Dynamic Programming. Lecture 7 Prof. Dr. Aydın Öztürk. Introduction. Dynamic Programming(DP) applies to optimi- zation problems in which a set of choices must be made in order to arrive at an optimal solution. As choices are made, subproblems of the same form arise.
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Introduction • Dynamic Programming(DP) applies to optimi- zation problems in which a set of choices must be made in order to arrive at an optimal solution. • As choices are made, subproblems of the same form arise. • DP is effective when a given problem may arise from more than one partial set of choices. • The key technique is to store the solution to each subproblem in case it should appear
Introduction (cont.) • Divide and Conquer algorithms partition the problem into independent subproblems. • Dynamic Programming is applicable when the subproblems are not independent.(In this case DP algorithm does more work than necessary)
Introduction (cont.) • Dynamic Programming algorithm solves every subproblem just once and then saves its answer in a table.)
Assembly-line Scheduling • If we are given a list of which stations to use in line 1 and which to use in line 2, then T(n)=Θ(n). • There are 2n possible ways to choose stations. Thus, the solution is obtained by enumarating all posible ways and computing how long each takes would take Ω(2n).
The steps of a dynamic programming • Characterize the structure of an optimal solution • Recursively define the value of an optimal solution • Compute the value of an optimal solution in a bottom-up fashion • Construct an optimal solution from computed information
Step 1:The structure of the fastest way through the factory • Consider the fastest possible way(fpw) to get from the starting point through station S1,j. • If j=1, there isonly one way thatthe chassis could have gone. • For j=2, 3, ..., n there are two choices: • The chassis could have come from S1, j-1and then directly to station S1, j. • The chassis could have come from S2, j-1and then directly to station S1, j-1.
Step 1:The structure of the fastest way through the factory (cont.) • First, suppose that the fastest way(fw) through station S1, j is throughstation S1, j-1. • The key observation is that the chassis must have taken a fw from the starting point through station S1, j-1.
Step 1:The structure of the fastest way through the factory (cont.) • Similarly, suppose that the fw through station S1, j is throughstation S2, j-1. • We observe that the chassis must have taken a fw from tha starting point through station S2, j-1.
Step 1:The structure of the fastest way through the factory (cont.) • We can say that for assembly-line scheduling, an optimal solution to a problem contains within it an optimal solution to sub-problems (finding the fw through either S1, j-1or S2, j-1). • We refer to this property as optimal substructure.
The fw through stations S1, j • The fw through station S1, j-1and then directly through station S1, j. or • The fw through station S2, j-1and then through station S1, j.
The fw through station S2, j • The fw through station S2, j-1and then directly through station S2, j. or • The fw through station S1, j-1and then through station S2, j.
Step 1:The structure of the fastest way through the factory (cont.) • To solve the problem of finding the fw through station j of either line, we solve the subproblems of finding the fw through station j-1 on both lines. • Thus, we can build an optimal solution to an instance of the assemly line scheduling problem by building on optimal solutions to subproblems
Step 2: A recursive solution • We define the value of an optimal solution recursively in tems of the optimal solutions to subproblems. • Let fi[j] denote the fastest possible time to get a chassis from the starting point throuhg station Si, j. • Let fi* denote the fastest time to get a chassis all the way through the factory.
Step 2: A recursive solution: An example Station S1,1 Station S1,2 Station S1,3 Station S1,4 Station S1,5 Station S1,6 9 3 4 Line 1 7 4 8 2 3 2 1 3 3 4 Chassis Enters 5 2 1 2 1 4 2 5 6 4 8 5 7 Line 2 Station S1,1 Station S1,2 Station S1,3 Station S1,4 Station S1,5 Station S1,6 j 1 2 3 4 5 6 j 2 3 4 5 6 f1[j] 9 18 20 24 32 35 1 2 1 1 2 l1[j] f*=38 l* =1 12 16 22 25 30 37 f2[j] 1 2 1 2 2 l2[j]
Step 2: A recursive solution (cont.) • fi[j] values give the values of optimal solutions to subproblems. • Let li[j] to be the line number 1 or 2, whose station j-1 is used in a fw throgh station Si, j. • We also define l* to be the line whose station nis used in a fw through the entire factory.
