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Topic: Bucky's puzzles (Read 49428 times)
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jnz
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My solution for Time-Division Multiplexing v3 (batedoc): giralab. The bottom half is recycled; the top half is new.
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Rene
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Also, has anybody else attempted the Even-Odd Partition(hard variation) tutusop? Those looking for a really tough, tough puzzle, well there you go. I am still planning to give it a go, but haven't found the time yet. In my first attempt I mixed up the hard and medium puzzles, and came up with a solution that solved neither of them 
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Bucky
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IT SEEMS That nobody has posted a solution to Medium either.
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That is the most ingenious method of solving an impossible puzzle that I have ever seen.
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Bucky
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...And a new Heavy puzzle, even though nobody's solved Medium yet. Since you solved One-Shot, the transbarrier communications group has faced budget cuts; The original three pathways have been reduced to two. Also, due to an increase in demand, you will need to solve several crates instead of just one. Fortunately, you have been provided with infinite communication in the downward direction. Two-Shot: dakolip
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That is the most ingenious method of solving an impossible puzzle that I have ever seen.
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jf
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IT SEEMS That nobody has posted a solution to Medium either.
Here is my solution to Even Odd Partition (medium): dadefuk -- but it's a SPOILER for the hard version ( tutusop), as I just reprogrammed my water clocks (from my previous solution pidazic).
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Bucky
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Here is my solution to Even Odd Partition (medium): dadefuk -- but it's a SPOILER for the hard version ( tutusop), as I just reprogrammed my water clocks (from my previous solution pidazic). I really like the simplicity of your switching mechanism. I think using a template copier misses half the point of the puzzle. Granted, it's a very clever form of template copier, but it's rather overkill because (other spoiler). Finally, I think that discarding the original 0 was a bad idea because it could have been pushed through to the other side to save some time, it counts up from 0 when counting down is quicker.(part of previous spoiler) Also, I've done a new, easier Even-Odd Partition puzzle that actually works as apparently intended. Even-Odd Partition Remix: bedibug.
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« Last Edit: February 02, 2007, 04:25:47 AM by Bucky »
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That is the most ingenious method of solving an impossible puzzle that I have ever seen.
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jf
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Bucky, those are valid points. I re-used the hard solution because it was easier than figuring out a new solution. Most of the work in the original solution was the switching mechanism, especially the reset-swap sequence. The "water clocks" can be programmed for just about any sequence, given enough room for the water.
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« Last Edit: February 02, 2007, 10:09:59 AM by jf »
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Rene
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Here is my solution to Two Shot ( dakolip): dytedom The only way I could see, was to multiply the times for two of the crates.
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jf
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Here is my solution to Two-Shot ( dakolip): lodygop. I came to the same conclusion as Rene, but I have a lot more machinery (much of which deals with initialization -- I wanted to solve in one pass, with crates, not barrels).
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Bucky
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There is a way to solve Two-Shot without outright multiplication. As a bonus challenge, solve it using the following communication scheme: Use copies of the third crate to trigger the gate (2 bits) and trigger each at an even or odd tick (2 bits).
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That is the most ingenious method of solving an impossible puzzle that I have ever seen.
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Rene
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I finally took the time to put my teeth into Even-Odd Partition (hard). What a very tough puzzle! Over the past weeks I have done some try-outs of different ideas, but nothing really stuck. Maybe because I did not have enough time, maybe because I was on the wrong track. But today I finally was able to push one idea through and solved it. Solution to Even-Odd Partition - hard ( tutusop): nyfycab I went for a programmable automata in the end, with a separate program for each side. The right side program just discards or sends back the first crate that comes out (the second crate is always send to the target matchers). The left side program decides among: send it to the right side (will add 2 to it), push it to the right side unchanged, or send it up to the target matchers.EDIT: and the solution for the medium. It is basically an adaptation of the hard one. I have made the right side more general - nicer, I think (would have worked for hard as well) Solution to Even-Odd Partition - medium ( sodices): butikyp The right side now programmatically selects the first crate or the second crate (or none) both for the target matchers and for back to the leftEDIT: And finally, the easy variant. It was maybe the wrong way around: from hard to easy, but it saved the trouble finding specific solutions for medium and easy. Solution to Even-Odd Partition - easy ( cubyfuh): nobytas This was just reprogramming of the medium solution
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« Last Edit: February 04, 2007, 09:00:24 PM by Rene »
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Rene
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I finally got around to solving one of your oldies as well: Timing is important ( sufebeb). As you say, the solution is..... odd. Anybody who has found a solution, will know why I don't post it here....
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Bucky
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I finally got around to solving one of your oldies as well: Timing is important ( sufebeb). As you say, the solution is..... odd. Anybody who has found a solution, will know why I don't post it here.... Just out of curiosity, how long did it take you to solve it?
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That is the most ingenious method of solving an impossible puzzle that I have ever seen.
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jf
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There is a way to solve Two-Shot without outright multiplication. As a bonus challenge, solve it using the following communication scheme: Use copies of the third crate to trigger the gate (2 bits) and trigger each at an even or odd tick (2 bits). Here is my solution for the bonus challenge: biburet. If you've read Bucky's spoiler, then you know that I did not use multiplication this time. It was quite a tight squeeze to get all the parts built. Ironically, sometimes the less space you have to work in, the more pieces you need to use. Very clever idea for the communication scheme, Bucky. I did have to serialize the left crates to avoid a potential collision at the gates when sensing two of the bits., which slows it down a bit, on top of the initialization overhead. It runs from 183 best-case to 244 worst-case ticks. Edit: I was able to shave that down to 174-235 ticks, with a slight optimization in gydabyt -- not that speed is the point to the solution. Edit again: I just realized that I did not use Bucky's communication scheme, exactly. I did not use the downward copier. Once one of the winches has been used to get a value up, you are reduced to a single winch, giving only 1 bit of communication based on value.
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« Last Edit: February 05, 2007, 09:33:59 PM by jf »
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Rene
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I finally got around to solving one of your oldies as well: Timing is important ( sufebeb). As you say, the solution is..... odd. Anybody who has found a solution, will know why I don't post it here.... Just out of curiosity, how long did it take you to solve it? I have been looking at it before, but put it away as "unsolvable". At the last attempt, it took me about half an hour to solve it. You basically just need to get the "aha" idea. After that it took me just a small effort to get the timing right. I think you could stare at it for days without seeing it, though.
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