## Wednesday, March 30, 2016

### Spanning set

Let $S = \{0,1,2,\dots,10\}$.

A set of numbers $A \subset S$ is called "spanning" if the set $\{ x, x+1, x+2, x+3 \mod 11 | x \in A \} = S$.

And a set of numbers $B \subset S$ is called "potentially spanning" if there is an integer $k$ such that the set $\{kb \mod 11 | b \in B \}$ is spanning.

Prove that every subset of $S$ with four elements are potentially spanning, and prove that there exists a subset of $S$ with three elements that is not potentially spanning.

Solution First, we show that there is a subset of three elements that is not potentially spanning.

Note that for a set to be spanning, it has to be $\{c,c+4, c+8\}$ for some $c$ (here we assume everything is modulo 11, so we don't write it for notational simplicity). This is because the maximum distance between two adjacent element is 4, so the distances between 3 elements must be 4+4+3 (or its rotation).

Now we claim that a set $\{x,y,z\}$ is potentially spanning if and only if there exists $r \neq 0$ and $s$ such that $\{x,y,z\} = \{r+s, 2r+s, 3r+s \mod 11\}$. First, clearly if a set is spanning, then all of its "rotations" are also spanning sets. Likewise, if a set is potentially spanning, then all of its rotations are also potentially spanning. So wlog we may assume that $s=0$. Then the set $\{r,2r,3r\}$ is potentially spanning because we can always take $k = 4r^{-1} \mod 11$ so that $\{kr,2kr,3kr\} = \{4,8,12=1\}$ a spanning set. The inverse modulo is guaranteed to exist because 11 is prime and $r \neq 0$.

Now the other direction. Suppose $\{x,y,z\}$ is potentially spanning, such that $\{kx,ky,kz\}$ is spanning. As we've established above, it must have form $\{c,c+4, c+8\}$ for some $c$. Thus we can take $r = 4k^{-1} \mod 11$ and $s = (c-4)k^{-1} \mod 11$ so that $\{k(r+s),2kr,3kr\} = \{4+ks,8+ks,12+ks\} = \{c,c+4,c+8\}$.

So we have shown that the potentially spanning sets are generated by (and only by) the number of $(r,s)$ pairs we can choose. There are 10 choices for $r$ and 11 for $s$, so there are at most 110 potentially spanning sets. We say at most because some choices of $r$ and $s$ may "collide" (result in the same spanning triplet). However, there are ${11 \choose 3} = 165$ possible triplets, so there are some triplets that are not generated by the $(r,s)$ construction above and thus not potentially spanning.

Now we show that a subset $B = \{b_1,b_2,b_3,b_4\}$ is always potentially spanning. Note that if any three of $b_i$s form a potentially spanning triplet then $B$ is potentially spanning. In particular, if $B$ contains an arithmetic sequence modulo 11 then $B$ is potentially spanning. So we assume that $B$ is free from arithmetic progression.

Now let $d_i = b_{i+1} - b_i$ (of course we mean $b_5 = b_1$. Then among all four $d_i$s, let $d$ be the maximal distance. If the maximal distance is 4, then we are done because $B$ itself is spanning. So we assume that the maximal distance is at least 5. Wlog, we may assume that this maximal distance is $d_4$ and that $b_1 = 0$. So this means $0=b_1 < b_2 < b_3 < b_4 = 11-d$ and $b_i$ must not contain arithmetic sequence. It's easy to see that the only possible combinations are:

Case 1: $d=5$, so $0=b_1 < b_2 < b_3 < b_4 = 6$. The only combinations are $\{0,1,4,6\}, \{0,1,5,6\},\{0,2,5,6\}$. All the other combinations contain an arithmetic sequence. But in the first two cases, we can take $k=8$ to get: $\{0,4,8,10\}, \{0,4,7,8\}$ which contains arithmetic sequence thus potentially spanning, and for the third case we can take $k=3$ to get $\{0,4,6,7\}$ which is spanning.

Case 2: $d=6$, so $0=b_1 < b_2 < b_3 < b_4 = 5$. The only combinations are $\{0,1,4,5\}$ and $\{0,2,3,5\}$. We can take $k=4$ to get $\{0,4,5,9\}$ and $k=2$ to get $\{0,4,6,10\}, both spanning sets. Case 3:$d=7$, so$0=b_1 < b_2 < b_3 < b_4 = 5$. The only combination is$\{0,1,3,4\}$. We can take$k=5$to get$\{0,4,5,9\}$, a spanning set. And there is no way to have$d > 7$and still fitting$b_2$and$b_3$without getting an arithmetic progression. So our proof is complete. ## Sunday, March 27, 2016 ### Various double counting problems These problems can all be solved by the double counting principle, that is, to count the same object two different ways and assert that those must match. Suppose$n > 4$. Let$A_1, A_2, \dots ,A_k$each be a set of$n$integers, where$k \leq 4^{n-1} / 3^n$. Show that we can color each integer with one of the four colors so that each set contains integers of 4 different colors. In a party attended by$n$people, there were a number of handshakes. A group of 10 persons is called "tight" if each person shook hands with the other 9. It is known that we cannot add another handshake (two people who didn't shake hands now shaking hands) without increasing the number of tight groups. Prove that the number of handshakes is at least$8n-36\$

In a class with many students, we choose 2016 committees among them. Each committee has 11 students, and each student may belong to multiple committees. Prove that there is a way to choose some of the students so that each committee has at least one chosen and one unchosen student.