A greedy algorithm is a simple, intuitive algorithm that is used in optimization problems. The algorithm makes the optimal choice at each step as it attempts to find the overall optimal way to solve the entire problem.

1. Welcome to Our Presentation
Presentation Topic: Greedy Algorithm

2. Group Members:
Name: Arifur Jaman Bappy ID: 153-15-607
Name: Mohammad Salim Hosen ID: 162-15-1044
Name: Md Rifat Rahman ID: 162-15-1049

3. What is Greedy Algorithm
A greedy algorithm is an algorithm that follows the
problem solving approach of making the locally optimal
choice at each stage with the hope of finding a global
optimum solution.

4. Pros and Cons of Greedy Algorithm
• Pros:
• Simple, Easy to Implement, Run Fast
• Cons:
• Very often they don’t provide global optimum solution.

5. When to use Greedy Algorithm
1. Greedy – choice property: When a global optimum
can be arrived by selecting local optimum
2. Optimal Substructure: An optimal solution to the
problem contains optimal solution to subproblems.

6. Greedy Algorithms We Will Discuss
• Greedy Coin Change Algorithm
• Greedy Fractional Knapsack Algorithm
• Greedy Huffman Algorithm

7. Greedy Algorithm to find Minimum number of Coins
Give
me $35
$5 $10 $20
You may give:
5+5+5+5+5+5+5
10+5+5+5+5+5
20+5+5+5
……………………

8. Greedy Coin Change Algorithm
• V = Given value to make change
• C = Largest coin no longer than V
• I = Number of Denomination
• T = Total Number of coins initially 0
1. while V>0
2. if C[i] <= V
3. num = V/C[i];
4. V = V-(num*C[i]);
5. T += num;
6. i = i – 1;

9. Greedy Coin Change Algorithm
Algorithm:
1. while V>0
2. if C[i] <= V
3. num = V/C[i];
4. V = V-(num*C[i]);
5. T += num;
6. i = i – 1;
Coins: $5, $10, $20
V = $35, i = 3, T = 0
Is $35 greater than 0 Yes
Is $20 less than or equal $35 Yes
num is $35/$20 = 1
Amount Left in V is $35 - $20 = $15
Total coin T is 0+1 = 1
i = 3-1 = 2

10. Greedy Coin Change Algorithm
Algorithm:
1. while V>0
2. if C[i] <= V
3. num = V/C[i];
4. V = V-(num*C[i]);
5. T += num;
6. i = i – 1;
Coins: $5, $10, $20
V = $15, i = 2
Is $15 greater than 0 Yes
Is $10 less than or equal $15 Yes
Num is $15/$10 = 1
Amount Left in V is $15 - $10 = $5
Total coin T is 1+1 = 2
i = 2-1 = 1

11. Greedy Coin Change Algorithm
Algorithm:
1. while V>0
2. if C[i] <= V
3. num = V/C[i];
4. V = V-(num*C[i]);
5. T += num;
6. i = i – 1;
Coins: $5, $10, $20
V = $5, i = 1
Is $5 greater than 0 Yes
Is $5 less than or equal $5 Yes
Num is $5/$5 = 1
Amount Left in V is $5 - $5 = $0
Total coin is 2+1 = 3
i = 1-1 = 0 So 3 Coin will be needed to make $35

12. Fractional Knapsack
• If we are allowed to take fractional amounts,
then this is the fractional knapsack problem.
• Goal: Choose items with maximum total benefit
but with weight at most total weight.

13. Fractional Knapsack
6 kg
Bronze – 3kg Silver – 2kg
Gold – 4kg
Platinum – 1kg
Diamond – 5kg

14. Fractional Knapsack
Item Bronze Silver Platinum Gold Diamond
Value ($) 25 20 15 40 50
Weight (kg) 3 2 1 4 5
Value/Weight 8.3 10 15 10 10
Item Taken Weight Value
Platinum 1 kg $15
Silver 2 kg $20
Gold 3 kg $30
Total: 3 Item 6 kg $65

15. Huffman Coding
Character Code Frequency Total Bits
A 000 10 30
E 001 15 45
I 010 12 36
S 011 3 9
T 100 4 12
P 101 13 39
Newline 110 1 3
TOTAL: 174
• Our Goal is to reduce total number of bits without losing any information.

16. Huffman Algorithm
Character Frequency
n 1
S 3
T 4
A 10
I 12
P 13
E 15
4
n S

17. Huffman Algorithm
Character Frequency
n 1
S 3
T 4
A 10
I 12
P 13
E 15
4
n S
8
T

18. Huffman Algorithm
Character Frequency
n 1
S 3
T 4
A 10
I 12
P 13
E 15
Node 18 18
4
n S
8
T
18
A

19. Huffman Algorithm
Character Frequency
n 1
S 3
T 4
A 10
I 12
P 13
E 15
Node 18 18
Node 25 25
I
25
P

20. Huffman Algorithm
Character Frequency
n 1
S 3
T 4
A 10
I 12
P 13
E 15
Node 18 18
Node 25 25
Node 33 33
4
n S
8
T
18
A
33
E

21. Huffman Algorithm
Character Frequency
n 1
S 3
T 4
A 10
I 12
P 13
E 15
Node 18 18
Node 25 25
Node 33 33
4
n S
8
T
18
A
33
E
I
25
P
58

22. Huffman Algorithm
Character Code
A 111
E 10
I 00
S 11001
T 1101
P 01
Newline 11000
4
n S
8
T
18
A
33
E
I
25
P
58
0
0 1
1
0 1
0 1
1
0
0 1

23. Result:
Character Code Frequency Total Bits
A 111 10 30
E 10 15 30
I 00 12 24
S 11001 3 15
T 1101 4 16
P 01 13 26
Newline 11000 1 5
TOTAL: 146
Total bits reduced to 146 from 176 without losses of any data.