Dynamic Programming

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Dynamic Programming

• Carrie Williams

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What is Dynamic Programming?

• Method of breaking the problem into smaller,
  simpler sub-problems
• Start with the smallest sub-problems and combine
  to produce a solution
• Similar to Divide and Conquer
• But uses a bottom-top approach

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When is it used?

• Most often used with Optimization Problems
• When the Brute Force approach becomes too time
  consuming
• Brute Force is when you find all possible
  solutions and compare them

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ExampleMatrix-Chain Multiplication

• Given A sequence of matrices (A1A2,,An) with Ai
  having dimension mi-1 x m
• Cost of a Solution The number of operations
  needed to compute A1A2An
• Optimal Solution The solution with minimal cost

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MCM Example

• We have a matrix sequence of (90,20,15,50,180)
• So we have matrices of the following sizes
• A1 90 x 20
• A2 20 x 15
• A3 15 x 50
• A4 50 x 180

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MCM Example Cont.

• We want to compute A1A2A3A4
• 5 possible ways
• (A1(A2(A3A4)))
• (A1((A2A3)A4))
• ((A1A2)(A3A4))
• ((A1(A2A3))A4)
• (((A1A2)A3)A4)
• Recall that the number of operations needed to
  compute an m x n matrix with a n x p is mnp

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MCM Example Cont.

• We could try all possible solutions and see what
  solution gives us the least cost
• OR
• We could use the method of dynamic programming
• Finding the optimal solution by finding the
  optimal solution of sub-problems

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Finding the Solution

• Start with the computation of two matrices
• Let Mi,j be the cost of the optimal solution
• Mi,jmin Mi,k Mk1,j mi-1mkmj
• iltkltj
• Now we must compute Mi,j, igt1, jltn, from the
  bottom up

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Finding the Solution cont.

• Using the previous example, we need to compute
  Mi,j for igt1, jlt4
• Mi,i0, for all i
• The solutions for 2 matrices are the following
• M1,2 27,000
• M2,3 15,000
• M3,4 135,000

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Chart of Optimal Solutions
4 3 2 1
1 27,000 0
2 15,000 0
3 135,000 0
4 0
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Computing the Cost of Three Matrices

• M1,3 min M1,2 M3,3 m0m2m3
• M1,1 M2,3
  m0m1m3 27,000 0 901550 94,500
• 0 15,000 902050 105,000
• M2,4 min M2,3 M4,4 m1m3m4
• M2,2
  M3,4 m1m2m4
• 15,000 0 2050180 195,000
• 0 135,000 2050180 315,000
• Minimum for M1,3 94,500 ((A1A2)A3)
• Minimum for M2,4 195,000 (A2(A3A4))

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Chart of Optimal Solutions
4 3 2 1
1 94,500 27,000 0
2 195,000 15,000 0
3 135,000 0
4 0
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Computing the Cost of Four Matrices

• M1,4 min M1,3 M4,4 m0m3m4
• M1,2 M3,4 m0m2m4
• M1,1 M2,4 m0m1m4
• 94,500 0 9050180 904,000
• 27,000 135,000
  9015180405,000
• 0 195,000 9020180 519,000
• Hence, the minimum cost is 405,000, which is
  ((A1A2)(A3A4))

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Chart of Optimal Solutions
4 3 2 1
1 405,000 94,500 27,000 0
2 195,000 15,000 0
3 135,000 0
4 0
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Conclusion

• We used the method of dynamic programming to
  solve the matrix-chain multiplication problem
• We broke the problem into three sub-problems
• The minimal cost of two matrices, then three
  matrices, and finally four matrices
• We found the solution by using the sub-problems
  optimal solutions