Question

In the figure below, which nodes (if any) will be pruned using alpha-beta pruning, assuming successor nodes are always explored left to right? (Select all that apply, or "None of the nodes will be pruned" if no nodes can be pruned.)

A MAX D MIN d d2 C3 bibi C2 14 5 2 6 4 2 12 8 3 TERMINAL CO

b1
		b3
		b3
		c1
		c2
		c3
		d1
		d2
		d3
		None of the nodes will be pruned

Answer 1

1. Conditions for alpha-beta pruning is α>=β.
2. Max turns only updates the value of alpha
3. Min turns only updates the value of beta.
4. Values of alpha and beta are passed to child nodes.
5. At the time of backtracking, node values of the tree are passed to parent nodes instead of values of alpha and beta.

α= -∞, β= +∞

МАX az B MIN di di d: by bi bz 14 5 4 12 8 TERMINAL

1. At node A the values of α= -∞, β= +∞, these alpha and beta are passed to node B.so now at node B the values are α= -∞, β= +∞.
2. At node B it is the turn of Min, where the value of β will be updated. So the value of Beta will be compared with.
   1. First terminal node of B. min (∞, 3) = 3. Now β = 3
   2. Second terminal node of B compared with update β. Min (3, 8) = 3
   3. Third terminal with updated β. Min(3, 12) = 3
   4. Now the value of 3 is replaced as value of Node B.
3. Now on backtracking we reach to node A. now it’s turn of Max. With all the available node values. Max (-∞, 3) = 3. Now the alpha value is 3 and beta is +∞.
4. Now alpha and beta values are passed to Node C. It’s the turn of Min so compare the beta value with
   1. Terminal node 2. Min (∞, 2) = 2. α = 3, β = 2, now α >= β .so the right node of the parent node will be pruned. C₂ is pruned.
   2. Now compare the Beta value with terminal node 6. Min (2, 6) is 2. α >= β. So the node 6(C₃) will be pruned.
   3. Now the value of Node C is changed to 2.
5. On backtracking to Node A. Alpha and Beta values are changed. It’s the turn of Max. So Max (3, 2) gives 3. Alpha = 3, Beta = +∞. These values are passed to Node D.
6. At Node D it’s the turn of Min. So the Beta value are compared as
   1. Terminal node 2. Min (+∞, 2) = 2. Alpha >= Beta so d₂ will be pruned.
   2. Terminal node 14. Min(2, 14) = 2. Alpha >= Beta so d₃ will be pruned.
   3. Finally the value of Node D set to 2
7. On Backtracking to Node A. Max(3,2,2) = 3.
8. Final tree of the problem is.
   МАХ 2 MIN di C bi b2 14 5 2 6 4 2 12 8 3. TERMINAL
   

The answer is nodes that are pruned is: c2, c3, d2, d3.