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Topic: Mastermind I (Read 949 times) |
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Einmaliger
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Posts: 4
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Mastermind I
« on: Mar 15th, 2007, 9:29am » |
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The search for "Mastermind" didn't yield any results, so I'll just start a new topic.
Here's my attempt of a solution:
In the first row, three pegs are correct, one is wrong. Suppose the green peg is the wrong one and the other three are correct. Then, the solution would contain red, yellow and brown, so in the second row (which contains all of these three colors, but in different positions) there would be at least three white keypegs. Since this isn't the case, our assumption was wrong and we know that the first peg of the solution is green.
Now suppose that the wrong peg in the first row is the brown one, and the first three are correct. Then, there would be at least three black keypegs in the thrid row. Thus, we know that the fourth peg of the solution is brown.
If the wrong peg in the first row is the red one, there would have to be at least three black keypegs in the fourth row. Thus, the second peg is red, and there is no yellow peg anywhere in the solution.
In the third line, we have two black and one white keypegs. One of the black ones stands for the green peg, the other one for the red one. Since the yellow peg is wrong, the white keypeg one can only represent the light blue peg.
Thus, the solution is green, red, light blue and brown (in this order).
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| « Last Edit: Mar 15th, 2007, 9:30am by Einmaliger » |
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Aryabhatta
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Re: Mastermind I
« Reply #1 on: Mar 15th, 2007, 9:52am » |
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Assuming that there is a unique solution, your solution checks out.
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