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POFast exact bigint factorial
 

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    <p>I have a fixed-point bignumber library and want to implement fast factorial with no precision loss.</p> <p>After some math tricks on paper I got this formula:</p> <pre><code>(4N)!=((2N)!).((2N)!).{ (2N+1).(2N+3).(2N+5)...(4N-1) }.(2^N)/(N!) </code></pre> <p>This is already pretty fast, and with some programming tricks the complexity nears <code>~ O(log(n))</code>.</p> <p>To be clear, my current implementation is this:</p> <pre><code>//--------------------------------------------------------------------------- longnum fact(const DWORD &amp;x,longnum &amp;h) // h return (x&gt;&gt;1)! to speed up computation { if (x==0) { h=1; return 1; } if (x==1) { h=1; return 1; } if (x==2) { h=1; return 2; } if (x==3) { h=1; return 6; } if (x==4) { h=2; return 24; } int N4,N2,N,i; longnum c,q; N=(x&gt;&gt;2); N2=N&lt;&lt;1; N4=N&lt;&lt;2; h=fact(N2,q); // get 2N! and N! c=h*h; for (i=(N2+1)|1;i&lt;=N4;i+=2) c*=i; c/=q; // c= ((2N!)^2)*T1 / N! for (i=N4+1;i&lt;=x;i++) c*=i; c.round(); c&lt;&lt;=N ; // convert 4N! -&gt; x!, cut off precision losses for (i=(N2+1)|1,N2=x&gt;&gt;1;i&lt;=N2;i++) h*=i; h.round(); // convert 2N! -&gt; (x/2)!, cut off precision losses return c; } //--------------------------------------------------------------------------- longnum fact(const DWORD &amp;x) { longnum tmp; return fact(x,tmp); } //--------------------------------------------------------------------------- </code></pre> <p>Now my question:</p> <ol> <li><p>Is there a <strong>fast way to obtain</strong> <code>N!</code> from this <strong>term:</strong> <code>T1 = { (2N+1).(2N+3).(2N+5)...(4N-1) }</code>?</p> <p>Already answered.</p></li> </ol> <p>So to be clear, I need to extract this <strong>unknown term:</strong></p> <pre><code>T2 = (4N)! / (((2N)!).((2N)!)) </code></pre> <p>so:</p> <pre><code>(4N)! = (((2N)!).((2N)!)).T2 </code></pre> <p>This would help a lot because then it would not be needed to compute <code>.../(N!)</code> for factorial.</p> <p>The <code>T1</code> term is always integer-decomposable to this:</p> <pre><code>T1 = T2 * N! </code></pre> <p>Finally, it hit me :) I have done a little program for primes decomposition of factorials and then suddenly all becomes much clearer:</p> <pre><code>4! = 2!.2!.(2^1).(3^1) = 24 8! = 4!.4!.(2^1).(5^1).(7^1) = 40320 12! = 6!.6!.(2^2).(3^1).(7^1).(11^1) = 479001600 16! = 8!.8!.(2^1).(3^2).(5^1).(11^1).(13^1) = 20922789888000 20! = 10!.10!.(2^2).(11^1).(13^1).(17^1).(19^1) = 2432902008176640000 24! = 12!.12!.(2^2).(7^1).(13^1).(17^1).(19^1).(23^1) = 620448401733239439360000 28! = 14!.14!.(2^3).(3^3).(5^2).(17^1).(19^1).(23^1) = 304888344611713860501504000000 32! = 16!.16!.(2^1).(3^2).(5^1).(17^1).(19^1).(23^1).(29^1).(31^1) = 263130836933693530167218012160000000 36! = 18!.18!.(2^2).(3^1).(5^2).(7^1).(11^1).(19^1).(23^1).(29^1).(31^1) = 371993326789901217467999448150835200000000 40! = 20!.20!.(2^2).(3^2).(5^1).(7^1).(11^1).(13^1).(23^1).(29^1).(31^1).(37^1) = 815915283247897734345611269596115894272000000000 </code></pre> <p>After analyzing the prime exponents of the <code>T2</code> term (the rest after half factorials ^ 2) I derive the formula for them:</p> <pre><code>T2(4N) = multiplication(i=2,3,5,7,11,13,17,...) of ( i ^ sum(j=1,2,3,4,5,...) of (4N/(i^j))-(2N/(i^j)) ) </code></pre> <ul> <li>where multiplication is through all <code>primes &lt;= 4N</code></li> <li>where sumation is until <code>i^j &lt;= 4N</code></li> </ul> <p>The problem is that the divisions <code>4N/(i^j)</code> and <code>2N/(i^j)</code> must be done in <strong>integer math</strong> so they <strong>cannot be simplified easily</strong>.</p> <p>So I have another question:</p> <ol> <li><p><strong>How can I compute this: <code>exponent(i) = sum(j=1,2,3,4,5,...) of (N/(i^j))</code> effectively?</strong></p> <p><code>i</code> is any prime where <code>i&lt;=N</code>. It should be easy.</p></li> </ol> <p>Now I calculate the exponent <code>e</code> for prime <code>i</code> inside the <code>T2(N)</code> term like this (but this is too complex for my taste):</p> <pre><code>for (e=0,a=N/i,b=(N&gt;&gt;1)/i;(a)||(b);e+=a-b-b,a/=i,b/=i); </code></pre> <p>... I will try implement <code>T2</code> into <code>fact(x)</code> and compare speeds ...</p>
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    1. COThis code seems really complicated. What's wrong with an O(n) loop?
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      1. USCarl Norum
    2. CO@CarlNorum thats exactly what I was thinking. *"After some meth tricks I got the formula [...] and with some programming tricks the complexity nears O(nlogn)"* `for(long long int i = 1 ; i <= n ; ++i) { n *= i; }` Whats wrong with the typical for loop O(n) implementation?
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      1. USManu343726
    3. COsorry my mistake it should be O(log(n)) this use subdivision of N to calculate 40! it use 20! and 10! , to compute 20! it use 10! and 5! ... and so on. to calculate 100! you need only 5 recursions instead of 99 in O(n) case
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