Subject: Re: card puzzle
Date: Fri, 24 Nov 2000 18:15:11 -0600 (CST)
From: Weiping Shi <wshi@ee.tamu.edu>
To: Shai Simonson <shai@stonehill.edu>
CC: "Dr. Melvin R. Mercer" <mercer@ee.tamu.edu>

Shai,

You can do it for 52+4 cards! But it's messy.

Let each suit contain 14 cards. Consider the
distribution of the 5 cards among 4 suits:

Case 1. 2-1-1-1.
Case 1.1. If two spades X and Y:
1 <= X-Y (or Y-X) <= 6 (mod 14), we are done.
Case 1.2. If two spades X and Y:
X-Y=Y-X=7 (mod 14), then at least two of the remaining 3
cards will be interval {X, X+1, ..., X+6}
or {Y, Y+1, ..., Y+6}. (If all three cards are
in the same interval, choose the highest two
in any pre-specified order. Let the two
chosen cards be i and j, and AWLG they are in {X, ..., X+6}.
Then show X as the first card.
We must have 0 <= min{i-j, j-i} <= 3 (mod 7).
We will show either i (or j) and use 4 of the
3!=6 possible permutations to identify the
offset j-i (or i-j). The 2 un-used permutation will
be used in case 3.

Case 2. 3-1-1 or 3-2 or 4-1.
This is what we just discussed in the last email.
Basically you can always pick two spades
that are not distance 7, and hide a spade that is
within distance 6 or less from the shown card.

Case 3. 2-2-1.
If the two spades are not distance 7 apart, then we are done.
If the two hearts are not distance 7 apart, we are also done.
Otherwise, we will show any spade as the first card,
followed by a permutation of spade, heart and diamond.
So we hide a heart.
To avoid confusing with case 1.2, we will use
the un-used 2 permutations in case 1.2 to identify
if the missing suit is heart or diamond. The value of the
missing card must be distance 7 from the shown card (heart)
in this case.

There is still some room for further improvement,
but it's getting really messy.

Weiping