/sci/ - Science & Math

>>2617300
P = NP

I don't know if it's that challenging, but I had fun with it.

integrate sin(x*x/2) dx from 0 to infinity

>>2617303
Unless P is zero or N is one, I doubt it.

why can't the value of money change?

>>2617307
Is that possible algebraically?

Was thinking more critical thinking than integration, but thanks.

What is the probability that the an answer chosen at random is the answer to this question?

a. 1/3
b. 1/3
c. 2/3

>>2617317
http://en.wikipedia.org/wiki/P_versus_NP_problem

There's a room of 8 people. One of them always tells the truth, the other 7 are flip floppers (they alternate between telling the truth and lying) You don't know whether they start off lying or start off telling the truth, and they aren't in-sync with each other (ie just because one starts off lying doesn't mean the others will and vice versa)

You can ask them 2 questions. Those 2 questions can be either directed at the same person or asked to two different people. Your goal is to figure out which one always tells the truth. What two questions do you ask, and to whom?

go for a math degree

>>2617341
I know, hence goofy face accompanying the comment to imply awareness of the ignorance of my response.

$.50x + $3y + $10z = $100

what are the values for x y and z

>>2617360

Needs additional constraints.

>>2617360
Isn't that incredibly arbitrary? I can choose any combination of those constants I want to make one hundred.

x=2
y=3
z=9

$.50x + $3y + $10z = $100

what are the values for x y and z

edit** i forgot xyz need to equal 100 units and cost $100

bump

This one's very simple to calculate mathematically, and has been discussed for many years, but it tends to puzzle people the first time around:

There are three doors, one of which contains a prize. You choose one, and then the host eliminates one of the two other choices. He follows it up by asking you if you'd like to switch your choice. What do you do?


Sorry if this is below you.

the original question involved a bunch of budgies
blue budgie costs 10$
green costs 3$
yellow costs 50cents

they want you to spend 100$
and have 100 budgies

i can only get 99 budgies for 100$

What is the probability that when you flip a coin three times, it comes as head at least once?

Ask one if there is one truthful person in the room.

If they say no, then just ask who the truthful one is.

If they say yes, ask which people will always flip flop.

Does that work?

>>2617403
This problem is still crazy even when I've looked at it mathematically a thousand times. You switch.

here is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

[Added, Feb 15: for the purposes of this logic puzzle, "highly logical" means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]

Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).

One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.

One evening, he addresses the entire tribe to thank them for their hospitality.

However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.

What effect, if anything, does this faux pas have on the tribe?

>>2617408
Because when they say yes, they are either now lying or staying truthful. The flip flopping ones will now be lying, so they will point out the one person who does not flip flop. The truthful person will simply point out the other seven.

>>2617405
1 - P(all 3 coins tails)

I love stat problems. A: 7/8

>>2617412
It always made sense to me, and I don't consider myself to be a mathematical genius.

>>2617395

To solve for n unknowns, you need at least n independent equations involving said unknowns.

>>2617422

this works out, yes and you got the right idea. good job :)

The answer I know of phrases the questions a little more elegantly, though, but you still get the idea

I am building a pyramid six blocks high. On each block I write a one- or two- digit number, such that the number on each block is the sum of the two blocks beneath it. What number do I write on the topmost block?

.....?
....XX
...XXX
..XXXX
.XXXXX
XXXXXX

> Edit: I mean triangle, not pyramid obviously

>>2617448
>>2617448
for this to work I think the bottom row can be all zeros and all other numbers will also be zeros.

I may have misunderstood the criteria though.

Solved earlier but you might not have been there.

>>2617448

too vague, needs more constraints

From an old riddle time on /b/;

You have three identically sized bags with an equal amount of identical candies inside. Each of the bags would naturally weigh 10kg. However, one bag has been poisoned, and weighs one additional gram.

