Data Structures and Algorithms
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1 Data Structures and Algorithms CS S-22 Dynamic Programming David Galles Department of Computer Science University of San Francisco
2 22-0: Dynamic Programming Simple, recursive solution to a problem Naive solution recalculates same value many times Leads to exponential running time
3 22-1: Fibonacci Numbers Calculating the nth Fibonacci number Fib(0) = 1 Fib(1) = 1 Fib(n) = Fib(n-1) + Fib(n-2)
4 22-2: Fibonacci Numbers int Fibonacci(int n) { if (n == 0) return 1; if (n == 1) return 1; } return Fibonacci(n-1) + Fibonacci(n-2);
5 22-3: Fibonacci Numbers Why is this solution bad? Recalculate values many times Many, many, times
6 22-4: Fibonacci Numbers Fib(6) Fib(5) Fib(4) Fib(3) Fib(2) Fib(2) Fib(1) Fib(1) Fib(0) Fib(3) b(1) Fib(0) Fib(2) Fib(1) Fib(4) Fib(1) Fib(0) Fib(3) Fib(2) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0)
7 22-5: Fibonacci Numbers Fib(7) Fib(6) Fib(5) Fib(5) Fib(4) Fib(4) Fib(3) Fib(3) Fib(3) Fib(2) Fib(1) Fib(2) Fib(2) Fib(1) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0) Fib(1) Fib(0) (1) Fib(0) Fib(2) Fib(3) Fib(4) Fib(1) Fib(0) Fib(2) Fib(1) Fib(3) Fib(2) Fib(1) Fib(0) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0)
8 22-6: Fibonacci Numbers Fib(8) Fib(7) Fib(6) Fib(5) Fib(5) Fib(4) Fib(4) Fib(3) Fib(3) Fib(3) Fib(2) Fib(1) Fib(2) Fib(2) Fib(1) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0) Fib(1) Fib(0) 1) Fib(0) Fib(2) Fib(3) Fib(4) Fib(1) Fib(0) Fib(2) Fib(1) Fib(3) Fib(2) Fib(1) Fib(0) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0) Fib(6) Fib(5) Fib(4) Fib(3) Fib(3) Fib(2) Fib(1) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0) Fib(2) Fib(4) Fib(1) Fib(0) Fib(3) Fib(2) Fib(2) Fib(1) Fib(1) Fib(0) Fib(1) Fib(0)
9 22-7: How Bad is Recalculation? Assume 2 GHz machine Add every cycle No time spent on recursive call overhead Lower bound on time required Fibonacci(100) will take:
10 22-8: How Bad is Recalculation? Assume 2 GHz machine Add every cycle No time spent on recursive call overhead Lower bound on time required Fibonacci(100) will take: 9087 Years
11 22-9: How Bad is Recalculation? Assume 2 GHz machine Add every cycle No time spent on recursive call overhead Lower bound on time required Fibonacci(100) will take: 9087 Years Fibonacci(200) will take:
12 22-10: How Bad is Recalculation? Assume 2 GHz machine Add every cycle No time spent on recursive call overhead Lower bound on time required Fibonacci(100) will take: 9087 Years Fibonacci(200) will take: millennia Well after heat death of the universe (100 trillion years)
13 22-11: Dynamic Programming Recalculating values can lead to unacceptable run times Even if the total number of values that needs to be calculated is small Solution: Don t recalculate values Calculate each value once Store results in a table Use the table to calculate larger results
14 22-12: Dynamic Programming To calculate Fibonacci(100), only need to calculate 101 values Fibonacci(n) can be calculated in time O(1) Assuming we have values for Fibonacci(n-1) and Fibonacci(n-2)
15 22-13: Dynamic Programming Create a table: FIB[] FIB[n] = nth Fibonacci number Fill the table from left to right Use old values in table to calculate new values
16 22-14: Faster Fibonacci int Fibonacci(int n) { int[] FIB = new int[n+1]; FIB[0] = 1; FIB[1] = 1; for (i=2; i<=n; i++) FIB[i] = FIB[i-1] + FIB[i-2]; } return FIB[n];
17 22-15: Dynamic Programming To create a dynamic programming solution to a problem: Create a simple recursive solution (that may require a large number of repeat calculations Design a table to hold partial results Fill the table such that whenever a partial result is needed, it is already in the table
18 22-16: World Series Two teams T 1 and T 2 T 1 will win any game with probability p T 2 will win any game with probability 1 p What is the probability that T 1 will win a best-of-seven series? Answer is not p : why not?
