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[KingZog3]

Another follow up on the -1/12 thing.

I've been watching some lectures online by the physicist Leonard Susskind.  Last week I watched the series on super symmetry and grand unification, now this week I'm going through the lectures on string theory and M theory.  In the lecture I just watched, he went through the -1/12 argument to deduce a 26 dimensional spacetime.  Some of you might find it interesting to watch physicists fumble through math 

The discussion begins at about 1:04:00

I've only watched part of it.. this all strikes me as something that is somehow tangential to "correct" but not complete.  Like they're doing computations that are kind of right but they're not using the right language.  The playing around and absorbing of infinities just doesn't sit right.

He's really going through some amazing hoops to do the sum of -1/12 thing.. though the pulling out an "infinite constant" thing is interesting.  But this is why I was never satisfied in my physics classes.  Things are never justified, just kind of.. done.

That's because you need the next, more advanced, physics to understand the current stuff you learn. And you keep getting more advanced until you reach the point where we don't have answer's yet.

Physics is always justified, or justifiable (until you reach the no answer stage). You just need to learn more to understand fully what you already learned.

[SirPeebles]

Yeah, but at least here it is clear that the sum is 1/epsilon^2 + -1/12 and not just -1/12.  In the lecture series, we've been viewing the string in an inertial reference frame with extremely large momentum along one axis so that the motion in the perpendicular plane is nonrelativistic.  As consequence, the energy has an extremely large additive constant on it, which he explains is the source of that 1/epsilon^2.  That is more satisfactory, in my mind.

[liopoil]

I don't have the education to totally understand this yet... but wow, cool.

[Witherweaver]

Right but.. could not a similar argument be made to make the sum A(x) + B, where A(x) goes to infinity and B is any number?  That epsilon seems arbitrary.. you could write 1/epsilon^2 as [1/epsilon^2 - k]+k for any constant k. 

[SirPeebles]

Right but.. could not a similar argument be made to make the sum A(x) + B, where A(x) goes to infinity and B is any number?  That epsilon seems arbitrary.. you could write 1/epsilon^2 as [1/epsilon^2 - k]+k for any constant k.

How do you propose to get 1/epsilon^2 - k in any natural way out of p^2?

[Witherweaver]

Epsilon is sqrt(1/(p^2+k)) instead of 1/p? 

I skipped ahead to his treatment of the sum, so maybe I'm missing some kind of organicness going on.

[Witherweaver]

Or, couldn't you take a function f_n(epsilon) with lim_{epsilon->0}f_n(epsilon) = 1 and consider the series

\sum{n=1}^\infty nf_n(\epsilon)

Then choose f_n so that the sums work out to give you <infinity>+<some fixed C>?  Or is e^{-epsilon*n} very special here?

[Titandrake]

This is a semi-old puzzle, but I recently both remembered it and found out the solution, so I might as well share it.

You and a partner have a deck of cards, and want to show off a neat trick. First, a third person chooses 5 cards from the deck of 52. You look at those cards, remove 1 of them, and place the other 4 on the table in any order. (You must place all 4 cards, I believe the puzzle is much easier if you don't have to.) Your partner must then say what card you removed from the 5.

Come up with a scheme that makes this trick possible.

[Kirian]

This is a semi-old puzzle, but I recently both remembered it and found out the solution, so I might as well share it.

You and a partner have a deck of cards, and want to show off a neat trick. First, a third person chooses 5 cards from the deck of 52. You look at those cards, remove 1 of them, and place the other 4 on the table in any order. (You must place all 4 cards, I believe the puzzle is much easier if you don't have to.) Your partner must then say what card you removed from the 5.

Come up with a scheme that makes this trick possible.

I assume you're not allowed to use positional cues?

Edit:  Oh, I just noticed that *I* get to pick out which of the five cards is discarded, not the third party.  That makes things much, much easier.

[Kirian]

So, for suits, there are 6 possible distributions in 5 cards:  5, 4-1, 3-2, 3-1-1, 2-2-1, 2-1-1-1.  No matter what, you can always remove a card that is the same suit as one of the four cards you play.  Play the same-suited card first, and your partner knows the suit of the unknown.

I'm trying to figure out ranks, give me a few minutes.

[Axxle]

While the third party is focusing on you playing the cards you mouth the card to your partner.

[eHalcyon]

Are positional cues not allowed then?

What about timing?

[Twistedarcher]

Kirian has a good method for the suit.

For the rank, there's enough ways to play the cards that you can assign a rank to 13 ways of playing out the cards.

Say 1 is the left most spot, 2 is next left most, etc., you can do something like:

1234 = 2
1243 = 3
1324 = 4
1342 = 5
1423 = 6
1432 = 7

A lot of memorization but possible.

Not sure of a way yet if you need to play them all in the same location.

Edit: you could use the position from your hand just as easily as the position on the table.

[eHalcyon]

Kirian has a good method for the suit.

For the rank, there's enough ways to play the cards that you can assign a rank to 13 ways of playing out the cards.

Say 1 is the left most spot, 2 is next left most, etc., you can do something like:

1234 = 2
1243 = 3
1324 = 4
1342 = 5
1423 = 6
1432 = 7

A lot of memorization but possible.

Not sure of a way yet if you need to play them all in the same location.

Edit: you could use the position from your hand just as easily as the position on the table.

There are many, many things you could do if you can use position to give information.  In order for this to be difficult, you would have to figure out a way to do it solely by the order that the cards are played.

[Twistedarcher]

Couldn't you just do the same with the ranks of the cards? The order you play them from lowest to highest can be substituted for left and right?

Ie, if you go lowest, second lowest, third lowest, highest, that can signal one rank, while going lowest, third lowest, second lowest, highest signals another rank.

