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Pythagorean triples/Java/Brute force primitives (view source)

Revision as of 20:45, 20 December 2011

814 bytes added ,  12 years ago
m
Somewhere in there I got rid of the need for 12 as a BigInteger, reformat
m (Extra notes and comments, move a comparison so it is only calculated if needed.)
m (Somewhere in there I got rid of the need for 12 as a BigInteger, reformat)
 
(13 intermediate revisions by the same user not shown)
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{{works with|Java|1.5+}}
This version brute forces primitive triple candidates and then scales them to find the rest (under the perimeter limit of course). Since it only finds the primitives mathematically it can optimize its candidates based on some of the properties [[wp:Pythagorean_triple#Elementary_properties_of_primitive_Pythagorean_triples|here]] -- namely that a and b have opposite evenness, only one of a and b is divisible by 3, only one of a and b is divisible by 4, c is always odd, and that a<sup>2</sup> + b<sup>2</sup> must be a perfect square (which [[wp:Square_number#Properties|don't ever end in 2, 3, 7, or 8]]). NotablyAfter using those rules to eliminate candidates for a,b pairs, it doesn'tchecks usethat a GCDand functionb toare checkcoprime. forSince primitivesmany (<code>BigInteger.gcd()</code>a,b usespair thecandidates binaryhave GCDalready algorithmbeen eliminated, [[wp:Computational_complexity_of_mathematical_operations#Number_theory|whichthis ischeck O(n<sup>2</sup>)]])actually speeds things up a little bit by letting the program skip some c loops. For a perimeter limit of 1000, it is about 35 times faster than [[Pythagorean triples#Java|the other brute force version]]. For a perimeter limit of 10000, it is about 517 times faster. It also does not markmarks the primitives.
 
It defines a <code>Triple</code> class which is comparable so it can be placed in a <code>SetTreeSet</code> for easy sorting and to remove duplicates (e.g.the thisGCD algorithmcheck findsshould [15,remove 20duplicates, 25] as a primitive candidate afterbut it's hadnice alreadyto been added by scaling [3, 4,make 5]sure). It also can scale itself by an integer factor.
 
Note: this implementation also keeps all triples in memory. Be mindful of large perimeter limits.
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import java.util.Set;
import java.util.TreeSet;
import static java.math.BigInteger.ONE;
 
import static java.math.BigInteger.*;
public class PythTrip2{
 
public static final BigInteger TWO = BigInteger.valueOf(2);
public class PythTrip{
private static BigInteger LIMIT = BigInteger.valueOf(100);
private static final BigInteger TWO = BigInteger.valueOf(2),
B3 = BigInteger.valueOf(3), B4 = BigInteger.valueOf(4),
public static class Triple implements Comparable<Triple>{
BigIntegerB7 a,= bBigInteger.valueOf(7), c,B31 peri;= BigInteger.valueOf(31),
B127 = BigInteger.valueOf(127), B191 = BigInteger.valueOf(191);
//change this to whatever perimeter limit you want;the RAM's the limit
public Triple(BigInteger a, BigInteger b, BigInteger c) {
private static BigInteger LIMIT = BigInteger.valueOf(100);
this.a = a;
 
this.b = b;
public static class Triple implements Comparable<Triple>{
this.c = c;
BigInteger a, b, c, peri = a.add(b).add(c);
boolean }prim;
 
public Triple(BigInteger a, BigInteger publicb, TripleBigInteger scale(longc, boolean kprim){
this.a = a;
return new Triple(a.multiply(BigInteger.valueOf(k)),
this.b = b;
b.multiply(BigInteger.valueOf(k)),
this.c = c.multiply(BigInteger.valueOf(k)));
}peri = a.add(b).add(c);
this.prim = prim;
@Override}
 
public boolean equals(Object obj) {
public Triple scale(long k){
if(obj.getClass() != this.getClass()) return false;
return new Triple trip = (Triplea.multiply(BigInteger.valueOf(k)obj;), b
return a .equalsmultiply(tripBigInteger.a) && b.equalsvalueOf(trip.bk) &&), c.equalsmultiply(trip.c);BigInteger
} .valueOf(k)), prim && k == 1);
}
 
@Override
@Override
public int compareTo(Triple o) {
public boolean equals(Object obj){
//sort by a, then b, then c
if(!aobj.equalsgetClass(o.a)) return!= athis.compareTogetClass(o.a);)
if(!b.equals(o.b)) return b.compareTo(o.b)false;
Triple trip = if(!c.equals(o.c)Triple) return c.compareTo(o.c)obj;
return a.equals(trip.a) && b.equals(trip.b) && return 0c.equals(trip.c);
}
 
@Override
public Stringint toStringcompareTo(Triple o){
return if(!a + ", " + b + ", " + c;.equals(o.a))
} return a.compareTo(o.a);
if(!b.equals(o.b))
}
return b.compareTo(o.b);
if(!c.equals(o.c))
private static Set<Triple> trips = new TreeSet<Triple>();
return c.compareTo(o.c);
return 0;
public static void addAllScales(Triple trip){
long k = 1;}
 
Triple tripCopy = new Triple(trip.a, trip.b, trip.c);
public String toString(){
while(tripCopy.peri.compareTo(LIMIT) < 0){
return a + ", " + b + ", " + c + trips.add(tripCopyprim ? " primitive" : "");
}
tripCopy = trip.scale(k++);
}
 
}
private static Set<Triple> trips = new TreeSet<Triple>();
public static void main(String[] args){
 
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();
//change this to whatever perimeter limit you want;the RAM's the limit
BigInteger peri2 = LIMIT.divide(BigInteger.valueOf(2)),
peri3 = LIMIT.divide(BigInteger.valueOf(3));
 
for(BigInteger aperi2 = ONE; aLIMIT.compareTodivide(peri3TWO), < 0; aperi3 = aLIMIT.adddivide(ONEB3)){;
 
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
b.compareTo(peri2) < 0; b = b.add(TWO)){
BigInteger//skip bbif =both or neither of a and b.multiply(b); are divisible by 3 and 4
//if(amod3 a^2 +== b^2 is not a perfect square then don't even test for c's.mod(B3).equals(ZERO)
if(aa || amod4 == b.addmod(bbB4).toStringequals(ZERO).matches(".*[2378]")) continue;
continue;
//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))
continue;
if(!a.gcd(b).equals(ONE))
continue;
BigInteger ab = a.add(b);
BigInteger aabb = aa.add(bb);
 
// c is always odd for primitives so if b is odd start at b+2
for(BigInteger c = b.add(b.and(ONE).equals(ONE)? TWO:ONE);
// otherwise c.compareTo(peri2) < 0; c = c.add(TWO)){b+1
for(BigInteger c = b.add(b.testBit(0) //if? a+b+cZERO >: periLimitONE); c
if(ab.add(c) .compareTo(LIMITperi2) >< 0; c = c.add(TWO)){
 
break;
}// 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 (compare == 0){
Triple prim = new Triple(a, b, c, true);
if(trips.add(prim)) primCount++;{
addAllScales(prim) primCount++;
addAllScales(prim);
}
}
}
}
}
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.");
+ " 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
Runtime: 22
Up to a perimeter of 100, there are 17 triples, of which 7 are primitive.</pre>
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