Moving from hexagon to hexagon, connected by the black lines, you are to form a string of numbers, starting at the number marked with an S, and finishing at the number marked with an F. Each number is made up of a certain amount of digits, and can only be used once. For example, the number (6,7,2) has three digits: 6, 7, and 2.

The first time you land on a hexagon, you use the top number on that hexagon, and if you revisit a hexagon, you’ll have to use the bottom number. Each number you visit on your journey from hexagon to hexagon must be “lower” than the one before, and numbers that have n digits can only be compared to numbers that have n – 1, n, or n + 1 digits by the following rules:

1) Moving from length n to length n + 1: simply append a new digit at the end of the number, while keeping the other digits the same. Example: you can move from (3,2) to (3,2,6); you cannot move from (5,4) to (5,2,7).

2) Moving from length n to length n: one and only one digit is to be reduced. Example: you can move from (3,7) to (3,2); you cannot move from (6,3) to (4,1).

3) Moving from length n to length n – 1: cut off the last digit and one digit is reduced. Example: you can move from (4,6,2) to (4,5); you cannot move from (2,1,7) to (2,1).

Here is an example puzzle:

Before I show the solution, let me give a bit of mathematical background on the basic idea for this puzzle: partially ordered sets.

Most of the number systems that we normally deal with are well-ordered. Basically, this means that any two numbers in our set can be compared to one another. In the counting numbers (1, 2, 3, 4 ...), I can say that 2 is less than 3, 7 is less than 12, and so on. I can pick any two numbers and figure out which one is smaller and which one is bigger (unless, of course, I pick the same number, in which case, the numbers are said to be equal).

In a partially ordered set, if I pick one number, I can compare that number to only some of the other numbers in the set. Say, for example, I create a partial ordering of the counting numbers in which you can only compare numbers with the same number of digits. Then, we could still say that 2 is less than 3, but 7 is no longer less than 12, since the two numbers cannot be compared to one another. 7 is not less than 12, and it’s not greater than 12 either; the two numbers simply cannot be compared.

But things can become even more complicated. Let’s change the previous example of partially ordered sets to say that if I have a number with n digits, then I can compare it with another number that has only n – 1, n, or n + 1 digits. So 2 is still less than 3, but now 7 is less than 12, since 7 has one digit and 12 has two. However 8 cannot be compared to 278, since 8 has one digit and 278 has three.

Or can it? It certainly cannot be compared directly, but say we compare both numbers to a number with two digits, like 67. 8 is less than 67, and 67 is less than 278. So we can make an indirect comparison. In fact, we can make an indirect comparison of all the counting numbers that don’t fall within our n – 1, n, n + 1 rule by building a sort of chain of numbers. This would imply that I can compare, either directly or indirectly, any two counting numbers, which would mean that I’m back to having a set that’s well-ordered.

This type of puzzle works the same way. You cannot move from a number like (6,4) to (2,1), unless a stepping stone number like (6,1) or (2,4) exists, just like the stepping stone number 67 we used to compare 8 to 278.

Following the highlighted yellow arrows below, you can trace the solution to the example above. The numbers which are used in the current step are blue, those which were used in a previous step are shaded grey. Notice how the finish hexagon must be landed upon twice: first to use up the top number, then to complete the puzzle.

Below is a chart of the example's problem space. This shows every possible move that can be made while

solving the puzzle. The correct solution is in blue, down the left side of the chart. Problem space charts for each Hexagonal Ordering puzzle are located on the Drawing Board.

Last updated: April 5, 2007

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