When optimising code, never guess, always measure. This page has been

Tagged As Software This is a truism that lots of people quote, but it can be hard to remember, especially in the heat of battle (as it were). Rather fortunately it came to mind just when needed, as I found something completely unexpected. I was writing a simple implementation of the Fermat difference of squares method of factoring. This involves writing the number to be factored as - you guessed it - the difference of two squares. If $n=a^2-b^2$ then $n=(a-b)(a+b)$ and we have a factorisation (provided we don't have $a-b=1$). Version 0 So the obvious thing to do is to take the square root of $n$, round it up to get a candidate for $a$, then look at $a^2-n$. If that's a square then we're done. def factor_fermat0(n): a = square_root_ceil(n) while True: d = a*a-n b = square_root_ceil(d) if b**2 == d: return a-b, a+b a += 1 Version 1 We can improve on this. We can increase $a$ until we have $a^2-b^2>n$, and then increase $b$ until $a^2-b^2<n$. If $n$ is odd then we'll get equality at some point, so we just continue until that happens, and we're done. def factor_fermat1( n ): a = square_root_ceil(n) b = square_root_ceil(a**2-n) while a**2-b**2 != n: while a**2-b**2 < n: a += 1 while a**2-b**2 > n: b += 1 return a-b, a+b Version 2 Now we notice that we are repeatedly computing the squares of $a$ and $b$, so we create variables to hold the squares and update them by remembering that $(x+1)^2=x^2+2x+1$. def factor_fermat2( n ): a = square_root_ceil(n) a2 = a**2 b = square_root_ceil(a**2-n) b2 = b**2 while a2-b2 != n: while a2-b2 < n: a2, a = a2+a+a+1, a+1 while a2-b2 > n: b2, b = b2+b+b+1, b+1 return a-b, a+b Version 3 We can improve that slightly further by having $c=2a+1$ and $d=2b+1$, and the code now becomes: def factor_fermat3( n ): a = square_root_ceil(n) a2 = a**2 c = 2*a+1 b = square_root_ceil(a**2-n) b2 = b**2 d = 2*b+1 while a2-b2 != n: while a2-b2 < n: a2, c = a2+c, c+2 while a2-b2 > n: b2, d = b2+d, d+2 return (c-d)/2, (c+d)/2-1 Let's see how we're doing These improvements all seem obvious and straight-forward, and we can check by doing some timing tests. The times, and indeed, the ratios of times, will vary according to the number being factored, so we can do two different numbers and see what the trend is. If you choose to analyse the algorithm then you'll see what's going on. However, we factored two different numbers: 4736906887651 =

1211141

x

3911111: Routine 0 1 2 3 Run 0 9.223 0.409 0.258 0.188 Run 1 9.138 0.408 0.258 0.188 Run 2 9.146 0.407 0.262 0.194 Run 3 9.149 0.416 0.258 0.188 Run 4 9.228 0.408 0.272 0.189 Run 5 9.187 0.407 0.258 0.188 Run 6 9.154 0.421 0.259 0.188 Run 7 9.157 0.407 0.259 0.189 47368008965429 =

3911111

x

12111139: Routine 0 1 2 3 Run 0 34.922 1.230 0.774 0.564 Run 1 34.276 1.214 0.775 0.562 Run 2 34.191 1.215 0.777 0.564 Run 3 34.198 1.214 0.776 0.564 Run 4 34.182 1.230 0.776 0.563 Run 5 34.226 1.244 0.775 0.563 Run 6 34.144 1.235 0.773 0.564 Run 7 34.104 1.233 0.775 0.565 We can see that the first routine is rubbish, but that's no surprise, as it's computing a square root every time through the loop. For the others we can see a clear improvement as we make those optimisations. It did surprise me just how big the improvement was from #2 to #3, but it was nice to see. Version 4 Then I made a small change to the code layout ready for another shot at an optimisation. Pretty uncontroversial - I ran the timing tests again to make sure there was no difference, using a larger number, and not bothering to use the reference "version 0" since that was so slow. def factor_fermat4( n ): a = square_root_ceil(n) a2 = a**2 c = 2*a+1 b = square_root_ceil(a**2-n) b2 = b**2 d = 2*b+1 while a2-b2 != n: while a2-b2 < n: a2 += c c += 2 while a2-b2 > n: b2 += d d += 2 return (c-d)/2, (c+d)/2-1 47368008965429 =

3911111

x

12111139: Routine 1 2 3 4 Run 0 1.225 0.774 0.562 0.622 Run 1 1.223 0.772 0.562 0.624 Run 2 1.238 0.776 0.567 0.625 Run 3 1.233 0.777 0.566 0.626 Run 4 1.229 0.769 0.563 0.624 Run 5 1.231 0.773 0.562 0.624 Run 6 1.225 0.771 0.563 0.625 Run 7 1.219 0.771 0.563 0.622 Wait - what? It's slower? How can that be? I'd've thought it would compile to exactly the same bytecode, and so execute in pretty much the same time. Just to be sure I ran it on a bigger number. 4736791755876611 =

39111113

x

121111147: Routine 1 2 3 4 Run 0 12.388 7.775 5.672 6.245 Run 1 12.215 7.767 5.660 6.256 Run 2 12.295 7.910 5.674 6.301 Run 3 12.196 7.749 5.645 6.247 Run 4 12.221 7.731 5.646 6.294 Run 5 12.227 7.750 5.638 6.232 Run 6 12.226 7.798 5.691 6.250 Run 7 12.164 7.753 5.646 6.264 There is no doubt: a2, c = a2+c, c+2

is faster than

a2 += c c += 2 That caught me by surprise, but it really does go to show: When optimising,

measure, measure,

measure. My guess is that in the first case the two evaluations and assignments can happen in parallel, and so may happen on different cores, whereas in the second case they must happen in serial, but that's just a guess. Maybe someone reading this knows for sure. Let me know. I've got more changes to make, but one thing is for sure - I'll be measuring the improvements as I go to make sure they really do make things better. References: https://en.wikipedia.org/wiki/Fermat%27s_factorization_method Comments The following comment was sent, but I don't know if the person is happy to be mentioned by name - if it's you, please let me know. However, they said this (paraphrased): Using the standard python interpreter the GIL would prevent a2, c = a2+c, c+2 from executing in parallel. Another big optimization would be to replace a**2 with a*a. Multiplication is much faster than power functions < 4. It is possible the interpreter does this automatically, but I don't think so. And to whomever it was - thank you for the response. <<<< Prev <<<< A Dog Called Mixture : >>>> Next >>>> Idle Thoughts About Pollard Rho ... You can follow me on Mathstodon.







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