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Douglas Zare
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Here is a construction of size $38$. Call the points cards and divide the deck into $4$ suits, so that $2$ suits have ranks $1,...,13$ and $2$ suits have ranks $1,...,12$. Every hand of $5$ cards must have at least $2$ cards in some suit. So, if we cover all possible pairs in each suit, the result will be a wheel.

The La Jolla Covering Repository says that $C(12,5,2)=9$ so we can cover all pairs within the short suits with $9$ hands each:

1  2  4  7  9 
1  3  4  6 12 
1  3  7 10 11 
1  5  6  8 12 
2  3  5  7  8 
2  6 10 11 12 
3  6  7  9 12 
4  5  8 10 11 
5  8  9 10 11 

Similarly, it says $C(13,5,2) \le 10$ so we can cover all pairs within the long suits with $10$ hands each:

1  2  3  4  5
1  6  7  8  9
1 10 11 12 13
2  3  6  7 10
2  3  8  9 11
4  5  6  7 11
4  5  8  9 10
2  3  4 12 13
5  6  7 12 13
1  8  9 12 13

The total is $38$ hands which intersect each hand of $5$ cards in at least $2$ cards. So, $L(50,5,5,2) \le 38$.


Edit: The best this method can do with the coverings known in the La Jolla Repository is $37$. That can be done by splitting a $50$ card deck into suits of sizes $(17, 11, 11, 11),$ $(13,13,13,11)$, or $(15, 13, 11, 11)$.

$C(17,5,2) \le 16$.

$C(15,5,2) \le 13$.

$C(13,5,2) \le 10$ as shown above.

$C(11,5,2) \le 7$.

Here is a construction of size $38$. Call the points cards and divide the deck into $4$ suits, so that $2$ suits have ranks $1,...,13$ and $2$ suits have ranks $1,...,12$. Every hand of $5$ cards must have at least $2$ cards in some suit. So, if we cover all possible pairs in each suit, the result will be a wheel.

The La Jolla Covering Repository says that $C(12,5,2)=9$ so we can cover all pairs within the short suits with $9$ hands each:

1  2  4  7  9 
1  3  4  6 12 
1  3  7 10 11 
1  5  6  8 12 
2  3  5  7  8 
2  6 10 11 12 
3  6  7  9 12 
4  5  8 10 11 
5  8  9 10 11 

Similarly, it says $C(13,5,2) \le 10$ so we can cover all pairs within the long suits with $10$ hands each:

1  2  3  4  5
1  6  7  8  9
1 10 11 12 13
2  3  6  7 10
2  3  8  9 11
4  5  6  7 11
4  5  8  9 10
2  3  4 12 13
5  6  7 12 13
1  8  9 12 13

The total is $38$ hands which intersect each hand of $5$ cards in at least $2$ cards. So, $L(50,5,5,2) \le 38$.

Here is a construction of size $38$. Call the points cards and divide the deck into $4$ suits, so that $2$ suits have ranks $1,...,13$ and $2$ suits have ranks $1,...,12$. Every hand of $5$ cards must have at least $2$ cards in some suit. So, if we cover all possible pairs in each suit, the result will be a wheel.

The La Jolla Covering Repository says that $C(12,5,2)=9$ so we can cover all pairs within the short suits with $9$ hands each:

1  2  4  7  9 
1  3  4  6 12 
1  3  7 10 11 
1  5  6  8 12 
2  3  5  7  8 
2  6 10 11 12 
3  6  7  9 12 
4  5  8 10 11 
5  8  9 10 11 

Similarly, it says $C(13,5,2) \le 10$ so we can cover all pairs within the long suits with $10$ hands each:

1  2  3  4  5
1  6  7  8  9
1 10 11 12 13
2  3  6  7 10
2  3  8  9 11
4  5  6  7 11
4  5  8  9 10
2  3  4 12 13
5  6  7 12 13
1  8  9 12 13

The total is $38$ hands which intersect each hand of $5$ cards in at least $2$ cards. So, $L(50,5,5,2) \le 38$.


Edit: The best this method can do with the coverings known in the La Jolla Repository is $37$. That can be done by splitting a $50$ card deck into suits of sizes $(17, 11, 11, 11),$ $(13,13,13,11)$, or $(15, 13, 11, 11)$.

$C(17,5,2) \le 16$.

$C(15,5,2) \le 13$.

$C(13,5,2) \le 10$ as shown above.

$C(11,5,2) \le 7$.

Source Link
Douglas Zare
  • 28k
  • 6
  • 90
  • 130

Here is a construction of size $38$. Call the points cards and divide the deck into $4$ suits, so that $2$ suits have ranks $1,...,13$ and $2$ suits have ranks $1,...,12$. Every hand of $5$ cards must have at least $2$ cards in some suit. So, if we cover all possible pairs in each suit, the result will be a wheel.

The La Jolla Covering Repository says that $C(12,5,2)=9$ so we can cover all pairs within the short suits with $9$ hands each:

1  2  4  7  9 
1  3  4  6 12 
1  3  7 10 11 
1  5  6  8 12 
2  3  5  7  8 
2  6 10 11 12 
3  6  7  9 12 
4  5  8 10 11 
5  8  9 10 11 

Similarly, it says $C(13,5,2) \le 10$ so we can cover all pairs within the long suits with $10$ hands each:

1  2  3  4  5
1  6  7  8  9
1 10 11 12 13
2  3  6  7 10
2  3  8  9 11
4  5  6  7 11
4  5  8  9 10
2  3  4 12 13
5  6  7 12 13
1  8  9 12 13

The total is $38$ hands which intersect each hand of $5$ cards in at least $2$ cards. So, $L(50,5,5,2) \le 38$.