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CS420 lecture ten BACKTRACK

CS420 lecture ten BACKTRACK. Solution vectors. In optimization problems, or more general in search problems, a set of choices are to be made to arrive at an optimum or solution.

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CS420 lecture ten BACKTRACK

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  1. CS420 lecture tenBACKTRACK

  2. Solution vectors • In optimization problems, or more general in search problems, a set of choices are to be made to arrive at an optimum or solution. • A solution can be written in the form of a vector X = (x1,x2,..,xn) and the search is directed by criterion P, defining feasible choices of X. • Often P can also decide whether a partial solution Xi=(x1,..,xi), in, is feasible, ie, there are extensions of Xi to a feasible solution. • We can either search for one solution, or we can search for all solutions.

  3. Constraints • Explicit constraints define the domain of the choice variables, eg xi>0, or xi=0 or 1, or li<xi<ui. • Implicit constraints specify that x1 .. xi satisfy P, ie relate the xi-s to each other. • The explicit constraints define a possible solution spacefor the problem • The implicit constraints narrow the solution space down to an answer space.

  4. Search Spaces • We concern us here with tree shaped spaces • A prefix x1 .. xi is a path from the root (the empty vector) with edges labeled x1 to xi • Children of xi form solution vectors of the form x1 .. xi+1.

  5. n Queens • Given an nxn chessboard and n queens, find all placements of the queens such that none of them attack each other • none of the queens occupy the same row, column, or diagonal. • As each queen occupies her own row, a (partial) choice vector X is a vector of integers from 1 to n, where xj is the column for the queen in row j. • One solution to the 4-queens problem is (2,4,1,3)

  6. 4 queens 2 4 1 3 solution Q Q Q Q

  7. Solution space for n-Queens • root has n children, 1 for each column • each of these has n-1 children, excluding its own column • next has n-2 children etc. • so the tree has n! nodes / states, many of these are infeasible • diagonal constraint • n! is a gross overestimate of the number of states we need to check

  8. Subset sum (SSS) • Given a set of integers m1 to mn and another integer M, find all subsets of mi-s such that their sum is M • Eg for m ={1,2,3}, M=3 we have two solutions {1,2} and {3}.

  9. Solution Vectors for SSS • One option: the indices of the chosen elements • then solution vectors have different length

  10. Solution Vectors for SSS X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 • One option: the indices of the chosen elements • then solution vectors have different length X3=3

  11. Alternative • Another, more regular, solution space has nodes with up to n choice values ci = 0 if mi is not in the subset and ci = 1 if it is. • In this case the leaves represent the subsets.

  12. C1= 1 C1= 0 Alternative C2= 0 C2= 0 C2= 1 C2= 1 C3= 0 C3= 1 C3= 0 C3= 0 C3= 1 C3= 1 C3= 1 C3= 0 • Another, more regular, solution space has nodes with up to n choice values ci = 0 if mi is not in the subset and ci = 1 if it is. • In this case the leaves represent the subsets.

  13. Backtrack • Backtrack walks depth first through the solution space • and uses the criterion P (sometimes called bounding function B) to prune the solution space to create the answer space.

  14. Iterative Backtrack • We are searching for all possible answers. • X will hold the (partial) solution vector (x1,x2,..,xi). • B is the bounding function. • Next(X) is the set of valid next states xi+1 from state xi , and therefore Bi+1(X[1..(i+1)) is true for all states in Next(X) and false for all other states. • We also need a function that decides whether a state is an answer state.

  15. IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } }

  16. k=1 x=[] cap = 3 X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  17. k=2 x=[1] feasible cap=2 X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  18. k=3 x=[1,2] answer path cap=0 print X=[1,2] X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  19. k=3 x=[1,2,3] infeasible backtrack X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  20. k=2 X=[1,3] infeasible backtrack X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  21. k=1 X=[2] feasible X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  22. k=2 X=[2,3] infeasible backtrack X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  23. k=1 X=[3] feasible X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  24. k=2 X=[3] cap=0 path to answer node print X=[3] X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  25. paths to answer node X=[1,2] X=[3] X1=1 X1=3 X1=2 X2=2 X2=3 X2=3 IBackTrack(n){ k = 1 while (k>0) { if (there remains an x in Next(X[1..(k-1)]) and Bk(X[1..k]) ) { if (X[1..k] is path to answer node) print X[1..k] // end if k++ // consider next set } else k-- // backtrack to previous set } } X3=3

