Dynamic Programming CISC4080, Computer Algorithms CIS, Fordham Univ. - - PowerPoint PPT Presentation
Dynamic Programming CISC4080, Computer Algorithms CIS, Fordham Univ. Instructor: X. Zhang Outline Introduction via example: Fibonacci, rod cutting Characteristics of problems that can be solved using dynamic programming More
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multiset: allow duplicate elements
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c1=1,2...min{n,k}{p[c1] + rn−c1}
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c1=1...min{n,k}{p[c1] + rn−c1}
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// for n > k
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How many times CutRod (2) is called? How about CutRod(1)?
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//return max. profit one can make with a rod of length n
CutRod (n, p[1…k])
CutRodHelper (n, p[1…k], r[]) 1. if r[n] >=0 return r[n] //if it has been calculated already 2. // no need to recalculate, return the stored result 3. 4. if n==0 return 0 //base case 5. //general case 6. curMax=0 7. for c1=1, 2, 3, … , min (n, k) 8. curProfit = p[c1] + CutRodHelper (n-c1) 9. curMax = max(curProfit, curMax) 10. 11. r[n]= curMax //save result in r[] for future reference 12. return curMax
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c1=1,2...min{n,k} {p[c1] + rn−c1}
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//return max. profit one can make with a rod of length n
CutRod (n, p[1…k])
CutRodHelper (n, p[1…k], r[]) 1. if r[n] >=0 return r[n] //if it has been calculated already 2. // no need to recalculate, return the stored result 3. 4. if n==0 return 0 //base case 5. //general case 6. curMax=0 7. for c1=1, 2, 3, … , min (n, k) 8. curProfit = p[c1] + CutRodHelper (n-c1) 9. curMax = max(curProfit, curMax) 10. 11. r[n]= curMax //save result in r[] for future reference 12. return curMax
21 Tracing with n=3 table r : c1 curProfit curMax
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given by array w[ ], v[ ]
weights <= W
10 + max value possible with 13-5
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What would be V13?
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with weights and values given by array w[ ], v[ ]
i such that w[i]<k {v[i] + Vk−w[i]}
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with weights and values given by array w[ ], v[ ]
i such that w[i]<k {v[i] + Vk−w[i]}
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How a problem’s optimal solution can be derived from
problem
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MSE(4)=4, array is A[4…4]
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Practice: 1) fill in ?? 2) How to find out starting and ending index of the max. subarray?
MSE(1)= A[1] MSE(i) = max (A[i], A[i]+MSE(i-1)); 2) Return max of MSE(i),
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int MaxSubArray (int A[1…n], int & start, int & end) { // Use array MSE to store the MSE(i) MSE[1]=A[1]; maxSubArray = MSE[1]; //use tabulation/bottom up to calculate MSE for (int i=2;i<=n;i++) { MSE[i] = ?? maxSubArray = max(maxSubArray, MSE[i]); } return maxSubArray; }
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int MaxSubArray (int A[1…n], int & start, int & end) { // Use array MSE to store the MSE(i) MSE[1]=A[1]; max_MSE = MSE[1]; for (int i=2;i<=n;i++) { MSE[i] = ?? if (MSE[i] > max_MSE) { max_MSE = MSE[i]; } } return max_MSE; }
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38 Sol to problem instance of size n Sol to problem instance of size n-1, n-2, … 1
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is the either the max. subarray ending at pos i-1 extended to pos i; or just A[i]
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1 2 3 4 5 6 7 8
to 9)
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1 2 3 4 5 6 7 8
LIS ending at pos 5
Let L(n) be length of LIS ending at n-th number
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Pos: 1 2 3 4 5 6 7 8
Let L(n) be length of LIS ending at n-th number. Recurrence relation: Note that the i’s in RHS is always smaller than the j
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Pos: 1 2 3 4 5 6 7 8
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22 34 18 30 76 1 3 19 80
1 2 3 4 5 6 7 8 9
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22 34 18 30 76 1 3 19 80
1 2 3 4 5 6 7 8 9
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22 34 18 80 76 1 3 19 20
1 2 3 4 5 6 7 8 9
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22 34 18 80 76 1 3 19 20
1 2 3 4 5 6 7 8 9
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22 34 18 80 76 1 3 19 20
1 2 3 4 5 6 7 8 9
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22 34 18 80 76 1 3 19 20
1 2 3 4 5 6 7 8 9
/*whether there is a subset of these numbers that add up to K, and output one such subset */
bool AddUpToK_Tabulation (int numbers[], int num_len, int K) { bool C[K+1][num_len+1]; // C[k][n]: can we add up to k using numbers[1…n] for n=0 to num_len C[0][n] = 0 // if K==0 return true for k=1 to K C[k][0] = false //fill in array row by row, left to right for k=1 to K for n=1 to num_len if C[k][n-1]==true C[k][n]=true //we can make k without last number //otherwise, can we include numbers[n] to make k? else if k==numbers[n] C[k][n] = true; else if k>numbers[n] and C[k-numbers[n]][n-1]==true C[k][n] = true; else //k<numbers[n] or cannot make k-numbers[n] using numbers[1…n-1] C[k][n] = false return C[K][num_len]; }
