Recent posts by Lebossle on Kongregate

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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

*Extension*bandwagon wooo

But just in case I’m submitting qwerty, prec, back, crystal and kad, for personal reasons.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Note: you could have submitted the answers too, so FoS you too.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Unless you think the exact next person is a mole, then both ypu and that person will send the same list.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

I think I have the exotic one, “mangosteen”

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Hopefully not too late

1. (2^y-1)^ x

2. No, because you can easily check that none of the possible sides for the smaller cubes (1 to 29) work.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Deadline is tomorrow, going by the title. But before nikeas gets any bright ideas, and since everyone already noticed my answer to 3 and didn’t have any objections to it, I’m submitting it.

3: There are many ways that could happen. One is if the first sphere had 1/4+1/2^6 of the lead, the second had 1/8+1/2^6, the third 1/16+1/2^6, …, the 16th 1/2^17+1/2^6, the 17th 1/2^18+1/2^9, the 18th 1/2^19+1/2^9, …, the 2^4+2^6th had 1/2^(2^4+2^6+1)1/2^9, the 2^42^6+1th had 1/2^(2^4+2^6+2)1/2^12, …. That is, the nth ball gets[1/2^(n1) of the lead, and then the first 2^4 balls get 1/2^6 of the lead each, the next 2^6 get 1/2^9 each, the next 2^8 get 1/2^12 each, …, the next 2^(2n) get 1/2^(3n) each.
This works because a sphere taking 1/2^(3n) of the lead needs k/2^(2n) of paint, for some fixed k, so each group, if not for the addition of 1/2^(n+1), would take 2^(2n)*k/2^(2n)=k paint, so the total would be k+k+k+k+…=infinity. Since the addition of lead to the spheres only increases the amount of paint they need, you still need infinite paint.
The trick is that the ratio paint/lead gets infinitely high as lead approaches 0, so if you make enough small enough balls you’ll take an unbounded amount of paint.

An nikeas, again, no, saying “you always take infinity paint” doesn’t work because if you put 1/2 of the lead on the first sphere, 1/4 on the second, 1/8 on the third, …, the total amount of paint you need is finite for the same reason the amount of lead you need is finite.

Edit: ok, idk how to make plus signals without textile deciding they are underline markers

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Problem 2 can be rewritten as “Find x<30 integer such that 30³+20*x³=y³ with y integer”. thing is, 30³+20*x³ is always a multiple of 10, so that means y³ is a multiple of 10, thus y is a multiple of 10, thus y³ is a multiple of 1000, thus 30³+20*x³ is a multiple of 1000, but since 30³ is a multiple of 1000 then 20*x³ is a multiple of 1000 too, so x³ is a multiple of 50, thus x is a multiple of 10, but since x is smaller than 30 then x is 10 or 20.

In case of your interpretation, you have 2^y-1 choices of subsets of houses for each dog, so the answer is (2^y-1)^x

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

1. is totally ambiguous, I’ve been trying to clear it up with Puls but he keeps on not answering my questions. I don’t know if the answer is 1, you just have to let the dogs free and they’ll find their way, or each dog can have (2^y-1) possible subsets of house, making the answer (2^y-1)^x, or each dog gets one house, making the answer y^x.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Originally posted by Helltank:

If there are infinite amount of spheres, you need an infinite amount of paint to paint them assuming that each sphere requires a non-zero amount of paint to be painted. I think this is obvious.

Nope, wrong, for the same reason an infinite amount of balls may require a finite amount of lead. 1/2+1/4+1/8+1/16+…=1, and anything that decreases exponentially does too.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Thing is, exactly how in Pulsaris’ problem you can’t say “I divide the lead into infinite balls of the same size” because there isn’t a number “infinitely close to zero”, you can’t say “infinite speed” because “infinity” is not a real number (that’s why you say the reals are the interval (-infinity,+infinity) → it’s OPEN in both sides because infinity IS NOT AN ELEMENT). Your path is a function of time → position, you can’t be in two places at the same time.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

EBWOP: actually, nvm, just save the first bucket to add some volume to all the other balls. So we had, sorted,
2^4 balls containing 1/2^6 of the lead each
2^6 balls containing 1/2^9 of the lead each
2^8 balls containing 1/2^12 of the lead each

Now we add 1/4 of the lead to the first ball, 1/8 to the second, 1/16 to the third, etc. It’s easy to see that the order is maintained, so they are strictly decreasing, and

BTW, if you want a challenge about infinities, try this: there is a pig somewhere in the 1×1 square in the center of a 3×3 square. This pig walks with speed at most 1/s. There is a Tazmanian Devil somewhere else in the 3×3 square, and he runs at any speed he wants (disconsider relativity). But the Devil is blind, and both Devil and pig are just points. Which route should the Devil take to catch the pig, no matter which route the pig takes, before the pig escapes the 3×3 square? (Got this from a class by Gugu)

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

3.

