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# CaptainEd

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1. ## Balancing Gold Coins

First, I define Restricted and Unrestricted coins, then describe and prove the algorithm for Restricted coins. Next, I request a slight tweak to the OP conditions--one additional coin known to be good. Then comes the definition of the procedure for Unconstrained coins, followed by the edge case (how many coins with only one weighing), and the formula for the Unconstrained case, since the OP requests the Unconstrained formula. Then come the first few values.
2. ## Balancing Gold Coins

cant do this compactly with iPhone. I'll answer sooon. (1) yes I'll make better demonstration about Restricted approach. (2) yes, I'll describe complete Unrestricted approach, which gives larger number
3. ## Balancing Gold Coins

sorry, my attempt to remove the post failed.
4. ## Balancing Gold Coins

[spoiler=answer maybe]in N weighings, can distinguish bad coin out of M=Sum(3^i), i=0...N-1 Can't always tell whether heavy or light[/spoiler] [spoiler=Background: restricted vs unrestricted]if you weigh coins in the balance scale, and the Left pan is heavy, then the coins in that pan are "restricted": each one is either Good or Heavy, while the coins in the other pan are restricted as either Good or Light. I abbreviate "heavy or good" as "H", and "light or good" as "L". We can distinguish among 3^N restricted coins in N weighings. [/spoiler] [spoiler=Strategy]However, we are starting with coins we know nothing about (and maybe one coin known to be good) So we will weigh a number of coins (p), leaving some of them (q) unweighed. Once you weigh the first batch, they become restricted, and you can resolve one of them in ceil(log base 3 (p)). If they balanced, then you have N-1 weighings to resolve the remaining q coins. So our initial task when approaching a set of coins is to choose p and q so that the number of weighings for p restricted coins equals number of weighings for q unrestricted coins. [/spoiler]

7. ## Count the Flags

Gavin, I don't see any rules saying where you can put the randomly generated bit. do we imagine that it can be placed into our finite memory? I like your suggestion that the random bit might contribute to a Roomba-like exploration :-) of course I don't see how yet.
8. ## Did you get more than me?

I'm missing something.
9. ## Interview Puzzle- "Fox in a hole"

I think that after we check 4, we know that fox is in an odd number right now, and will move to an even one that night.
10. ## Interview Puzzle- "Fox in a hole"

Don't watch the video!!! it is a good video, it clearly states the problem, but then it gives the answer! (Clearly stated as well) if you watch the video, stop after the problem statement. I enjoy the problem, and I appreciated the ingenious solution, nice job!
11. ## Mouse & Cheese Cube

My analysis was incomplete. If you skip the Center X, then a string of alternating C and F (with E between) is constrained by the number of E. With 12 E, at most 13 (C,F) can be interleaved with them. And, because there are only 6 F, that means the upper limit on a path would use 7 Cs and 6Fs. So as Rocdocmac says, if it's possible without touching X, it must start and end with C: CEFECEFECEFECEFECEFECEFEC. Either way, with or without the X cell, 25 appears to be the upper bound, and Thalia found one.
12. ## Mouse & Cheese Cube

(Sorry I don't know how to make spoiler on iPhone, but it seems Thalia is the winner, so I'm not surprising anyone) Here's why 25 is max: Lets label a corner C, Edge Center E, Face Center F, and the hidden Center X. There are 12 E, 8C, 6F, and 1X. Because of the required 90 degree turn at each cell, CE can only be followed by EF (or Exit). All possible moves are: CE : EF FE : EC or EF EC: CE EF : FE or FX FX : XF FE : EC Notice only F can reach X, so the Sequence of moves must include FXF, if the Center is to be reached at all. Also notice that once you have moved from C to E, you can only move from E to F. For each C, you can pass through one F. (In CEFE sequence). But while there are 8 C, there are only 6 F, so the total sequence can have only 6 C, and the maximum is only 25. Good news is, there are enough E to join all the C and F, so the sequence could look like ECEF ECEF ECEF X FECE FECE FECE A path that avoids X would have to have an E between two Fs, which would waste an E, leaving only 24.
13. ## Show 4 divides this algebraic expression.

I hope this is clear: Oops! Plasmid beat me to it on the 8:13 version of this puzzle.
14. ## Celebrity Interview

Bonanova, cute puzzle! Pickett, great solve! Thank you both, that was fun to watch!
15. ## Celebrity Interview

Can't do spoilers on my phone.
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