基于数论变换的大整数乘法的性能测试

        看来之前所说的有些错误的地方，这个算法只能处理到800位的乘法。之后做了一个测试，即比较用数论变换方法和用普通方法做大整数乘法的时间消耗。如下图，表示了程序运行所花的时间随着问题规模的增加而增加的曲线。其中红色的曲线是普通乘法，而黑色的曲线则是基于数论变换的大整数乘法。可以看到当问题规模增大时，我们的算法的时间性能就体现出来了。不过不够完美的是，基于数论变换的方法，只能处理到800多位（81n<=65537）。

        当然，跟Java库中的BigInteger类中的乘法相比，我的实现代码就慢了很多。主要是BigInteger中不是用十进制来保存的，而是用了更大的进制（256还是65536来着？也记不清了），因此它的效率是来自于算法的常数因子比较小。在可测的范围之内都比我们的算法效率要高。这也可以说是理论与实际之间的一些差距吧。或者说BigInteger中的乘法是综合各方面考虑的权衡。



测试的代码如下：

1.Test.java

package NumericTransformation;import java.awt.Graphics;import javax.swing.JFrame;import javax.swing.JPanel;public class Test extends JPanel {/** * @param args */public static void main(String[] args) {// TODO Auto-generated method stubJFrame jFrame=new JFrame();jFrame.setBounds(0, 0, 800, 800);jFrame.add(new Test());jFrame.setVisible(true);jFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);}@Overridepublic void paint(Graphics g) {// TODO Auto-generated method stubsuper.paint(g);//g.drawString("this is our first draw string function call", 10, 10);for(int i=10;i<800;i+=5){try {new NumericTransformation().testOnce(i, i, g);} catch (Exception e) {// TODO Auto-generated catch blocke.printStackTrace();}}}}