19 22-17: World Series Calculate the probability that T 1 will win the series, given T 1 needs to win x more games, and T 2 needs to win y more games PT1win(x,y) The probability that P 1 will win a best-of-seven series is then PT1win(4,4) The probability that P 1 will win a best-of-seven series, if P 1 has already one 2 games, and P 2 has won 1 game is then PT1win(2,3)
20 22-18: World Series Base cases: What is PT1win(0,x)?
21 22-19: World Series Base cases: What is PT1win(0,x)? 1! T 1 has already won! What is PT1win(x,0)?
22 22-20: World Series Base cases: What is PT1win(0,x)? 1! T 1 has already won! What is PT1win(x,0)? 0! T 1 has already lost!
23 22-21: World Series Recursive Case: PT1win(x,y) If T 1 wins the next, game, then the probability that T 1 will win the rest of the series is
24 22-22: World Series Recursive Case: PT1win(x,y) If T 1 wins the next, game, then the probability that T 1 will win the rest of the series is PT1win(x-1,y) If T 1 loses the next game, then the probability that T 1 will win the rest of the series is:
25 22-23: World Series Recursive Case: PT1win(x,y) If T 1 wins the next, game, then the probability that T 1 will win the rest of the series is PT1win(x-1,y) If T 1 loses the next game, then the probability that T 1 will win the rest of the series is PT1win(x,y-1)
26 22-24: World Series Recursive Case: PT1win(x,y) If T 1 wins the next, game, then the probability that T 1 will win the rest of the series is PT1win(x-1,y) If T 1 loses the next game, then the probability that T 1 will win the rest of the series is PT1win(x,y-1) Probability that T 1 will win is p Probability that T 1 will lose is 1 p What then is PT1win(x,y)?
27 22-25: World Series Recursive Case: PT1win(x,y) If T 1 wins the next, game, then the probability that T 1 will win the rest of the series is PT1win(x-1,y) If T 1 loses the next game, then the probability that T 1 will win the rest of the series is PT1win(x,y-1) Probability that T 1 will win is p Probability that T 1 will lose is 1 p PT1win(x,y) = p * PT1win(x-1,y) + (1 p) * PTwin(x,y-1)
28 22-26: World Series float PT1win(int x, int y, int p) { if (x == 0) return 1; if (y == 0) return 0; } return p * PT1win(x-1,y,p) + (1 - p) * PT1win(x, y-1,p);
29 22-27: World Series float PT1win(int x, int y, int p) { if (x == 0) return 1; if (y == 0) return 0; } return p * PT1win(x-1,y,p) + (1 - p) * PT1win(x, y-1,p); Just like Fibonacci, recalculating values exponential time How many total values do we need to calculate for PT1win(n,n)?
30 22-28: World Series float PT1win(int x, int y, int p) { if (x == 0) return 1; if (y == 0) return 0; } return p * PT1win(x-1,y,p) + (1 - p) * PT1win(x, y-1,p); Just like Fibonacci, recalculating values exponential time How many total values do we need to calculate for PT1win(n,n)? n 2
31 22-29: World Series P[x,y] = # of games required for T 1 to win, if T 1 needs to win x more games, and T 2 needs to win y more games. P[0,x] = 1 for all x > 0 P[x,0] = 0 for all x > 0 P[x,y] = p P[x 1,y]+(1 p) P[x,y 1] Need to fill out the table such that when we need a partial value, it has already been computed
32 22-30: World Series P[x,y] y x
33 22-31: World Series P[x,y] y x
34 22-32: World Series P[x,y] y x
35 22-33: World Series P(x,y) p =
36 22-34: World Series P(x,y) p =
37 22-35: World Series P(x,y) p =
38 22-36: World Series P(x,y) p =
39 22-37: World Series P(x,y) p =
40 22-38: World Series P(x,y) p =
41 22-39: World Series P(x,y) p =
42 22-40: World Series P(x,y) p =
43 22-41: World Series P(x,y) p =
44 22-42: World Series P(x,y) p =
45 22-43: World Series P(x,y) p =
46 22-44: World Series P(x,y) p =
47 22-45: World Series P(x,y) p =
48 22-46: World Series P(x,y) p =
49 22-47: World Series P(x,y) p =
50 22-48: World Series P(x,y) p =
51 22-49: World Series P(x,y) p =
52 22-50: World Series P(x,y) p =
53 22-51: World Series P(x,y) p =
54 22-52: Sequences & Subsequences A sequence is an ordered list of elements <A, B, C, B, D, A, B> A subsequence is a sequence with some elements left out; Subsequences of <A, B, C, B, D, A, B> <B, B, A> <A, B, C> <B, D, A, B> <C>
55 22-53: Sequences & Subsequences A sequence is an ordered list of elements <A, B, C, B, D, A, B> A subsequence is a sequence with some elements left out NON-Subsequences of <A, B, C, B, D, A, B> <D, A, C> <A, B, B, C> <C, A, D> <B, D, B, A>
56 22-54: Common Subsequences Given two sequences S 1 and S 2, a common subsequence is s subsequence of both sequences <A, B, C, B, D, A, B>, <B, D, C, A, B, A> Common Subsequences: <B, C, A> <B, D> <B, A, B> <B, C, B, A>
57 22-55: LCS Longest Common Subsequence Need not be unique <A, B, C, B, D, A, B>, <B, D, C, A, B, A> <B, C, B, A> <B, D, A, B>
58 22-56: LCS Given the sequences: <A, B, A, B, B> <B, C, A, B> LCS must end in B. Why?
