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COSC 3101A - Design and Analysis of Algorithms 7

COSC 3101A - Design and Analysis of Algorithms 7. Dynamic Programming Assembly-Line Scheduling Matrix-Chain Multiplication Elements of DP. Many of these slides are taken from Monica Nicolescu, Univ. of Nevada, Reno, monica@cs.unr.edu. Dynamic Programming.

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COSC 3101A - Design and Analysis of Algorithms 7

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  1. COSC 3101A - Design and Analysis of Algorithms7 Dynamic Programming Assembly-Line Scheduling Matrix-Chain Multiplication Elements of DP Many of these slides are taken from Monica Nicolescu, Univ. of Nevada, Reno, monica@cs.unr.edu

  2. Dynamic Programming • An algorithm design technique (like divide and conquer) • Divide and conquer • Partition the problem into independent subproblems • Solve the subproblems recursively • Combine the solutions to solve the original problem COSC3101A

  3. Dynamic Programming • Applicable when subproblems are not independent • Subproblems share subsubproblems • A divide and conquer approach would repeatedly solve the common subproblems • Dynamic programming solves every subproblem just once and stores the answer in a table • Used for optimization problems • A set of choices must be made to get an optimal solution • Find a solution with the optimal value (minimum or maximum) • There may be many solutions that return the optimal value: an optimal solution COSC3101A

  4. Dynamic Programming Algorithm • 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 COSC3101A

  5. Assembly Line Scheduling • Automobile factory with two assembly lines • Each line has n stations: S1,1, . . . , S1,n and S2,1, . . . , S2,n • Corresponding stations S1, j and S2, j perform the same function but can take different amounts of time a1, j and a2, j • Entry times e1 and e2and exit times x1 and x2 COSC3101A

  6. Assembly Line • After going through a station, can either: • stay on same line at no cost, or • transfer to other line: cost after Si,j is ti,j , j = 1, . . . , n - 1 COSC3101A

  7. Assembly Line Scheduling • Problem: what stations should be chosen from line 1 and which from line 2 in order to minimize the total time through the factory for one car? COSC3101A

  8. 1 0 0 1 1 1 2 3 4 n 0 if choosing line 2 at step j (= 3) 1 if choosing line 1 at step j (= n) One Solution • Brute force • Enumerate all possibilities of selecting stations • Compute how long it takes in each case and choose the best one • Problem: • There are 2n possible ways to choose stations • Infeasible when n is large COSC3101A

  9. 1. Structure of the Optimal Solution • Let’s consider the fastest way possible to get from the starting point through station S1,j • If j = 1: determine how long it takes to get through S1,1 • If j ≥ 2, have two choices of how to get to S1, j: • Through S1, j - 1, then directly to S1, j • Through S2, j - 1, then transfer over to S1, j COSC3101A

  10. 1. Structure of the Optimal Solution • Suppose that the fastest way through S1, j is through S1, j – 1 • We must have taken a fastest way from entry through S1, j – 1 • If there were a faster way through S1, j - 1, we would use it instead • Similarly for S2, j – 1 COSC3101A

  11. Optimal Substructure • Generalization: an optimal solution to the problem find the fastest way through S1, j contains within it an optimal solution to subproblems: find the fastest way through S1, j - 1 or S2, j - 1. • This is referred to as the optimal substructure property • We use this property to construct an optimal solution to a problem from optimal solutions to subproblems COSC3101A

  12. 2. A Recursive Solution • Define the value of an optimal solution in terms of the optimal solution to subproblems • Assembly line subproblems • Finding the fastest way through station j on both lines, j = 1, 2, …, n COSC3101A

  13. 2. A Recursive Solution (cont.) • fi[j] = the fastest time to get from the starting point through station Si,j • j = 1 (getting through station 1) f1[1] = e1 + a1,1 f2[1] = e2 + a2,1 COSC3101A

  14. S1,j-1 S1,j a1,j-1 a1,j t2,j-1 a2,j-1 S2,j-1 2. A Recursive Solution (cont.) • Compute fi[j] for j = 2, 3, …,n, and i = 1, 2 • Fastest 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) COSC3101A

  15. 2. A Recursive Solution (cont.) • f* = the fastest time to get through the entire factory f* = min (f1[n] + x1, f2[n] + x2) COSC3101A

  16. 2. A Recursive Solution (cont.) e1 + a1,1 if j = 1 f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j) if j ≥ 2 e2 + a2,1 if j = 1 f2[j] = min(f2[j - 1] + a2,j ,f1[j -1] + t1,j-1 + a2,j) if j ≥ 2 COSC3101A

