? It is unknown if P = NP, but most researchers believe that P and NP are not the same class 1 million dollar prize for proving that P = NP or P != NP http://en.wikipedia.org/wiki/P_versus_NP_problem
- 6. P vs NP (2) Conclusion: the problems in NP P should be more difficult than the problems in P For these problems, we cannot find a solution in polynomial time at this moment in time Maybe we shall find in the future, especially if someone manages to prove that P = NP Therefore, at this moment in time we could separate between problems in: P – less difficult NP P – more difficult
- 7. Polynomial Reduction Given 2 decision problems, A1 and A2 Remember: A decision problem is a problem that has only two possible outputs: yes/no We say that problem A1 can be reduced in polynomial time to problem A2 (A1 P A2) if: There exists a polynomial time algorithm F that transforms any input data for A1, x, into an input data for A2, F(x) A1(x) == yes A2(F(x)) == yes A1(x) == yes => A2(F(x)) == yes A1(x) == no => A2(F(x)) == no Image source: http://homepages.ius.edu/rwisman/C455/html/notes/Chapter34/NP-Completeness.htm
- 8. Polynomial Reduction (2) Thus, we can use any solution to A2 to solve A1 If we have a solution for A2, we also have a solution for A1 If A1 P A2, then we say that problem A1 is easier or at most as difficult as A2 This looks a bit strange, doesn’t it ? From the algorithm complexity point of view See next slide why
- 9. Polynomial Reduction (3) Solve A1(x) x2 = F(x) RETURN SolveA2(x2) From the algorithm point of view, it seems that the algorithm for solving A1 has a greater complexity than the one for solving A2 Complexity(SolveA1) = Complexity(F) + Complexity(SolveA2) However, we are interested from the classes of problems point of view: We can use the best algorithm for solving A2 and then use F to solve A1 If A2 P => A1 P If A2 NP => A1 P or NP (If A2 NP P => A1 P or NPP) Therefore, A1 is always easier or as at most as difficult as A2
- 10. Polynomial Reduction (4) Another property of polynomial reduction: If A1 NP => A2 NP (If A1 NP P => A2 Because A2 cannot be easier than A1 NPP) Therefore, we shall use polynomial reduction to highlight that a problem is more difficult than another one w.r.t. classes of problems
- 11. NP-hard and NP-complete A problem Q is called NP-hard if: For It is more difficult or at least as difficult as any problem in NP However, Q may not even be in NP! There are problems even more difficult that those in NP! They are NP-hard Q1 NP: Q1 P Q A problem Q is called NP-complete if It is NP-hard and it is in NP These are the most difficult problems in NP!
- 12. NP-hard and NP-complete (2) NP-hard and NP-complete are also classes of problems A possible graphical representation is: Image source: http://en.wikipedia.org/wiki/NP-complete
- 13. NP-complete Problems Graph Clique Graph Vertex Cover Given a undirected graph G(V, E). Is there a clique of size k in G? Given a undirected graph G(V, E). Is there a vertex cover of size k ? Quick Info: A clique is a subset of vertices V’ V such that for any v1, v2 V’ there is an edge (v1, v2) E[G] A vertex cover is a subset of vertices V’ V such that for any edge (v1, v2) E[G], v1 V’ and/or v2 V’ At least an endpoint of any edge in the graph is covered in V’
- 14. NP-complete Problems (2) Graph Coloring N-Queens Problem Hamiltonian Cycle Travelling Salesman Problem Minesweeper Task scheduling Etc. A lot of interesting problems are NP-hard (some of them are NP-complete)
- 15. Conclusions There are a lot of problems that are very difficult to solve (NP-hard) There is no polynomial time solution for them, at least at this moment in time We need a method for solving them Simple solution: backtracking (with heuristics)
- 16. Backtracking Useful to solve difficult problems: Many optimization problems Combinatorial problems Problems for which you want to know all the solutions NP-complete problems Backtracking improves the brute-force “generate and test” solution for a problem Generate and test Generate all possible solutions After a solution is final, test if it is correct
- 17. Generate and Test Example: k-clique for a graph G(V, E) 1. Generate all combinations of k vertices 1.a. Choose a value from V for the 1st vertex in the clique 1.b. Choose a value from V for the 2nd vertex in the clique … 1.?. Choose a value from V for the kth vertex in the clique 2. Test if the generated solution is correct 2.a. If it is correct and you are looking for only 1 solution, then stop 2.b. Else continue generating the next solution
