第四章 Dynamic Programming 技术

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第四章 Dynamic Programming 技术. 邹权(博士) 计算机科学系. 4. 1 Introduction. Fibonacci number F ( n ). 1if n = 0 or 1 F ( n -1) + F ( n -2) if n > 1. F ( n ) =. Pseudo code for the recursive algorithm: F ( n ) 1 if n =0 or n =1 then return 1 2 else return F ( n -1) + F ( n -2). - PowerPoint PPT Presentation

Text of 第四章 Dynamic Programming 技术

  • Dynamic Programming

  • 4.1 IntroductionFibonacci number F(n)F(n) = 1if n = 0 or 1F(n-1) + F(n-2)if n > 1Pseudo code for the recursive algorithm:F(n)1if n=0 or n=1 then return 12elsereturn F(n-1) + F(n-2)

    n012345678910F(n)1123581321345589

  • The execution of F(7)F7F6F5F5F4F3F3F2F1F0F2F1F0F1F2F1F0F2F1F0F2F1F0F4F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1F1

  • The execution of F(7)Computation of F(2) is repeated 8 times!F7F6F5F5F4F3F3F2F1F0F2F1F0F1F2F1F0F2F1F0F2F1F0F4F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1F1

  • The execution of F(7)Computation of F(3) is also repeated 5 times!F7F6F5F5F4F3F3F2F1F0F2F1F0F1F2F1F0F2F1F0F2F1F0F4F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1F1

  • The execution of F(7)Many computations are repeated!!How to avoid this?F7F6F5F5F4F3F3F2F1F0F2F1F0F1F2F1F0F2F1F0F2F1F0F4F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1F1

  • Idea for improvementMemorization Store F1(i) somewhere after we have computed its value Afterward, we dont need to re-compute F1(i); we can retrieve its value from our memory.F1(n)1 if v[n] < 0 then2 v[n] F1(n-1)+F1(n-2)3 return v[n]Main()1 v[0] = v[1] 12 for i 2 to n do3 v[i] = -14 output F1(n)

  • Look at the execution of F(7)v[0]

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    11-1-1-1-1-1-1

  • Look at the execution of F(7)v[0]

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    11-1-1-1-1-1-1

  • Look at the execution of F(7)v[0]

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    112-1-1-1-1-1

  • Look at the execution of F(7)v[0]

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    112-1-1-1-1-1

  • Look at the execution of F(7)v[0]

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    1123-1-1-1-1

  • Look at the execution of F(7)v[0]

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    1123-1-1-1-1

  • Look at the execution of F(7)F1F0v[0]

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    11235-1-1-1

  • Look at the execution of F(7)F1F0F2F1F0F1F7F6F5F5F4F3F3F2F1F0F2F1F2F1F0F2F1F0F4F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1v[0]

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    11235-1-1-1

  • Look at the execution of F(7)F1F0F2F1F0F1F7F6F5F5F4F3F3F2F1F0F2F1F2F1F0F2F1F0F4F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1v[0]

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    112358-1-1

  • Look at the execution of F(7)F1F0F2F1F0F2F1F0F2F1F0F3F1F1F7F6F5F5F4F3F3F2F1F0F2F1F4F4F3F3F2F1F0F2F1F0F2F1F0F1F1v[0]

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    112358-1-1

  • Look at the execution of F(7)F1F0F2F1F0F2F1F0F2F1F0F3F1F1F7F6F5F5F4F3F3F2F1F0F2F1F4F4F3F3F2F1F0F2F1F0F2F1F0F1F1v[0]

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    11235813-1

  • Look at the execution of F(7)F1F0F2F1F0F2F1F0F2F1F0F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1F1F7F6F5F5F4F3F3F2F1F0F2F1F4F4v[0]

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  • Look at the execution of F(7)F1F2F1F0F2F1F0F2F1F0F4F3F3F3F2F1F0F2F1F0F2F1F0F1F1F1F1F7F6F5F5F4F3F3F2F1F0F2F0F1F4v[0]

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    1123581321

  • Can we do even better?ObservationThe 2nd version still make many function calls, and each wastes times in parameters passing, dynamic linking, ...In general, to compute F(i), we need F(i-1) & F(i-2) onlyIdea to further improveCompute the values in bottom-up fashion.That is, compute F(2) (we already know F(0)=F(1)=1), then F(3), then F(4)This new implementation saves lots of overhead.F2(n)1 F[0] 12 F[1] 13 for i 2 to n do4F[i] F[i-1] + F[i-2]5 return F[n]

  • Recursive vs Dynamic programmingRecursive version:F(n)1if n=0 or n=1 then return 12elsereturn F(n-1) + F(n-2)Dynamic Programming version:F2(n)1 F[0] 12 F[1] 13 for i 2 to n do4F[i] F[i-1] + F[i-2]5 return F[n]TooSlow!Efficient!Time complexity is O(n)

  • Summary of the methodologyWrite down a formula that relates a solution of a problem with those of sub-problems. E.g. F(n) = F(n-1) + F(n-2).Index the sub-problems so that they can be stored and retrieved easily in a table (i.e., array)Fill the table in some bottom-up manner; start filling the solution of the smallest problem.This ensures that when we solve a particular sub-problem, the solutions of all the smaller sub-problems that it depends are available.For historical reasons, we call such methodologyDynamic Programming.In the late 40s (when computers were rare), programming refers to the "tabular method".

