BrainDen.com - Brain Teasers

# EventHorizon

VIP

523

1. ## Arc length = area

I'm sure I know what BMAD was saying. Yes, ignore the units / squared units difference. The problem doesn't involve a circle though. You need to find a function f(x) such that the area under the curve between two points is the same as the arc length between those two points along the function. Here the arc length is like the length of a string representing the function cut at the two points. How to find the arc length of a function: Restatement of problem using previous spoiler: One answer is simple, the other is less so.

3. ## An interesting limit

A quick check: The whole #!:

7. ## Miscellaneous logic puzzle from exam

I looked at that for a while. Though it's not much, here's what I got...
8. ## Tiling a hexagon

This puzzle is not quite solved yet. So here's my attempt
9. ## Help Bob find Alice's P(x)\$%^

Alice is so kinky...
10. ## Cutting pizza

Proof for (3)
11. ## Help me please to solve

I installed the plus4 app, and it is just a mastermind clone. +x means you have x numbers in the right spot, and -x means you have x numbers right, but they are in the wrong spots. It could be that a guess would result in +2-2, meaning that you have all the right numbers, but two are in the wrong spots.
12. ## Amoeba evacuation puzzle

So the best you can do is 00 00 or 0 0 ... 0 0 000...00

14. ## Amoeba evacuation puzzle

Looks like I misunderstood the initial configuration. I assumed the x's were already amoeba and not just space to vacate.
15. ## Amoeba evacuation puzzle

Example 2 is not looking good. Perhaps it is unsolvable....
16. ## A Boggle-like Challenge

Yay for integer sequences (found using your 19355 )... Time to look at 3-regular graphs... which aren't as simple and quite possibly don't have this property. So it looks like we basically just need to find how many 8-regular graphs there are with 26 labeled vertices. It'll be off by a little, but will be a good estimate for N.

18. ## A Boggle-like Challenge

N is not going to be that simple to calculate. Any given adjacency you find could have multiple possible cycles that could be found within it (So it may incorrectly be counted multiple times). Adjacencies don't need to be as neatly organized as curr3nt's answer was.
19. ## A Challenged Flight Deck

I think the bit debt idea will always have situations that produce contradictions, so I'm thinking it should be discarded. I looked over plainglazed's 38, and I think it works. Nice. Edit: I got another 38. It's slightly more complicated (1 more branch needed), but is based on yours, so I won't bother posting it. It uses 6of8's and 5of8's in combinations instead of groups of 5's in a bigger group of 20.
20. ## A Challenged Flight Deck

I didn't really explain it too well. Here's a (hopefully) better attempt to show that with 3 bits of information, you can get 4 of the next 5 cards right and still end up with 2 bits after A couple examples Example 2 shows that this method does not work as I thought it did, since I reached a contradiction. So back to the drawing board... hopefully something from this can be salvaged. I think it may still work, but that I can't 'borrow' quite as many bits as I thought. I think there needs to be a 1 bit buffer at all times. Hopefully that solves the issue, but I'll need to look into it.
21. ## Dicey Permutations

So, after reading your post and looking at the fewest for 3 again. I thought of one configuration to try. Since testing a configuration with my code is fast, I threw it into my code and...

23. ## A Challenged Flight Deck

I wondered where I could get that extra bit... v v v v v doesn't work
24. ## A Challenged Flight Deck

I've got a strategy that gets 38 (or more) on 99.8% of decks, and needs one more bit on the rest... one bit... grr. Edit: Hey... 500th post! v v v v doesn't work Edit 2: I got a strategy that guarantees 38. I'm going to flush out this idea before posting (hoping I can get 39).
25. ## A Challenged Flight Deck

×

• #### Activity

• Riddles
×
• Create New...