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Minimum possibility -minimax algorithm:
Finally, we want to put the scores on the top edges in the tree. So there is over again a choice. Whenever, in this case, we have to remember that player two is making the choices, and they will act in order to minimisation the score one of that player 1 gets. So there is, in the case when player one chooses the 3 card, player 2 will choose the 7 to reduces the score player 1 can get. So there is choice, we choose the minimum possibility of the three to put on the edges at the top of the tree as follows:
In the direction of choose the correct first card, well there would player one purely looks next to the topmost edges of the final tree and chooses the one with the highest score. In fact in this generally, there is alternate choosing the 7 will certification player one scores 10 in this game we can assuming that player one chooses according to such the minimax strategy just to move 2, but - importantly - making no assumptions worry for how player two will choose.
Prove the equations A + A‾ .B + A.B‾ = A + B using the Boolean algebraic theorems ? Ans. The equation is A + A‾.B + A.B‾ = A + B L.H.S. = A + A‾ .B + A.B‾ = (A + A.B‾) + A‾.B
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