ProjectEuler problem 35

[Tiberiu Gal]

"You'll skip 70% of checks..."
that's true for the Circular check ... but in the whole app, the 
most time is spent finding primes.


On Wednesday, 16 May 2012 at 14:01:19 UTC, Andrea Fontana wrote:
> Not that big gain?
>
> If you check for circulars numbers from 0 to 1.000.000 you can 
> skip intervals from 200.000 to 299.999, from 400.000 to 699.999 
> and from 800.000 to 899.999 (and other sub-intervals, of 
> course).
>
> You'll skip 70% of checks...
>
> On Wednesday, 16 May 2012 at 13:14:37 UTC, Tiberiu Gal wrote:
>> Good point there, unfortunately it's not that big a gain in 
>> this particular instance.
>>
>> thank you
>>
>> On Wednesday, 16 May 2012 at 12:46:07 UTC, Andrea Fontana 
>> wrote:
>>> What about some logic optimizations?
>>>
>>> F.E. if number contains one of these digit 0,2,4,6,8,5 is not 
>>> circular of course ..
>>>
>>>
>>> On Wednesday, 16 May 2012 at 09:26:45 UTC, Tiberiu Gal wrote:
>>>> hi
>>>>
>>>> many claim their code solves the problem in order of ms ( 
>>>> c/pascal/haskell code)
>>>>
>>>> my D code takes ~1.5  seconds.
>>>> Can it be made faster ( without pointers )?
>>>> it runs the same with or without caching the primes, and 
>>>> most of the time it spends finding primes; isCircularPrime 
>>>> can be optimized a bit, obviously, but it's not the 
>>>> bottleneck.
>>>>
>>>> thanks
>>>>
>>>> ===========================================
>>>> module te35;
>>>>
>>>> import std.stdio, std.math, std.conv;
>>>>
>>>> const Max = 1_000_000;
>>>>
>>>> byte[Max] primes;
>>>>
>>>> void main() {
>>>> 	primes[] = -1;
>>>> 	int cnt;
>>>> 	foreach(i; 2 .. Max) {
>>>> 		//writefln("is %s prime ? %s ", i, i.isPrime);
>>>> 		if( i.isPrime && i.isCircularPrime) {
>>>> 			cnt++;
>>>> 			//writefln("\t\tis %s, circular prime ? %s ", i, 
>>>> i.isCircularPrime);
>>>> 		}
>>>> 	}
>>>>
>>>> 	writeln(cnt);
>>>> 	
>>>> }
>>>>
>>>> bool isPrime(int n) {
>>>> 	byte x = 0;
>>>> 	if( ( x = primes[n]) != -1) return (x == 1);
>>>>
>>>> 	if( n < 2 && n > 0 ) {
>>>> 		primes[n] = 0;
>>>> 		return true;
>>>> 	}
>>>> 	
>>>> 	//int s = cast(int) (sqrt( cast(real) n) ) + 1;
>>>> 	for(int i=2; i*i < n + 1; i++) {
>>>> 		if( n %i == 0 ) {
>>>> 			primes[n] = 0;
>>>> 			return false;
>>>> 		}
>>>> 	}
>>>> 	
>>>> 	primes[n] = 1;
>>>> 	return true;
>>>>
>>>> }
>>>>
>>>> bool isCircularPrime( int n) {
>>>> 	
>>>> 	auto c = to!string(n);
>>>> 	for(int i ; i < c.length; i++){
>>>> 		c = c[1 .. $] ~ c[0];
>>>> 		if( !(to!int(c) ).isPrime )
>>>> 			return false;
>>>> 	}
>>>> 	return true;
>>>> }