/sci/ - Science & Math

slope of red triangle <span class="math">\neq[/spoiler] slope of blue triangle

>>2528785
>>2528789
yeah yeah we all know that, said so myself in my post (OP here), but could you get some nice riddles in here like the "Blue Eyes" one I posted?
there's lots of math puzzles online but most of them really suck or are too easy. there used to be some good ones on here in the past, let's get the good riddles going!

>>2528800
what is the eanswer to the blue eyes riddle, OP?

>>2528807
http://xkcd.com/solution.html

>>2528821
thanx :)

bump, come on guys, it's been exlusively religion/outer space colonies/transhumanism and other bullshit on /sci/ for the past couple of days, let's get some good ol' math puzzles in here!!

What is the greatest number of queens one can fit on a chess-board without any of them being able to take any of the other queens?

>>2528847
>implying outer space colonies/transhumanism threads are a bad thing

Though I would like to see more math-related threads replace the troll threads

>>2528855
8,
better one is how many do you need to attack all the fields.

I've been trying to follow the proof that Euler's totient/phi function is multiplicative. Anyone got a good proof?

>>2528878
Correct, but can you place them?
Here is a template you can use.

>>2528900

>>2528878
5 i think.

>>2528935
correct

bump for more, singularity threads are getting boring, there must be more good logic/math puzzles around like the blue eyes one

Goldbachs conjecture.

>>2529536
fuck yeah singularity

>>2529544

how about no

Without reading the solution, any ideas on the blue eyes problem? It's already known that the Guru can see someone with blue eyes before she said it, because the problem states everyone can see everyone's eyes. The only new information the Guru brings is the fact that she said she can see someone with blue eyes. Which seems useless.

for those of you who have looked at the solution, is it legitimate? i don't want to look at it and spoil the answer but i also don't want to waste my time if it isn't legitimate. thanks bros

the solution is legit, think induction

>>2529727
yeh, seems to check out. but i have no idea why the guru is even needed, the same thing would happen even if he didn't exist/didn't say aything, because everybody can see everybody elses eye colour.

>>2529597
Answer to blue eyes problem:
(no one posted the problem, so I'm assuming it's the one I think it is)
If there are n people with blue eyes, they leave on day n. Think of it like this:
If you're the only person with blue eyes, you look around and see no one else with blue eyes. So it must be you and you leave immediately.
If you see someone else with blue eyes and they leave immediately, it's because they figured out they have blue eyes, via the above process. If they don't leave immediately, it must be because they see someone else with blue eyes. If you look around and see no one else with blue eyes, then it must be you. So both of you have blue eyes and leave on the second day.

repeat this process indefinitely.

>>2528896
Someone (probably you) posted this question on a separate thread and it was answered there.

For which regular polygons P is the following true:
Infinite copies of P, all identical in size, can be placed on a plane such that every point on the plane is covered by a polygon, but there is no area of overlap.
For example, infinite squares can cover a plane without overlap, but infinite regular 35245-gons (of the same size) cannot.

Bonus points: prove your answer.

>>2529740
Wrong.
Think of the case where there are only two people on the island, both with blue eyes. They would just sit there forever, each thinking that the other guy's eyes were blue, but having no idea what color their own eyes were.

When the guru says "I see someone with blue eyes," then they can each wait and see what the other does that night (in response to that statement), and deduce that their own eyes are blue the next day.

>>2529778

Triangles, squares, hexagons, their angles are all factors of 360.

>>2529778
triangles

that was retarded

>>2529794
The guy above you is correct.

>>2529778
>>2529785
>>2529794
>>2529806
Wow, just wait until you guys hear about Platonic solids...

>>2529820
The only regular polyhedron that can fill R^3 using "blocks" of the same size is a cube.

>>2529740
the guru is absolutely needed. it isn't "common knowledge" that there is someone with blue eyes before the guru announces it. i'm saying "common knowledge" because I'm referring to the logical notion of common knowledge which is needed to be able to solve the puzzle. look it up and realize the guru is definitely needed!

>>2529740
Because induction doesn't work without a base.
None of the deductions can be made without the first case, and the first case can't occur if the guru doesn't say anything

>>2529833
Proof of this, or link to proof?

>>2529863
This isn't a formal proof, but there are only five regular polyhedra, so it's not too hard to look at four of them and see it's not possible.

>>2528855
depends if you are alowed other pieces, the answer would be 56

>>2529848
It IS common knowledge. Everyone can plainly see that someone has blue eyes.

I reasoned that all blue-eyed people on the island would leave on the first night. I didn't account for the hypothetical possibility of any given person to have a different eyecolour.

So I was mistaken.