  • Dynamic programming VS Divide-and-conquerDivide-and-conquer method 1. Subproblem is independent. 2. Subproblem is solved repeatedly.Dynamic programming (DP) 1. Subproblem is not independent. 2. Subproblem is just solved once.Common: Problem is partitioned into one or more subproblem, then the solution of subproblem is combined.DP reduces computation by Solving subproblems in a bottom-up fashion.Storing solution to a subproblem the first time it is solved.Looking up the solution when subproblem is met again.

  • 4.2 Assembly-line schedulingEndSi[j] :The jth station on line i ( i=1 or 2) .ai[j] :The assembly time required at station Si[j]. ei :Entry time. xi: Exit time.

  • One SolutionBrute forceEnumerate all possibilities of selecting stationsCompute how long it takes in each case and choose the best oneProblem:

    There are 2n possible ways to choose stationsInfeasible when n is large

  • fi[j] = the fastest time to get from the starting point through station Si[j]Let f * denote fastest time to get a chassis all the way through the factory.

    j = 1 (getting through station 1) f1[1] = e1 + a1[1] f2[1] = e2 + a2[1]f * = min( f1[n]+x1 , f2[n]+x2 )e1e2a1[1]a1[2]a1[3]a2[1]a2[2]a2[3]a1[n-1]a1[n]a2[n-1]a2[n]x1x2t1[n-1]t2[n-1]BeginLine 1Line 2t1[1]t2[1]t1[2]t2[2]

  • 2. A Recursive Solution (cont.)Compute fi[j] for j = 2, 3, ,n, and i = 1, 2Fastest way through S1[j] is either:fastest way through S1[j-1]then directly through S1[j], or f1[j] = f1[j - 1] + a1[j]fastest way through S2[j-1], transfer from line 2 to line 1, then through S1[j] f1[j] = f2[j -1] + t2[j-1] + a1[j]

    f1[j] = min(f1[j - 1] + a1[j], f2[j -1] + t2[j-1] + a1[j])a1[j]a1[j-1]a2[j-1]t2[j-1]S1[j]S1[j-1]S2[j-1]

  • The recursive equation

    For example, n=5:

    Solving top-down would result in exponential running timef1[j]f2[j]12345f1[5]f2[5]f1[4]f2[4]f1[3]f2[3]2 times4 timesf1[2]f2[2]f1[1]f2[1]

  • 3. Computing the Optimal SolutionFor j 2, each value fi[j] depends only on the values of f1[j 1] and f2[j - 1]Compute the values of fi[j]in increasing order of j

    Bottom-up approachFirst find optimal solutions to subproblemsFind an optimal solution to the problem from the subproblemsf1[j]f2[j]12345

  • Additional InformationTo construct the optimal solution we need the sequence of what line has been used at each station:li[j] the line number (1, 2) whose station (j - 1) has been used to get in fastest time through Si[j], j = 2, 3, , nl* - the line whose station n is used to get in the fastest way through the entire factoryl1[j]l2[j]2345

  • Step 3: Computing the fastest wayDPFastestWay(a, t, e, x, n)1 f1[1] e1 + a1[1]; f2[1] e2 + a2[1]2 for j 2 to n do3 if f1[j - 1] + a1[j] f2 [j - 1] + t2[j-1] + a1[j] then4 f1[j] f1[j - 1] + a1[j]5 l1[j] 16 else f1[j] f2 [j - 1] + t2[j-1] + a1[j] 7 l1[j] 28 if f2[j - 1] + a2[j] f1[j - 1] + t1[j-1] + a2[j] then9 f2[j] f2[j - 1] + a2[j] 10 l2[j] 211 else f2 [j] f1[j - 1] + t1[j-1]+ a2[j] then 12 l2[j] 113 if f1[n] + x1f2[n] + x2 then14 f* f1[n] + x115 l* 116 else f* f2[n] + x2 17 l* 2Running time: (n)

  • For exampleBeginEnd123456jf1[j]f2[j]f *=389121816202224253230353723456jl1[j]l2[j]l *=12111112222

  • 4. Construct an Optimal SolutionPrintStations(l, n)1 i l* 2 print line i , station n3 for j n downto 2 do4 i li[j]5 print line i , station j - 1line 1, station 6line 2, station 5line 2, station 4line 1, station 3line 2, station 223456jl1[j]l2[j]l *=12111112222line 1, station 1

  • 24327944885475233421213612BeginEndThe fastest assembly wayf *=38

  • 4.3 Matrix-chain multiplication Problem: Given a chain A1, A2, . . . , An of n matrices, where for i = 1, 2, . . . , n , matrix Ai has dimension pi-1 pi, fully parenthesize the product A1 A2...An in a way that minimizes the number of scalar multiplications. A1 A2 Ai Ai+1 An p0 p1 p1 p2 pi-1 pi pi pi+1 pn-1 pn A product of matrices is fully parenthesized if it is either a single matrix or the product of two fully parenthesized matrix pr