  17. 3. Computing the Optimal Solution f* = min (f1[n] + x1, f2[n] + x2) f1[j] = min(f2[j - 1] + a2,j ,f1[j -1] + t1,j-1 + a2,j) • Solving top-down would result in exponential running time 1 2 3 4 5 f1[j] f1(1) f1(2) f1(3) f1(4) f1(5) f2[j] f2(1) f2(2) f2(3) f2(4) f2(5) 2 times 4 times COSC3101A

  18. increasing j 3. Computing the Optimal Solution • For 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 approach • First find optimal solutions to subproblems • Find an optimal solution to the problem from the subproblems 1 2 3 4 5 f1[j] f2[j] COSC3101A

  19. increasing j Additional Information • To 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, …, n • l* - the line whose station n is used to get in the fastest way through the entire factory 2 3 4 5 l1[j] l2[j] COSC3101A

  20. Example e1 + a1,1, if j = 1 f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j) if j ≥ 2 1 2 3 4 5 f1[j] 9 18[1] 20[2] 24[1] 32[1] f* = 35[1] f2[j] 12 16[1] 22[2] 25[1] 30[2] COSC3101A

  21. 4. Construct an Optimal Solution Alg.: PRINT-STATIONS(l, n) • i ← l* • print “line ” i “, station ” n • for j ← ndownto 2 • do i ←li[j] • print “line ” i “, station ” j - 1 line 1, station 5 line 1, station 4 line 1, station 3 line 2, station 2 line 1, station 1 1 2 3 4 5 f1[j]/l1[j] 9 18[1] 20[2] 24[1] 32[1] l* = 1 f2[j]/l2[j] 12 16[1] 22[2] 25[1] 30[2] COSC3101A

  22. Dynamic Programming Algorithm • Characterize the structure of an optimal solution • Fastest time through a station depends on the fastest time on previous stations • Recursively define the value of an optimal solution • f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j) • Compute the value of an optimal solution in a bottom-up fashion • Fill in the fastest time table in increasing order of j (station #) • Construct an optimal solution from computed information • Use an additional table to help reconstruct the optimal solution COSC3101A

  23. Matrix-Chain Multiplication Problem: given a sequence A1, A2, …, An, compute the product: A1  A2  An • Matrix compatibility: C = A  B colA = rowB rowC = rowA colC = colB A1  A2  Ai  Ai+1  An coli = rowi+1 COSC3101A

  24. Matrix-Chain Multiplication • In what order should we multiply the matrices? A1  A2  An • Parenthesize the product to get the order in which matrices are multiplied • E.g.:A1  A2  A3 = ((A1  A2)  A3) = (A1  (A2  A3)) • Which one of these orderings should we choose? • The order in which we multiply the matrices has a significant impact on the cost of evaluating the product COSC3101A

  25. rows[A]  cols[A]  cols[B] multiplications k k MATRIX-MULTIPLY(A, B) ifcolumns[A]  rows[B] then error“incompatible dimensions” else fori  1 to rows[A] do forj  1 to columns[B] doC[i, j] = 0 fork  1 to columns[A] doC[i, j]  C[i, j] + A[i, k] B[k, j] j cols[B] j cols[B] i i = * B A C rows[A] rows[A] COSC3101A

  26. Example A1  A2  A3 • A1: 10 x 100 • A2: 100 x 5 • A3: 5 x 50 1. ((A1  A2)  A3): A1  A2 = 10 x 100 x 5 = 5,000 (10 x 5) ((A1  A2)  A3) = 10 x 5 x 50 = 2,500 Total: 7,500 scalar multiplications 2. (A1  (A2  A3)): A2  A3 = 100 x 5 x 50 = 25,000 (100 x 50) (A1  (A2  A3)) = 10 x 100 x 50 = 50,000 Total: 75,000 scalar multiplications one order of magnitude difference!! COSC3101A

  27. Matrix-Chain Multiplication • Given a chain of matrices A1, A2, …, An, where for i = 1, 2, …, n matrix Ai has dimensions pi-1x 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 x p1 p1 x p2 pi-1 x pi pi x pi+1 pn-1 x pn COSC3101A

  28. 1. The Structure of an Optimal Parenthesization • Notation: Ai…j = Ai Ai+1 Aj, i  j • For i < j: Ai…j = Ai Ai+1 Aj = Ai Ai+1 Ak Ak+1  Aj = Ai…k Ak+1…j • Suppose that an optimal parenthesization of Ai…j splits the product between Ak and Ak+1, where i  k < j COSC3101A

