Game Theory a problem which is a bit similar to the Impossible Puzzle

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  • #1
rbpl
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Game Theory a problem which is a bit similar to the "Impossible Puzzle"

From numbers 1 to 10, two integers X, and Y (not necessarily distinct) are chosen by a referee . The referee informs secretly to Joe the integer U where U = X + Y . The referee informs secretly to Bob the integer V where V = X^2 + Y^2. Before the referee makes his choice, he explains this rule to both players in their common presence. The players take turns to guess the numbers X; Y selected. In case a player is not sure of the numbers he can say "I don't know". Then it is the turn for the opponent. Suppose the game starts with Bob. Suppose he says, "I don't know" and immediately Joe says, I know. When will this happen?

How do I get started?

I thought about the problem it is a bit similar to the "Impossible Puzzle" but the main idea of the "Impossible Puzzle" is that you get the sum and the product here I have X^2 + Y^2 instead of X*Y. However to point of such an exercise is to find the actual values.

I know that when you add X and Y where they are between 1 and 10 you will get 20 different solutions, but when you add X^2 and Y^2 you get more than 20 solution thus it should be easier to find the solution since you have the answer for V.

The part that throws me off the most is the question "When will this happen?"
 

Answers and Replies

  • #2
hgfalling
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Well, there may be more than one solution, so you need to identify all the solutions.
 
  • #3
rbpl
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There are 100 combinations for X + Y and 100 combinations for X^2 + Y^2. Since U and V could be any of the results we can assume that the correct solution is one of the 100 combination for 100 combinations.

Whatever the result is for X + Y the guy has to check numbers for at least 1 result and at most 10.

Whatever the result is for X^2 + Y^2 the guy has to check numbers for at least 1 result and at most 4.

Since this is how many times the results repeat themselves if we calculate all of the combinations.

Am I correct?
 
  • #4
mistermath
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49+1 = 50
25+25 = 50

I bet this is the only time you get the same # that's up to you to figure out. However, this shows you one scenario that you're looking for.

If you get V = 50, Bob doesn't know which one of the answers worked, the other guy will know V = 50 so he will look at U, if U = 8 then x = 7, y = 1. If U = 10 then x = y = 5.

If you play with this, unless there are other possible senarios where this works, x = y = 5 is the only answer.

Thinking about it further, x must equal y or else how would Joe know for sure. (Like is x = 7 y = 1 or is it y = 7, x = 1?)
 
  • #5
rbpl
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Thank you guys for your help.

Mistermath that totally makes sense, when I was trying to see the pattern in the results for x^2 + y^2 I made a stupid miscalculation and for every 1^2 * 10^2... and 10^2 * 1^2... I wrote 100... instead of 101... which made other numbres just as attractive as 50. Thanks again.
 
  • #6
I like Serena
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I believe there are a few more solutions:
25+25=50
49+1=50
49+16=65
64+1=65
49+36=85
81+4=85
Since Bob does not know, it must be one of these.
Corresponding sums are: 10, 8, 11, 13, 9, 11 respectively.
So with U=11 Joe wouldn't know either.
But since he does, it must be one of the other solutions.
That is, the solutions are (5,5), (7,1), (7,6), (8,1).
 
  • #7
mistermath
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I believe there are a few more solutions:
25+25=50
49+1=50
49+16=65
64+1=65
49+36=85
81+4=85
Since Bob does not know, it must be one of these.
Corresponding sums are: 10, 8, 11, 13, 9, 11 respectively.
So with U=11 Joe wouldn't know either.
But since he does, it must be one of the other solutions.
That is, the solutions are (5,5), (7,1), (7,6), (8,1).

The problem with what you've said (or maybe you did not finish) is of the final 4 choices:
(5,5), (7,1), (7,6), (8,1) it could also have been: (5,5), (1,7), (7,6), and (1,8)

the only one that works perfectly is (5,5). Since Joe must not only know what the two numbers are, but he must know the actual value of x and the actual value of y.
 
  • #8
I like Serena
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The problem with what you've said (or maybe you did not finish) is of the final 4 choices:
(5,5), (7,1), (7,6), (8,1) it could also have been: (5,5), (1,7), (7,6), and (1,8)

the only one that works perfectly is (5,5). Since Joe must not only know what the two numbers are, but he must know the actual value of x and the actual value of y.

I deduced that the number pairs must be unordered, because otherwise the puzzle would be unsolvable. (5,5) in this case would not be a proper solution, because Bob would almost never know the answer. That is, if for instance Bob would have V=52, the numbers could be X=6 and Y=4 or they could be X=4 and Y=6. Joe would still have U=10, but would not know which solution to select.
 
  • #9
mistermath
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If Joe knows the answer, then the answer must be one of: (1,1), (2,2), (3,3),.. because otherwise he could get the x and y part incorrect.

1^2+1^2 = 2
2^2+2^2 = 8
3^2+3^2 = 18
4^2+4^2 = 32

etc.. so we have as the only possibilities for V to be:
2, 8, 18, 32, 50, 72, 98, 128, 162, 200

If V is any of these, Bob knows right away what x and y are... except if V = 50. 50 has more than 1 representation: (5,5), (1,7), (7,1) so he says he does not know and then Joe's job is easy because he has the sum.

What you did was assume the x,y are unordered because what if V is 52? Well, like you said, V couldn't have been 52 because then Joe wouldn't know the answer; but Joe knew the answer so contradiction and V = 52 is not a possibility.
 
  • #10
I like Serena
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If Joe knows the answer, then the answer must be one of: (1,1), (2,2), (3,3),.. because otherwise he could get the x and y part incorrect.

1^2+1^2 = 2
2^2+2^2 = 8
3^2+3^2 = 18
4^2+4^2 = 32

etc.. so we have as the only possibilities for V to be:
2, 8, 18, 32, 50, 72, 98, 128, 162, 200

If V is any of these, Bob knows right away what x and y are... except if V = 50. 50 has more than 1 representation: (5,5), (1,7), (7,1) so he says he does not know and then Joe's job is easy because he has the sum.

What you did was assume the x,y are unordered because what if V is 52? Well, like you said, V couldn't have been 52 because then Joe wouldn't know the answer; but Joe knew the answer so contradiction and V = 52 is not a possibility.

Joe would only know that U=10. Joe has no knowledge of V, and as far as Joe is concerned V could be 50, 52, or some other number. In other words, Joe would not know whether (x,y) is (5,5), (6,4), (4,6) or some other combination. This is a contradiction because Joe is supposed to know the answer.
 

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