Puzzle challenges!

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Re: Puzzle challenges!

Postby Ravana » September 9th, 2017, 8:27 pm

Learning haskell
Code: Select all
coin :: Int -> Int
coin n
  | n <= 1 = 0
  | otherwise = 1 + coin (div (n + 2) 3)
Code: Select all
*Test> coin 10
3
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Re: Puzzle challenges!

Postby Samonella » September 10th, 2017, 3:37 am

My guess:
edit:
Curiosity about Ravana's answer:
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Re: Puzzle challenges!

Postby Ravana » September 10th, 2017, 10:46 am

About Curiosity
That +2 is needed because div truncates float to get integer. So +2 when dividing by 3 results in rounding up.
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Re: Puzzle challenges!

Postby Caladbolg » September 10th, 2017, 2:37 pm

Correct, both of you :D It's log_3 (N) rounded up in general case.
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Re: Puzzle challenges!

Postby Samonella » September 10th, 2017, 3:32 pm

Ooh I see, very clever.

I have one more puzzle, though it's pretty similar to a couple of the others in this thread.

You are in a room with 20 bags of coins, containing 500 each. All of the coins are worthless except those in one particular bag, which are valid currency for recruiting units. The coins are impossible to tell apart, except that the fake ones weigh exactly 1 newton each and the real ones weigh 1.001.

Also in the room is a spring scale, which tells you the exact weight of whatever you measure (in newtons). However, after one use the scale breaks and becomes inaccurate. Typical.

You are only allowed to leave the room carrying 500 coins, so how can you use your single measurement to identify the bag that's valuable?

edit: Winner, Inky!
Last edited by Samonella on September 10th, 2017, 6:53 pm, edited 1 time in total.
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Re: Puzzle challenges!

Postby Ravana » September 10th, 2017, 5:04 pm

However, after one use the scale breaks and becomes inaccurate. Typical.

Consistently inaccurate?
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Re: Puzzle challenges!

Postby Inky » September 10th, 2017, 5:38 pm

Samonella wrote:which are valid currency for recruiting units

Haha, love it!!!

Also, the fact that the scale works once means it already works more often than the typical scale I've come across :whistle:

I think this should work:
Spoiler:
Commented playthroughs with screenshots:
Let's Play Dead Water, Let's Play Invasion from the Unknown and Let's Play After the Storm (in progress)
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Re: Puzzle challenges!

Postby Samonella » September 10th, 2017, 6:52 pm

Ravana wrote:Consistently inaccurate?
No, completely worthless.

Inky's answer was right, congratulations on your 500 gold!
The last few months have been nothing but one big, painful reminder that TIMTLTW.

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Re: Puzzle challenges!

Postby Inky » September 10th, 2017, 7:41 pm

Yeah, think of all the ruffians that could buy!

Here's an interesting question (that totally stumped me when I first encountered it):

The spider and the ant

A giant spider is attached to a cave wall with a 1 meter silk thread. An ant starting at the cave wall is trying to reach the spider by crawling along the thread (don't ask why, maybe it just wants to say hi).

The ant has crawled 1 cm (= .01 meters) along the thread, when suddenly the spider jumps forward by 1 meter (these things can jump really far, it is how they always come out of nowhere to maul your units in campaigns), stretching the thread to 2 meters (the ant maintains its relative position on the thread, so now instead of 1 cm it is 2 cm from the wall). Undaunted, the ant crawls another 1 cm along the thread, but then the spider jumps another meter, stretching the thread to 3 meters.

Assuming this continues indefinitely (so ignoring stuff like lifespans or whether silk can actually stretch), will the ant ever reach the spider?

Hint:

How about we do the same thing as for the other yes/no riddle: just make a guess as to what you think will happen and if you want you can give your reasoning for it, then after a few people guess (or if nobody posts for a while) I'll say which answer is correct. :)
Commented playthroughs with screenshots:
Let's Play Dead Water, Let's Play Invasion from the Unknown and Let's Play After the Storm (in progress)
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Re: Puzzle challenges!