To aid in your quest to eat as much candy as possible, you are given a precision scale, but with one catch; you may only use the scale ONCE. You may only place items on the scale at the same time, and remove them at the same time (or in other words, you can't put all three bags on the scale at the same time and remove them one at a time to find the poisoned bag, or some variation on the idea).

How do you figure out which bag holds the poisoned candy?

develop a complex model that predicts stock movements perfectly and gathers all the information required to make these perfect predictions. There's your problem. You have a lifetime to solve it. Get to it.

>>2617458
My fault, the actual puzzle requires that every number be different (and so therefore zero is not allowed)

So congrats on the first correct solution, but can you solve it in hard mode?

>>2617478
>>2617478

720?

>>2617485
Has to be a one or two digit number

>>2617489
I'm assuming it also has to be natural numbers only, or there would be a large number of solutions using integers or negative numbers.

>>2617489

well then I'd say 99 since it has to be pretty big

>>2617469
Possible? Instincts say put two on scale, but logic fails me.

>>2617412
Only if you know that the host has to offer you the chance to switch regardless of whether you've picked the prize door.

>>2617521
It's possible.

>>2617469

but two bags on the scale. If one weighs more than the other, that bag is poisoned. If they both weigh the same, neither is poisoned

>>2617521
Assuming the total weight from that was over 20 kg, then you randomly eat one candy. If you die, it is poison, if you live, the remaining one is poison.

>>2617338
Can someone see if this is right? I think it's asking what is the probability of any question predicting it's own probability, which is zero, since there are two one thirds and one two thirds.

tl;dr Is answer zero?

>>2617545
I see it as a self-referential paradox. If you say 1/3, then the answer is 2/3; if you say 2/3, then the answer is 1/3.

>>2617537
When I say scale, I mean the bathroom weighing scale type scale, not the "Lady Justice" balancing scale.
>>2617539
You have to find the answer without dying.
>>2617545

>>2617415
I can never get questions like this.
Similar problem, solution unknown:

Room filled with one hundred blue heads. The participants are aware that there are two colors in the experiment, blue and green. If one becomes aware that one is blue, then he must leave.

At regular intervals, a light it switched off and on again. During the dark interval, those who know they have blue heads must leave.

Each only knows that nintey nine others are blue and that they themselves might be green or blue.

How many lights off does it take to vacate the room?

>>2617569
The problem with this type is the fact that any information gathered by one person is equally applicable to any other person. Therefore they all must leave at the same time, or all never leave.

I suspect a paradox somewhere in them.

>>2617459
Can some one confirm this answer?
r = 0.8571428571

>>2617557
Is it possible?

>>2617574
I'll work on it.

I am my nephew's sister, but I am also a man.

How is this possible?

>>2617582
I already said it was.

>>2617598
You're transgender.

>>2617598
you are lady gaga

>>2617610
Not transgendered.

>>2617569

If you had green on your head, no one else would leave the room because they saw the green head. Therefore, you would leave the room because no one else did.

On the contrary, if someone leaves the room it can be assumed they left because they saw no one else with green on their head. Thus, you can conclude you have blue on your head.

>>2617621
I don't follow...

>>2617621
Everyone has blue heads. Therefore, I don't think it's possible to deduce.
Remember: >>2617572

>>2617469

Put it on the scale, if the last digit is an odd number, there is poison.

Because of this logic:
Let All bags = 1gram
Poison bag = 2gram
3 bags in total therefore total grams = 4g

For poison total must be
3g

For non poisoned
must be 2g

Therefore
Safe = even
Poison = odd

>>2617598
Nephew and man are words with singular definitions, but sister is often used in honorary situations.

Both the uncle and his nephew are in a sisterhood somehow, and are thus "each other's sisters".

>>2617598
You're someone who gets the words brother and sister confused easily.

>>2617641
The poison total is 1 gram, which means two bags weigh exactly 10,000 grams each while one bag weighs 10,001 grams.

I'm a little surprised that people are having such a hard time, I've posted this riddle on /sci/ before and two people arrived at the correct answer within 10 minutes.

