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X m ; y) := x 1 ∆(x 1 + y, x 2 , . . , x m ) + x 2 ∆(x 1 , x 2 + y, . . , x m ) + · · · + x m ∆(x 1 , . . , x m + y). 8. WHERE TO GO FROM HERE ? 27 Then g(x 1 , . . , x m ; y) = x1 + · · · + x m + m 2 y ∆(x 1 , . . , x m ). Proof: Observe that g is a homogeneous polynomial of degree one more than the degree of ∆(x 1 , . . , x m ). Swapping x i and x j changes the sign of g (since ∆ can be seen to be alternating). Hence if we substitute x i = x j = x then g becomes 0. It follows that x i − x j divides g.

1 DEFINITION. A sunflower with k petals and core Y is a family of sets, along with a set Y , such that | | = k, and for all A, B ∈ with A = B we have A∩ B = Y . Moreover, the sets A \ Y , which we call the petals, are nonempty. 2 LEMMA (Sunflower Lemma). Let be a family of sets, each of size s. (k − 1)s then contains a sunflower with k petals. Proof: We prove the result by induction on s. If s = 1 then | | > k − 1, so contains at least k size-1 sets. These form a sunflower (with Y = ). Fix s ≥ 2, and suppose the result holds for all smaller values of s.

K − 1)s−1 sets. Consider x := {A \ {x} : A ∈ , x ∈ A}. By induction we find a sunflower with k petals in this family. Adding x to each member gives the desired sunflower of . The Sunflower Lemma has found a number of applications in computer science. As we did in the chapter on Ramsey theory, we may wonder what the best possible bound is that guarantees a sunflower with k petals. Denote this number by f (s, k). (k − 1)s + 1. The upper bound was just proven; for the lower bound, consider the family of SDRs of a collection of s pairwise disjoint size-(k − 1) sets A1 , .