>>2529781
thanks

>>2529833
lern-2-octahedron

>>2529876
Well, five regular convex polyhedra. But obviously it's even harder for a concave polyhedron.

>>2529891
Show me how to do it with octahedra.

>>2529889
Even if there are only blue eyes and brown eyes, you're still wrong.

>>2528900

easy as hell, since you can only put one piece into a row and there are eight rows, you put one into each row a few spaces apart so they can't attack eachother horizontally/vertically/diagonally

You have twelve marbles. They are all identical except that one marble weighs either a tiny bit more or a tiny bit less than the others. You do not know whether it is heavier or lighter. You have a scale that can detect any difference in weight. However, the scale does not tell you how much something weighs; you just put objects on either side and the heavier side goes down.

Using the scale three times or less, how can you determine which marble is different from the others?

Note 1: You do not know if the unusual marble is heavier or lighter!
Note 2: This is possible. In fact, I know of two solutions, so if you try to prove that it's impossible, you're a troll.

There are five people. Four of them are "togglers" and one is a "truthteller." The truthteller always answers questions truthfully. If you ask a toggler a question, and you have not previously asked them anything, then they will randomly either answer you honestly or lie. However, if you ask them a second question, the truth value of their second answer will not be the truth value of their first answer (ie they toggle between truth and lying).

You get to ask two questions, divided as you choose between the five people. You can ask two people each one question, or ask both questions to the same person. Your two questions do not need to be the same.

How do you determine the truthteller?

Let (a,b,c) be natural numbers.
Assume (a^4) + (b^4) = (c^4).

Prove or disprove:
Among a,b, and c, an odd number of them are multiples of five.

>>2529988
That's dumb. Fermat's Last Theorem says the assumptions are impossible.

>>2529932
1. Put 6 marbles on both sides of scale
2. One side drops/raises , Put 3 from the dropping/raising side on both sides
...Then wat

bump

>>2528935
>>2530011
the numbers are 0, 1 and 1

>>2530013
no, you do 4 on each side and leave 4 off
if they are equal you split the 4 off the scale then the 2 you get from that
OR if they are not equal split the 4 on the scale in step 1 and continue from there.

>>2529932
put two marbles on each side.

[if they are balanced] the other two, put one on each side. They will not be balanced. Replace one of these with one of the original 4 marbles.

[if they're not] remove one marble from each side. If this is now balanced, replace one of the marble with one of the two you removed.

should be able to tell from there

>>2530077
Dude, what.

>>2530077
But since you do not know if the odd one out is heavier or lighter, how do you determine which side has the unequal one and still keep it within 3 tries?

>>2530084
The second one uses the scale 4 times.

>>2530107
yeah, I read it as 6 marbles to start with for some reason. Gotta rethink slightly

>>2530105
well, then, switch 4 with 5 and see if that works, if I had this infront of me I could solve it but I only have to work it in my mind.

5 marbles on each side

1. If equal, Put the 2 left on and problem solved
2. If unequal, I dunno lol

bump

>>2530084
What if you weigh 4 and 4; they're unbalanced; you remove 1 from each side; they're still unbalanced? You now have one weighing left.

>>2529956
ask every on of them 'are you speaking truth'. All will say yes. Then ask 'did you lie to me'. If they said truth the first time, they'll say no, if they lied first time, they'll say yes. Dump all that say yes, since your guy says truth all the time.

Rinse and repeat 'till you have only one person left.

>>2530145
You only get to ask two questions total. NOT two per person.

>>2530145
>Full Retard

>>2529956

A. Asksome one if they are the truthteller:
If Yes GOTO B, if No GOTO C
B. Ask the same person who is the truth teller (no matter what they will have to answer truthfully).
C. Ask the same person who is not the truth teller (they have to lie and their answer will be the truth teller)

>>2530191
ding ding ding ding!

>>2530145
>trollface.tiff

>>2528855
There is this programming quiz problem called "8 queens" where you need to compute all the legal arrangements for 8 queens.

>>2530095
If you're confused then just imagine the simple version with only 3 marbles. With 1 weighting you can determine which one has a different weight:

Leave 1 aside, and put 1 on each side of the scale.

This method easily scales up to arbitrary N. Each weighting reduces N by a factor of 3.

>>2530396
>Each weighting reduces N by a factor of 3

Only if you know whether the unusual marble is heavier or lighter, which you don't, in this case.

>>2530447
Oh sorry, confused it with the "one marble is heavier" puzzle.

who wants me to tell the solution to the marble puzzle? anyone still interested in solving it by themselves or has everyone given up?

>>2529932
im lazy, but does it involve putting 6 on 1 side and six on the other?