>>2617643
Ding ding ding.
A winrar are you.

>>2617469
do I have a pet or small child for testing purposes?

>>2617469
I'd put two randomly selected bags onto the 'use once only' scale.
Best case scenario : Exactly 20.000 Kg of candy for me.
Worst case: Scale reads 20.001 Kg, I take the remaining bag home. 10Kg is still quite a lot!

>>2617693
Nope, sorry bro.

>>2617643
I don't have much respect for puzzles that are based on intentional abuse of semantics.

Pic closely related, though it's not quite the same thing.

>>2617696
You only have a 33% chance of discovering the poisoned bag that way. The solution doesn't involve a scenario where the player has to say "one of these two bags is poisoned".

>>2617659
Is this a "think outside the box" puzzle like arranging six matches into triangles?

>>2617706
Hey, don't yell at me. I'm just the guy that got the answer.

>>2617574
Correct, r=6/7

>>2617723
I can't really say it's the same, but it does involve thinking outside the box.

>>2617729
More precisely, does it only involve the scale?

>>2617729
Take 1, 2, and 3 candies out of each bag respectively (take 1 out from bag one, 2 out from bag 2, and 3 out from bag 3) and weigh all the bags. The one extra gram divided in a certain way (too lazy to do any math) will tell you which bag is poisoned.

Do I win a cookie?

>>2617469
Ok ..do this:
>>2617696
and for the second situation mix the candy with the poison in a bag .... now one can clearly say which bad has poison in it!

>>2617729
Since it's getting late, I could give the answer if people want.

>>2617729
Well my first thought is to put half of one bag and one other complete bag on the scale. 15kg plus either 0g, 0.5g, or 1g will identify the poisoned load. But I dunno if you're allowing me such accurate halving.

>>2617729

More precisely, does it only involve using the scale, and then only in the conventional manner?

Frigging alpha in my captcha screwed me up.

Poison bag problem:
Take 1 candy out of bag 1, 2 candies out of bag 2, 3 candies out of bag 3. With the one extra gram divided equally (supposedly) the difference will be noticeable.

Do I win?

>>2617760
Winrar!
right?

I'm okay with getting the answer.
Assuming the scale is the only thing being used, and is only being used in the conventional way.

>>2617764
Please do

>>2617760
This guy got it. Here's your cookie.

>>2617781
>>2617760
Argh, posting to /sci/ took forever for some reason. Sorry bout the double post.

>>2617796
Same thing happened to me.

>>2617801
Same here, 4chan's servers are fucking up again.

>>2617621
i don't understand. the original riddle says if one discovers he's blue. that would mean anyone who saw the green head would leave, which could presumably take only one extinguishing of the lights, no? this question can't really be answered without knowing the field of view for the heads/some silly constraint.

>>2617415
everyone dies? they all find out their eye types. if someone wants to kill himself but gets told he doesn't have to kill himself, he then knows he has brown eyes. and if they're staunchly religious, i'm picturing a scenario where they all want to commit suicide, but individually they arrive at the position i just suggested, where they find out whether they have brown or blue eyes.

>>2617813
They all know someone has to die. Can they allow the one person to slip by unharmed? No one knows their eye colour. They are merely informed one exists.

Anyone got some fresh material out there?

>>2617801
>>2617806
Well at least I'm not alone.

>>2617792
OMNOMNOMNOMNOM

Good riddle, liked the twist where it was a regular type scale. It'd have to be mighty precise though.

I always liked these kinds of puzzles but there are far too many where it's a matter of "statistics" despite the fact that a simpler answer exists or the person is messing around with semantics.

Prime example:
You've wandered into a cave with 3 chests. Inside the chests are 2 silver coins, 2 gold coins, and 1 gold and 1 silver coin. You don't know which chest is which however. Reaching into a chest, you pull out a gold coin. What is the chance (or probability) that the next coin is also gold?

