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Analysis of Algorithms CS 477/677

Analysis of Algorithms CS 477/677. Instructor: Monica Nicolescu Lecture 12. Dynamic Programming. An algorithm design technique for optimization problems (similar to divide and conquer) Divide and conquer Partition the problem into independent subproblems

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Analysis of Algorithms CS 477/677

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  1. Analysis of AlgorithmsCS 477/677 Instructor: Monica Nicolescu Lecture 12

  2. Dynamic Programming • An algorithm design technique for optimization problems (similar to divide and conquer) • Divide and conquer • Partition the problem into independentsubproblems • Solve the subproblems recursively • Combine the solutions to solve the original problem CS 477/677 - Lecture 12

  3. Dynamic Programming • Used for optimization problems • Find a solution with the optimal value (minimum or maximum) • A set of choices must be made to get an optimal solution • There may be many solutions that return the optimal value: we want to find one of them • Applicable when subproblems are not independent • Subproblems share subsubproblems CS 477/677 - Lecture 12

  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 CS 477/677 - Lecture 12

  5. 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 again and again  the problem has overlapping subproblems CS 477/677 - Lecture 12

  6. 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 • Times to enter are e1 and e2and times to exit are x1 and x2 CS 477/677 - Lecture 12

  7. Assembly Line • After going through a station, the car can either: • stay on same line at no cost, or • transfer to other line: cost after Si,j is ti,j , i = 1, 2, j = 1, . . . , n-1 CS 477/677 - Lecture 12

  8. Assembly Line Scheduling • Problem: What stations should be chosen from line 1 and what from line 2 in order to minimize the total time through the factory for one car? CS 477/677 - Lecture 12

  9. S1,j-1 S1,j a1,j-1 a1,j t2,j-1 a2,j-1 S2,j-1 1. Structure of the Optimal Solution • Let’s consider all possible ways to get from the starting point through station S1,j • We 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 CS 477/677 - Lecture 12

  10. 2. A Recursive Solution • f* = the fastest time to get through the entire factory • fi[j] = the fastest time to get from the starting point through station Si,j f* = min (f1[n] + x1, f2[n] + x2) CS 477/677 - Lecture 12

  11. 2. A Recursive Solution • 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 CS 477/677 - Lecture 12

  12. S1,j-1 S1,j a1,j-1 a1,j t2,j-1 a2,j-1 S2,j-1 2. A Recursive Solution • Compute fi[j] for j = 2, 3, …,n, and i = 1, 2 • Fastest way through S1, j is either: • the way through S1, j - 1 then directly through S1, j, or f1[j - 1] + a1,j • the way through S2, j - 1, transfer from line 2 to line 1, then through S1, 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) CS 477/677 - Lecture 12

  13. 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] CS 477/677 - Lecture 12

  14. increasing j 4. Construct the Optimal Solution • We need the information about which 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 number (1, 2) whose station n has been used to get in fastest time through the exit point 2 3 4 5 l1[j] l2[j] CS 477/677 - Lecture 12

  15. Compute initial values of f1 and f2 FASTEST-WAY(a, t, e, x, n) • f1[1] ← e1 + a1,1 • f2[1] ← e2 + a2,1 • for j ← 2to n • do if f1[j - 1] + a1,j ≤ f2[j - 1] + t2, j-1 + a1, j • then f1[j] ← f1[j - 1] + a1, j • l1[j] ← 1 • else f1[j] ← f2[j - 1] + t2, j-1 + a1, j • l1[j] ← 2 • if f2[j - 1] + a2, j ≤ f1[j - 1] + t1, j-1 + a2, j • then f2[j] ← f2[j - 1] + a2, j • l2[j] ← 2 • else f2[j] ← f1[j - 1] + t1, j-1 + a2, j • l2[j] ← 1 Compute the values of f1[j] and l1[j] Compute the values of f2[j] and l2[j] CS 477/677 - Lecture 12

  16. FASTEST-WAY(a, t, e, x, n) (cont.) • if f1[n] + x1 ≤ f2[n] + x2 • then f* = f1[n] + x1 • l* = 1 • else f* = f2[n] + x2 • l* = 2 Compute the values of the fastest time through the entire factory CS 477/677 - Lecture 12