You're are standing before two gates. One leads to heaven, the other leads to Hell, but you don't know what is behind either door. There are two gatekeepers. One of them always tells the truth and the other always lies, but you don't know which is the honest one or which is the liar.

You can only ask one question to one of them in order to find the way to heaven. What is the question?

>>2530544
Old problem is old. Ask one what the other would say. Then go the other way.

>>2529932
Divide the 12 marbles into 3 groups, and weight two of the groups. If the scale is equal, to got subpuzzle A. Otherwise go to subpuzzle B.

Subpuzzle A: 4 marbles with one odd marble within, PLUS 8 standard marbles. You have 2 weightings left.

For the first weighting: put one marble aside, and weight the other 3 along with a standard marble.
If the scale is even, the one set aside is different.
If the scale is tilted, then exchange 1 unknown marble from each side. If the scale doesn't move, then the only unmoved marble is odd. If the scale moved, the move to subpuzzle C.

Subpuzzle C: You have 2 unknown marbles and 10 standard marbles. You have 1 weighting left.
youshouldbeabletosolvethis.jpg

I'll still working on subpuzzle B. Gimme 2 more minutes.

I remember a couple that aren't terribly common.

---------------------------------------------------------------
THE THREE SISTERS:

You are to marry one of three sisters (because you are the prince of some fancy kingdom and you gotta choose a princess, let's say.) There's the oldest sister, who always tells the truth. There's the youngest sister who always lies. Then there's the middle sister, who will randomly speak the truth or lie - It's impossible to tell beforehand.

Now, you wouldn't mind marrying the oldest sister. She always speaks the truth and would make a nice wife. You also wouldn't mind marrying the youngest sister - Since you know she always lies, you can at least deduce the truth from that, so it shouldn't be too bad. But under no circumstances do you want to marry the middle sister!

The three sisters are presented to you, and since you have never met them before, you don't know who is who. You are allowed to ask one of the sisters a single question. What do you ask in order to make sure that you will always be able to choose a suitable wife?

---------------------------------------------------------------

The answer can be deduced logically, nothing fancy is going on. The sisters aren't omniscient but will answer to the best of their ability. The sisters know each other (because they're sisters, durr).

>>2530572
R U A FEMALE

Solution continued from >>2530558

Subpuzzle B: You have an uneven scale with 4 marbles on each side, with 4 standard marbles as weights. You have 2 weightings left.

For the first weighting: Remove one marble from one side and remove 2 marbles from the other side. Place a standard marble to "balance" out the numbers on each side. Exchange 3 of the "suspect" marbles currently on the scale. So now you have 3 marbles set aside, and 3 on each side of the scale.

Now we arrive at a trifucation: either the scale is 1) balanced, or 2) the scale is unblanced but identical to the previous weighing, or 3) scale is balanced and different from the initial weighting.

Case 1) If the scale is balanced, the 3 marbles set aside is "suspect". Continue to subpuzzle D.

Case 2) The scale is unbalanced and identical to the previous weighting, which means 2 of the unmoved marbles currently on the scale is "suspect", go to subpuzzle C.

Case 3) One of the 3 marbles you exchanged was suspect, continue to subpuzzle D.

And the second one, which I think I read on /sci/.

---------------------------------------------------------------
THE LIONS AND THE SHEEP:

On a deserted island far away in the pacific live 100 lions and 1 sheep. The lions are all-knowing and perfectly logical beings. But as all living things they need to eat. The following is true:

- A lion can either decide to eat grass, or eat the sheep.
- If a lion eats the sheep, it turns into a sheep (and may thus be eaten by other lions).
- Lions can survive indefinitely by eating only grass.
- Lions prefer meat over grass, but will only eat the sheep if they are sure that they in turn won't be eaten by another lion.

The question is simple. Does the sheep get eaten?

---------------------------------------------------------------

>>2530615
no

>>2530615
Sounds like a paradox. No lion would eat the sheep because he could then get eaten. But if no lion will eat the sheep, then any lion could eat the sheep, because no lion would eat him afterward.

>>2530615
Can the lions divide the sheep between them so that they all become the sheep and therefore cannot eat each other?

>>2530615
even number of lions: sheep does not get eaten
odd number of lions: sheep gets eaten

>>2530646
Naw, there can only be one sheep at any time.

>>2530659
Motivate.

>>2530659
Yeah, this.

>>2530611 here
I'm stuck on subpuzzle D. Crap.

>>2530615
Suppose 1 lion + 1 sheep:
Eat.

Suppose 2 lions + 1 sheep:
Whoever eats sheep will get eaten, therefore both lions eat grass forever.

