This is a game/challenge for one player. Start by drawing an n-by-n grid on paper. (n could be about 5 to 10, maybe.)

Next, either use a pre-existing integer sequence (such as {a(k)}, where a(n) = the number of divisors of n), or randomly pick the integers in the sequence.

If randomly picking the sequence, first write down n^2 numbers in a list before continuing to the next part of the game.

It might help to first write down the numbers even if you are using a preexisting sequence. Write down n^2 numbers, just in case they are needed.

Next, place the integers into the grid, starting at any square, and then placing each number -- in order by the order of the indexes (a(1), then a(2), then a(3)...etc), each number in any empty square horizontally, vertically, or diagonally adjacent to the last square filled with a number.

(a(k) is always next to a(k-1).)

Continue doing this until you cannot place any more integers (because there are no empty squares next to where you last put a number).

Next, starting at any square with a number in it, draw a path of connected line segments from square to adjacent square -- adjacent and in the direction of either vertical, horizontal, or diagonal -- such that each number drawn to is greater than or equal to the number in the previous square of the path.

(The numbers of the path never descend.)

The path must not visit any square more than once. But two diagonal segments of the path may cross.

Move until you can't move anymore. (The last square visited by the path will not be bordered by any unvisited square with a number >= the value in the last square.)

Your score is the number of squares your path visits.

Note: I realize that you could use the all-1's sequence, say, and score a perfect n^2 points each time, but that wouldn't be much fun.

As an easy challenge to myself I used the first 16 terms of the number-of-divisors sequence (1,2,2,3,2,4,2,4,3,4,2,6,2,4,4,5)

and a 4-by-4 grid. I got a top score of 13 (in several ways).

Can you do better?

Thanks,

Leroy Quet

## Wednesday, July 22, 2009

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## 1 comment:

I found a 14-point solution (counting all squares visited, including the starting square) using the number-of-divisors sequence and a 4-by-4 grid.

Can anyone get 15 or 16 points? Is it even possible to do so?

Leroy

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