# Puzzle about $4$-tuples less than $(x,x,x,x)$

Let $X$ be a $4$-tuple of matching entries, so $X = (x, x, x, x)$. Let $C$ be a collection of $4$-tuples such that $(a,b,c,d) \in C \implies a+b+c+d \leq 4x$ where $X$ must be an element of $C$, and no two $4$-tuples in $C$ have the same $a$,$b$,$c$, or $d$ value.

Let $f(X)$ be defined as the maximum possible cardinality of $C$. How would you find a closed form expression for $f(X)$?

So for example, if $X = (10,10,10,10)$, a possible maximal set $C$ would be

$$C = \{ (10,10,10,10), (1,13,13,13), (13,1,12,12), (12,12,1,11), (11,11,11,1), (2,9,9,9),\\ (9,2,8,8), (8,8,2,7), (7,7,7,2), (3,6,6,6), (6,3,5,5), (5,5,3,4), (4,4,4,3) \}\;,$$

where $|C| = 13$.

• This is my first post, any feedback on wording / formatting would be appreciated. I suspect the answer is something like f(x) = (4x-1)/3, perhaps with some ugly floors and ceilings, but I'd need some proof that the greedy example for C I've provided actually gives a C of maximum cardinality. – Otay Jan 22 '17 at 7:59
• You should include your own thoughts at solving this in your post. Since you are a new user and your question sounds like an exercise that has been given to you, this post reads like "I don't want to do my homework. Do it for me." – Mike Pierce Jan 22 '17 at 8:04
• I actually constructed this question after playing a video game that ranks players by four individual stats, then calculates an overall ranking. So this problem answers what the worst case overall ranking is for a player with balanced individual ranks. It's just for fun. (Is there a tag for that?) – Otay Jan 22 '17 at 8:33
• If I understand correctly, the C you provided is not of maximum cardinality. Couldn't I replace $(9, 2, 8, 8)$ with $(9, 14, 8, 8)$, replace $(8, 8, 2, 7)$ with $(8, 8, 14, 7)$, and then insert $(14, 2, 2, 14)$? That would bump the cardinality up to 14. – Archr Jan 22 '17 at 16:34
• Yeah! You're right its not max cardinality. – Otay Jan 23 '17 at 3:02