Puzzle Time

jon-nyc

Yeah that’s a cheat. You have the order of the cards. That’s all.

Klaus

I'm not quite there yet, but I think I'm on the right track. That said, maybe I'm making this too complicated.

:::

Let's give the card names from 1 to 52.

The selected cards are c-1 to c-5, whereby c-1 < c-2 < c-3 < c-4 < c-5.

He removes c3.

Now George proceeds as follows:

Let's call the 24 different orders in which four cards can be arranged s-1 to s-24.

Case 1: c-2 and c-4 < 26.

In this case, c-3 must be between 1 and 26, but not c-2 or c-4. This means, at most 24 possibilities remain. If c-3 was the n'th card (not counting c-2 and c-4), choose order s-n.

Case 2: c-2 and c-4 >= 26.

This case is symmetrical to case 1.

Case 3: c-2 < 26 and c-4 >= 26.

Here I'm stuck.

:::

It would all be easy if somehow one additional bit of information could be transmitted, which would bump up the 24 sorting orders to 48, which would be enough (since only 48 cards remain).

Klaus

Ah, got it!

jon-nyc

I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.

Axtremus

@jon-nyc said in Puzzle Time:

I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.

:::

On that point, I’m thinking the selection of that card has to be suite based. Since there are only four suites and George gets five cards, it is guaranteed that at least two cards will come from the same suite.

:::

Klaus

Got it now, I think.

Here's how to get that extra bit of information out of the card choice.

:::

Now consider the colors (spades, clubs etc.) of the cards.

The notation 4.1 means that there is one color that shows up four times and 1 color that shows up once.

For the five cards there are these possibilities.

a) 4.1
b) 3.2
c) 3.1.1
d) 2.2.1
e) 2.1.1.1

For four cards, there are these possibilities:

i) 4.0
ii) 3.1
iii) 2.2
iv) 2.1.1
v) 1.1.1.1

From each of a) to e) you can choose a card to reach either one of i), iii) or v), or one of ii) and iv).

Let's call i), iii) and v) the "zero" case and ii) and iv) the "one" case.

There's the extra bit. Together with the 24 sorting orders I can then encode a number between 1 and 48.

:::

jon-nyc

How would you do the encoding of the 48? What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?

Ax - you’re on the right track. Or a right track at least.

Klaus

@jon-nyc said in Puzzle Time:

What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?

That part's easy.

From the 52 cards, subtract the 4 cards that were handed to Cats. Sort them. Name them 1 to 48.

The number encoding is: Name the 24 sorting orders s1 to s24. If you transmit 0 via the method I described above, sorting s-n stands for card n. If you transmit 1, it stands for 24+n.

jon-nyc

Right but the need to choose a particular card to implement your ‘bit’ could conflict with your ability to represent the number, couldn't it?

Klaus

Yes, but I was kind of hoping you wouldn't notice.

jon-nyc

Lol.

There’s a much easier way which people could easily do in their head in real time to make the trick work.

Horace

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Considering 5 random distinct numbers from 1 to 52, the magician could choose either the highest or the lowest number to keep secret, while ordering the remaining four to identify a number between 1 and 24, since there are 24 orderings of four distinct things A,B,C, and D where A is the lowest number, B the second lowest, etc. The magician would choose either the highest or lowest of the 5 numbers, whichever one was within 24 of the remaining highest number, considering 52 to roll over to 1 as numbers increase. This should always be possible. The assistant would identify the hidden number as the indicated offset from the highest of their four cards, rolling over from 52 to 1 if necessary.

:::

jon-nyc

:::

Yes that should work.

A conceptually similar solution that is easier to compute in your head in ‘magic trick time’:

Based in Ax’s observation that there is always at least one suit with two cards within the 5.

George chooses one of the two cards of the same suit, puts the other on top of the four he hands to Cats.

With the other three cards he can communicate a number between 1-6. He use that number to indicate how much higher the reserved card is than the top card, allowing for the cycle from K->A->2.

Example. He puts the queen of clubs on top with the reserved card being the 5. He indicates 6 with the other three cards. Cats knows that from Q you count K,A,2,3,4,5 so the reserved card is a 5.

What if the reserved card was a six? In that case George would have reserved the Q and handed Cats the 6 on top of the deck, still indicating 6 with the other three cards. (7,8,9,10,J,Q). In other words, by strategically choosing between the two cards of that suit, the additional 3 cards can convey enough information.

:::

Horace

Yeah that's conceptually identical but for the range being 1-13 rather than 1-52. But with some practice I don't think either is particularly difficult to compute in magic trick time. To identify the number between 0 and 23 based on the order of 4 cards, let hte first card indicate which quartile the number is in, then the second card indicate which tri-cile of that, and the third card which of the two remaining numbers it is.

Klaus

@Horace said in Puzzle Time:

The magician would choose either the highest or lowest of the 5 numbers, whichever one was within 24 of the remaining highest number

Ah yes, that was the puzzle piece I was missing.

Horace

jon where do you get these puzzles? Or do you invent them?

jon-nyc

Mostly from a mailing list from a university ‘math circle‘ we were involved with in NC

Horace

Today it is my responsibility to present a game during a department zoom meeting that we can all participate in. I'm going to give them this puzzle, but I'll have them try to solve it as a group rather than individually and competitively. Should be fun. Maybe I'll be surprised and they'll come up with a solution quickly.

jon-nyc

jon-nyc

I've since learned that this very method (using top card as suit indicator and three cards as identifier) was used by a magician from the 30s. Probably invented by him.

The New Coffee Room

Puzzle Time