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Pythagorean triples/Java/Brute force primitives: Difference between revisions
Pythagorean triples/Java/Brute force primitives (view source)
Revision as of 20:45, 20 December 2011
, 12 years agoSomewhere in there I got rid of the need for 12 as a BigInteger, reformat
m (With the GCD check we can be sure that we have found a primitive, so we can mark them (negligible speed change)) |
m (Somewhere in there I got rid of the need for 12 as a BigInteger, reformat) |
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import java.util.Set;
import java.util.TreeSet;
import static java.math.BigInteger.*;
public class PythTrip{
//change this to whatever perimeter limit you want;the RAM's the limit
private static BigInteger
public static class Triple implements Comparable<Triple>{
BigInteger a, b, c, peri;
boolean prim;
public Triple(BigInteger a, BigInteger b, BigInteger c, boolean prim){
this.a = a;
this.b = b;
this.c = c;
peri = a.add(b).add(c);
this.prim = prim;
}
public Triple scale(long k){
return new Triple(a.multiply(BigInteger.valueOf(k)), b
.multiply(BigInteger.valueOf(k)), c.multiply(BigInteger
.valueOf(k)), prim && k == 1);
}
@Override
public boolean equals(Object obj){
if(obj.getClass() != this.getClass())
return false;
Triple trip = (Triple) obj;
return a.equals(trip.a) && b.equals(trip.b) && c.equals(trip.c);
}
@Override
public int compareTo(Triple o){
if(!a.equals(o.a))
return a.compareTo(o.a);
if(!b.equals(o.b))
return b.compareTo(o.b);
if(!c.equals(o.c))
return c.compareTo(o.c);
return 0;
}
public String toString(){
return a + ", " + b + ", " + c + (prim ? " primitive" : "");
}
}
private static Set<Triple> trips = new TreeSet<Triple>();
public static void addAllScales(Triple trip){
long k = 2;
Triple tripCopy = trip.scale(k++);
while(tripCopy.peri.compareTo(LIMIT) <= 0){
trips.add(tripCopy);
tripCopy = trip.scale(k++);
}
}
public static void main(String[] args){
long primCount = 0;
long start = System.currentTimeMillis();
BigInteger peri2 = LIMIT.divide(TWO), peri3 = LIMIT.divide(B3);
for(BigInteger a = B3; a.compareTo(peri3) < 0; a = a.add(ONE)){
BigInteger aa = a.multiply(a);
boolean amod3 = a.mod(B3).equals(ZERO);
boolean amod4 = a.mod(B4).equals(ZERO);
//b is the opposite evenness of a so increment by 2
for(BigInteger b = a.add(ONE); b.compareTo(peri2) < 0; b = b
//if a^2+b^2 isn't a perfect square, don't even test for c's
BigInteger aabb = aa.add(b.multiply(b));
if((aabb.and(B7).intValue() != 1)
&& (aabb.and(B31).intValue() != 4)
&& (aabb.and(B127).intValue() != 16)
&& (aabb.and(B191).intValue() != 0))
if(!a.gcd(b).equals(ONE))
continue;
BigInteger ab = a.add(b);
// c is always odd for primitives so
// otherwise
for(BigInteger c = b.add(b.testBit(0) ? ZERO :
// if a+b+c > periLimit
if(ab.add(c).compareTo(LIMIT) > 0) break;
int compare = aabb.compareTo(c.multiply(c));
// if a^2 + b^2 != c^2
if(compare < 0){
break;
}else if
Triple prim = new Triple(a, b, c, true);
if(trips.add(prim)){
primCount++;
addAllScales(prim);
}
}
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}
}
for(Triple trip : trips){
System.out.println(trip);
}
System.out.println("Runtime: " + (System.currentTimeMillis() - start));
System.out.println("Up to a perimeter of " + LIMIT + ", there are "
+ trips.size() + " triples, of which " + primCount
+ " are primitive.");
}
}</lang>
Output:
<pre>3, 4, 5 primitive
5, 12, 13 primitive
6, 8, 10
7, 24, 25 primitive
8, 15, 17 primitive
9, 12, 15
9, 40, 41 primitive
10, 24, 26
12, 16, 20
12, 35, 37 primitive
15, 20, 25
15, 36, 39
16, 30, 34
18, 24, 30
20, 21, 29 primitive
21, 28, 35
24, 32, 40
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