59 22-57: LCS Given the sequences: <A, B, A, B, B> <B, C, A, B> LCS must end in B. Length of LCS: 1 + lengthlcs (<A, B, A, B>, <B, C, A>)
60 22-58: LCS Given the sequences: <A, B, A, B> <B, C, A> The last element in the LCS must be: not B not A
61 22-59: LCS Given the sequences: <A, B, A, B> <B, C, A> The last element in the LCS must be: not B not A Length of LCS: Maximum of: lengthlcs (<A, B, A>, <B, C, A>) lengthlcs (<A, B, A, B>, <B, C>)
62 22-60: LCS Pseudo-Code LCS(Seq1, Seq2) { if (Seq1 is empty) (Seq2 is empty) return 0; if (last elem in Seq1 = last elem in Seq2) return 1 + LCS(Seq1 - last element, Seq2 - last element) return MAX(LCS(Seq1 - last element, Seq2), LCS(Seq1, Seq2 - last element)) }
63 22-61: LCS Pseudo-Code LCS(int x, int y, String S1, String S2) { if ((x == 0) (y == 0)) return 0; } if (S1.charAt(x-1) == S2.charAt(y-1)) return 1 + LCS(x-1, y-1, S1, S2); else return MAX(LCS(x-1, y, S1, S2), LCS(x, y-1, S1, S2));
64 22-62: LCS Pseudo-Code LCS(int x, int y, String S1, String S2) { if ((x == 0) (y == 0)) return 0; } if (S1.charAt(x-1) == S2.charAt(y-1)) return 1 + LCS(x-1, y-1, S1, S2); else return MAX(LCS(x-1, y, S1, S2), LCS(x, y-1, S1, S2)); Requires exponential time in (x+y)
65 22-63: LCS For x,y: Total number of subproblems
66 22-64: LCS For x,y: Total number of subproblems (x + 1) * (y + 1) (O(x y))
67 22-65: LCS Create a table T T[i,j] = LCS(i, j, S1, S2) T[x,0] = 0 T[0,x] = 0 T[x,y] = 1 + T[x-1,y-1] if S1[x] = S2[y] MAX(T[x-1,y], T[x,y-1]) otherwise
68 22-66: LCS 0 A 1 B 2 C 3 B 4 D 5 A 6 B 7 B D C A B A
69 22-67: LCS B D C A B A A 1 0 B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
70 22-68: LCS B D C A B A A B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
71 22-69: LCS B D C A B A A B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
72 22-70: LCS B D C A B A A B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
73 22-71: LCS B D C A B A A B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
74 22-72: LCS B D C A B A A B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
75 22-73: LCS B D C A B A A B 2 0 C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
76 22-74: LCS B D C A B A A B C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
77 22-75: LCS B D C A B A A B C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
78 22-76: LCS B D C A B A A B C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
79 22-77: LCS B D C A B A A B C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
80 22-78: LCS B D C A B A A B C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
81 22-79: LCS B D C A B A A B C 3 0 B 4 0 D 5 0 A 6 0 B 7 0
82 22-80: LCS B D C A B A A B C B 4 0 D 5 0 A 6 0 B 7 0
83 22-81: LCS B D C A B A A B C B 4 0 D 5 0 A 6 0 B 7 0
84 22-82: LCS B D C A B A A B C B 4 0 D 5 0 A 6 0 B 7 0
85 22-83: LCS B D C A B A A B C B 4 0 D 5 0 A 6 0 B 7 0
86 22-84: LCS B D C A B A A B C B 4 0 D 5 0 A 6 0 B 7 0
87 22-85: LCS B D C A B A A B C B 4 0 D 5 0 A 6 0 B 7 0
88 22-86: LCS B D C A B A A B C B D 5 0 A 6 0 B 7 0
89 22-87: LCS B D C A B A A B C B D 5 0 A 6 0 B 7 0
90 22-88: LCS B D C A B A A B C B D 5 0 A 6 0 B 7 0
91 22-89: LCS B D C A B A A B C B D 5 0 A 6 0 B 7 0
92 22-90: LCS B D C A B A A B C B D 5 0 A 6 0 B 7 0
93 22-91: LCS B D C A B A A B C B D 5 0 A 6 0 B 7 0
94 22-92: LCS B D C A B A A B C B D A 6 0 B 7 0
95 22-93: LCS B D C A B A A B C B D A 6 0 B 7 0
