We have $A_0,A_1,\dots,A_5$ sets, with the same length $|A_0|=\cdots=|A_5|=5$. We know that

  • $A_i\cap A_{i\pm 1}=\varnothing$
  • $|A_i\cap A_{j}|=2$, when $j\neq i\pm 1$ and $j\neq i$
  • for all different $i,j,k,l$, if $A_i\cap A_j\neq\varnothing$ and $A_k\cap A_l\neq\varnothing,$ we have $A_i\cap A_j\neq A_k\cap A_l$

Here we suppose that $A_0=A_6,$ $A_{-1}=A_5,$ $A_i=A_{i ~\mathrm{mod}(6)}$.

How can find length of $A_0\cup A_1\cup\cdots A_6$ set?

If we calculate it directly, by hand we will get 14, but I am interested in algorithmic way to find solution, because this task can be generalized so that: we have $A_0,A_1,\dots,A_n$ set, $|A_0|=\cdots=|A_n|=a_n$ and second will become $|A_i\cap A_{j}|=b_n.$

  • $\begingroup$ These conditions can't hold. Consider $i=k=0$, $j=l=2$. By the second condition $\left|A_0\cap A_2\right|=2$, so $A_0\cap A_2\neq\varnothing$. Hence by the third condition $A_0\cap A_2 \neq A_0\cap A_2$, a contradiction. $\endgroup$ – Christoph Dec 11 '15 at 12:31
  • $\begingroup$ @Christoph Perhaps the third condition is missing the assumption that $\{i,j\}\not=\{k,l\}$. $\endgroup$ – Michael Burr Dec 11 '15 at 12:35
  • $\begingroup$ I don't know. Also the second condition would imply that $\left|A_i\right| = \left|A_i \cap A_i\right| = 2\neq 5$. So it is probably also missing the assumption $i\neq j$. $\endgroup$ – Christoph Dec 11 '15 at 12:37
  • $\begingroup$ I made few changes, now it should be hold conditions $\endgroup$ – Giorgi Dec 11 '15 at 14:28

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