Alternate Title: What is the biggest binary counter you can squeeze onto the board?

Assuming an infinite supply of balls, design a device that counts (in binary) a single color of balls and stops when it cannot count any higher. What is the largest number of balls you can count using a standard-size board?

I found a few solutions, with varying degrees of faithless to the prompt and to the rules of Turing Tumble:

This is the highest I could count without breaking any rules. As in the workbook, I use the convention that a bit pointing left = 0 and pointing right = 1.

This time Iâ€™m counting blue balls. This solution technically satisfies the prompt, as long as you donâ€™t mind red balls in your blue balls. Every 16th blue ball will cause the lower four bits to overflow, which is carried over to the upper four bits by a red ball. Thus, 16 red balls will be require to count 255 blue balls. Incidentally, the total number balls collected at the bottom will be equal to the count read off the full 8-bit register plus the count read off the upper four bits (e.g. if the register reads 01100100, there are 100 blue balls + 6 red balls = 106 balls the bottom).

This is where things get sketchy. To make this work, you need a real board so that (a) the 512th ball will be captured against the side of the board and (b) you can tape of the blue chute so that all the balls trigger the red lever.

For this to work, you again need to tape over the blue chute. Also, the uppermost for bits will need to be read as the *opposite* of the direction they are pointing (I used gear bits for these as a reminder).