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1. Advanced Artificial
Intelligence
MSCS 1
Lecture 5

2. Greedy search
• Greedy search is a variation of the A* algorithm, where g(node) is set
to zero, so that only h(node) is used to evaluate suitable paths. In this
way, the algorithm always selects the path that has the lowest
heuristic value or estimated distance (or cost) to the goal.
• Greedy search is an example of a best-first strategy.
• Greedy-search methods tend to be reasonably efficient, although in
the worst case, like depth-first search, it may never find a solution at
all.

3. Knapsack problem
• The knapsack problem is an interesting illustration of the use of
greedy search algorithms and their pitfalls. The fractional knapsack
problem can be expressed as follows:
• A man is packing items into his knapsack. He wants to take the most
valuable items he can, but there is a limit on how much weight he can
fit in his knapsack. Each item has a weight wi and is profit pi. He can
only fit a total weight of W in his knapsack.

4. Knapsack problem
• The items that he wants to take are things that can be broken up and still
retain their value (like flour or milk), and he is The items that he wants to
take are things that can be broken up and still retain their value (like flour
or milk), and he is able to take fractions of items. Hence, the problem is
called the fractional knapsack problem.
• In solving this problem, a greedy-search algorithm provides the best
solution.
• The problem is solved by calculating the value per unit weight of each item:
pi/wi, and then taking as much as he can carry of the item with the
greatest
value per unit weight. If he still has room, he moves on to the item with the
next highest value per unit weight, and so on.

5. Fractional Knapsack problem
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Items N = 7
Bag capacity M = 15

6. Fractional Knapsack problem
• Now we have a bag with capacity of 15. lets suppose 15 kg.
• Now we want to transfer this bag from one location to another.
• We are facing this problem in daily life.
• Now problem is to filling of the objects in a container.
• If you put all the objects in the bag with the capacity, there is no
problem.
• But the objects weight is more than the bag capacity, here the
problem start.

7. Fractional Knapsack problem
• We should maximize the profit.
• So this problem is optimization problem.
• Can we apply the greedy method to solve this problem?
• Yes, if it is suitable.
• So what are the constraints here?
• The bag capacity is 15, so all the objects which we want to transfer
should be <= 15.

8. Fractional Knapsack problem
• For this problem we have many solutions.
• The optimal result is to gain a maximum profit.
• X = {x1, x2, x3, x4, x5, x6, x7}
• We can say this problem is a fractional problem
• 0 ≤ X ≤ 1

9. Fractional Knapsack problem
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3
• Items n = 7
• Bag capacity m = 15

10. Fractional Knapsack problem
• Profit of each item per kg is
• X1 = 5, x2=1.3, x3=3, x4=1, x5=6, x6=4.5, x7=3
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3
Items n = 7
Bag capacity m = 15

11. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3
Items n = 7
Bag capacity m = 15

12. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

13. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

14. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

15. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

16. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

17. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

18. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

19. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

20. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

21. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

22. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

23. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3

24. Fractional Knapsack problem
• Now we select the items on the basis of their profit
Objects 0 1 2 3 4 5 6 7
Profits p 10 5 15 7 6 18 3
Weights w 2 3 5 7 1 4 1
Max profit p/w 5 1.3 3 1 6 4.5 3
Items n = 7
Bag capacity m = 15

25. A* algorithm for searching
• Informed searching algorithm
• A* Search algorithm is one of the best and popular technique used in
path-finding and graph traversals.
• Informally speaking, A* Search algorithms, unlike other traversal
techniques, it has “brains”. What it means is that it is really a smart
algorithm which separates it from the other conventional algorithms.
• And it is also worth mentioning that many games and web-based
maps use this algorithm to find the shortest path very efficiently
(approximation).

26. A* algorithm for searching
• A* algorithms are similar to best-first search but use a somewhat
more complex heuristic to select a path through the tree. The best-
first algorithm always extends paths that involve moving to the node
that appears to be closest to the goal, but it does not take into
account the cost of the path to that node so far.
• The A* algorithm operates in the same manner as best-first search
but uses the following function to evaluate nodes:
• f(n) = g(n) + h(n)

27. A* algorithm for searching
• g(n)--- actual cost from start node to n.
• h(n)--- estimation cost from n to goal node.

28. A* algorithm for searching
• S is start node and G is goal node.
• A* is an admissible algorithm which mean that it will give an optimal
solution.

29. A* algorithm for searching
• We are taking a directed graph in this example. We also can take a
tree to solve this problem. From S to B, 4 is actual value and in red
color 12 is heuristic value.

30. A* algorithm for searching
• Explore S, SB, SC.

31. A* algorithm for searching
• We explore SCE, SCD, SBF, SBE

32. A* algorithm for searching
• We explore SCE, SCD, SBF, SBE

33. A* algorithm for searching
• We explore SCDEG

34. A* algorithm for searching
• Time complexity
• T(C) = O(v+E)
• In artificial intelligence point of view
• O(b^d)
• Where b is the branch factor, d is the depth.