Suppose 3 lions + 1 sheep:
One smart-ass lion eats the sheep, because he knows from above that he won't get eaten.

Suppose 4 lions + 1 sheep:
Knowing the above, no lions will eat the sheep because he knows If he eats the sheep, he will get eaten per above.

So it's even-odd just like >>2530659 said.

There is a party with P people. For each pair of people, they have shaken hands either once or not at all. Let Q be a subset of P, such that each person in Q has shaken an odd number of hands.

Is |Q| even or odd? Prove your answer.

>>2530752
Yeah, that's the correct lion solution. Now figure out the sisters.

>>2530825
Sorry, Q is a subset of P containing ALL people that have shaken an odd number of hands.

>>2530837
this is easy:
we can disregard the people who don't shake hands. each handshake connects two people, thus the sum of each person's number of handshakes over all people is twice the number of handshakes. but since each person only shakes hands once, the sum of each person's number of handshakes over all people is equal to the number of people shaking hands. by the previous this is equal to twice the number of handshakes and thus even. we are done.

>>2530572
Sister solution:

Sister A, B and C - Unknown ages.

Pull sister A aside and ask her which of sister B and C is younger. Pick this one.

Reason:

Truth sister will answer the lying sister is youngest, you marry lying sister

Lying sister will say truth sister is youngest (since technically older than middle sister), you marry the truth sister.

Doesn't matter what the unknown sister answers, you aren't marrying her.

enemas...

no wait these are real riddles

>>2530544
Ask them, "is the other guy lying?"

If yes, then go through that door.  If no, then go through the other door.

>>2531392
Both would answer yes, telling you nothing.

Ask one of them "Do you both tell the truth?"; The truth teller will say no, the liar will say yes.

Assuming you know which gatekeeper is assigned to which door (liar to hell, truth teller to heaven) you will know which door to enter. OP for this riddle did not specify if the gatekeepers are statically assigned as such so not sure if this is the correct answer.

buhlump

six people play a game, each of them rolls a die and doesn't look at it, then each person looks at the die of each other person, then each person offers a guess as to what numbe ris on their die, if at least one person guesses right the group wins, what strategy can the group employ to maximize their chance of winning?

>>2532386
oooh I knew a similar one with prisoners...let me think

>>2534087

is the answer enemas?

captcha: but uderp

i shit you not

>>2532386
I may be wrong, but isn't it impossible to do any better than guessing? The dice rolls are independent, so no matter how much you know about anyone else's roll, it doesn't affect the odds of guessing your own, and as there is no way to transfer information by guessing nor a way to relate the guesses to increase their chances of guessing correctly 'together', I can't see how you can do anything clever here. A hint, please?

I think it's something related to Hamming Codes? I read before a variation where the guess if the person is wearing Blue or Red hat.

>>2532386
Could a person use his guess to try to convey information to the other players?
Person [A] will see everybody elses dice and should guess at a number he has seen.
Everybody else then guesses at this exact same number. Somebody has the number, so the team wins according to the rules.

>>2534184
If we've heard the same problem, then the condition there is that the players must all guess correctly or all guess incorrectly in order to win. There they can collaborate in advance to do better than random guessing because their victories aren't independent - winning for me is a win only if you also win, and if you lose, then I must lose to win as well and all that. In this problem there's no such thing, but obviously if you are to guess 'separately' in the sense that first player 1 guesses and then player 2 and everybody can hear each other and so on, then obviously you can communicate and score a trivial win.

>>2528772
NO NO NO

I've seen this "solution" to the blue eyes problem, and it is wrong. It is essentially the same flaw as with the proof that "all horses are the same color" argument.

>>2534311
that is, you need to establish three base cases in order for the induction argument to work.

>>2529597
The information is useful. Think what would happen if there was only one blue-eyed person on the island. I didn't look at the solution but I know this puzzle so it's cheating (a little)

>>2534311
>>2534314
you are wrong, only one base case is needed here.
lrn2induction undergrad

>>2534314
are you claiming that the inductive step from 2 people to 3 doesn't work? well then please elaborate why, because it obviously does.

"The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:"
Meaning: The Guru speaks once in all their endless years on the island
"The Guru is allowed to speak once (let's say at noon). On one day in all their endless years on the island, standing before the islanders, she says the following:"
Intended meaning: The Guru speaks every day at noon, in one of these days in all their endless years...

I seriously think that punctuation was bad there.

Btw, she could have said "I see 100 people with blue eyes" and, as anyone who has blue eyes would know that there are other 99 people with blue eyes, so the only one left would have to be you wouldn't them all leave the island?

>>2534403
ok nevermind, I got it, there is no error in the riddle, my bad