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Topic: TTTPPP's Puzzles (Read 23726 times)

TTTPPP

A list of the puzzles I've made: *Note that for the three "Most Significant Bewilderment" puzzles you are not meant to be able to put pieces into the mechanism (eg. dropping an initial barrel down to get a 1 crate out). I only remembered to prevent this in the third level. Enjoy!


« Last Edit: September 06, 2007, 10:54:26 AM by TTTPPP »

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jf

For the Doubling Chamber: dinysuk. Rene's tunupyv is impressive for its simplicity.



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TTTPPP

I wish there was a slow motion button so I could see how Rene's solution works! I'll just think of it as magic.
JF's is also far simpler than my solution, and I love the use of 11+1 = 1, 22+2 = 2, etc.



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TTTPPP

New puzzle  Much Ado About Nothing vivevop. I think this is quite an obvious idea for a puzzle, so I wouldn't be surprised if there's been a similar one before. So far I've got a solution which works about 66% of the time padigyk, and can in theory solve Much Ado About Nothing 16  the same thing with 16 ? crates (i.e. the main puzzle is Much Ado About Nothing 5). Can anyone manage a partial solution better than this, or a solution for Much Ado About Nothing 15, 14, etc.


« Last Edit: July 27, 2007, 10:49:26 AM by TTTPPP »

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TTTPPP

New puzzle: Nudge Nudge pibanyd  Solution noribynI was trying to make a much harder puzzle, but came across this as a subpuzzle of it! I'm not even sure the harder puzzle was possible though!



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hatkirby

Solution to Nudge Nudge by yours truly! Nudge Back  zutyzek



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If you take four, and then add four, and then take away, add, subtract, add, add, multiply by, subtract, raise to the power of, divide by, and add four, and then find the reciprocal, what do you get? Of course, it's obviously 0.058823529411764705882352941176471! Duhhh!
Four is the answer to all!



Rene

For Nudge Nudge ( pibanyd): docilinI am still working on Most Significant Bewilderment Level 1. That one is possible but hard. I have done the math, and now need to do the implementation work. Most Significant Bewilderment Level 3 is theoretically possible, but I do not think there is enough space to make the solution. Even the math was very hard to figure out. For Much Ado About Nothing: there have been similar puzzles. For example,  use 0, 1, 3, and 9 to match a random crate  dubynod)  use three random crates between 05 to make a 0  cannot find that one anymore But I haven't seen this one before. I think it should be theoretically possible with 4 random crates to always make a 0, but I didn't come any further than you did at the moment. Use zeros directly, and subtract duplicates from each other


« Last Edit: August 06, 2007, 12:21:05 PM by Rene »

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TTTPPP

It's not possible to solve Much Ado About Nothing 4, as you may be given the crates 1,2,4 and 8. I've come up with improvements on my previous method, but haven't yet been able to implement them. It would solve MAAN9.



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Rene

And finally, my solution to Most Significant Bewilderment Level 1 ( himaxot): varobolStill looking at 2 and 3, but I am afraid they are probably not possible.



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jnz

For "Waiting for a fad to pass" ( nylabuv): gogoxyr. That was tricky. I don't think I've seen a puzzle like that before.



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TTTPPP

Good solution jnz! I was hoping there'd be some different ways of solving it, but I could only find mine. I'm guessing the longest running solution in the space given is by changing the +++++ in jnz's solution to ++++. Then it even solves "02C"! The aim of the puzzle was really to see how long something could be made to run in the space, so if you move the 1 crate about to form a different solution I'd still be interested to see it!



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jnz

The aim of the puzzle was really to see how long something could be made to run in the space
Ok, how about pocubeg? Solves "8E8"



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