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Exhaustive Search

Exhaustive Search. Brute Force Methods. guarantee best fitness value is found feasible for ‘small’ data sets only. SAT satisfiability problem. set of propositions prop[n] fitness function logical expression based on propositions boolean fitness(prop)

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Exhaustive Search

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  1. Exhaustive Search

  2. Brute Force Methods • guarantee best fitness value is found • feasible for ‘small’ data sets only

  3. SAT satisfiability problem • set of propositions prop[n] • fitness function • logical expression based on propositions boolean fitness(prop) • try all 2n combinations of T/F for propositions

  4. boolean fitness(boolean b[]) //fitness function boolean [] prop = new boolean[n]; boolean satisfied = sat(n-1); boolean sat (int index) // find a solution { if (index==0) //base case { prop[0] = true; if fitness(prop) return true; prop[0] = false; return fitness(prop); } prop[index] = true; // recursive if (sat(index-1)) return true; prop[index] = false; return sat(index-1); } ... ... 2 1 0 ..T ..F .TT .FT .TF .FF TTT FTT TFT FFT TTF FTF TFF FFF

  5. efficiency without risk • pruning the tree: • suppose fitness(prop) is (p0 \/ ~p1) /\ (~p2) partial evaluation ... ... nodes can be ignored: fitness can not be true 2 1 0 ..T ..F .TT .FT .TF .FF TTT FTT TFT FFT TTF FTF TFF FFF

  6. TSP travelling salesman • directed edges, complete graph: exhaustive search is the permutation problem enumerate all n! permutations of n distinct items (cities) {0,2,3,…,n-1} rank in path array D[i][j] 0 1 2 3 1 0 3 1 4 2 3 2

  7. TSP travelling salesman O(nn) input(V) // number of cities (Vertices) int[V] rIP // rankInPath, initialized to 0==not ranked visit(1) // generate visiting sequences void visit(int rank) { for (city = 0 to V-1) if (rIP[city] == 0) // not yet visited { rIP[city] = rank; if (rank == V) fitness(rIP) else visit(rank+1) rIP[city] = 0 } } fitness(int[] p) calculate pathlength from D[i][j] if bestpath, save p

  8. efficiency without risk • fix first city  O((n-1)n-1) • use sets instead of searching array  O((n-1)!) • keep partial fitness values • reduce cost of fitness evaluation • apply branch and bound BUT… still O(en)

  9. Variations on TSP • undirected edges: D[i][j] == D[j][i] • incomplete graph: some D[i][j] = null • Euclidean distances (on a plane) • cities are located on plane at (xi,yi) • D[i][j] is computed from coordinates: D[i][j] = D[j][i] = sqrt((xi-xj)2 + (yi-yj)2)  other data structures, efficiencies

  10. Continuousproblem spaces Where is height of land? 1. what scale to sample? x[0,1], y[0,1] interval length: 0.1: 100 data points 0.01: 10,000 0.001: 1,000,000 --- y  --- x 

  11. Continuousproblem spaces Where is height of land? 1. what scale to sample? x[0,1], y[0,1] interval length: 0.1: 100 data points 0.01: 10,000 0.001: 1,000,000 --- y  --- x 

  12. Continuousproblem spaces Where is height of land? 2. constraints – ignore water fewer data points but constraints must be tested --- y  --- x 

  13. Continuousproblem spaces Where is height of land? 3. where to locate sample --- y  --- x 

  14. Continuousproblem spaces - NLP Non-Linear Programming problems Typical problems are functions of multiple variables over domains of each Maximize f(x1,x2,x3,…,xn) for x1  D1, x2  D2, x3  D3,…, xn  Dn NLP problems are NP complete *Linear Programming problems are polynomial solvable O(nk)

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