1 / 37

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

nero-hampton
Download Presentation

Analysis of Algorithms CS 477/677

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related