  29. Optimal Substructure Ai…j= Ai…k Ak+1…j • The parenthesization of the “prefix” Ai…kmust be an optimal parentesization • If there were a less costly way to parenthesize Ai…k, we could substitute that one in the parenthesization of Ai…j and produce a parenthesization with a lower cost than the optimum  contradiction! • An optimal solution to an instance of the matrix-chain multiplication contains within it optimal solutions to subproblems COSC3101A

  30. 2. A Recursive Solution • Subproblem: determine the minimum cost of parenthesizing Ai…j = Ai Ai+1 Ajfor 1  i  j  n • Let m[i, j] = the minimum number of multiplications needed to compute Ai…j • Full problem (A1..n): m[1, n] COSC3101A

  31. 2. A Recursive Solution (cont.) • Define m recursively: • i = j: Ai…i = Ai m[i, i] = 0, for i = 1, 2, …, n • i < j: assume that the optimal parenthesization splits the product Ai Ai+1 Aj between Ak and Ak+1, i  k < j Ai…j= Ai…k Ak+1…j m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj Ai…k Ak+1…j Ai…kAk+1…j COSC3101A

  32. pi-1pkpj m[i, k] m[k+1,j] 2. A Recursive Solution (cont.) • Consider the subproblem of parenthesizing Ai…j = Ai Ai+1 Ajfor 1  i  j  n = Ai…k Ak+1…j for i  k < j • m[i, j] = the minimum number of multiplications needed to compute the product Ai…j m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj min # of multiplications to compute Ai…k min # of multiplications to compute Ak+1…j # of multiplications to computeAi…kAk…j COSC3101A

  33. 2. A Recursive Solution (cont.) m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj • We do not know the value of k • There are j – i possible values for k: k = i, i+1, …, j-1 • Minimizing the cost of parenthesizing the product Ai Ai+1 Aj becomes: 0if i = j m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}if i < j ik<j COSC3101A

  34. Reconstructing the Optimal Solution • Additional information to maintain: • s[i, j] = a value of k at which we can split the product Ai Ai+1 Aj in order to obtain an optimal parenthesization COSC3101A

  35. 3. Computing the Optimal Costs 0if i = j m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}if i < j ik<j • How many subproblems do we have? • Parenthesize Ai…j for 1  i  j  n • One problem for each choice of i and j • A recurrent algorithm may encounter each subproblem many times in different branches of the recursion  overlapping subproblems • Compute a solution using a tabular bottom-up approach  (n2) COSC3101A

  36. 3. Computing the Optimal Costs (cont.) 0if i = j m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}if i < j ik<j • How do we fill in the tables m[1..n, 1..n] and s[1..n, 1..n] ? • Determine which entries of the table are used in computing m[i, j] • m[i, j] = cost of computing a product of j – i – 1 matrices • m[i, j] depends only on costs for products of fewer than j – i – 1 matrices Ai…j= Ai…k Ak+1…j • Fill in m such that it corresponds to solving problems of increasing length COSC3101A

  37. second first m[1, n] gives the optimal solution to the problem 3. Computing the Optimal Costs(cont.) 0if i = j m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}if i < j ik<j • Length = 0: i = j, i = 1, 2, …, n • Length = 1: j = i + 1, i = 1, 2, …, n-1 1 2 3 n n j 3 Compute rows from bottom to top and from left to right In a similar matrix s we keep the optimal values of k 2 1 i COSC3101A

  38. Example: min {m[i, k] + m[k+1, j] + pi-1pkpj} m[2, 2] + m[3, 5] + p1p2p5 m[2, 3] + m[4, 5] + p1p3p5 m[2, 4] + m[5, 5] + p1p4p5 k = 2 m[2, 5] = min k = 3 k = 4 1 2 3 4 5 6 6 5 • Values m[i, j] depend only on values that have been previously computed 4 j 3 2 1 i COSC3101A

  39. 2 0 25000 2 1 0 7500 5000 0 Example 1 2 3 Compute A1  A2  A3 • A1: 10 x 100 (p0 x p1) • A2: 100 x 5 (p1 x p2) • A3: 5 x 50 (p2 x p3) m[i, i] = 0 for i = 1, 2, 3 m[1, 2] = m[1, 1] + m[2, 2] + p0p1p2 (A1A2) = 0 + 0 + 10 *100* 5 = 5,000 m[2, 3] = m[2, 2] + m[3, 3] + p1p2p3 (A2A3) = 0 + 0 + 100 * 5 * 50 = 25,000 m[1, 3] = min m[1, 1] + m[2, 3] + p0p1p3 = 75,000 (A1(A2A3)) m[1, 2] + m[3, 3] + p0p2p3 = 7,500 ((A1A2)A3) 3 2 1 COSC3101A