Step 3: Computing the fastest times We can write a simple recursive algorithm based on the equations for f1[j] and f2[j] to compute the fw through the factory. However its running time is exponential. Let ri( j ) be the number of references made tofi [ j ] in a recursive algorithm:
Step 3: Computing the fastest times(cont.) • We observe that for each value of fi[j] depends only on the values of f1[j-1] and f2[j-1]. • By computing the fi[j] values in order of increasing station numbers we can compute the fw through the factory in time.
Matrix-Chain multiplication (cont.) • Matrix multiplication is assosiative, and so all parenthesizations yield the same product. • For example, if the chain of matrices is then the product can be fully paranthesized in five distinct way:
Matrix-Chain multiplication MATRIX-MULTIPLY (A,B) if columns[A] ≠ rows[B] then error “incompatible dimensions” else for i←1 to rows[A] do for j←1 to columns[B] do C[i,j]←0 for k←1 to columns[A] do C[ i,j ]← C[i,j]+A[i,k]B[k,j] return C
Matrix-Chain multiplication (cont.) • The problem: Given a chain of nmatrices, where matrix Ai has dimension pi-1xpi, fully paranthesize the product in a way that minimizes the number of scalar multiplications.
Matrix-Chain multiplication (cont.) • One way to express B(x) in closed form is Expanding into a Taylor series about x=0 we get Comparing the coefficients of x we finally show that
Matrix-Chain multiplication (cont.) Step 1: The structure of an optimal paranthesization(op) • Find the optimal substructure and then use it to construct an optimal solution to the problem from optimal solutions to subproblems. • Let Ai...jwhere i ≤ j, denote the matrix product Ai Ai+1 ... Aj • Anyparenthesization of Ai Ai+1 ... Aj must split the product between Ak andAk+1 fori ≤ k < j.
Matrix-Chain multiplication (cont.) The optimal substructure of the problem: • Suppose that an op of Ai Ai+1 ... Aj splits the productbetween Akand Ak+1 then the paranthesization of the subchain Ai Ai+1 ... Ak within this parantesization of Ai Ai+1 ... Aj must be an op of Ai Ai+1 ... Ak
Matrix-Chain multiplication (cont.) Step 2: A recursive solution: • Let m[i,j] be the minimum number of scalar multiplications needed to compute the matrix Ai...jwhere 1≤ i ≤ j ≤ n. • Thus, the cost of a cheapest way to compute A1...nwould be m[1,n]. • Assume that the op splits the product Ai...jbetween Akand Ak+1.wherei ≤ k <j. • Then m[i,j] =Theminimum cost for computing Ai...k and Ak+1...j+ the cost of multiplying these two matrices.
Matrix-Chain multiplication (cont.) To help us keep track of how to constrct an optimal solution we define s[ i,j] to be a value of kat which we can splitthe product Ai...jto obtain an optimal paranthesization. That is s[ i,j] equals a value k such that
Matrix-Chain multiplication (cont.) Step 3: Computing the optimal costs It is easiy to write a recursive algorithm based on recurrence for computing m [i,j]. But the running time will be exponential!...
Matrix-Chain multiplication (Contd.) MATRIX-CHAIN-ORDER(p) n←length[p]-1 for i←1 to n do m[i,i]←0 for l←2to n do for i←1 to n-l+1 do j←i+l-1 m[i,j]←∞ for k←ito j-1 do q←m[i,k] + m[k+1,j]+pi-1pkpj if q < m[i,j] then m[i,j] ←qs[i,j] ←k return m and s
Matrix-Chain multiplication (cont.) Step 4: Constructing an optimal solution An optimal solution can be constructed from the computed information stored in the table s[1...n, 1...n]. We know that the final matrix multiplication is The earlier matrix multiplication can be computed recursively.
Matrix-Chain multiplication (Contd.) PRINT-OPTIMAL-PARENS (s, i, j) 1 if i=j • then print “Ai” • else print “ ( “ • PRINT-OPTIMAL-PARENS (s, i, s[i,j]) • PRINT-OPTIMAL-PARENS (s, s[i,j]+1, j) 6 Print “ ) ”
