I was helping my son with his math the other night and we hit a question called The Magic Box. You are given a 3x3 square and the digits 3,4,5,6,7,8,9,10,11, and
are expected to find a way of placing them such that each row, each column, and each diagonal adds up to 21.
I'm a programmer, so my first thought was, hey, I'll make a recursive algorithm to do this. The previous question, measuring 4 quarts when you have a 3 quart measure and a 5 quart measure, was solved due to insights remembered from Die Hard With A Vengeance, so clearly, I'm not coming at these questions from the textbook.
With a little more insight, I solved it. 7 is a third of 21, and it the center of an odd-numbered sequence of numbers, so clearly, it is meant to be the center. There is only one way you can use 11 on a side, with 4 and 6, so that a center column or row will be 3 7 11. If you know that column and the 4 11 6 row, you know at least this:
And because you know the diagonals, you know that it'll be
And you only have 5 and 9 left, and they're easy to just plug in
So, that's the logical way to to solve it. Clearly, order isn't important; it could be reversed on the x and y axis and still be a solution. But, once the thought of solving with a recursive algorithm came into my head, I could not leave well enough alone. So, here's a recursive program that finds all possible solutions for this problem.
I'm a programmer, so my first thought was, hey, I'll make a recursive algorithm to do this. The previous question, measuring 4 quarts when you have a 3 quart measure and a 5 quart measure, was solved due to insights remembered from Die Hard With A Vengeance, so clearly, I'm not coming at these questions from the textbook.
With a little more insight, I solved it. 7 is a third of 21, and it the center of an odd-numbered sequence of numbers, so clearly, it is meant to be the center. There is only one way you can use 11 on a side, with 4 and 6, so that a center column or row will be 3 7 11. If you know that column and the 4 11 6 row, you know at least this:
. 3 . . 7 . 4 11 6
And because you know the diagonals, you know that it'll be
8 3 10 . 7 . 4 11 6
And you only have 5 and 9 left, and they're easy to just plug in
8 3 10 9 7 5 4 11 6
So, that's the logical way to to solve it. Clearly, order isn't important; it could be reversed on the x and y axis and still be a solution. But, once the thought of solving with a recursive algorithm came into my head, I could not leave well enough alone. So, here's a recursive program that finds all possible solutions for this problem.
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