96 22-94: LCS B D C A B A A B C B D A 6 0 B 7 0
97 22-95: LCS B D C A B A A B C B D A 6 0 B 7 0
98 22-96: LCS B D C A B A A B C B D A 6 0 B 7 0
99 22-97: LCS B D C A B A A B C B D A 6 0 B 7 0
100 22-98: LCS B D C A B A A B C B D A B 7 0
101 22-99: LCS B D C A B A A B C B D A B 7 0
102 22-100: LCS B D C A B A A B C B D A B 7 0
103 22-101: LCS B D C A B A A B C B D A B 7 0
104 22-102: LCS B D C A B A A B C B D A B 7 0
105 22-103: LCS B D C A B A A B C B D A B 7 0
106 22-104: LCS B D C A B A A B C B D A B 7 0 1
107 22-105: LCS B D C A B A A B C B D A B
108 22-106: LCS B D C A B A A B C B D A B
109 22-107: LCS B D C A B A A B C B D A B
110 22-108: LCS B D C A B A A B C B D A B
111 22-109: LCS B D C A B A A B C B D A B
112 22-110: Memoization Can be difficult to determine order to fill the table We can use a table together with recursive solution Initialize table with sentinel value In recursive function: Check table if entry is there, use it Otherwise, call function recursively Set appropriate table value return table value
113 22-111: LCS Memoized LCS(int x, int y, String S1, String S2) { if ((x == 0) (y == 0)) T[x,y] = 0; return 0; if (T[x,y]!= -1) return T[x,y]; if (S1.charAt(x) == S2.charAt(y)) T[x,y] = 1 + LCS(x-1, y-1, S1, S2); else T[x,y] = MAX(LCS(x-1, y, S1, S2), LCS(x, y-1, S1, S2)); return T[x,y]; }
114 22-112: Fibonacci Memoized int Fibonacci(int n) { if (n == 0) return 1; if (n == 1) return 1; if (T[n] == -1) T[n] = Fibonacci(n-1) + Fibonacci(n-2); } return T[n];
115 22-113: Making Change Problem: Coins: 1, 5, 10, 25, 50 Smallest number of coins that sum to an amount X? How can we solve it?
116 22-114: Making Change Problem: Coins: 1, 4, 6 Smallest number of coins that sum to an amount X? Does the same solution still work? Why not?
117 22-115: Making Change Problem: Coins: d 1, d 2, d 3,..., d k Can assume d 1 = 1 Value X Find smallest number of coins that sum to X Solution:
118 22-116: Making Change Problem: Coins: d 1, d 2, d 3,..., d k Can assume d 1 = 1 Value X Find smallest number of coins that sum to X Solution: We can use any of the coins d i whose value is less than or equal to X We then have a smaller subproblem: Finding change for value up to X d i. How do we know which one to chose? Try them all!
119 22-117: Making Change Problem: Coins: d 1, d 2, d 3,..., d k Can assume d 1 = 1 Value X Find smallest number of coins that sum to X Solution: C[X] = smallest number of coins required for amount X What is the base case? What is the recursive case?
120 22-118: Making Change C[X] = smallest number of coins required for amount X, using coins d 1,d 2,d 3...d k Base Case: C[0] = 0 Recursive Case: C[X] = min 1 i n 1+C[X d i] (where d n is the largest coin X)
121 22-119: Making Change d 1 = 1,d 2 = 4,d 3 =
122 22-120: Making Change d 1 = 1,d 2 = 4,d 3 =
123 22-121: Making Change d 1 = 1,d 2 = 4,d 3 =
124 22-122: Making Change d 1 = 1,d 2 = 4,d 3 =
125 22-123: Making Change d 1 = 1,d 2 = 4,d 3 =
126 22-124: Making Change d 1 = 1,d 2 = 4,d 3 =
127 22-125: Making Change d 1 = 1,d 2 = 4,d 3 =
128 22-126: Making Change d 1 = 1,d 2 = 4,d 3 =
129 22-127: Making Change d 1 = 1,d 2 = 4,d 3 =
130 22-128: Making Change d 1 = 1,d 2 = 4,d 3 =
131 22-129: Making Change d 1 = 1,d 2 = 4,d 3 =
132 22-130: Making Change Given the table, can we determine the optimal way to make change for a given value X? How?