[Kirian]

Couldn't you just do the same with the ranks of the cards? The order you play them from lowest to highest can be substituted for left and right?

Ie, if you go lowest, second lowest, third lowest, highest, that can signal one rank, while going lowest, third lowest, second lowest, highest signals another rank.

That breaks down if there was a full house, for instance...

[Axxle]

Couldn't you just do the same with the ranks of the cards? The order you play them from lowest to highest can be substituted for left and right?

Ie, if you go lowest, second lowest, third lowest, highest, that can signal one rank, while going lowest, third lowest, second lowest, highest signals another rank.

It works if you also assign value to suits, but then Kirian's way of indicating suit doesn't work.

[eHalcyon]

Couldn't you just do the same with the ranks of the cards? The order you play them from lowest to highest can be substituted for left and right?

Ie, if you go lowest, second lowest, third lowest, highest, that can signal one rank, while going lowest, third lowest, second lowest, highest signals another rank.

It works if you also assign value to suits, but then Kirian's way of indicating suit doesn't work.

I've been thinking along this line.  You can have an overall sorting for all 52 cards (e.g. sort by rank with Ace high, break ties by suit with Diamond low, then Clubs, then Hearts and Spades high).  Ignoring the first card, which indicates suit, you have the 3 remaining cards which can be labelled low, med and high.  Then the order of those cards can convey rank.

The problem is that 3 cards will only get you 3! = 6 permutations, which is just under half of the 13 we need.  4 cards gives 24, but then we lose the suit.

Is there a way to involve the initial suit card without making any assumptions about the 5 cards you are given?

Edit: One more thing to consider -- since the first card you play is of the same suit as the removed card, that is actually one card you don't have to account for.  You only have to worry about 12 ranks, because the rank of the first card you play is automatically excluded.  Hmmm... you could have a convention like "always pick the higher card of the pair"... but that doesn't help narrow it down if the pair includes the very highest card.

[eHalcyon]

OK, so Kirian noted the possible distributions for suits.

5; 4-1; 3-2; 3-1-1; 2-2-1; 2-1-1-1

What are the possible distributions that can be seen by the guesser?

4; 3-1; 3-1 or 2-2; 2-1-1; 2-1-1; 1-1-1-1

If all 4 remaining cards are the same suit, then you can use all 4 cards to convey rank because it doesn't matter which is first.  That's trivial now.  You can do something similar from a 3-1 hand.  You are playing 3 cards of the same suit, so you can actually choose the order for them. 

The tightest case where this still works is from a 3-1-1 start.  You take one of the three matching, then you have 2-1-1 left.  Which of those 2 do you play first as the suit indicator?  That doubles the 6 permutations of the 3 other cards, which is enough to cover 12 ranks.

So now we just have to figure out what to do if we get a 2-2-1 or a 2-1-1-1 start.  It could be a different rank indication system because the guesser will be able to see that it was a 2-2-1 start (the pair that is revealed does not include the first suit indicating card) or a 2-1-1-1 start (the cards that are revealed are all different suits).

[Tables]

I think I have the solution. It's probably wrong (because I only spent about 5 minutes thinking ) but I'll spoiler tag in case it's right.

The basic idea is we need the card we put down as our suit card to not only indicate suit, but also always be picked specifically depending on the other number. The way we do this is as follows: We first get our 'suit pair' and we choose the card which is at most 6 higher than the other to set aside, assuming numbers wrap around. So e.g. if we had a 9 and a 2, then we set aside the 2 (as wrapping around we count from 9: 10, J, Q, K, A, 2). Then we play the 9 first, and then we just need permutations of the other cards to indicate a number from 'count up 1' to 'count up 6', which we can easily define as there are 6 permuations.

[Twistedarcher]

Ok, so we need a way to get from 6 permutations to 12. So basically just need one more choice to make.

[Titandrake]

Solution by Tables is correct. If you want a hint instead of getting spoiled by the entire solution, then worrying about the distribution of suits isn't important. All you need is that one suit has at least 2 cards.

[Tables]

As an example of the above method, let's say we number cards from lowest to highest then by suit alphabetically (so we'd go 2 Clubs, 2 Diamonds, 2 Hearts, 2 Spades, 3 Clubs, ... A Hearts, A Spades). Number the three cards (one is set aside, and one is the suit card) 1, 2, 3 from lowest to highest. Then for simplicity, we'll say S, 1, 2, 3 is lowest (add 1), then S, 1, 3, 2 (add 2), then S, 2, 1, 3 (add 3), then S, 2, 3, 1 (add 4), then S, 3, 1, 2 (add 5), and finally S, 3, 2, 1 (add 6).

So now I've made an actually complete method. Here's an example in action. I'm given H2, S8, CK, H8, C3. I have 2 Hearts and 2 Clubs, but I'm going to go with Hearts for my suit (semi-arbitrarily). H8 is six more than H2, so that's the one I set aside. H2 is my suit card. I want to indicate add 6, so I need the 'largest' permuation, 3, 2, 1. That happens to be CK, then S8, then C3. So I lay out for my partner: H2, CK, S8, C3.

My partner sees this and the H2 immediately tells him the hidden card is a Heart, from 3-8. He looks at the permuation of other cards and sees it's in decreasing order. He knows to add 6, so he announces the hidden card is H8 and is correct.

Phew. I'm more convinced of my method's correctness now.

PPE: Titan just confirmed it's correct, but still here's an example.

[eHalcyon]

Nicely done.  The wrapping around was the key I missed.  Very clever.

[Kirian]

Ah yes, I missed the wrapping around as well, I was going with my start on suits, and following that with OK, what you we do with 4 of a kind, full house, etc.  Then I got down to pairs and realized it was starting to fall apart.