Sly's Puzzles

[TTTPPP]

I agree that it's only solvable probabilistically.

A 'proof':

There are a finite number of solutions to the top left section, and given any solution for the top left section there is a finite minimum time that it will finish in.
Therefore there is a time T which is the longest minimum time for a solution for the top left to finish.  Therefore no matter what arrangement is in the top left section it will be solved with non-negligible probability in time T.

Therefore any complete solution to the puzzle must have the other two sections solved by time T+t (where t is the time for the trapdoors to close).

There are no solutions for the bottom left section as long as the random crate only produces 0s.  There is a non-negligible chance that nothing but 0s will be produced in the bottom right section for time T+t.

Therefore there are no complete solutions to the puzzle.

[sly]

Interesting math, you're right, it won't solve all the time, but you can get close.

I added some wires.

Synchronize 2 : jedixyg

[jnz]

Wires!  That's what I'm talking about!
For Synchronize 2 (jedixyg): xefoxep

adept: your Synchronize solution (jofytar) seems to be a tick off when the random crate on the right is 0.

[sly]

I forgot a component.

Synchronize 3 : gonygef

[Rene]

For Zero-base (rycecux): xoxozyx

About Dyslexia (nexifud): I already used almost all the available space for TTTPPP's "Much Ado About Nothing 9". If the random crate in Dyslexia happens to be '0', then it is equivalent to MAAN 9, so I would judge this puzzle unrealistically difficult.

[jnz]

About Dyslexia (nexifud): I already used almost all the available space for TTTPPP's "Much Ado About Nothing 9". If the random crate in Dyslexia happens to be '0', then it is equivalent to MAAN 9, so I would judge this puzzle unrealistically difficult.

Obfuscated hint: what if you ignore the barrel that comes out?
For Synchronize 3 (gonygef): is it possible to get a 1 through the monster's mouth?

[Rene]

About Dyslexia (nexifud): I already used almost all the available space for TTTPPP's "Much Ado About Nothing 9". If the random crate in Dyslexia happens to be '0', then it is equivalent to MAAN 9, so I would judge this puzzle unrealistically difficult.

Obfuscated hint: what if you ignore the barrel that comes out?

Your hint does not really help. You still need to get the original crate out. The only way to do that is with a '0' crate, so you still need to come up with a '0' crate. That makes the problem equivalent MAAN 9 again. And with the puzzle in the middle of the field, the solution is going to be very awkwardly shaped.
If you're seeing something that I'm missing, then let me know.

[jnz]

If you're seeing something that I'm missing, then let me know.

No, that's what I had in mind.  I thought it reduced it from "unrealistically difficult" to merely "extremely difficult".  Of course, you don't see me attempting it...

[Bucky]

I've got an alternate idea that might work most of the time for Dyslexia.  I don't have the space to complete the design as shown, but it may be possible to get something working most of the time with the same algorithm. (The demo machine has many things wrong with it and halts after producing a zero crate.  Ideally, instead of filtering for 0s we'd use the method shown to acquire an n-1 barrel and use them to filter directly for the target value.  There ARE cases where it won't work, but it should do fine most of the time.

Dislexia fragment: sogigid

EDIT:  I think I have the special cases worked out.   If the target crate is 0, the n-1 method won't work, but we can detect that and alter the traffic flow accordingly.  Also, it will only fail to produce a 0 if we have odd crates and an even target, or a 0, 4, 8 or C target.  We can usually solve this by extracting two crates every 16 machine cycles and adding them together.    Now I just need Rene to figure out how to hook everything up.  I figure it has at least a 95% chance of working.

[adept]

adept: your Synchronize solution (jofytar) seems to be a tick off when the random crate on the right is 0.

I can't believe I didn't test for that-  This one seems to work for synchronize gebylyb

Synchronize 2 : lecigeh

[Rene]

I've got an alternate idea that might work most of the time for Dyslexia.  I don't have the space to complete the design as shown, but it may be possible to get something working most of the time with the same algorithm. (The demo machine has many things wrong with it and halts after producing a zero crate.  Ideally, instead of filtering for 0s we'd use the method shown to acquire an n-1 barrel and use them to filter directly for the target value.  There ARE cases where it won't work, but it should do fine most of the time.

Dislexia fragment: sogigid

EDIT:  I think I have the special cases worked out.   If the target crate is 0, the n-1 method won't work, but we can detect that and alter the traffic flow accordingly.  Also, it will only fail to produce a 0 if we have odd crates and an even target, or a 0, 4, 8 or C target.  We can usually solve this by extracting two crates every 16 machine cycles and adding them together.    Now I just need Rene to figure out how to hook everything up.  I figure it has at least a 95% chance of working.

Normally I would jump to a challenge like that  But, unfortunately, I can still see some holes in your reasoning. You have a problem if the target crate is even. If it is even, then the 9 crates need to contain a multiple of the target crate, in order for it to work. E.g. if the target crate is 8, then the 9 crates need to contain either a 0 or an 8. Randomly adding two of the crates together has a low chance of producing 0 or 8, and is therefore not a viable approach. You need to carefully select which crates to add together (or subtract from each other) in order to produce the right value.
Therefore, I need some more of your creative thinking before we can construct a solution.

[Bucky]

Normally I would jump to a challenge like that  But, unfortunately, I can still see some holes in your reasoning. You have a problem if the target crate is even. If it is even, then the 9 crates need to contain a multiple of the target crate, in order for it to work. E.g. if the target crate is 8, then the 9 crates need to contain either a 0 or an 8. Randomly adding two of the crates together has a low chance of producing 0 or 8, and is therefore not a viable approach. You need to carefully select which crates to add together (or subtract from each other) in order to produce the right value.
Therefore, I need some more of your creative thinking before we can construct a solution.

I'm aware of this.  I didn't make the case that it worked all the time.  It'd only work 95% of the time as written, failing mostly on a '0' and '8'.  Unfortunately, I can't see how to solve it with case 0 without having a MAAN 9 solution on the side.

With the other cases, however, we can deal with the issue of 4, C and 8 at the same time as the even crate problem simply by pulling out a crate every 16 cycles and solving for that crate in the same manner as the original.  By the pigeonhole principle, two of the crates have the same value mod 8.  This neatly solves every case but 0, and usually solves 0 as well, leaving us with better than a 97% chance.

[sly]

Wow, I had no idea Dyslexia would be that difficult. I didn't see Much Ado About Nothing until after I made this level and I was thinking the same thing that either two crates are the same for zero or two crates added together equal zero. This led me to think this was possible with all numbers and left it as is. But now that I think about it, it doesn't work that way. Oh and I really like the 'machine gun reloading' effect you made Bucky.

I spent forever making this new puzzle and I kept thinking of ways to make it difficult. Unfortunately, I only got this, which is fairly easy. And then I couldn't figure out how to make the top two crates be switchable. btw: 7s are kinda the default input number

Rubik's Rectangle: xycusud

Oh, and, please I beg of you to make this puzzle better and harder, and smaller, if possible. You're free to mod it.

[jnz]

For Rubik's Rectangle (xycusud): linexis.  I cheated.  Also, you forgot to lock the space next to the 8 barrel but I didn't use that.
Edit: here's a non-cheat version: gosifed
Edit: here's a proof-of-concept top-swapper: budeses

[Rene]

Here is another cheat solution to Rubiks Rectangle (xycusud): bupigun