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Alice and Bob
#11
You can even do this with up to 124 unique cards. (in the deck)

There is only one way to do this trick with exactly 124 cards. With 52 cards, I tried to find out how many ways, but I couldn't figure out what 6497400! was.
Peace cannot be obtained without war. Why? If there is already peace, it is unnecessary for war. If there is no peace, there is already war."

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#12
I still don't see how it's possible (I don't want to ruin it by looking up a solution). This is the best I can think of:

Let say the cards in the deck have been assigned numbers 1 to 52.

Volunteer randomly picks 5 cards.

Alice always shows the 4 lowest numbered cards to Bob. This way Bob knows the fifth card is higher than the highest card he's show. The worst case scenario is the first 4 cards get picked - then there are still 48 possibilities. Best case scenario is #51 and #52 getting picked - Bob is shown #51 and knows the fifth has to be #52.

Alice doesn't have to show them to Bob in the same order they were picked, right? Maybe it could mean something different to Bob depending on whether he's first shown the highest of the 4, or the lowest of the 4, etc.
4 * 3 * 2 * 1 = 24 different ways to order the 4 cards. So in the worst case scenario this could cut it down to 48 / 24 = 2 possibilites.

And that's where I get stuck. Can I have a hint?
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#13
I was looking for solutions, but they were all a little too geometrical for me. But here is proof that there is at least one solution:

The number of combinations of 4 cards shown from 52 cards (in one of 24 orderings) is:
(52 choose 4) * 24.

The number of combinations of 5 unsorted cards picked from 52 cards is (52 choose 5).

You can see that these are vastly different amounts. For 124 cards, they are equal, indicating one unique solution.

Numerically, I'm still stumped as to how to solve this problem however...
Peace cannot be obtained without war. Why? If there is already peace, it is unnecessary for war. If there is no peace, there is already war."

Visit www.neobasic.net to see rubbish in all its finest.
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#14
Hi there,

I thought it would be more convincing to split up the solver,
so that Alice & Bob are two independant entities,
communicating only through a little shared show-file.

You'll find them here:
http://home.graffiti.net/vspickelen:graf.../Cards.zip

Regards,
vspickelen
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#15
Quote:So Alice selects one same-suit card and hides the other,
then uses a permutation of the remaining 3 cards to
encode their distance.
Wait, I thought the volunteer picks all five cards!

You mean the volunteer picks only one card, and Alice gets to pick the other four?
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#16
I hearby declare vspickelen the winner of this challenge. Special thanks goes to Agamemnus for helping me explain it Wink

Vspickelen, nicely done. That's the same solution I thought of. Not exactly the same but based on the same principles.

Sterling: (WARNING: spoilers ahead!)
No, the volunteer chooses five cards but Alice gets to decide which card to make the guessed card. She also doesn't need to hide the other suited (signal) card. It just makes it more difficult for the audience to figure out the trick. (BTW, what he meant by "hide" the other card is that by the sum of the shown cards Bob would be able to determine the location of the same-suited card. For example, if the sum of the four cards divided by four has a remainder of 0, then make the signal card the first card. If the remainder is 1, then make it the second, etc.)

She could, for example, order the cards so that the guessed card always has the same suit as the first card. It's still a bit tricky to solve since now there are only three cards to make a permutation with. (WARNING: more spoilers! Stop here if you want to solve the rest of the problem on your own.) Three cards could only represent six combinations and there are twelve possible cards to guess from. But Alice can choose which of the two suited cards is the guessed card. Now she just orders them so that the following is true:

card1's suit = card5's suit

and

(card1's value + the number shown by the combination [1 through 6]) MOD 13 = card5's value

(To keep things simple, values could be assigned like: A=1, 2=2, 3=3,.... ,J=11, Q=12, and K=0)
Voila!

Quote:You can even do this with up to 124 unique cards. (in the deck)
:o Wow. I didn't know that. Since there wasn't any information left over after my solution, I assumed 52 was the max. That get's me thinking...
hat were we arguing about again?
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#17
Thanks,

I'm in high feathers with my honours,
Cheerz!

vspickelen
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#18
On second thought, I looked at a simpler case of 2 shown cards shown and 3 cards drawn and I don't see how this is possible.. turns out that you just get several different results for every 2 cards.
Peace cannot be obtained without war. Why? If there is already peace, it is unnecessary for war. If there is no peace, there is already war."

Visit www.neobasic.net to see rubbish in all its finest.
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#19
The 2-card 'message space' is out of proportion to the 52-card deck.
It's far too small to encode the whole 3-card 'uncovered hand space'.
They match only if hand! / (deck - hand + 1) >= 1.

Alie & Ben have recently learned a brilliant strategy from prof. Berlekamp.
Their divination abilities are boundless now. I presented them decks with 8, 27, 52, 124 and even 725 cards.
A fifth card symbol called 'tips' was coined specially for this occasion.

http://home.graffiti.net/vspickelen:graf...ardspp.zip

vspickelen
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#20
But you already won the challenge! Now you're just showing off. (I'm kidding of course Big Grin) It looks interesting. I'll read it more thoroughly when I have time.
hat were we arguing about again?
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