133 22-131: Making Change Given the table, can we determine the optimal way to make change for a given value X? How? Look back through table, determine which coin was used to get the smallest number of coins (examples) We could also store which coin we used to get the smallest number of coins
134 22-132: Making Change d 1 = 1,d 2 = 4,d 3 = d 1 (1) d 1 (1) d 1 (1) d 2 (4) d 2 (4) d 3 (6) d 3 (6) d 2 (4) d 2 (4) d 3 (6)
135 22-133: Matrix Multiplication Quick review (on board) Matrix A is i j Matrix B is j k # of scalar multiplications in A B?
136 22-134: Matrix Multiplication Quick review (on board) Matrix A is i j Matrix B is j k # of scalar multiplications in A B? i j k
137 22-135: Matrix Chain Multiplication Multiply a chain of matrices together A B C D E F Matrix Multiplication is associative (A B) C = A (B C) (A B) (C D) = A (B (C D)) = ((A B) C) D = A ((B C) D) = (A (B C)) D
138 22-136: Matrix Chain Multiplication Order Matters! A : ( ),B : ( ), C : ( ),D : (100 1) ((A B) C) D Scalar multiplications: A (B (C D)) Scalar multiplications:
139 22-137: Matrix Chain Multiplication Order Matters! A : ( ),B : ( ), C : ( ),D : (100 1) ((A B) C) D Scalar multiplications: 2,010,000 A (B (C D)) Scalar multiplications: 30,000
140 22-138: Matrix Chain Multiplication Matrices A 1,A 2,A 3...A n Matrix A i has dimensions p i 1 p i Example: A 1 : 5 7, A 2 : 7 9, A 3 : 9 2, A 4 : 2 2 p 0 = 5,p 1 = 7,p 2 = 9,p 3 = 2,p 4 = 2 How can we break A 1 A 2 A 3... A n into smaller subproblems? Hint: Consider the last multiplication
141 22-139: Matrix Chain Multiplication M[i,j] = smallest # of scalar multiplications required to multiply A i... A j Breaking M[1, n] into subproblems: Consider last multiplication (use whiteboard)
142 22-140: Matrix Chain Multiplication M[i,j] = smallest # of scalar multiplications required to multiply A i... A j Breaking M[1, n] into subproblems: Consider last multiplication: (A 1 A 2... A k ) (A k+1... A n ) M[1,n] = M[1,k]+M[k +1,n]+p 0 p k p n In general, M[i,j] = M[i,k]+M[k +1,j]+p i 1 p k p j What should we choose for k? which value between i and j 1 should we pick?
143 22-141: Matrix Chain Multiplication Recursive case: M[i,j] = min i k<j (M[i,k]+M[k+1,j]+p i 1 p k p j What is the base case?
144 22-142: Matrix Chain Multiplication Recursive case: M[i,j] = min i k<j (M[i,k]+M[k+1,j]+p i 1 p k p j What is the base case? for all i M[i,i] = 0
145 22-143: Matrix Chain Multiplication M[i,j] = min i k<j (M[i,k]+M[k +1,j]+p i 1 p k p j In what order should we fill in the table? What to we need to compute M[i,j]?
146 22-144: Matrix Chain Multiplication M[i,j] = min i k<j (M[i,k]+M[k +1,j]+p i 1 p k p j In what order should we fill in the table? What to we need to compute M[i,j]?
147 22-145: Matrix Chain Multiplication M[i,j] = min i k<j (M[i,k]+M[k +1,j]+p i 1 p k p j What about the lower-left quadrant of the table?
148 22-146: Matrix Chain Multiplication M[i,j] = min i k<j (M[i,k]+M[k +1,j]+p i 1 p k p j What about the lower-left quadrant of the table? Not Defined
149 22-147: Matrix Chain Multiplication Matrix-Chain-Order(p) n # of matrices for i 1 to n do M[i,i] 0 for l 2 to n do for i 1 to n l+1 j i+l 1 M[i,j] for k i to j 1 do q M[i,k]+M[k +1,j]+p i 1 p k p j if q < M[i,j] then M[i,j] = q S[i,j] = k
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