![7
Min Coin Problem (4) – Brute-Force
Algorithm-2: min-coin(coins, m) : a
Input:
Coins: set of coins
m: the number of Rupees to make with coins.
Returns:
a: the minimum number of coins used to make m.
[To be written on board].](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-7-320.jpg)
![13
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0
R 0
A 0
S 0
H 0
|s1|=m
|s2|=n
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
Define map[n+!, m+1] matrix
Let map[0, :]=0 and map[:, 0]:=0](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-13-320.jpg)
![14
Longest Common Subsequence (4)
X B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0
A 0
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-14-320.jpg)
![15
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0
A 0
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-15-320.jpg)
![16
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1
A 0
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
S1[i]=s[j]](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-16-320.jpg)
![17
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1 1
A 0
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
S1[i]!=s[j]333](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-17-320.jpg)
![18
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1 1 1 1 1
A 0
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-18-320.jpg)
![19
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1 1 1 1 1
A 0 0 1
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-19-320.jpg)
![20
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1 1 1 1 1
A 0 0 1 2
S 0
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-20-320.jpg)
![21
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1 1 1 1 1
A 0 0 1 2 2 2 2
S 0 0 1 2 2 2 2
H 0
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-21-320.jpg)
![22
Longest Common Subsequence (4)
B R A N X H
0 0 0 0 0 0 0
C 0 0 0 0 0 0 0
R 0 0 1 1 1 1 1
A 0 0 1 2 2 2 2
S 0 0 1 2 2 2 2
H 0 0 1 2 2 2 3
LCS(s1, s2)=3
i=1 to m
j=1
to
n
If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1
Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-22-320.jpg)
![32
Applications of Greedy Algorithm (3)
Merging Sorted Lists
Input: n sorted arrays
A [1], A[2], A[3], …., A[n]
Problem: To merge all the sorted arrays into one sorted array as fast as possible.
E.g., Merge Sort](https://image.slidesharecdn.com/daa-lecture-10-11-250220182434-5ac86d32/85/Design-and-Analysis-of-Algorithm-Lecture-pptx-32-320.jpg)
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Design and Analysis of Algorithm-Lecture.pptx
- 1. Design and Analysis of Algorithms (DAA) LECTURE 10 DYNAMIC PROGRAMMING APPLICATION DR. HAMID H. AWAN
- 2. 2 Review of lecture 09 Introduction to Dynamic Programming Hallmarks of Dynamic Programming Fibonacci Series
- 3. 3 Contents of the lecture Topics Minimum Coin Problem The Longest Common Sub-string Pre-requisite knowledge C/C++ programming Mathematical induction Theorem proving taxonomy
- 4. 4 Min Coin Problem Suppose your are given some coins. Each coin HAS A WORTH OF Rs 1, Rs 4 or Rs 5. That is coins={1, 4, 5} Now you are asked to make Rs 13 with the help of given coins. How will you make Rs 13? Can you determine the combination which requires minimum number of coins to make Rs 13?
- 5. 5 Min Coin Problem (2) Greedy Approach Take the largest coins first to reach as closer as possible with minimum number of coins. Solution: 5+5+1+1+1=13 Total coins usedL: 5 Brute Force approach: Perform an exhaustive search to find the most optimum result. It guarantees best solution. It turns out that 5+4+4=13 requires only 3 coins to make Rs 13. Better than greedy approach, but at the cost of??
- 6. 6 Min Coin Problem (3) – Brute-Force
- 7. 7 Min Coin Problem (4) – Brute-Force Algorithm-2: min-coin(coins, m) : a Input: Coins: set of coins m: the number of Rupees to make with coins. Returns: a: the minimum number of coins used to make m. [To be written on board].
- 8. 8 Min Coin Problem (4) – Dynamic Programming Algorithm-3: min-coinDP(coins, m, mem) : a Input: Coins: set of coins m: the number of Rupees to make with coins. Mem: memorization array Returns: a: the minimum number of coins used to make m.
- 9. 9 Min Coin Problem (5) – Dynamic Programming Time and space complexity? T(m)=O(m) S(m)=O(m)
- 10. 10 Longest Common Subsequence Consider two strings S1=“ABCD” S2=“ACFD A, C and D are common LCS (s1, s2):=“ACD” |LCS(s1, s2)|:=3 Lines can’t overlap. |s1| is not necessarily equal to | s2| A B C D A C F D
- 11. 11 Longest Common Subsequence (2) Brute-Force Approach Every character of s1 may have to be compared with every character of s2 in worst case. If length of s1 is m and that of s2 is n then: S
- 12. 12 Longest Common Subsequence (3) The Dynamic Programming approach Example: Consider s1=“BRANXH” and s2=“CRASH” B R A N C H C R A S H B R A N X H C R A S H |s1|=m |s2|=n
- 13. 13 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 R 0 A 0 S 0 H 0 |s1|=m |s2|=n i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j]) Define map[n+!, m+1] matrix Let map[0, :]=0 and map[:, 0]:=0
- 14. 14 Longest Common Subsequence (4) X B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 A 0 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 15. 15 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 A 0 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 16. 16 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 A 0 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j]) S1[i]=s[j]
- 17. 17 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 A 0 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j]) S1[i]!=s[j]333
- 18. 18 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 1 1 1 A 0 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 19. 19 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 1 1 1 A 0 0 1 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 20. 20 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 1 1 1 A 0 0 1 2 S 0 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 21. 21 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 1 1 1 A 0 0 1 2 2 2 2 S 0 0 1 2 2 2 2 H 0 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 22. 22 Longest Common Subsequence (4) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 1 1 1 A 0 0 1 2 2 2 2 S 0 0 1 2 2 2 2 H 0 0 1 2 2 2 3 LCS(s1, s2)=3 i=1 to m j=1 to n If s1[i]==s2[j] then map[I, j]:=map[i-1, j-1]+1 Else map[I, j]:=argmax (map[i, j-1], map[i-1, j])
- 23. 23 Longest Common Subsequence (5) B R A N X H 0 0 0 0 0 0 0 C 0 0 0 0 0 0 0 R 0 0 1 1 1 1 1 A 0 0 1 2 2 2 2 S 0 0 1 2 2 2 2 B 0 1 1 2 2 2 2
- 24. 24 Longest Common Subsequence (6) Time Complexity? T(m, n)= O(m+n) Space Complexity? S(m, n)=O(mn)
- 25. 25 Thank you…! End of Lecture 04
- 26. Design and Analysis of Algorithms (DAA) LECTURE 11 GREEDY ALGORITHM DR. HAMID H. AWAN
- 27. 27 Review of lecture 10 Applications of Dynamic Programming Min Coins Problem Longest Common Subsequent Problem
- 28. 28 Contents of the lecture Topics Greedy Algorithm The knapsack problem The shortest path finding problem Pre-requisite knowledge C/C++ programming Mathematical induction Theorem proving taxonomy
- 29. 29 The Greedy Approach Greedy Algorithm A technique to build a complete solution by making a sequence of best “selection” steps. Selection depends on actual problem Focus is on “What is best step from this point”.