Many ways he could do it, one way is to divide the lead in buckets containing a geometric progression of volumes, like the first has 1/2 of the lead, the second 1/4, third 1/8, …, then for each bucket you make it into balls that need around the same amount of paint total for each bucket. If you do the math, if you have 1/n of the lead, you want to divide it into n² balls of equal radius to balance it out. This would work, if it wasn’t for the restriction that the balls have strictly decreasing radius, but you can easily add some epsilons around to do it and having the total value still be infinity.

Do I need to calculate a nice formula or is this enough? Anyone that had the first semester of pure maths in college, or has any mathematical analysis knowledge, would agree that this is enough.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

2.

30³ + 20 * 1³ = 27020, cubic root of that is 30.0074
30³ + 20 * 2³ = 27160, cubic root of that is 30.0591
30³ + 20 * 3³ = 27540, cubic root of that is 30.1987
30³ + 20 * 4³ = 28280, cubic root of that is 30.4668
30³ + 20 * 5³ = 29500, cubic root of that is 30.8987
30³ + 20 * 6³ = 31320, cubic root of that is 31.5215
30³ + 20 * 7³ = 33860, cubic root of that is 32.3516
30³ + 20 * 8³ = 37240, cubic root of that is 33.3941
30³ + 20 * 9³ = 41580, cubic root of that is 34.644
30³ + 20 * 10³ = 47000, cubic root of that is 36.0883
30³ + 20 * 11³ = 53620, cubic root of that is 37.7088
30³ + 20 * 12³ = 61560, cubic root of that is 39.4851
30³ + 20 * 13³ = 70940, cubic root of that is 41.3965
30³ + 20 * 14³ = 81880, cubic root of that is 43.4236
30³ + 20 * 15³ = 94500, cubic root of that is 45.5488
30³ + 20 * 16³ = 108920, cubic root of that is 47.7569
30³ + 20 * 17³ = 125260, cubic root of that is 50.0346
30³ + 20 * 18³ = 143640, cubic root of that is 52.3711
30³ + 20 * 19³ = 164180, cubic root of that is 54.7571
30³ + 20 * 20³ = 187000, cubic root of that is 57.1848
30³ + 20 * 21³ = 212220, cubic root of that is 59.6479
30³ + 20 * 22³ = 239960, cubic root of that is 62.1412
30³ + 20 * 23³ = 270340, cubic root of that is 64.6602
30³ + 20 * 24³ = 303480, cubic root of that is 67.2011
30³ + 20 * 25³ = 339500, cubic root of that is 69.7611
30³ + 20 * 26³ = 378520, cubic root of that is 72.3374
30³ + 20 * 27³ = 420660, cubic root of that is 74.9279
30³ + 20 * 28³ = 466040, cubic root of that is 77.5308
30³ + 20 * 29³ = 514780, cubic root of that is 80.1445

So there are no integers there.

tldr: the side of the smaller squares has to be a multiple of 10 (because the final cube has volume multiple of 10), so it’s either 10 or 20, and they both don’t work.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Each DOG must have at least one DOGHOUSE? Don’t you mean the other way aroud?

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Am I the only one that got Pulsaris’s public announcement of who the moles are?

 
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Topic: Forum Games / [Game-Related] Town of Salem.

All Any is pretty balanced, most of the time. Some unlikely setups that happened to me:

No evil roles
Some games later, a game with 1 exe, 1 amne, 1 arso and the rest town, forgot to screenie
4 SK’s

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

EBWOP: My eggy

By the way, I already said once that Puls was the host ever, and I have to agree with cocoa that this game is proving that point again. But I don’t see why not keep on playing for this easy FG win.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

I’m very good at maths

Phew, got here fast enough. Easter coming soon.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Lol, that would be literally impossible to get with an anagram solver.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Since Kad didn’t tell his letter, here’s the list:
LEGISLATORSHIP
PHRASEOLOGISTS
POSTINDUSTRIAL
PROSTAGLANDINS
PSYCHOSURGICAL
SUPERDIPLOMATS
So now we should vote between what is the word we think is most probable, or if we think we’re not going to make it (in that case we would submit “BCEFJMNKQVWXYZ”)

 
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Topic: Forum Games / [Game-Related] Town of Salem.

Jailor is a unique role…

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

6 words with 2 letters mismatching and 1 letter missing now. Kad, please?

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

Also, none of those 10 words in my list have both a ‘o’ and a second ‘s’, so this just got the best guesses list up to 26 words.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

This is very basic with programming. With the letters so far and a random wordlist from google there are 10 words with 5 letters mismatching (that is, the 4 missing plus 1 wrong one). And there are only 3 words with all different letters in the wordlist: dermatoglyphic, ambidextrously and troublemakings.

BTW, I think that if we can’t get a good guess at the word we should just throw 14 letters that were not claimed together just to screw up the liars.

Not posting the list here just in case the ones who didn’t post are the moles and won’t bother computing what it is. But if anyone has a better wordlist it’d be great.

 
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Topic: Forum Games / [TASK 9] -- The Moles -- [DL: 19 April 2200 UTC+8]

You only want to lie if you think that we are not going to get it right, and if you think you’re not going to survive the quiz. Very few people should get into that second category, so I’m pretty sure we’r going to get enough letters to get the word right.

Mine is ‘r’