CaptainEd
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Everything posted by CaptainEd
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Nobody, you hit the key--solving adjacent groups in 5 rather than 6 is a worthy and achievable goal. I'm even more strongly convinced that 8 is the maximum required for the W problem. (I imagine D being far, far away...) edit: oops, I'm wrong, result of #3:1 yields 12-5 OR 3-8 OR 4-9. Still checking...
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Now that I try to work on Y-san's question about minimized expectation,
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What's wrong with Palascosas' answer? As I read it, the worst case only requires 52 lookers:
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CaptainEd is quietly reading and learning from you guys. I'm clueless. I didn't even understand plasmid's solution, until bushindo gave it his stamp of approval. I'll feel satisfied once I understand it thoroughly.
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I offer the rails for Remnar's train of thought:
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If not, we've got to think deeper...
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Well, I couldn't agree with "poorly", especially as it was the first X, which led you to the correct word: chalky Your turn!
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swoops--__ steals--oo scores--o trying--o better--__ poorly--xo
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Cute, Palmer! Yes, it's best not to leave one's pets out unattended for too long... swoops--__ steals--oo scores--o
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Fun, indeed! I've got one with 6 letters.
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I'll bet scsw got over his ennui when he imagined the torture he'd give us solvers with this word!
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Thank you, Phatfingers.
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Yes, it's the details of creating those chains where my math errors caused problems. Actually, either P1 or P2 could take both responsibilities. Thanks for the interesting problem!
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Thanks, Phatfingers, -Yes, this dataset fails. -Yes, I think you did the swaps inaccurately, at a one-off location, but it fails according to (my interpretation of ) my instructions, as well. -No, the attempt wasn't foiled by having no way of knowing where the other number was living--P1 gets to see exactly half, which means P1 (actually us, as the puppeteers) knows exactly what's in the other half, and exactly where they will go once P2 has sorted them. The problem is, at this point, we're doing software-level bookkeeping (let's see, should that be (C+1/2) or (1+C/2)?) without my actually writing and testing the code. I apologize for that. I'll do the code and get it right (RSN), but here is the main fact, which I suspect can be agreed upon without my specifying the precise locations:
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Edit: sorry, gang, I'm having trouble copying cells from an Excel spreadsheet and preserving columns. My answer (in post 17) stands, but it might be hard to follow without seeing examples. I'll get them here somehow...sometime...
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OK, third time's a charm.
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It looks to me like the OP says that, after each prisoner has taken a turn, there must be one slip in each jar. (Also, note we don't have anybody's names, or need them, OP says there are only numbers in jars) Separately, my recent solution has a problem
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drum-roll please....Palmerc7 with "paradiddle". Good job, have fun
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schlepping--oooo sanguinely--xxxo varicocele--xxxxo xenogeneic--oo manometers--xoo protostele--xxxo zabaglione--xxxoo scathingly--xxo rhinestone--xoo waveringly--xxxoo deodorants--ooooo farcically--xxxoo xenografts--ooo valorously--xxo phlegmatic--xoooo skillfully--xo melancholy--xxo hootenanny--oo sanguinous--xx zygomorphy--oo tangential--xoooo cacography--xooo
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schlepping--oooo sanguinely--xxxo xenogeneic--oo manometers--xoo zabaglione--xxxoo scathingly--xxo rhinestone--xoo waveringly--xxxoo deodorants--ooooo farcically--xxxoo xenografts--ooo valorously--xxo phlegmatic--xoooo skillfully--xo melancholy--xxo hootenanny--oo sanguinous--xx zygomorphy--oo tangential--xoooo cacography--xooo I've been to hootenannies, definitely a real world.
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schlepping--oooo sanguinely--xxxo xenogeneic--oo manometers--xoo zabaglione--xxxoo scathingly--xxo rhinestone--xoo waveringly--xxxoo deodorants--ooooo farcically--xxxoo xenografts--ooo valorously--xxo phlegmatic--xoooo skillfully--xo melancholy--xxo sanguinous--xx tangential--xoooo
