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# Some Like It Hot

## Question

The final flight of the Arctic Circle of Logician's return home was finally uneventful. Warm thoughts as they neared Miami International Airport inspired one last quick challenge from one of the few logicians not asleep: "Let's define the hierarchy of suits as Diamonds < Clubs < Hearts < Spades < Diamonds and that of the ranks as Ace < 2 < 3 ... < Queen < King < Ace. I'll pick a card at random (suit and rank) and you must guess it. If your guess is either correct or one greater or less than my card in suit or rank correct in suit and one greater or less in rank or correct in rank and one greater or less in suit, I'll respond 'Hot' otherwise I'll say, 'Cold'. What is the fewest number of guesses that guarantees you can determine my card?"

EDIT

## Recommended Posts

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...I'm going to say 9. There are only 10/52 spaces (card combinations) that you will get 'cold' due to the suit thing, so once you get a 'cold' you know what the suit is. Simplest way of doing it is to try the same number with all four suits. Worst case if you get a hot # (not suit) and end up with 4 'hot's, then you move to a 'cold' distance away and find your suit, which takes a max of 3, and then try the #'s at +/- 2 of your original number, which takes another max of 2 to determine the #.

If you get the 'cold' suit on your 4th try with your original #, it only takes a max of 5 more moves (tries) to find the number.

Edited by Yoruichi-san
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yeah, looks like you are reading the problem correctly and got that puzzle correct. gonna edit the OP to hopefully correctly describe the puzzle i had intended.

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yeah, looks like you are reading the problem correctly and got that puzzle correct. gonna edit the OP to hopefully correctly describe the puzzle i had intended.

Yeah...that's what I originally thought the problem was...until I read the OP again..

...haven't had the time to solve it yet, but it's like playing battleship on a looping (periodic boundary conditions) 4X13 board with cross shaped bombs ;P

And thank you for making the suit order almost like bridge, but not quite...just to make it confusing for bridge players *cough*

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the edit does make a little better explanation as to why the "looping" - especially of the suits. when i was working this puzzle out, had been doing C<D<H<S but when i added the looping concept thought for some reason the colors should alternate. in any event, doubt this one will trump you.

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I can be sure of touching the card within 12. If it's in the first 6, it'll take up to 3 more to name the card. If it's in the second 6, it'll take up to 2 more to name the card. Therefore, my worst case is 14.

Let's call the cards C(s,r), where s is a suit from 1...4 (DCHS) and r is a rank from 1...13 (A...K)

Method: pick a card, any card, call it C1 = C(s,r)

Now establish a sequence of cards C2...c12 as follows Ci+1 = Ci(s+1,r+2)

Procedure: test C1, c2,..., Ci until you get a Hot. Let's call this Ci = C(s,r)

if i<7: test C(s+1,r+1)

----if Hot, test C(s,r+1),

--------if Hot, Announce C(s,r+1)

--------if Cold, Announce C(s+1,r)

----if Cold, test C(s-1,r-1)

--------if Hot, test C(s,r-1)

------------if Hot, Announce C(s,r-1)

------------if Cold, Announce c(s-1,r)

--------if Cold, Announce C(s,r)

if i>6: test C(s,r+1)

----if Hot, test C(s,r+2)

--------if Hot, Announce C(s,r+1)

--------if Cold, Announce C(s,r)

----if Cold, Announce C(s,r-1)

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Takes 13 tests to be sure of getting a Hot. If it's in the first 7, it will take from 1 to 3 more tests to name the card. If it's in the second 5, it will take 1 or 2 more tests. If it's in the 13th, it will take 1 more test.