>Sephiroth eflogr
Really captcha?

>>2617834
I could probably pull some riddles from Marilyn vos Savant's column.

Consider a medical diagnosis (cancer) in which there are two alternative hypotheses: (1) that the patient has a particular form of cancer and (2) that the patient does not. The available data is from a particular laboratory test with two possible outcomes:
(positive) and (negative).
We have prior knowledge that over the entire population of people only .008 (.8%) have this disease.
Furthermore, the lab test is only an imperfect indicator of the disease. The test returns a correct positive result in only 98% of the cases in which the disease is
actually present and a correct negative result in only 97% of the cases in which the disease is not present. In other cases, the test returns the opposite result.

Suppose we now observe a new patient for whom the lab test returns a positive result.What is the probability that the patient has cancer?

>>2617569
I want the answer to this. In the variants I know the people are told that there is at least one blue/green head. Solving for 100 people is reduced by one light switch to solving for 99 people, which is reduced by one light flick to solving for a room of 98 people and so on, but to solve the endpoint for a room with 2 people needs the people to be told that there is at least one green/blue head. No?

So I want the answer please.

>>2617853
Are you pulling from the same chest?

>>2617853
Yeah, I actually messed up the riddle some when I posted it, because solving it requires the player to account for the weight of the bag itself, before he counts every piece of candy in a bag to determine what an individual piece of candy weighs.

Let's just say an empty bad is magical and completely weightless.

>>2617881
I posted it and I have no idea what the answer is, I suspect it's either one, two, or never.

Also, see
>>2617572
Any information or assumption for one person can be applied to ALL people, so any action one takes they will ALL take.

>>2617853
Probability is 2/3 that next coin is gold.

>>2617853
33% chance.

>>2617901
>>2617910
Same chest both times?

>>2617886
Yes, you have to pull from the same chest.

>>2617901
This is correct but I have a big issue with it. In other versions of the riddle the chest has drawers to it. In this case order could matter in counting either of the two gold coins from the all gold chest. From simply a chest however, order shouldn't matter. You either pulled from the gold or silver chest and the next coin is either gold or silver. I really feel the answer should be 50% given the wording of the problem.

>>2617921
Same chest
>>2617910
I do believe it's actually 2/3.
Though I could be remembering the answer incorrectly.

Someone solve it.

>>2617923
Well you either picked your coin from the second chest or the third chest. If you picked from the second chest, you will get a gold next, if you picked from the third chest you won't. So what is the chance that the gold you picked came from the second chest? 2/3.

>>2617939
The question as I originally read it didn't ask about the chance of you pulling from a certain chest, just coins. I understand what you're trying to get across and you've done a better job than most others in that regard but I still say it should be 50%. Two chests and you've commited to one. What're the odds you get that other gold coin choosing from two chests? Should be 50%.

Meh, that's just me I guess.

>>2617931
It should be only one third, because you can't exclude the chest with silver without a new mathematical set (so it's not 50%) and there is only one scenario out of the three possibles where you would pull a second gold coin from the chest.

If we were drawing from a different chest it would be an entirely different matter.

>>2617939
Thank you, I just understood it.

>>2617898
Well it sounds a lot like a similar but to my mind at least much more solvable problem, where the answer goes like this:

Of the twenty people, at least one has a blue head. If there were only one blue head, he will know after one light because he can only see green heads. So one light down, one person walks out, game ends.
If there are two blue heads, then those people only see one blue head, and when that head doesn't walk out after the first light, they know that that blue head did not see only green heads - and since he can't see any other blue heads, he concludes that he also has a blue head. Two light switches, two people walk out, game ends. Three people, three light switches. One hundred people, one hundred light switches.

Only, it's all built on that piece of information that everyone knows there is at least one blue head. Without that, you can have a blue head looking at all green heads and not walking out, so the game just don't work.

Did you misstate that kind of problem, or is this just a harder one I don't get?

