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Author

Topic: TTTPPP's Puzzles (Read 27411 times)


Rene



« Last Edit: August 28, 2007, 11:18:13 PM by Rene »

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TTTPPP

The "Under and Over" series were easier than I'd planned, so I've currently given up trying to come up with situations which require distinctly different uppipe draws. Well done Rene with MAAN9, I've been struggling to get a solution to fit in the window for quite a while now! I came up with one which would work if the play area was about 5 or 6 times wider! Also nice work with MAAN16  I had thought it must be possible to make an elegant solution to find pairs, but resorted to the brute force tree sort instead. I can't see a good approach to MAAN puzzles with less crates than 9.



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TTTPPP

Right, a new puzzle. Apologies for the fact that I can't solve it! "Rubik'son" bagabugI came up with an idea of trying to implement a Rubik's cubeesque puzzle, which only allowed permutations of the crates. The spoiler below gives a description of how to use the machine (which I wouldn't really consider part of the puzzle). The machine outputs a random permutation of the numbers 1,2,3 and 4. This is the target to match using the four crates on the left. The lefthand side takes four different inputs A,B,C and D. These affect the crates as follows: An A will permute the crates from 1,2,3,4 to 4,3,2,1 (i.e. a reversal). A B will permute the crates from 1,2,3,4 to 4,1,2,3 (i.e. a rotation). A C will permute the crates from 1,2,3,4 to 4,3,1,2. A D will push the crates into the solution spot.
My progress so far: I have made a solution for the first crate, using only B barrels as input to the machine. I'm fairly sure this is the wrong way to go about a solution  maybe it's better to fix the 1 crate, then fix the 2 crate, etc.



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Bucky

I got further than TTTPPP did. I almost didn't fit what I have, and am not relishing handling the remaining logic in the remaining space. I need to, for every barrel header barrel, set it to move a B from beginning to end of each group that follows and has a larger header barrel. Then I need to add "AD" at the end. Partial solution: kysycytEDIT:I'm even closer now. I just need to add the needed B to the start of the next word to pass through. Partial solution: xylevoh.EDIT: Sooo close! lazedad


« Last Edit: September 05, 2007, 11:25:52 PM by Bucky »

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That is the most ingenious method of solving an impossible puzzle that I have ever seen.



jnz

For Rubik'son ( bagabug): kyrunadStrategy: Rotate the given pattern left until the 1 is in the leftmost position, keeping track of the number of rotations. This leaves us with 6 possibilities (1234, 1243, 1324, 1342, 1423, 1432). Hash these possibilities down to a single barrel and then use it as input to a lookup table to select some permutations that will produce the 1xxx pattern. Then push in the Bs we saved from the first step to rotate the pattern into the proper final position, followed by a D to flush the machine.


« Last Edit: September 06, 2007, 01:50:16 AM by jnz »

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TTTPPP

jnz, that is quite brilliant! I've filled a couple of sheets of paper with diagrams and equalities, and was nowhere near that solution. Bucky, I'm afraid I still don't understand quite what you're up to! I was trying to use your barrel pusher from the right hand side, and have that for the first 3 barrels, and then have a second one for the fourth barrel. This involves coming up with 8 preset sequences of input barrels, which 'magically' produce the desired ordering. At this point I gave up and went to bed!



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jnz

jnz, that is quite brilliant! I've filled a couple of sheets of paper with diagrams and equalities, and was nowhere near that solution.
Why thank you. I had a page of notes in Notepad, an Excel spreadsheet ( to find a collisionfree hash function (4x+2y+z)), and a Perl script ( to produce the entries in the lookup table).



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Rene

Rubik'son is a very original and well designed puzzle; it really reminds of Rubik's Cube. I liked it very much. For Rubik'son ( bagabug): gavidekMy strategy: I used a Rubiks type of solution as well. Each of the ABC patterns that I used is designed to swap only two digits and leave the other two intact. So, the first digit that comes out is looked up in a table, and produces the pattern that swaps that digit into the '1' position. The same swap is also done for the digits that the machine produces: an '1' that comes out later is replaced by the remembered digit. Subsequently, a similar thing is done for the second and third digit, producing the patterns that swap the second and third digits into place. BTW: the Rubicon identifier for the puzzle is quite funny: "bagabug" EDIT: just looked at jnz's solution. Pretty smart. It is faster than mine, partly because you permutate three barrels at a time and use the efficient rotation operation.


« Last Edit: September 07, 2007, 04:12:43 PM by Rene »

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jnz

Rene: I'd originally thought about doing something similar to your solution but then I realized that I'd have to keep track of which crates I had already swapped. That seemed complicated so I moved on to other ideas. You've managed to handle that problem in a quite simple and compact manner though. Kudos!



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TTTPPP

Another nice solution from Rene as well! I enjoyed the challenge of making this puzzle, so I'm glad it was interesting.



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Gameman

for waiting for a fad to pass: banyrad



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Yugioh duel; i play exodia, you lose. Nope, I play Rubicon contractor! *gets run over by dozer pushing a 4 crate*



Bucky

for waiting for a fad to pass: banyradThis fails to solve. Doubling Chamber vs. Template Copier: xygetyh


« Last Edit: November 06, 2007, 08:05:21 AM by Bucky »

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That is the most ingenious method of solving an impossible puzzle that I have ever seen.



jf

For Rubik'son ( bagabug): kurened. I've been working on this one for quite some time, on and off, as I have found time to spend on it. That was quite an interesting puzzle! Now that I've completed my solution, I've looked at the spoilers and the other solutions to compare. I came up with the same general design as Rene, implemented differently, and it works like this: The left side handles a string of three barrels to solve the first three crates. If you solve three crates the fourth must end up in the right place. barrel 1 (for 1st crate) if (1) then nothing because the 1 crate is in the right spot already, if (2) then ABCD>BACD (swap 1st and 2nd), if (3) then ABCD>CBAD (swap 1st and 3rd), if (4) then ABCD>DBCA (swap 1st and 4th); barrel 2 (for 2nd crate) if (1) then nothing  another barrel will move the crate if (2) then nothing because the 2 crate is in the right spot, if (3) then ABCD>ACBD (swap 2nd and 3rd), if (4) then ABCD>ADCB (swap 2nd and 4th); barrel 3: (for 3rd crate) if (1 or 2) then nothing  another barrel will move the crate if (3) then nothing because the 3 crate is in the right spot, if (4) then ABCD>ABDC (swap 3rd and 4th)
The right side processes the first three barrels out of the machine. When a 1 is seen, a copy of the first barrel out replaces it. When a 2 is seen, a copy of the second barrel (which itself could be a copy of the first barrel) replaces it. For example, if 1 2 3 4 comes out, then 1 2 3 4 is passed to the left (and only 1 2 3 are used). If 4 3 2 1 comes out, then 4 3 3 4 is passed to the left. If 3 1 2 4 comes out, then 3 3 3 4 passes left. This simple repeating of the 1 and 2 positions remembers the state of swapped barrels for the left side.I was also working on a full table lookup of all 24 sequences, which I'll finish up as I have time.


« Last Edit: January 03, 2008, 03:15:54 AM by jf »

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jf

Here is my alternate design for Rubik'son ( bagabug): logilygThis solution uses a fulltable lookup, where the table contains all solution sequences ahead of time. This was also quite difficult. I kept alternating between this design and my other design as I ran into problems with each.



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