BrainDen.com - Brain Teasers

# CaptainEd

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7

1. ## Coloring a grid

Nice compact proof, Bonanova, thank you!

Part b

4. ## 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.
5. ## 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
6. ## Balancing Gold Coins

sorry, my attempt to remove the post failed.
7. ## 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]

10. ## 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.
11. ## Did you get more than me?

I'm missing something.
12. ## 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.
13. ## 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!
14. ## 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.
15. ## 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:
16. ## 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.
17. ## Celebrity Interview

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

Can't do spoilers on my phone.
19. ## The lion and the tamer, Part 2

The more I tried to simulate Plasmid's argument, the more powerful it felt. While T is moving North, it doesn't matter how close L is to T, or how tiny the angle theta off vertical, with a sufficiently small stepsize, L will waste a portion of each step moving South, widening the angle theta, until eventually | T - L | > stepsize.
20. ## Folding paper

You're stronger than I am; I edit the post, removing everything, and then I put in some quiet, gentle, lame stuff.
21. ## Folding paper

@BMAD I can't prove this is the largest @bonanova: in my picture ( representing real folds on real paper ) a "tail" lapped over the Southeast corner, adding an extra two edges beyond the ones identified by your argument. I presume that they become part of the overall envelope of the figure, leading to my answer, greater than yours.
22. ## Folding paper

My guess for Most sides:
23. ## Folding paper

Another interesting puzzle! Thanks, BMAD! Here's a question from me: when you say "repeat this process", do you mean "pick any corner and bring that corner to the middle of the paper"? Or do we bring the corner to any place at all that we can choose (not just the absolute center of the paper)?
24. ## Escaping the train

I like Charlie's argument; I think Charlie earns the BGS,
25. ## Who wants a dollar

Rollie, your friend is "logical" and "competitive". But is your friend "greedy" as well? Does your friend prefer making more money than you? or more money, period? You suggested cooperating with your friend what if you...
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