>>2617976
I think mines just a variation on yours that more confusing and less solvable. Maybe after exactly one hundred switches, everyone will leave? By the same logic. Or one hundred and one.

>>2617939
Actually, ignore my post, I'm pretty sure this guy is right. I wasn't excluding the silver chest but I was excluding the mixed chest. It's a 2/3 chance, it's similar to the three doors riddle.

>>2618003
Confirming, that guy is right. On the first draw, there were six possible outcomes, with a 50% chance of drawing a gold coin.

Since we drew a gold coin, we had picked one of three scenarios, Gg, Gg, and Gs. On the second draw we had a 2/3 chance of drawing a gold coin.

You are presented with two indistinguishable envelopes containing some money. You are further informed that one of the envelopes contains twice as much money as the other. You may select any one of the envelopes and you will receive the money in the selected envelope. When you have selected one of the envelopes at random but not yet opened it, you get the opportunity to take the other envelope instead. Should you switch to
the other envelope?

This is apparently a paradox, heads up.

>>2618030
Sounds like this riddle >>2617403 but with a flat 50% chance.

Probably missing something obvious, though.

The switching argument: One line of reasoning proceeds as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is

(1\2) 2A + (1\2) (A\2) = (5\4)A

8. This is greater than A, so I gain on average by switching.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

Bump, letters is too complicated. Anything else?

>3. The other envelope may contain either 2A or A/2.
There's a flaw in the math. That would mean one letter has 4 times as much money as the other.

You have a parcel with a message inside.

You need to send it to your friend without anyone else reading the message. It is possible to place key-locks on such parcels so that only the person with the key may open the parcel, but each person has only one lock and one key.

The only medium for communication is the parcel.

The only means of transporting parcels is through the post, and all unlocked parcels are opened and screened for messages and keys (if they discover and duplicate your key, your lock is useless).

Given the means and limitations of transport and security, construct a method for sending messages that is consistently secure.

Summary: How do you send a private message to someone through public channels while ensuring that privacy is maintained during transit given only locks and keys (where unsecured keys sent through transit result in the end of privacy)?

Put your lock on the parcel, and mail the parcel. Write "place your lock on this parcel so it has both our locks and mail it back to me" on the outside of the parcel. Get the parcel back, unlock your lock, and mail it back. Now it only has your friend's lock, and only he can unlock it

>>2618110
You forget A is not invariant

>>2617300
heres a mind fucker. how is humor an evolutionarily advantageous trait?

94 yellow 1 green 5 blue?
>>2617404

>>2618176

nice

for gold coin one
pr(2nG)= prob that u picked a gold from chest 2 =2/6 2 golds, 6 possible coins
pr(2)=1/3, 3 chests, 1 number 2 chest
Pr(2|G)=pr(2nG)/Pr(G)
=(1/3)/(1/2)
=2/3

bump for interest

>>2618057
This has been solved for a few months now. Basically the argument is that the probability distribution function for various moneys in the envelop is not constant, and you have extra information not represented by your probability function.

Effectively, you know there's some $ amount where there's NO FUCKING WAY the other has double. So for some parts of the real number line, your function is actually 100% guaranteed to be envelope and 1/2 envelope, not 50% that and 50% envelope and double envelope.

>>2618184
that's not a mind fucker
humor gets people to like you, more likely to help you out when you're in danger of dying
humor makes bitches wet, which increases your likelihood of cumslinging at a useful target

Arrange six matches into triangles.

>>2618152

Public key cryptography. In a real world example like yours, you would send an unlock padlock (only you had the key to open it) to the person you would like to send stuff to. Then that person would put the message/object inside a box and lock it with your padlock, which obviously only you could open since only you have the key. And vice-versa, when you want to send stuff to the other person you send your open padlock to them.

>>2620577
TETRAHEDRON BITCH

>>2620577
Very overused.

See:
>>2617723

>>2617448
Can someone explain this to me?
And was there ever a solution found?