Question:
I have list of numbers, $\{1,2...1000\}$. This has ${1000 \choose 2}=499500$ unique pairs.
If I pick 2 numbers at a time from the list, I can pick all unique pairs in 499500 tries.
If I pick 8 numbers at a time from the list, what is the lowest number of tries in which I can pick all the unique pairs?
Repeating of pairs is allowed.
example: with smaller numbers : 8 (instead of 1000 numbers), 4 (instead of 8 picks)
list - 1,2,3,4,5,6,7,8; unique pairs - 28 (8c2).
If I pick 2 numbers at a time from the list, it will take me 28 tries to pick all the unique pairs.
If I pick 4 numbers at a time, it will only take 6 tries to pick all the unique pairs
try 1 - 1,2,3,4 - covers these pairs, 1,2; 1,3; 1,4; 2,3; 2,4; 3,4
try 2 - 5,6,7,8 - covers these pairs, 5,6; 5,7; 5,8; 6,7; 6,8; 7,8
try 3 - 1,2,5,6 - covers these pairs, 1,5; 1,6; 2,5; 2,6; (1,2; 5,6; - repeating)
try 4 - 3,4,7,8 - covers these pairs, 3,7; 3,8; 4,7; 4,8; (3,4; 7,8; - repeating)
try 5 - 1,2,7,8 - covers these pairs, 1,7; 1,8; 2,7; 2,8; (1,2; 7,8; - repeating)
try 6 - 5,6,3,4 - covers these pairs, 5,3; 5,4; 6,3; 6,4; (5,6; 3,4; - repeating)
Divide into 8 piles of size 127, including a little repetition. Rename them as ordered pairs from $(1,1)$ to $(8,127)$.
With $127^2$ octets, namely $O_{ab}=\{(i,ai+b\mod127)\}$ you can connect any numbers from different piles. You then need to connect numbers within a pile.
Divide each $127$ pile into eight piles of $17$, with some repetition, and you need $17^2$ octets for each $127$ pile.
You now have $64$ piles of $17$, and all numbers in different piles have been done.
Within a pile of $17$, each number must be in three octets, which is $51$ places needed, or at least seven octets. I think that is possible, but I've only covered $17$ by using eight octets.
That gives a total of $127^2+8×17^2+64×8=18953$ octets. The obvious minimum of ${1000\choose2}/{8\choose2}$ is $17840$