Let's call the cards C(s,r), where s is a suit from 1...4 (DCHS) and r is a rank from 1...13 (A...K)

Method: pick a card, any card, call it C1 = C(s,r)

Now establish a sequence of cards C2...c13 as follows:

Ci+1 = Ci(s+1,r+2)

Procedure: test C1, c2,..., Ci until you get a Hot. Let's call this Ci = C(s,r)

if i<8: test C(s+1,r+1)

----if Hot, test C(s,r+1),

--------if Hot, Announce C(s,r+1)

--------if Cold, Announce C(s+1,r)

----if Cold, test C(s-1,r-1)

--------if Hot, test C(s,r-1)

------------if Hot, Announce C(s,r-1)

------------if Cold, Announce c(s-1,r)

--------if Cold, Announce C(s,r)

if i in 8...12: test C(s,r+1)

----if Hot, test C(s,r+2)

--------if Hot, Announce C(s,r+1)

--------if Cold, Announce C(s,r)

----if Cold, Announce C(s-1,r)

if i = 13: test C(s,r+1)

----if Hot, Announce C(s,r)

----otherwise, Announce C(s-1,r)

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Yeah, my method was slightly less elegant, but yielded the same result for max.

I like Cap's method as it basically 'climbs' or spirals the looping grid with the 5 square crosses (roll the grid into a long cylinder), leaving a zigzag ribbon which is taken on the second iteration in groups of 3 (the max possible to take of the remaining configuration with a single cross).

I did C(s+2, r+1), in pairs, and then moved up 3 rows, leaving two one square 'holes' for the first 8 crosses, used 9 and 10 to 'fill out' the top 3 rows and then moved down to test the pairs of holes for 11-13. If it was a full cross (10 or below), it would take a max 3 more moves to determine the number. If it was a pair of holes, only one more move is needed, for a max of 14.

Edited by Yoruichi-san
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Now that Y-san has the same max, I feel more confidence. Arigatoo gozaimasu!

I admire that you were able to get a larger number of full crosses early, so your Mean Time To First Hot is lower by a fraction. Now I see that, if we ignore the geometry, and just get to grab 5 cards each time, it would still take 11 tests to be sure of getting a Hot, so our taking 13 doesn't feel all so wasteful.

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hey hey you two - think you can know after fewer guesses.

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Err...desolee, mon capitaine...

Actually, my method gives a max of 13, not 14...doh...you don't need to test the last pair, if it's not one of the other 50 squares, it has to be one of the remaining 2, so you use the 13th test to find which one.

I don't think there's a more efficient packing of the crosses, though (I think holes of 2 near eachother are the best you can do), and it makes sense in terms of the math, ie 5*10=50,+3 to figure out which one of the 5.

Edit: And I don't think you need the 13th test either? You don't need to get a 'hot' to figure out which it is, you can assume the last group is 'hot' by the process of elimination.

Edited by Yoruichi-san
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And so it remains a team effort. But your arrangement wins by one test, I believe.

Your 13th test distinguishes the last pair, so no 14th test, agreed.

My 13th test distinguishes my only pair, so no 14th test. Unfortunately, my 12th test touches 3, so it takes two more to find which one for sure.

This is why your solution ending in 3 pairs is better than mine, ending in a trio and a pair.

Good job, Y-san. Hontoo ni utsukushii desu!

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right on - had a minor brainurism when first reading Y-san's solution. couldn't quite visualize her layout and quickly accepted her claim for fourteen. my solution for thirteen is similar in concept but slightly different in layout.

good work you two. thanks for playing

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Thanks for the pretty puzzle!

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And so it remains a team effort. But your arrangement wins by one test, I believe.

Your 13th test distinguishes the last pair, so no 14th test, agreed.

My 13th test distinguishes my only pair, so no 14th test. Unfortunately, my 12th test touches 3, so it takes two more to find which one for sure.

This is why your solution ending in 3 pairs is better than mine, ending in a trio and a pair.

Good job, Y-san. Hontoo ni utsukushii desu!

Actually, I redrew your solution, and I think if you zig-zag your crosses instead of spiraling (i.e. C(s,r), C(s+1,r+2). C(s, r+4), etc), you end with 3 pairs as well.

Edit: Lol...great minds think alike and reply at the same time, it appears...yeah, thanks for the puzzle, and the brainurism was mine . I actually liked both versions of the puzzle .

Edited by Yoruichi-san

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