  40. l =3 l = 2 35*15*5=2625 10*20*25=5000 m[3,4]+m[5,5] + 15*10*20 =750 + 0 + 3000 = 3750 m[3,5] = min m[3,3]+m[4,5] + 15*5*20=0 + 1000 + 1500 = 2500 COSC3101A

  41. 4. Construct the Optimal Solution • Store the optimal choice made at each subproblem • s[i, j] = a value of k such that an optimal parenthesization of Ai..j splits the product between Ak and Ak+1 • s[1, n] is associated with the entire product A1..n • The final matrix multiplication will be split at k = s[1, n] A1..n = A1..s[1, n] As[1, n]+1..n • For each subproduct recursively find the corresponding value of k that results in an optimal parenthesization COSC3101A

  42. 4. Construct the Optimal Solution(cont.) • s[i, j] = value of k such that the optimal parenthesization of Ai Ai+1 Aj splits the product between Ak and Ak+1 • A1..n = A1..s[1, n] As[1, n]+1..n 1 2 3 4 5 6 6 • s[1, n] = 3  A1..6 = A1..3 A4..6 • s[1, 3] = 1  A1..3 = A1..1 A2..3 • s[4, 6] = 5  A4..6 = A4..5 A6..6 5 4 3 j 2 1 i COSC3101A

  43. 4. Construct the Optimal Solution (cont.) PRINT-OPT-PARENS(s, i, j) if i = j then print “A”i else print “(” PRINT-OPT-PARENS(s, i, s[i, j]) PRINT-OPT-PARENS(s, s[i, j] + 1, j) print “)” 1 2 3 4 5 6 6 5 4 j 3 2 1 i COSC3101A

  44. Example: A1  A6 ( ( A1 ( A2 A3 ) ) ( ( A4 A5 ) A6 ) ) s[1..6, 1..6] 1 2 3 4 5 6 PRINT-OPT-PARENS(s, i, j) if i = j then print “A”i else print “(” PRINT-OPT-PARENS(s, i, s[i, j]) PRINT-OPT-PARENS(s, s[i, j] + 1, j) print “)” 6 5 4 j 3 2 1 P-O-P(s, 1, 6) s[1, 6] = 3 i = 1, j = 6 “(“ P-O-P (s, 1, 3) s[1, 3] = 1 i = 1, j = 3 “(“ P-O-P(s, 1, 1)  “A1” P-O-P(s, 2, 3) s[2, 3] = 2 i = 2, j = 3 “(“ P-O-P (s, 2, 2)  “A2” P-O-P (s, 3, 3)  “A3” “)” “)” i … COSC3101A

  45. Elements of Dynamic Programming • Optimal Substructure • An optimal solution to a problem contains within it an optimal solution to subproblems • Optimal solution to the entire problem is built in a bottom-up manner from optimal solutions to subproblems • Overlapping Subproblems • If a recursive algorithm revisits the same subproblems over and over  the problem has overlapping subproblems COSC3101A

  46. Optimal Substructure - Examples • Assembly line • Fastest way of going through a station j contains: the fastest way of going through station j-1 on either line • Matrix multiplication • Optimal parenthesization of Ai Ai+1 Aj that splits the product between Ak and Ak+1 contains: an optimal solution to the problem of parenthesizing Ai..k and Ak+1..j COSC3101A

  47. Discovering Optimal Substructure • Show that a solution to a problem consists of making a choice that leaves one or more similar problems to be solved • Suppose that for a given problem you are given the choice that leads to an optimal solution • Given this choice determine which subproblems result • Show that the solutions to the subproblems used within the optimal solution must themselves be optimal • Cut-and-paste approach COSC3101A

  48. Parameters of Optimal Substructure • How many subproblems are used in an optimal solution for the original problem • Assembly line: • Matrix multiplication: • How many choices we have in determining which subproblems to use in an optimal solution • Assembly line: • Matrix multiplication: One subproblem (the line that gives best time) Two subproblems (subproducts Ai..k, Ak+1..j) Two choices (line 1 or line 2) j - i choices for k (splitting the product) COSC3101A

  49. Parameters of Optimal Substructure • Intuitively, the running time of a dynamic programming algorithm depends on two factors: • Number of subproblems overall • How many choices we look at for each subproblem • Assembly line • (n) subproblems (n stations) • 2 choices for each subproblem • Matrix multiplication: • (n2) subproblems (1  i  j  n) • At most n-1 choices (n) overall (n3) overall COSC3101A

  50. Readings • Chapter 15 COSC3101A

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