The final puzzle I'll be discussing from the MatheMagic book is 'Magic Hexagons'. For this one, I'll cover the basic puzzle from the book, and then look at extending it (until I come unstuck).
The puzzle is borderline trivial:
Place the numbers 1-7 in this hexagon so that the three lines through the central hexagon all have the same total.
The method is straighforward: place the number in the middle of the range (4) in the central position, and then pair the others around it so that the smallest is with the largest (1 and 7), then the next smallest with the next largest (2 and 6), and finally the last pair (3 and 5).
However, extending this (in the same way as I did with the triangles in my previous post) is not trivial at all. There are many solutions for the regular hexagon with side length = 3, but I took my calculations along a slightly different path and came completely unstuck.
Consider this hexagonal grid, with irregular sides: is it possible to populate them so that all the straight-line paths with length four have the same sum?
The issue is that the numbers in dark blue (in this example, 4 and 7) are included in three separate lines (the vertical, and two diagonal lines) and swapping numbers in and out of these hexagons affects three different lines (and the ones you're swapping into). The numbers in the pale blue hexagons are also considered in two different lines (the two diagonal lines that intersect) and again there's too much juggling to do. In the simple example, and in the regular 3x3 hexagon puzzle, each number only contributes to one line and swapping with other numbers only affects the start and destination lines. In my puzzle, there are between two and six lines being affected (depending on the location of the two numbers being swapped).
So, is it possible to solve this puzzle. Short answer: no, I don't think it is.
The puzzle is borderline trivial:
Place the numbers 1-7 in this hexagon so that the three lines through the central hexagon all have the same total.
However, extending this (in the same way as I did with the triangles in my previous post) is not trivial at all. There are many solutions for the regular hexagon with side length = 3, but I took my calculations along a slightly different path and came completely unstuck.
Consider this hexagonal grid, with irregular sides: is it possible to populate them so that all the straight-line paths with length four have the same sum?
The five lines shown here with arrows indicate the lines (each containing four hexagons) that I will attempt to balance.
First attempt: place the smaller numbers (1, 2) in the central hexagons, and those at the middle of the range (6, 7) then try to pair up the numbers around the outside of the shape.
First attempt: place the smaller numbers (1, 2) in the central hexagons, and those at the middle of the range (6, 7) then try to pair up the numbers around the outside of the shape.
Not a great start, but I was able to make some improvements by changing the numbers in the four central hexagons. I still have 2 in the central area, but have included 4 instead of 1, and 8 instead of 6. Still working on keeping the mid-range and low-end numbers in the middle.
The issue is that the numbers in dark blue (in this example, 4 and 7) are included in three separate lines (the vertical, and two diagonal lines) and swapping numbers in and out of these hexagons affects three different lines (and the ones you're swapping into). The numbers in the pale blue hexagons are also considered in two different lines (the two diagonal lines that intersect) and again there's too much juggling to do. In the simple example, and in the regular 3x3 hexagon puzzle, each number only contributes to one line and swapping with other numbers only affects the start and destination lines. In my puzzle, there are between two and six lines being affected (depending on the location of the two numbers being swapped).
So, is it possible to solve this puzzle. Short answer: no, I don't think it is.
Love it!
ReplyDeleteI found several solutions for your puzzle. Here is one of them
ReplyDeleteI found a solution to your puzzle:
Referring to the diagram above with the blue and green "core", replace the
4 with a 7, the 8 with a 2, the 2 with a 3, and the 7 with an 8. Starting
from the 12 at the top, going clockwise, replace the outer cells with the
following numbers: 1, 9, 6, 14, 13, 12, 11, 10, 5, and 4. The resulting
grid will have all 5 arrows add up to 28.