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22 34 18 80 76 1 3 19 20
1 2 3 4 5 6 7 8 9
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/* Output max. value, and objects to select @param W: given weight capacity, >=0 @ param n: # of objects, n>=0 @param w, v: weights and values @return max value achievable from obj [1…n] */
Knapsack_Norepeat (W, w, v, n) { if W==0 or n == 0 //base case return 0 //general case: reduce to smaller problems? //Q: include n-th obj or not, which one yields larger value? if (w[n] > W) //it’s possible to include n-th obj V1 = Knapsack_Norepeat (W-w[n], w, v, n-1) + V[n] //if we include n-th obj V2 = Knapsack_Norepeat (W, w, v, n-1) //if we don’t include n-th obj return max(V1, V2) else //impossible to include n-th obj // only one option: do not include n-th obj return Knapsack_Norepeat (W, w, v, n-1) }
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and n
number of objects
Knapsack_Norepeat (W, w, v, n) { initialize V[W+1][n+1] to store solutions to subproblems // V[w][n] is the max value achievable with weight capacity w and obj 1…n fill V with -1 Knapsack_Norepeat_helper (W, w, v, n, V); } Knapsack_Norepeat_helper (W, w, v, n, V) { if (V[w][n]>=0) // If the problem has been solved before…. return V[w][n] if (w==0 or n==0) V[w][n]= 0 return V[w][n] //general case: reduce to smaller problems? //Q: include n-th obj or not, which one yields larger value? if (w[n] > W) //it’s possible to include n-th obj V1 = Knapsack_Norepeat_helper (W-w[n], w, v, n-1, V) + V[n] V2 = Knapsack_Norepeat_helper (W, w, v, n-1, V) V[w][n] = max (V1, V2) return V[w][n] else //impossible to include n-th obj // only one option: do not include n-th obj V[w][n] = Knapsack_Norepeat_helper (W, w, v, n-1, V) return V[w][n]; }
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Running time analysis? T(w,n) = O (nm)
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*/
13.
15. 16.} Worst-case running time: O(n2m)
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If we ignore last element in X
If we ignore last element in Y
/* Return the longest subsequence of X, Y @param X: is a string @param Y: is a string @return the longest common subsequence of X and Y*/ LCS (X, Y) { m=|X|, n=|Y| if (|X| == 0 or |Y|==0) return “ “; //empty string //general case // can we match last letters of X and Y? if X[m] == Y[n] return LCS(Xm-1,Yn-1)+X[m] //concatenated with last letter else // Xm and Yn cannot be both in LCS s1 = LCS(Xm-1,Yn) s2 = LCS(Xm, Yn-1) return longer of s1 and s2 }
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X = 〈A, B, D, E〉 Y = 〈Z, B, E〉 X = 〈A, B, D, G〉 Y = 〈Z, B, D〉
Three Questions? Here we need to calculate prefix (Xm-1, Yn-1), and pass them to recursive calls
/* Return the longest subsequence of X, Y @param X : is a string @param Y: is a string @return the longest common subsequence of Xm and Yn*/ LCS (X, Y, m, n ) { if (m== 0 or n==0) return “ “; //empty string //general case // can we match last letters of X and Y? if X[m] == Y[n] return LCS(X,Y, m-1, n-1)+X[m] //concatenated with last letter else // Xm and Yn cannot be both in LCS s1 = LCS(X, Y, m-1, n ) s2 = LCS(X, Y, m, n-1) return longer of s1 and s2 } // keep X, Y as they are, use parameters to specify prefix length // subproblem’s size is given by m and n, at least one dimension is decreased
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X = 〈A, B, D, E〉 Y = 〈Z, B, E〉 X = 〈A, B, D, G〉 Y = 〈Z, B, D〉
Three Questions?
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C[2,3] C[2,4] C[3,3] C[3,4]
Y A B C B D A B
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C[3,4]= length of LCS (X3, Y4) = Length of LCS (BDC, ABCB) 3rd row, 4th column element
Memoization: After computing a solution to a subproblem, store it in a table. Subsequent calls check the table to avoid redoing work. LCS (X, Y) create and initialize table c[m][n] to -1 LCS_helper(X, Y, |X|, |Y|, c) LCS_helper(X, Y, i, j, c) if c[i, j] = NIL // LCS(i,j) has not been solved yet then if i==0 or j==0 c[i,j]=0 return 0 if x[i] = y[j] c[i, j] ←LCS_helper(x, y, i–1, j–1, c) + 1 else c[i, j] ←max{ LCS_helper(x, y, i–1, j, c), LCS_helper(x, y, i, j–1,c) } else return c[i,j]
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C[2,3] C[2,4] C[3,3] C[3,4]
Y A B C B D A B
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0 1 2 3 4 5 6 7
C[3,4]= length of LCS (X3, Y4) = Length of LCS (BDC, ABCB) 3rd row, 4th column element
C[2,3] C[2,4] C[3,3] C[3,4]
Y A B C B D A B
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C[3,4]= length of LCS (X3, Y4) = Length of LCS (BDC, ABCB) 3rd row, 4-th column element
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Output A Output B Output C Output B
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Dimension of A, B, and A x B?
Total (scalar) multiplication: 4x2x3=24
Total (scalar) multiplication: n2xn1xn3
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# of multiplication: p3p4p5+ p2p3p5+ p1p2p5
Find the one with minimal multiplications?
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