Simple Proof for A-H Array Puzzle

[ramsey2879]

I was doing a brain treaser where the letters A-H had to be put into an array 2 by 4 such that no letter is adjacent or diagonal to a letter one unit different. That is a C can't be above, below, to one side of, or diagonal from a B or D. I solved the problem but the thought occurred to me that using modular logic it can easily be proven that the H has to be above,below,to one side of, or diagonal to the A. In other words, it is immposible to put the numbers 1-8 into an 2 by 4 square box array, such that where each every set of boxes that touch either by a side
or corner the difference between the 2 boxes mod 8 is more than 1. Does anyone care to give a simple proof?

 

[ramsey2879]

I was doing a brain treaser where the letters A-H had to be put into an array 2 by 4 such that no letter is adjacent or diagonal to a letter one unit different. That is a C can't be above, below, to one side of, or diagonal from a B or D. I solved the problem but the thought occurred to me that using modular logic it can easily be proven that the H has to be above,below,to one side of, or diagonal to the A. In other words, it is immposible to put the numbers 1-8 into an 2 by 4 square box array, such that where each every set of boxes that touch either by a side
or corner the difference between the 2 boxes mod 8 is more than 1. Does anyone care to give a simple proof?

Proof:let the two rows of boxes be numbered 1-2 and the 4 boxes on each row be numbered 1-4. If the boxes are filled with numbers 0-7, each number except the 0 and seven have two neighboring numbers which differed by 1, but the numbers in boxes (1,2) and (2,2) both touch all boxes but (1,4) and (2,4). Mod 8 each number has two neighbors, if the numbers in boxes (1,2) and (2,2) share a neighbor, there are still three numbers but only two boxes to fill. Thus it is impossible to fill the 8 boxes with numbers 0-7 so that no two touching boxes have a neighboring numbers in the boxes. But mod 9, 0 is not a neighbor to 7, thus 0 and 2 can be placed in boxes (1,2) and (2,2) and the two neighbors, 1 and 3 can be placed in boxes (1,4) and (2,4). Likewise, 7 and 5, can be placed in boxes (2,3) and (1,3) and the two neighbors, 4 and 6 can be placed in boxes (1,1) and (2,1). 0 corresponds to "A" and 7 corresponds to "H".

 

[Norwegian]

Assume it is possible. Then every 2x2 square must be filled with numbers, no two of which are consecutive mod 8. Since the square contains 4 numbers, and the 4 gaps between the numbers are all at least 2, and sum to 8, we conclude that all the numbers in the square have the same parity, which then also holds for the entire array, and we have a contradiction.

 

[ramsey2879]

Assume it is possible. Then every 2x2 square must be filled with numbers, no two of which are consecutive mod 8. Since the square contains 4 numbers, and the 4 gaps between the numbers are all at least 2, and sum to 8, we conclude that all the numbers in the square have the same parity, which then also holds for the entire array, and we have a contradiction.

In other words, a 2X2 square must have only the numbers 0,2,4,6 or 1,3,5,7 since there is always a 3 gap difference between the smallest and largest number. However, when you note that there are 3 such 2 by 2 squares in a 2 by 4 array and only 2 possible sets of 4, then you have a contridiction. Thanks

 

[blue_raver22]

I commend you for solving the A-H array puzzle and for applying modular logic to prove that the H must be adjacent or diagonal to the A. Your observation is correct, and I agree that it would be impossible to put the numbers 1-8 into a 2 by 4 array without violating this rule.

To provide a simple proof, we can start by considering the possible positions of the A and H in the array. Since the array is 2 by 4, there are only 8 possible positions for the A and H, as shown below:

1. A _ _ _
2. _ A _ _
3. _ _ A _
4. _ _ _ A
5. _ _ _ _ H
6. _ _ _ H _
7. _ _ H _ _
8. _ H _ _ _

Now, let's assume that the H is not adjacent or diagonal to the A. This means that in each of the 8 possible positions, the H must be at least 2 units away from the A. However, since the array is only 2 units wide, this is not possible. Therefore, the H must be adjacent or diagonal to the A in order to satisfy the rules of the puzzle.

In other words, if we try to place the H more than 1 unit away from the A, there will always be at least one empty space in between them. This leaves us with only 7 possible positions for the H, all of which are adjacent or diagonal to the A.

In conclusion, your use of modular logic is a clever and effective way to prove that the H must be adjacent or diagonal to the A in the A-H array puzzle. Well done!