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VashaNatasha [74]
3 years ago
7

Prove the following assertion: For every game tree, the utility obtained by MAX using minimax decisions against a suboptimal MIN

will never be lower than the utility obtained playing agains an optimal MIN. Can you come up with a game tree in which MAX can do still better using a suboptimal strategy against a suboptimal MIN?

Computers and Technology
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

From the given diagram, consider a MIN node whose children are terminal nodes, if MIN plays  

suboptimal. MIN will never be lower than the utility obtained playing against an optimal MIN  

MIN will always select a move having minimax utility greater than or equal to the move that is  

predicted by the minimax that is the MIN-played optimal value.  

Then the MIN node's value is increased to MAX. This is done by induction.  

One can do better than the minimax strategy, if the suboptimal play is predicted by MIN.

If MIN always falls for certain for certain kind of trap and losses, then setting up a trap guarantees a win.

Explanation:

See attached picture also.

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Write a loop that fills an array int values[10] with ten random numbers between 1 and 100. Write code for two nested loops that
zalisa [80]

Answer:

#include <iostream>

#include <cstdlib>

using namespace std;

int main() {

   int[] array = new int[10];

   int index = 0;

   while(index < array.size()){

           int number = (rand() % 100) + 1;

           for (int i = 0; i < 1; i++) {

               array[index] = number;

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           }

     }

}

Explanation:

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