- 30. 30 Applications of Greedy Algorithm Sorting Merging sorted lists Knapsack Minimum Spanning Trees Huffman encoding
- 31. 31 Applications of Greedy Algorithm (2) Sorting Select the minimum element in the list and move it to the beginning. E.g., selection sort, insertion sort. How much is a greedy sorting algorithm optimal?
- 32. 32 Applications of Greedy Algorithm (3) Merging Sorted Lists Input: n sorted arrays A [1], A[2], A[3], …., A[n] Problem: To merge all the sorted arrays into one sorted array as fast as possible. E.g., Merge Sort
- 33. Dynamic Programming vs. Greedy Algorithms Dynamic programming We make a choice at each step The choice depends on solutions to subproblems Bottom up solution, from smaller to larger subproblems Greedy algorithm Make the greedy choice and THEN Solve the subproblem arising after the choice is made The choice we make may depend on previous choices, but not on solutions to subproblems Top down solution, problems decrease in size 33
- 34. 34 Dynamic Programming vs. Greedy Algorithms (2) Dynamic Programming = Brute- Force + Greedy Algorithm
- 35. 35 Optimization problems An optimization problem is one in which you want to find, not just a solution, but the best solution A “greedy algorithm” sometimes works well for optimization problems A greedy algorithm works in phases. At each phase: You take the best you can get right now, without regard for future consequences You hope that by choosing a local optimum at each step, you will end up at a global optimum
- 36. 36 The Knapsack Problem The 0-1 knapsack problem A thief robbing a store finds n items: the i-th item is worth vi dollars and weights wi pounds (vi, wi integers) The thief can only carry W pounds in his knapsack Items must be taken entirely or left behind Which items should the thief take to maximize the value of his load? The fractional knapsack problem Similar to above The thief can take fractions of items
- 37. 37 0-1 Knapsack - Dynamic Programming P(i, w) – the maximum profit that can be obtained from items 1 to i, if the knapsack has size w Case 1: thief takes item i P(i, w) =vi *wi+ P(i - 1, w-wi) Case 2: thief does not take item i P(i, w) = P(i - 1, w)
- 38. 38 The Fractional Knapsack Problem Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight Goal: Choose items with maximum total benefit but with weight at most W. If we are allowed to take fractional amounts, then this is the fractional knapsack problem. In this case, we let xi denote the amount we take of item i Objective: maximize Constraint: S i i i i w x b ) / ( S i i W x
- 39. 39 Example Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight Goal: Choose items with maximum total benefit but with weight at most W. Weight: Benefit: 1 2 3 4 5 4 ml 8 ml 2 ml 6 ml 1 ml $12 $32 $40 $30 $50 Items: Value: 3 ($ per ml) 4 20 5 50 10 ml Solution: • 1 ml of 5 • 2 ml of 3 • 6 ml of 4 • 1 ml of 2 “knapsack”
- 40. 40 The Fractional knapsack algorithm The greedy algorithm: Step 1: Sort pi /wi into nonincreasing order. Step 2: Put the objects into the knapsack according to the sorted sequence as possible as we can. Example Capacity =20 S.No Weight Price 1 18 25 2 15 24 3 10 15
- 41. 41 The Fractional knapsack algorithm Solution p1 /w1 = 25/18 = 1.32 p2 /w2 = 24/15 = 1.6 p3 /w3 = 15/10 = 1.5 Sort in the descending order that will select the one item # 2 and half item number three and none from first item so Optimal solution: x1 = 0, x2 = 1, x3 = 1/2 value 31.5
- 42. 42 Shortest paths on a special graph Problem: Find a shortest path from v0 to v3 . The greedy method can solve this problem. The shortest path: 1 + 2 + 4 = 7.
- 43. 43 Shortest paths on a multi-stage graph Problem: Find a shortest path from v0 to v3 in the multi-stage graph.
- 44. 44 Solution of the above problem dmin(i,j): minimum distance between i and j. This problem can be solved by the dynamic programming method. d m i n ( v 0 , v 3 ) = m i n 3 + d m i n ( v 1 , 1 , v 3 ) 1 + d m i n ( v 1 , 2 , v 3 ) 5 + d m i n ( v 1 , 3 , v 3 ) 7 + d m i n ( v 1 , 4 , v 3 )