2.NumericTransformation.java

package NumericTransformation;import java.awt.Color;import java.awt.Graphics;import java.math.BigInteger;import java.util.Random;import java.util.Scanner;/** * 实现了基于数论变换的大整数乘法 * @author wwf * */public class NumericTransformation {public static final int M=65537;public static final int R=3;//3 是65537的一个本原根，也可以说是生成元public static final int RP=21846;public static final int MAX_POW=65536;/** * @param args * @throws Exception  */public static void main(String[] args) throws Exception {// TODO Auto-generated method stubScanner in=new Scanner(System.in);/*System.out.println("input two integer numbers : ");String numa=in.next();String numb=in.next();System.out.println("the result of multiplication is : "+new NumericTransformation().multiply(numa, numb));*/}void testOnce(int length,int position,Graphics g) throws Exception{long time1=0,time2=0;System.out.println("calculate");for(int i=0;i<500;i++){Random rand=new Random();StringBuffer sba=new  StringBuffer();StringBuffer sbb=new StringBuffer();for(int j=0;j<length;j++){sba.append((rand.nextInt(10)+20)%10);sbb.append((rand.nextInt(10)+20)%10);}String sa=sba.toString();String sb=sbb.toString();long original=System.currentTimeMillis();multiply(sa, sb);time1+=System.currentTimeMillis()-original;original=System.currentTimeMillis();int[] result=new int[sba.length()+sbb.length()+1];for(int j=0;j<sba.length();j++){for(int k=0;k<sbb.length();k++){result[j+k]+=(sba.charAt(j)-'0')*(sbb.charAt(k)-'0');}}for(int j=0;j<result.length-1;j++){result[j+1]+=result[j]/10;result[j]%=10;}time2+=System.currentTimeMillis()-original;}g.setColor(Color.BLACK);g.fillRect(position, (int) (500-time1/10), 5, 5);g.setColor(Color.red);g.fillOval(position, (int) (500-time2/10), 5, 5);}/** * 这个是程序对外的接口 * @param a * @param b * @return * @throws Exception */public String multiply(String a, String b) throws Exception{if(a.startsWith("-")&&b.startsWith("-")){return multiply(a.substring(1), b.substring(1));}else if(((!a.startsWith("-")&&(!b.startsWith("-"))))){long [] pa=new long[a.length()];long [] pb=new long[b.length()];for(int i=0;i<a.length();i++){char c=a.charAt(i);if(!Character.isDigit(c)){throw new Exception("number format error!!!");}pa[i]=c-'0';}for(int i=0;i<b.length();i++){char c=b.charAt(i);if(!Character.isDigit(c)){throw new Exception("number format error!!!");}pb[i]=c-'0';}long [] presult=this.numMultiply(pa, pb);StringBuffer result=new StringBuffer();for(long i:presult){result.append(i);}while(result.charAt(0)=='0'&&result.length()>1){result.delete(0, 1);}return result.toString();}else if(a.startsWith("-")){return "-"+multiply(a.substring(1), b);}else{return "-"+multiply(a, b.substring(1));}}/** * a和b是两个乘数的十进制表示形式(两个乘数都是非负整数) * @param a * @param b * @return * @throws Exception */private long[] numMultiply(long []a,long[] b) throws Exception{if(a.length*81>M&&b.length*81>M){throw new Exception("number too large to perform FFT based Multiplication");}long[] result=polyMultiply(a, b);/*for(int i=0;i<result.length;i++){System.out.print(result[i]+" ");}System.out.println();*/for(int i=result.length-1;i>=0;i--){if(result[i]>10){result[i-1]+=result[i]/10;result[i]%=10;}}/*for(int i=0;i<result.length;i++){System.out.print(result[i]+" ");}System.out.println();*/return result;}/** * 多项式乘法 * @param pa * @param pb * @return */private long[] polyMultiply(long [] pa,long[]pb){/* * expand pa and pb */int dim=1;while(dim<pa.length+pb.length){dim*=2;}long [] tmp=new long[dim];for(int i=0;i<pa.length;i++){tmp[tmp.length-i-1]=pa[pa.length-i-1];}pa=tmp;tmp=new long[dim];for(int i=0;i<pb.length;i++){tmp[tmp.length-i-1]=pb[pb.length-i-1];}pb=tmp;//System.out.println("dim = " + dim);long r=1;for(long i=0;i<MAX_POW/dim;i++){r=(r*R)%M;}//System.out.println("in polyMultiply : r = "+r);long[] psa=transform(pa,r);long[] psb=transform(pb,r);long[] psp=new long[psa.length];for(int i=0;i<psp.length;i++){psp[i]=(psa[i]*psb[i])%M;}long rr=1;for(int i=0;i<MAX_POW/dim;i++){rr=(rr*RP)%M;}long[] result=transform(psp,rr);for(int i=0;i<result.length;i++){result[i]/=dim;}return result;}/** * 变换 * 假设pa的长度是n=2^t * @param pa * @return */private long[] transform(long[] pa,long r){if(pa.length==1){long[] result=new long[1];result[0]=(r*pa[0])%M;//System.out.println("in transform : ");//System.out.println("r = "+r);//System.out.println("pa : "+pa[0]);//System.out.println("result : "+result[0]);return result;}//System.out.prlongln("r = "+r);long []pa1=new long[pa.length/2];long []pa2=new long[pa.length/2];for(int i=0;i<pa.length/2;i++){pa1[i]=pa[2*i];pa2[i]=pa[2*i+1];}//System.out.println("r*r%M = "+(long)((long)r*r)%M);long[] ps1=transform(pa1,(long)(((long)r)*r)%M);long[] ps2=transform(pa2,(long)(((long)r)*r)%M);long[]result=new long[pa.length];long tmpr=r;for(int i=0;i<result.length/2;i++){result[i]=(tmpr*ps1[i]+ps2[i])%M;result[i+result.length/2]=((-tmpr*ps1[i]+ps2[i])%M+M)%M;tmpr=(tmpr*r)%M;}//System.out.println("in transform");//System.out.println("r = "+r);//System.out.println("pa : ");/*for(int i=0;i<pa.length;i++){System.out.print(pa[i]+" ");}System.out.println();System.out.println("result : ");for(int i=0;i<result.length;i++){System.out.print(result[i]+" ");}System.out.println();*/return result;}}