of the card.) The upshot of this is that the maximizer wins the overall game if and only if the card in question is a club. A formal analysis proceeds similarly, labelling the nodes as follows:

![0_image_0.png](0_image_0.png)

Note, incidentally, that the maximizer's ability to change the card does not help him win the game.

Now suppose that we apply deep pruning to this game. The ply four node is one where the minimizer can force a value of at most …, suggesting that the siblings of the bottom …
node can be pruned. But doing so produces the following tree:
$ V $

![0_image_1.png](0_image_1.png)

$\mathbf{d}$. 
If the maximizer reaches ply 3, he can win by changing the card to a club if need be.

Of course, the minimizer won't let the maximizer reach ply 3; at ply 2, he'll move left sset to the maximizer wins only if the card is a diamond. That means that the maximizer wins at the root just in case the card is either a club or a diamond.

A partial graph of the values for this game is as follows:

![0_image_2.png](0_image_2.png)

where we have included the crucial fact that x ∧ y = 0 if x ̸ = y (since the minimizer knows the card) and ♥ V ♥ = 1 because the maximizer can invoke his special rule.  Other least upper bounds are not shown in the diagram. The maximizing function V moves up the figure; the minimizing function ∧ moves down.

The deep prune fails because we can't "push" the value ♠ Λ 0 past the ♡ to get to the near the root. Somewhat more precisely, the problem is that

$$\heartsuit=\heartsuit\vee(\clubsuit\wedge0)\neq\left(\heartsuit\wedge\clubsuit\right)\lor\left(\heartsuit\wedge0\right)=0$$

This suggests the following: