itt math jokes

[ZemZem]

Not a math joke, but:

[Lance Masayoshi]

Not a math joke, but:

lol

Also Imaginary Numbers are real

[Makaze]

Logic is close enough, right?

A man's wife tells him to go to the store to buy a loaf of bread. She adds, "If they have eggs, buy a dozen."

The man comes home with twelve loaves of bread.

[buttmuncher.ops]

[Ansem]

too easy 3/10

[RandomX2]

http://www.buzzfeed.com/babymantis/20-spectacularly-nerdy-math-jokes-1opu

my favorite is "shit just got real"

LOL, number 10 is actually pretty usable.

@Makaze: Nice, I love logic jokes. I was gonna post the old "3 logicians walk into a bar" one, but since everyone knows that I'll just leave this here. It's a difficult riddle instead of a joke, and I couldn't solve it alone, but it's damn satisfying when you get the answer.

[Ansem]

^10/10 would definitely use in real life. I looked up the solution thinking I had gotten it right, but like the puzzle said, its not simple.

[NTG]

Help I can't distinguish my cup* from my donut**

*or mug

**or doughnut

Edited April 15, 2014 by NTG

[Kriemhild]

@Makaze: Nice, I love logic jokes. I was gonna post the old "3 logicians walk into a bar" one, but since everyone knows that I'll just leave this here. It's a difficult riddle instead of a joke, and I couldn't solve it alone, but it's damn satisfying when you get the answer.

I think I figured out the answer, but how do I actually check if it's correct?

[Ansem]

Google

[Kriemhild]

Google

But that would spoil the correct answer if I'm wrong...

[Aleph]

I just spoiled myself and as it turns out it involves induction, which I would not have guessed any time soon.

Then again my brain is already fried trying to write a function in C that returns true if and only if a supplied line segment intersects with a given cylinder

[Fruity Insanity]

I hardly get anything in this thread.

My math level isn't high enough, apparently.

[RandomX2]

Since there's three of us with the answer now, you can always post your guess (if you haven't googled it yet) and we'll tell you if you're right.

@Fruity Insanity: Do you even grind bro?

[Acacia Sgt]

I like this thread. The Pascal one has to be my most favorite so far. lol

Speaking of that riddle... hahaha, it's funny how once finding the answer it all becomes easy to understand... then again, it did said it was simply a use of logic. Not that the logic itself is simple, but you know what I mean...

Edited April 16, 2014 by Acacia Sgt

[MRDRHAWK]

I was gonna make one up but then I realized it was just gonna be a shitty tangent

[Kriemhild]

Alright, here's my guess at the blue eyes riddle:

[spoiler=Blue Eyes Guess]All of the people with blue eyes leave the island together, at the 100th night.

Explanation: A blue-eyed person, B1, will see 99 people with blue eyes. To him, there are two possibilities: there are 99 blue-eyed people, or 100 blue-eyed people. B1 deduces that IF there are 99 blue-eyed people, then each of those blue-eyed people (e.g. B2) would deduce two possibilities: there are 98 blue-eyed people, or 99 blue-eyed people. B1 knows that in this b=99 hypothetical scenario, B2 would deduce that IF there are 98 blue-eyed people, then each of those 98 blue-eyed people (e.g. B3, but this set also includes B1) would deduce two possibilities: there are 97 blue-eyed people, or 98 blue-eyed people.

The chain goes on. B99 deduces that there are either 2 blue-eyed people or one. IF there's only one, then B100 would deduce that there is either 1 person with blue eyes, or nobody. Before the Guru spoke, both possibilities would work and nobody leaves the island due to uncertainty. But after the Guru spoke, B100 would know that the latter is impossible. Therefore, B100 would know that he actually has blue eyes and will leave the island on the first night. But after the first night, B100 doesn't leave (b is not truly equal to 1), so B99 would now know that the b = 1 is impossible. Therefore, there are only two people with blue eyes and both will leave the island on the second night. This doesn't happen, so B98 knows that b cannot be equal to 2, and so, all three people with blue eyes will leave the island on the third night.

The chain continues until we go back to B1. On the 99th night, nobody left the island. So B1 knows that b is not equal to 99. B1 will now correctly deduce that b is equal to 100. All blue-eyed people will arrive at the same conclusion. They will all deduce that they have blue eyes and will leave on the 100th night.

I probably couldn't explain that well, but if anybody else arrived at the same solution, they might understand what I mean.

[Acacia Sgt]

Alright, here's my guess at the blue eyes riddle:

[spoiler=Blue Eyes Guess]All of the people with blue eyes leave the island together, at the 100th night.

Explanation: A blue-eyed person, B1, will see 99 people with blue eyes. To him, there are two possibilities: there are 99 blue-eyed people, or 100 blue-eyed people. B1 deduces that IF there are 99 blue-eyed people, then each of those blue-eyed people (e.g. B2) would deduce two possibilities: there are 98 blue-eyed people, or 99 blue-eyed people. B1 knows that in this b=99 hypothetical scenario, B2 would deduce that IF there are 98 blue-eyed people, then each of those 98 blue-eyed people (e.g. B3, but this set also includes B1) would deduce two possibilities: there are 97 blue-eyed people, or 98 blue-eyed people.

The chain goes on. B99 deduces that there are either 2 blue-eyed people or one. IF there's only one, then B100 would deduce that there is either 1 person with blue eyes, or nobody. Before the Guru spoke, both possibilities would work and nobody leaves the island due to uncertainty. But after the Guru spoke, B100 would know that the latter is impossible. Therefore, B100 would know that he actually has blue eyes and will leave the island on the first night. But after the first night, B100 doesn't leave (b is not truly equal to 1), so B99 would now know that the b = 1 is impossible. Therefore, there are only two people with blue eyes and both will leave the island on the second night. This doesn't happen, so B98 knows that b cannot be equal to 2, and so, all three people with blue eyes will leave the island on the third night.

The chain continues until we go back to B1. On the 99th night, nobody left the island. So B1 knows that b is not equal to 99. B1 will now correctly deduce that b is equal to 100. All blue-eyed people will arrive at the same conclusion. They will all deduce that they have blue eyes and will leave on the 100th night.

I probably couldn't explain that well, but if anybody else arrived at the same solution, they might understand what I mean.

That's pretty much it. It's basically everybody doing an elimination process of everybody else also eliminating who else is eliminating who has already realized there is more blue-eyed than what they see (if this doesn't make sense then I just don't know how to word it properly). It all boils down to "Group of size X leave on the Xth night."

[Ansem]

Topic title to nerd trivia y/y

[Makaze]

Alright, here's my guess at the blue eyes riddle:

[spoiler=Blue Eyes Guess]All of the people with blue eyes leave the island together, at the 100th night.

Explanation: A blue-eyed person, B1, will see 99 people with blue eyes. To him, there are two possibilities: there are 99 blue-eyed people, or 100 blue-eyed people. B1 deduces that IF there are 99 blue-eyed people, then each of those blue-eyed people (e.g. B2) would deduce two possibilities: there are 98 blue-eyed people, or 99 blue-eyed people. B1 knows that in this b=99 hypothetical scenario, B2 would deduce that IF there are 98 blue-eyed people, then each of those 98 blue-eyed people (e.g. B3, but this set also includes B1) would deduce two possibilities: there are 97 blue-eyed people, or 98 blue-eyed people.

The chain goes on. B99 deduces that there are either 2 blue-eyed people or one. IF there's only one, then B100 would deduce that there is either 1 person with blue eyes, or nobody. Before the Guru spoke, both possibilities would work and nobody leaves the island due to uncertainty. But after the Guru spoke, B100 would know that the latter is impossible. Therefore, B100 would know that he actually has blue eyes and will leave the island on the first night. But after the first night, B100 doesn't leave (b is not truly equal to 1), so B99 would now know that the b = 1 is impossible. Therefore, there are only two people with blue eyes and both will leave the island on the second night. This doesn't happen, so B98 knows that b cannot be equal to 2, and so, all three people with blue eyes will leave the island on the third night.

The chain continues until we go back to B1. On the 99th night, nobody left the island. So B1 knows that b is not equal to 99. B1 will now correctly deduce that b is equal to 100. All blue-eyed people will arrive at the same conclusion. They will all deduce that they have blue eyes and will leave on the 100th night.

I probably couldn't explain that well, but if anybody else arrived at the same solution, they might understand what I mean.

Why doesn't the same logic apply to any other eye colors? It seems to me that B100 could also be a brown eyed person making the same deduction. A brown eyed person would see 100 blue eyed people, but likewise, they would not know that there are not 101 blue eyed people.

Why do only the blue eyed people leave the island?

Edited April 17, 2014 by Makaze

[Acacia Sgt]

Why doesn't the same logic apply to any other eye colors? It seems to me that B100 could also be a brown eyed person making the same deduction. A brown eyed person would see 100 blue eyed people, but likewise, they would not know that there are not 101 blue eyed people.

Why do only the blue eyed people leave the island?

Because of the same blue-eyed people. They know the blue eyed people can tell who has or not non blue eyes except for themselves, so once each know they have blue eyes and leave, the others conclude that by leaving, then they each know that, at the very least, they don't have blue eyes. Basically, the same logical process the blue-eyed use to see if they have blue eyes or not.

Edited April 17, 2014 by Acacia Sgt

[Makaze]

Because of the same blue-eyed people. They know the blue eyed people can tell who has or not non blue eyes except for themselves, so once each know they have blue eyes and leave, the others conclude that by leaving, then they each know that, at the very least, they don't have blue eyes. Basically, the same logical process the blue-eyed use to see if they have blue eyes or not.

What keeps them from making the exact same decuction as the blue eyed people? According to the logic here, all of the people, regardless of eye color, would conclude that they had blue eyes.

[Acacia Sgt]

What keeps them from making the exact same decuction as the blue eyed people? According to the logic here, all of the people, regardless of eye color, would conclude that they had blue eyes.

Because the blue eyed people reach the conclusion first, and leave.

For example, let's say only 1 person had blue eyes. Everybody but him would think that there is either 1 (the blue eyed person they see) or 2 (the blue eyed person and themselves) blue eyed people. The blue eyed, seeing none, would think there is either none or 1 (himself). Since he knows the Guru saw a blue eyed person, then he, who sees none, concludes he is the one, and so leaves. After this everybody concludes that if the single blue eyed person left, then it's because he didn't saw anybody with blue eyes, and so concluded that he must be the only one if the Guru saw someone with Blue eyes. Since he left, then they conclude that they don't have blue eyes.

This process is applied with more complexity as the number of blue eyed people increase, but the conclusion is the same. Once the blue eyed people realize they have them and leave, the rest can conclude that they don't have blue eyes, since otherwise the blue eyed people wouldn't have left.

Edited April 17, 2014 by Acacia Sgt

[Makaze]

Because the blue eyed people reach the conclusion first, and leave.

For example, let's say only 1 person had blue eyes. Everybody but him would think that there is either 1 (the blue eyed person they see) or 2 (the blue eyed person and themselves) blue eyed people. The blue eyed, seeing none, would think there is either none or 1 (himself). Since he knows the Guru saw a blue eyed person, then he, who sees none, concludes he is the one, and so leaves. After this everybody concludes that if the single blue eyed person left, then it's because he didn't saw anybody with blue eyes, and so concluded that he must be the only one if the Guru saw someone with Blue eyes. Since he left, then they conclude that they don't have blue eyes.

This process is applied with more complexity as the number of blue eyed people increase, but the conclusion is the same. Once the blue eyed people realize they have them and leave, the rest can conclude that they don't have blue eyes, since otherwise the blue eyed people wouldn't have left.

I see. So each blue eyed person needs to wait for [(Perceived # of Blue Eyes) + 1] days.

All of the other blue eyed people do the same. Thus, when no one leaves on the (Perceived # of Blue Eyes)th day, they know that they, too, have blue eyes.

I would never have figured that out on my own.

Edited April 17, 2014 by Makaze

[euklyd]

Got it. Thanks to Makaze to confirming that I wasn't making shitty logical leaps.

[4/17/14 3:21:06 PM] Makaze: What is the answer to the original riddle, then?

[4/17/14 3:21:46 PM] Euklyd:

[4/17/14 3:16:41 PM] Euklyd: they leave on D[number of blue people]

[4/17/14 3:16:55 PM] Euklyd: And all at the same time.

[4/17/14 3:21:49 PM] Euklyd: therefore D100

[4/17/14 3:21:52 PM] Euklyd: all the blue

[4/17/14 3:22:32 PM] Euklyd: Is there something else that I need to answer?

[4/17/14 3:23:24 PM] Makaze: Nope.

Not sure if I would have figured it out entirely on my own, but I wasn't being led to the solution at all, so I'm happy.

I'd say the thing that gave it away was Hextator's post about "it involves induction."