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    copied!<p>I like the answer and anlysis by Artelius. Especially the proposal to use </p> <blockquote> <p>f(x,y) = x + y*K</p> </blockquote> <p>for some constant K is interesting and I'd like to just add a few more thoughts. What I'm doing here is not new, but very closely related to the <a href="http://www.brpreiss.com/books/opus4/html/page214.html" rel="noreferrer">Fibonacci hashing</a>, which I think has been proposed by Knuth.</p> <p>If we are using 64-bit integers then a collision f(x1, y1) = f(x2, y2) means</p> <blockquote> <p>0 = (dx + dy * K) mod 2<sup>64</sup>,</p> </blockquote> <p>where dx = x1 - x2 and dy = y1 - y2. This is the same as</p> <blockquote> <p>K = -dx*dy<sup>-1</sup> mod 2<sup>64</sup>,</p> </blockquote> <p>where dy<sup>-1</sup> is the modular inverse modulo 2<sup>64</sup>. If we want to choose the K such that f(x1, y1) != f(x2, y2) whenever the differences dx and dy are both small then we have to choose K such that </p> <blockquote> <p>K = -dx*dy<sup>-1</sup> mod 2<sup>64</sup>,</p> </blockquote> <p>has no solution such that both dx and dy are small. This can be achieved for example by choosing K close to phi * 2<sup>64</sup>, where phi = (sqrt(5)-1)/2 is the golden ratio. The golden ratio has a very special continued fraction expansion, i.e. in a certain sense it is a number that is hard to approximate well with a fraction.</p> <p>Hence, for 64-bit unsigned integers the following functions could be used</p> <blockquote> <p>f(x,y) = x + y * 11400714819323198485;</p> </blockquote> <p>or equivalently when using signed 64-bit integers</p> <blockquote> <p>f(x,y) = x - y * 7046029254386353131;</p> </blockquote>
 

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