Harder Version of Challenge #39 from the book


#1

In Puzzle #39 from the book, the challenge is to generate the repeating pattern rBBBrBBBrBBBrBBB, which is 3 blue (B) followed by 1 red ®, 3 blue, 1 red, repeating until the board runs out of marbles.

In this version, the challenge is the generate a similar pattern but with 4 blue followed by 1 red (repeating until out of marbles). It will look like rBBBBrBBBBrBBBBrBBBB,

Use the minimal number of parts and favor simple parts (ramps, bits, crossovers) over more “expensive” parts such as gears and gear bits.

I can provide a solution upon request.


#2

17 parts. I’m not including a screenshot to avoid spoilers. Only one of those crossovers is really needed, but using several saves on the number of ramps we need.


#3

Nice job! You did it in fewer parts than I did, so I think your solution is the preferred one. It also has a very clean and efficient layout. I like to look at not only the total parts count, but also the “cost” of the parts. I think crossovers and ramps are the cheapest parts while gears and gear bits are the most expensive parts. Your solution not only uses fewer total parts, it also uses predominantly cheap parts. Another way to judge a solution is by the path length travelled by the balls. I think your solution is great in that respect too. Thanks for sharing.


#4

I just realised that I’m missing one ramp in the bottom centre slot. I’ll update the link later.


#5

Let’s take this up a notch. Try doing BBBBBrBBBBBr… I already have a solution.


#6

Design a method that works for any N, assuming you’ve got enough board space and parts.
Here is one solution, but likely not the most efficient in terms of parts and real estate - in fact - I could only fit N=4 on the largest emulator board.

Note 1 - for some reason when this loads, a gear ends up out of place for some reason, so you’ll need to fix it. Should be obvious.

Note 2 - start with the blue lever. This solves the puzzle with blues separating many reds, instead of the opposite. I’m too lazy to redo it the other way.