Postby Ravana » September 10th, 2017, 8:30 pm

WolframAlpha
says it takes 1.509*10^43 jumps for ant to reach spider.

Personally only managed to calculate up to 10^8 before Exception: stack overflow
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Re: Puzzle challenges!

Postby Inky » September 10th, 2017, 9:56 pm

Well wolfram alpha is one way to do it! :D :whistle:
Ravana's answer is correct! 8)

Answer
Yes; the ant's first cm is 1/100 of the thread length, then the second cm is 1/2 * 1/100 of the total length...the Nth cm is 1/N*1/100 of the thread length. So at the Nth step the portion of the thread the ant has covered is 1/100 * (1 + 1/2 + 1/3 + ... + 1/N) of the thread and we want to know whether this sum can surpass 1 is N if large enough.

In fact it can, because the sum 1 + 1/2 + 1/3 + ... + 1/N can become larger than any number you want if you add up enough terms (since it consists of 2 terms each > 1/4, followed by 4 terms > 1/8, then 8 terms > 1/16 and so on, where each group sums to > 1/2). This means no matter what the length of the original thread and the speeds of the spider and ant were, the ant would still reach the spider!

It just takes really long... as Ravana said, the value of N that makes this series sum to 100 is 1.509*10^43. (You could also approximate sum[t=1 to N] 1/t by integral[t=1 to N] 1/t dt = ln(N), to see that the time needed is on the order of e^100 = 2.67*10^43).
Commented playthroughs with screenshots:
Let's Play Dead Water, Let's Play Invasion from the Unknown and Let's Play After the Storm (in progress)
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Re: Puzzle challenges!

Postby skeptical_troll » September 10th, 2017, 10:30 pm

just thought it was nice noticing that this riddle it's a pretty good model of light (the ant) travelling in a Universe (the silk thread) with zero acceleration and decreasing expansion rate (Hubble parameter ~ 1/N). So if one uses the analogy to solve the riddle...
Spoiler:

Probably over-killing, but since it was already solved I thought it was worth adding some variations :mrgreen:
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Re: Puzzle challenges!

Postby Samonella » October 8th, 2017, 8:43 pm

I came across another puzzle that is quite a bit like one from earlier in the thread, but this one is a little harder.

A band of 20 pirates has managed to steal a chest of treasure containing 100 gold coins, and it is time to divide up the booty. However, the pirate code specifies a very unusual method for doing so. First, the captain gets to propose a plan for dividing the gold; his plan might be that they divide it up equally between all 20 pirates, or he might suggest that he gets all the gold for himself, or one coin for everyone except the first mate, who gets the rest, or any other combination he wants. The whole crew (including the captain) will then vote on whether or not the plan passes. It does, as long as 50% or more of the pirates vote yes. Otherwise, the captain will walk the plank, and the first mate will then become captain and can propose a plan of his own. The process continues, following a predetermined hierarchy through the crew, until a plan passes.

Being pirates, every member of the crew is purely interested in getting as much gold as possible (without dying, of course). Additionally, they're quite a rough bunch even for pirates, so if one happened to know it would get the same amount of gold regardless of whether it votes yes or no, it will vote no just for the fun of making someone walk the plank. And finally, they know each other very well, and are all perfect logicians.

So the question is, what plan should the captain propose in order to get as much gold as possible without risking his own life?
The last few months have been nothing but one big, painful reminder that TIMTLTW.

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Re: Puzzle challenges!

Postby WTrawi » October 13th, 2017, 1:43 pm

I've heard this riddle before and solved it, but now I can't remember the answer. I will think about it because I love this one :D
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Re: Puzzle challenges!

Postby Ravana » October 13th, 2017, 4:11 pm

#rank where #x = on which day #x is captain.

#1=91:#3=1:#5=1:#7=1:#9=1:#11=1:#13=1:#15=1:#17=1:#19=1

Spoiler:
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