Challenge: how high can you count?


#1

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?


#2

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.


#3

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?


#4

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).


#5

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


#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


#7

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


#8

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.


#9

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.