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January 2

Srinivasa Ramanujan's deathbed theory solved after 100 years

Ramanujan the Indian mathematician's deathbed theory has been solved recently which I think is important and should be added in the article. But, the question is "how" and "where". I asked Favonian here and also posted in article's talk page. Favonian thinks, we need some sources that make more of the actual math and less of the dreams and divine inspiration. So, do I. The thing is very much complicated (starting with a dream ends with black hole). Any suggestion? --Tito Dutta (talk) 01:47, 2 January 2013 (UTC)[reply]

One doesn't solve a theory. The news is that another of Ramanujan's conjectures has been confirmed. —Tamfang (talk) 23:16, 26 June 2013 (UTC)[reply]

Concave quadrilateral requirement?

Resolved

Given 4 edge lengths (a,b,c,d) in *order* it is relatively easy to tell whether a convex simple quadrilateral exists with those lengths (test to see if any lengths are greater than the sum of the other 3), is there an easy way to calculate whether a concave simple quadrilateral exists with those edge lengths in order?Naraht (talk) 14:54, 2 January 2013 (UTC)[reply]

If a≤b≤c≤d then a simple concave one exists. Draw sides a and b in a straight line (so they form a 180° angle). Then draw in the other two sides to form a convex quadrilateral. Finally, tweak the interior angle at the vertex of the two shortest sides so it is slightly greater than 180°. I'll try to think about whether this generalizes to a≤c≤b≤d and b≤a≤c≤d, which I think would cover all cases. Duoduoduo (talk) 17:02, 2 January 2013 (UTC)[reply]

In general, with side sequence a, b, c, d, pick any two adjacent sides s1 and s2, and compare the sum of their lengths to the sum of s3 and s4. Say s1+s2<s3+s4. Then draw s1 and s2 in a straight line, and complete the convex quadrilateral by drawing in s3 and s4. Then tweak the 180° angle between s1 and s2 to make it a little more than 180°. Done. This covers all cases except when a+b=c+d and b+c=d+a, which upon subtracting one from the other gives a-c=c-a hence a=c, and which upon adding gives b=d -- thus a parallelogram is the only one that can't be distorted into a concave simple quadrilateral. Duoduoduo (talk) 17:31, 2 January 2013 (UTC)[reply]

Thank you, that is neat and clean.I'd figured out that it couldn't be done with a rectangle, but hadn't figured out how much larger the set was.Naraht (talk) 20:26, 2 January 2013 (UTC)[reply]

It's not that much "larger" a set at all (when looking at the lengths only). Parallelogram and rectangle are the same except for the angles. - ¡Ouch! (^{hurt me} / _{more pain}) 12:44, 8 January 2013 (UTC)[reply]

my hat, your hat, quantum entanglement, and OTP

Many here know the puzzle of how a team of hat-guessers can improve their performance of *both* (two out of two) guessing their own hat color correctly, if each is wearing a hat that is either black or white with 50/50 probability and cannot see their own hat color, only their opponent's.

You would think that if A is randomly guessing his own, then he has a 50% chance of getting it right, and if B is randomly guessing his own, then he has a 50% chance of getting it right, and observing their opponent "doesn't help", since they're predicting their own - not their opponent's. The chances they're both right then would be 25%. Yet they can coordinate ahead of time and a strategy then exists that get's the pair to *both* be right 50% of the time. In fact, a strategy exists for them to *never* both be right!!!!

The spoiler is below.

To *never* both be right, A always guesses for his own hat color the color he sees on his partner, B. Meanwhile, B always guesses the opposite of what he sees on A. Then they cannot both be right.

To both be right 50% of the time they must coordinate their guesses by guessing for their own hat color the hat color they see on their opponent.

So far, all this is known. Now we turn to quantum entanglement. Say I perform quantum entanglement on two sets of bits, AB... and CD... (the bit series continues).

How can we use the effect above to get a counterintuitive result? 91.120.48.242 (talk) 15:44, 2 January 2013 (UTC)[reply]

I assume by OTP you mean one-time pad? Does that mean you already know the answer - if not, why is the one-time pad relevant, and why do you think there is a counterintuitive result? 130.88.99.231 (talk) 17:04, 2 January 2013 (UTC)[reply]

You need to define precisely which version of the hat guessing game you're talking about. The version I know of involves 3 people, with hats, two of which are the same color and one different (and the participants don't know it it's two white and one black or vice-versa). They can each see the hats on the other two, but not their own, and must try to guess their own. They can not communicate with each other, and know when the others guess, but not what they guessed or if they were correct. Therefore:

1) The person who sees 2 hats of the same color on the other two knows that he has the opposite color, and guesses so.

2) Each of the other two, knowing that the first one would only have guessed if he saw two identical colored hats on the remaining players, know that the color of their hat must match the other remaining player, and so guess.

Thus, the 2nd part is an introduction to game theory, where the 2nd and 3rd guessers base their guesses on the behavior of the 1st guesser, which is assumed to be rational. StuRat (talk) 17:53, 2 January 2013 (UTC)[reply]

Looks to me that he defined his version very precisely, and it wasn't yours. Take a look below the spoiler. Duoduoduo (talk) 18:45, 2 January 2013 (UTC)[reply]

He said each person has a 50% probability of having a hat of each color, but I'm not clear on whether that means they must each have the same color or opposite colors. StuRat (talk) 19:35, 2 January 2013 (UTC)[reply]

Evidently each is given a hat of which the colour is determined by a fair coin toss. The two coin tosses are independent. This fits perfectly with the outcomes described according to the strategies used. The challenge only comes with the quantum case, but the setup is a little less obvious here. Presumably we can entangle the first pair of bits (A and C) in a singlet state (e.g. antiparallel spins), the second pair etc. Then if each can observe the other's "hat" (spin) and guess the opposite as its own, then both guesses will be 100% right. But I fail to see the counterintuitiveness in this, so I presume it must be some more interesting setup intended. — Quondum 20:06, 2 January 2013 (UTC)[reply]