  26. Recursive back track • global variable X[1:n], initial call RBackTrack(1) RBackTrack(k) { for each X[k] with Bk(X[1..k])==true if (X[1..k] is a path to an answer state) print(X[1..k] RBackTrack(k+1) }

  27. n-queens • X[1:k] is a partial solution with X[i] the column number (1..n) of a queen in row i 1:k. • When we place queen k+1 we need to check that she does not occupy any of the already occupied columns (that's easy) and any of the already occupied diagonals: • two queens in (i,j) (row i, column j) and (p,q) occupy the same diagonal if (i-j) == (p-q) or (i+j)==(p+q), which is equivalent with |i-p|==|j-q|.

  28. iterative n-queens with global partial solution X Place(k) for (i = 1 to k-1) if (X[i]==X[k] or // same columns | X[i]-X[k] | == |i-k| ) // same diagonal return(false); // end for return(true)

  29. nQueens(n) { X[1] = 0 //place first queen "in front of" row 1 k=1 while (k>0) { X[k]++ // move queen k to next column while (X[k]<=n and not(Place(k)) { X[k]++ } if (X[k] <= n) // found valid spot for queen k if (k==n) print(X) else {k++; X[k]=0} // start next row else k-- // backtrack }

  30. Sum of Subsets • Two ways to represent the search space for subset sum. • Here we consider the regular case with choices xi=0 (do not take object i) and xi=1 (take object i) at level i. • A choice for the bounding function Bk =

  31. Simplifying / Strengthening assumption for SSS • From hereon we assume that the objects are sorted in non-decreasing order, because this allows us to strengthen our observations.

  32. Weight gathered so far • In some state X[1..k] at level k the total weight gathered by objects 1 to kis either M, and we have found a solution and thus we don't need to go deeper in that state... • or the weight gathered is less than M, and then X[1..k] cannot lead to an answer node if

  33. Weight gathered so far • In some state X[1..k] at level k the total weight gathered by objects 1 to kis either M, and we have found a solution and thus we don't need to go deeper in that state... • or the weight gathered is less than M, and then X[1..k] cannot lead to an answer node if • WHY?

  34. A better bound • Original bound plus previous observation leads to a better bound: not too little and not too much

  35. Recursive SSS • s is the weight gathered by the objects chosen so far, and thus needs to be initialized at 0 • r is the total weight of the remaining objects, and thus needs to be initialized at the sum SumW of all the weights. • kis the current 1-based level. If we take object k we can simplify the bound test because the sum of the weights of the chosen objects (now including k) plus remaining objects has not changed and is still >= M • The initial call is SumOfSub(0,1,SumW).

  36. No degenerate problems We assume that W[1]<=M and SumW>=M, otherwise we have a degenerate problem.

  37. SumOfSub(s,k,r) { // two possible next states X[k] = 1 //1: take object k, we can: Bk-1 = true if (s+W[k]==M) print X[1..k] // subset found else (if s+W[k]+W[k+1]<=M) // Bk = true SumOfSub(s+W[k],k+1,r-W[k]) // do not take object k, // now we have to check whether the rest // of the objects still can reach M and // the lightest remaining object is not too heavy if (s + r - W[k] >= M // still can reach M and s+W[k+1] <=M // rest objects not too heavy ) SumOfSub(s,k+1,r-W[k]) }

  38. SumOfSub(s,k,r) { // two possible next states X[k] = 1 //1: take object k, we can: Bk-1 = true if (s+W[k]==M) print X[1..k] // subset found else (if s+W[k]+W[k+1]<=M) // Bk = true SumOfSub(s+W[k],k+1,r-W[k]) // do not take object k, // now we have to check whether the rest // of the objects still can reach M and // the lightest remaining object is not too heavy if (s + r - W[k] >= M // still can reach M and s+W[k+1] <=M // rest objects not too heavy ) SumOfSub(s,k+1,r-W[k]) } Notice that the algorithm does not check k<=n. This is implicit in the fact that when SumOfSub is called s<M and s+r>M, hence r>0 and thus k<=n.

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