- 18. Alternative View of a Problem Most of these problems can be transformed into the following problem: He have a set of n variables: V1, …, Vn Each of these variables have a specified domain: There are a set of constraints that should be respected by the final solution: One value for each domain should be assigned to each variable in the final solution Vi Dom(Vi) = Domi[1..ki]; each domain has ki possible values to choose from Constraints for a single variable (E.g.: V2 != 3) Constraints between two variables (E.g.: V2 != V3 or V1+V3 = 2) Other kind of constraints We need to determine a value for each variable that is part of the domain of that variable and this instantiation respects all the defined constraints Constraints Satisfaction Problem (CSP)
- 19. Example – k-clique We have k variables: V1, …, Vk – the vertices of the k-clique Each variable can take values from all the vertices of the graph G(Vertices[n], Edges[m]) Dom(V1) = … = Dom(Vk) = {1, …, n} Considering the vertices of the graph are labeled from 1..n Constraints: Vi != Vj for all 1 <= i < j <= k (Vi, Vj) Edges for all 1 <= i < j <= k
- 20. Generate and Test – Revisited For the new formulation of the problems GenerateAndTest(Vars, Domains, Constraints) FOR (Vars[1] in Domains[1]) FOR (Vars[2] in Domains[2]) … FOR (Vars[n] in Domains[n]) CheckConstraints(Vars, Constraints) Complexity: (k1 * k2 * … * kn) where ki = size(Domains[i]) If k1 = k2 = … = kn = k => (kn) Exponential complexity
- 21. Generate and Test – Recursive We can write easily a recursive solution Same complexity as the previous algorithm GenerateAndTestRecursive(Vars[1..n], Domains, Constraints, k) IF (k == n + 1) CheckConstraints(Vars, Constraints) ELSE FOR (i = 1; i <= size(Domains[k]); i++) Vars[k] = Domains[k][i] GenerateAndTestRecursive(Vars[1..n], Domains, Constraints, k+1) Initial call: GenerateAndTestRecursive(Vars, Domains, Constraints, 1)
- 22. Solution Tree Root level: No variable is assigned First level: First variable is assigned with all possible values from the domain Last level: The last variable is assigned Complexity: generated by all the levels in the tree! Depends on the height – d Depends on the (average) branching factor – b O(bd) More details on whiteboard
- 23. Problems with G&T A correct solution = a solution that is consistent w.r.t. all the constraints that are checked Some inconsistencies appear while building the solution Why only check for consistency when the solution is final? Also check the consistency of partial solutions If a partial solution is not consistent, abandon it And assign the next value in the domain to the current variable, if any left If no values are left in the domain of the current variable, go back to the previous one and continue This is called backtracking
- 24. Backtracking Improvement of G&T that checks for the consistency of the partial solutions Thus, search in the solution tree is pruned We can further improve the search by using heuristics => the complexity for finding the correct solution is reduced We cannot reduce the height of the tree But we can reduce the average branching factor of the pruned solution tree However, usually backtracking is still O(bd) for the worst case We cannot guarantee it to have a lower complexity
- 25. Backtracking – recursive scheme We can devise a recursive scheme for most problems solvable using backtracking BKTRecursive(Vars[1..n], Domains, Constraints, k) IF (k == n + 1) PrintSolution(Vars) ELSE // when no next value exists, the index is reset to the first value WHILE (ExistsNextValue(Vars[k], Domains[k])) Vars[k] = NextValue(Vars[k] , Domains[k], Constraints) IF (CheckConstraints(Vars, Constraints, k)) BKTRecursive (Vars[1..n], Domains, Constraints, k+1) PrintSolution, ExistsNextValue, NextValue, CheckConstraints are problem and method depending
- 26. Remarks Initial call: Because consistency is verified after choosing each value for a variable in the partial solution, it means that the final solution is also consistent Therefore, just print it ExistsNextValue, NextValue – method dependent BKTRecursive(Vars, Domains, Constraints, 1) Simple to implement for usual backtracking, just iterate through the domain array for each variable until reaching the end More complex for using BKT with heuristics CheckConstraints, PrintSolution – problem dependent One of the only things that change from problem to problem
- 27. Backtracking – iterative scheme We can also devise an iterative scheme for most problems solvable using backtracking BKTIterative(Vars[1..n], Domains, Constraints) k=1 WHILE (k <= n +1) IF (k == n + 1) PrintSolution(Vars) k-CONTINUE // when no next value exists, the index is reset to the first value WHILE (ExistsNextValue(Vars[k], Domains[k])) Vars[k] = NextValue(Vars[k] , Domains[k], Constraints) IF (CheckConstraints(Vars, Constraints, k)) k++ BREAK IF (!ExistsNextValue(Vars[k], Domains[k])) k--
