Editing Double hashing (section)

=== Triple hashing ===
Adding a quadratic term <math>i^2,</math><ref name=Kirsch08>{{cite journal
 |title=Less Hashing, Same Performance: Building a Better Bloom Filter
 |first1=Adam |last1=Kirsch  |first2=Michael |last2=Mitzenmacher |authorlink2=Michael Mitzenmacher
 |journal=Random Structures and Algorithms |volume=33 |issue=2 |pages=187–218
 |date=September 2008 |doi=10.1002/rsa.20208 |citeseerx=10.1.1.152.579
 |url=https://www.eecs.harvard.edu/~michaelm/postscripts/rsa2008.pdf
}}</ref> <math>i(i+1)/2</math> (a [[triangular number]]) or even <math>i^2 \cdot h_3(x)</math> ('''triple hashing''')<ref>Alternatively defined with the triangular number, as in Dillinger 2004.</ref> to the hash function improves the hash function somewhat<ref name=Kirsch08/> but does not fix this problem; if:
: <math>h_1(y) = h_1(x) + k \cdot h_2(x) + k^2 \cdot h_3(x),</math>
: <math>h_2(y) = -h_2(x) - 2k \cdot h_3(x),</math> and
: <math>h_3(y) = h_3(x).</math>
then
: <math>\begin{align}
h(k-i, y) &= h_1(y) + (k - i) \cdot h_2(y) + (k-i)^2 \cdot h_3(y) \\
          &= h_1(y) + (k - i) (-h_2(x) - 2k h_3(x)) + (k-i)^2 h_3(x) \\
          &= \ldots \\
          &= h_1(x) + k h_2(x) + k^2 h_3(x) + (i - k) h_2(x) + (i^2 - k^2) h_3(x) \\
          &= h_1(x) + i h_2(x) + i^2 h_3(x) \\
          &= h(i, x). \\
\end{align}</math>