# Challenge: how high can you count?

Using just board and pieces as it ships, release 1 blue, 1 red, 2 blue, 1 red, 3 blue, 1 red, 4 blue, 1 red, … n blue, 1 red. So output as it appears on output ramp looks like:

. . . . bbbbrbbbrbbrb

What is the highest that you can count to in this manner?

Whatever your solution, it can’t count arbitrarily high - since a simple argument shows that you cannot continue the pattern indefinitely without more and more pieces and board space.
But, do you have a structure and approach that is systematic enough that you could count arbitrarily high if you had more parts and more board space?

Addendum: I was able to count to 3 here. But after getting to three, it continues with some extra balls before repeating, so the repeating pattern looks like this: rbbbrbbrbbbrbbrb. It would be nice if I could repeat just the part that counts to three (i.e., rbbbrbbrb). Can anybody get a count up to four?

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This will count to 4 and repeat itself nicely (pull on the blue lever). This technique can be used to count to arbitrary power of 2 given sufficiently large board.

I’ve looked at your solution but I have to say, I don’t know how to extend your structure to higher powers of 2. Do you mind elaborating on that?

This one counts to 8 (but it needs a large board). It is also structured so it can be grown indefinitely (a more compact version is possible without structuring it this way).

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Ah yeah, I see it now, very cool. Is it feasible to modify this structure so it loops after, say, counting up to 6?

Terminating it when some value is reached is easy (just count blue balls). Making it loop with non-power-of-2 period is harder, some sort of reset functionality has to be added

Very nice!! I imagine this could also be “inverted” to create a “count-down” timer as in “10, 9, 8, …,2, 1, blast-off!”

Actually, by using two blue balls per count, it is possible to go to 8 (and keep looping) on a standard board (but more pieces are needed than a single set contains). It’s a nice puzzle to get to 8. Click here if you want to see a solution.

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This is a nice challenge! I managed to count up to six on a normal-sized board with just the normal set of parts. The difficulty was not so much the logic, but getting to a set-up that fits. My solution is not periodic. After reaching six the sequence terminates. So the result is bbbbbbrbbbbbrbbbbrbbbrbbrb

Here’s my solution.

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