- 28. Example: n-Queens Problem Given a table of chess size n x n, find a possible positioning of n queens such that none of the queens attack themselves Or find all possible positionings n = 8 usual chess table n = 1 => 1 solution n = 2, 3 => 0 solutions n = 4 => 2 solutions … n = 8 => 92 solutions … n = 25 => 2,207,893,435,808,352 solutions
- 29. n-Queens Problem A possible solution for n = 8 (from Wikipedia)
- 30. n-Queens Problem Three possible approaches First approach n2 variables – one for each position on the table Each variable has the domain {0, 1} if a queen is placed on that particular position Complexity: O(2n*n) Branching factor for each node: 2 Height of tree: n*n Second approach n variables – one for each queen Each variable has the domain {1, …,n2} – the position on the table for each queen Complexity: O(n2*n) Branching factor: n2 Height of tree: n
- 31. n-Queens Problem CheckConsistency1(Vars[1..n*n], Domains, k) FOR (i = 1..k-1) rowi = (i – 1) / n columni = (i - 1) % n rowk = (k – 1) / n columnk = (k – 1) % n IF (rowi == rowk || columni == columnk || abs(rowi -rowk) == abs(columni - columnk)) IF (Vars[i] == 1 AND Vars[k] == 1) // we already have a queen on the same row // or the same column or the same diagonal RETURN false RETURN true
- 32. n-Queens Problem CheckConsistency2(Vars[1..n], Domains, k) FOR (i = 1..k-1) rowi = (Vars[i] – 1) / n columni = (Vars[i] - 1) % n rowk = (Vars[k] – 1) / n columnk = (Vars[k] – 1) % n IF (rowi == rowk || columni == columnk || abs(rowi -rowk) == abs(columni - columnk)) // we already have a queen on the same row // or the same column or the same diagonal RETURN false RETURN true
- 33. n-Queens Problem Third approach Idea: the queens cannot be placed on the same row! n variables – one for the each position of queen i on row i (i=1..n) Each variable has the domain {1, …,n} – the column where the queen is placed on each row Complexity: O(nn) Branching factor for each node: n Height of tree: n The position of each queen would be (i, Vars[i]) i = 1..n
- 34. n-Queens Problem CheckConsistency3(Vars[1..n], Domains, k) FOR (i = 1..k-1) rowi = i columni = Vars[i] rowk = k // always rowk != rowi columnk = Vars[k] IF (rowi == rowk || columni == columnk || abs(rowi - rowk) == abs(columni - columnk)) // we already have a queen on the same row // or the same column or the same diagonal RETURN false RETURN true
- 35. Example: Graph Coloring Problem Given an undirected graph G(V, E), can we color each vertex of the graph using k colors such that any two vertices joined by an edge have different colors ? (u, v) E : color[u] != color[v] Modeling the problem: N = |V| variables – one for each vertex Each domain has k values: {1, …, k} – the color of each vertex Complexity: O(kn) Height: n Branching factor: k
- 36. Graph Coloring Problem CheckConsistency(Vars[1..n], Domains, k) FOREACH (v IN Adj[k]) // each vertex in the list of adjacency of vertex k IF (v < k AND color[k] == color[v]) RETURN false RETURN true
- 37. Improvements for Backtracking How can we improve backtracking ? How we model the problem Use of heuristics in order to reduce the average branching factor Look at n-queens Variables that have a smaller domain, should be instantiated firstly Variables that have the most constraints, should be instantiated firstly Etc. Forward Checking: more on whiteboard Advanced: Arc-Consistency (AC) algorithms
- 38. Heuristics for Backtracking Minimum Remaining Values (MRV) - Choose the variable that has the least number of valid values in its domain Combined with Forward-Checking Most Constraining Variable (MCV) – Choose the variable that has the most constraints with previous variables Least Constraining Value (LCV) – Choose the value that imposes the least number of constraints on the remaining variables Mainly useful if we want to find a solution, not all of them More info here: http://www.ai.kun.nl/aicourses/bki212a/slides/AISP-CSPch5.pdf
- 39. Conclusions There are some problems that are very difficult For these problems it’s ok to use backtracking But solvable using exponential algorithms NP-complete Maybe with heuristics However, for optimization problems maybe instead of backtracking, it may sometimes be useful to find an approximate solution and not the optimum one In polynomial time
- 40. References CLRS – Chapter 36 http://ww3.algorithmdesign.net/sample/ch13-np.pdf http://www.cs.cmu.edu/~avrim/451/lectures/lect1030.pdf http://faculty.ksu.edu.sa/YAlohali/Documents/ch3Constraint%20Satisfaction%20Problems1.pptx Advanced: Backtracking and Constraint Satisfaction Problems: http://kti.mff.cuni.cz/~bartak/downloads/CPschool05notes .pdf

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