  17. 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] CS 477/677 - Lecture 12

  18. 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 CS 477/677 - Lecture 12

  19. Matrix-Chain Multiplication Problem: given a sequence A1, A2, …, An of matrices, 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 CS 477/677 - Lecture 12

  20. Matrix-Chain Multiplication • In what order should we multiply the matrices? A1  A2  An • Matrix multiplication is associative: • 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 overall cost of executing the entire chain of multiplications CS 477/677 - Lecture 12

  21. 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] CS 477/677 - Lecture 12

  22. Example A1  A2  A3 • A1: 10 x 100 • A2: 100 x 5 • A3: 5 x 50 • ((A1  A2)  A3): A1  A2 takes 10 x 100 x 5 = 5,000 (its size is 10 x 5) ((A1  A2)  A3) takes 10 x 5 x 50 = 2,500 Total: 7,500 scalar multiplications 2. (A1  (A2  A3)): A2  A3 takes 100 x 5 x 50 = 25,000 (its size is 100 x 50) (A1  (A2  A3)) takes 10 x 100 x 50 = 50,000 Total: 75,000 scalar multiplications one order of magnitude difference!! CS 477/677 - Lecture 12

  23. 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 xpn CS 477/677 - Lecture 12

  24. 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 CS 477/677 - Lecture 12

  25. 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 CS 477/677 - Lecture 12

  26. 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] • i = j: Ai…i = Ai m[i, i] = • 0, for i = 1, 2, …, n CS 477/677 - Lecture 12

  27. pi-1pkpj m[i, k] m[k+1,j] 2. A Recursive Solution • Consider the subproblem of parenthesizing Ai…j = Ai Ai+1 Ajfor 1  i  j  n = Ai…k Ak+1…j for i  k < j • Assume that the optimal parenthesization splits the product Ai Ai+1 Aj at k (i  k < 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 CS 477/677 - Lecture 12

  28. 2. A Recursive Solution 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}ifi < j ik<j CS 477/677 - Lecture 12

  29. 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 subproblem for each choice of i and j 1 2 3 n  (n2) n j 3 2 1 CS 477/677 - Lecture 12 i

  30. 3. Computing the Optimal Costs 0if i = j m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}ifi < j ik<j • How do we fill in table m[1..n, 1..n]? • Determine which entries of the table are used in computing m[i, j] Ai…j= Ai…k Ak+1…j • Fill in m such that it corresponds to solving problems of increasing length CS 477/677 - Lecture 12

  31. second first m[1, n] gives the optimal solution to the problem 3. Computing the Optimal Costs 0if i = j m[i, j]= min {m[i, k] + m[k+1, j] + pi-1pkpj}ifi < j ik<j • Length = 1: i = j, i = 1, 2, …, n • Length = 2: j = i + 1, i = 1, 2, …, n-1 1 2 3 n n j 3 Compute elements on each diagonal, starting with the longest diagonal. In a similar matrix s we keep the optimal values of k. 2 1 i CS 477/677 - Lecture 12

  32. 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 CS 477/677 - Lecture 12

  33. 2 0 25000 2 1 0 7500 5000 0 Example min {m[i, k] + m[k+1, j] + pi-1pkpj} 1 2 3 Compute A1  A2 A3 • A1: 10 x 100 (p0x p1) • A2: 100 x 5 (p1x p2) • A3: 5 x 50 (p2x 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 CS 477/677 - Lecture 12

  34. 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 1 2 3 n n j 3 2 1 i CS 477/677 - Lecture 12

  35. 4. Construct the Optimal Solution • 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..k Ak+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 1 2 3 n n j 3 2 1 i CS 477/677 - Lecture 12

  36. 4. Construct the Optimal Solution • s[i, j] = value of k such that the optimal parenthesization of Ai Ai+1 Aj splits the product between Ak and Ak+1 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 CS 477/677 - Lecture 12

  37. Readings • Chapter 15 CS 477/677 - Lecture 12 37

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