找兩素數p和q
取n=p*q
取t=(p-1)*(q-1)
取任何壹個數e,要求滿足e<t並且e與t互素(就是最大公因數為1)
取d*e%t==1
這樣最終得到三個數: n d e
設消息為數M (M <n)
設c=(M**d)%n就得到了加密後的消息c
設m=(c**e)%n則 m == M,從而完成對c的解密。
註:**表示次方,上面兩式中的d和e可以互換。
在對稱加密中:
n d兩個數構成公鑰,可以告訴別人;
n e兩個數構成私鑰,e自己保留,不讓任何人知道。
給別人發送的信息使用e加密,只要別人能用d解開就證明信息是由妳發送的,構成了簽名機制。
別人給妳發送信息時使用d加密,這樣只有擁有e的妳能夠對其解密。
rsa的安全性在於對於壹個大數n,沒有有效的方法能夠將其分解
從而在已知n d的情況下無法獲得e;同樣在已知n e的情況下無法
求得d。
<二>實踐
接下來我們來壹個實踐,看看實際的操作:
找兩個素數:
p=47
q=59
這樣
n=p*q=2773
t=(p-1)*(q-1)=2668
取e=63,滿足e<t並且e和t互素
用perl簡單窮舉可以獲得滿主 e*d%t ==1的數d:
C:\Temp>perl -e "foreach $i (1..9999){ print($i),last if $i*63%2668==1 }"
847
即d=847
最終我們獲得關鍵的
n=2773
d=847
e=63
取消息M=244我們看看
加密:
c=M**d%n = 244**847%2773
用perl的大數計算來算壹下:
C:\Temp>perl -Mbigint -e "print 244**847%2773"
465
即用d對M加密後獲得加密信息c=465
解密:
我們可以用e來對加密後的c進行解密,還原M:
m=c**e%n=465**63%2773 :
C:\Temp>perl -Mbigint -e "print 465**63%2773"
244
即用e對c解密後獲得m=244 , 該值和原始信息M相等。
<三>字符串加密
把上面的過程集成壹下我們就能實現壹個對字符串加密解密的示例了。
每次取字符串中的壹個字符的ascii值作為M進行計算,其輸出為加密後16進制
的數的字符串形式,按3字節表示,如01F
代碼如下:
#!/usr/bin/perl -w
#RSA 計算過程學習程序編寫的測試程序
#watercloud 2003-8-12
#
use strict;
use Math::BigInt;
my %RSA_CORE = (n=>2773,e=>63,d=>847); #p=47,q=59
my $N=new Math::BigInt($RSA_CORE{n});
my $E=new Math::BigInt($RSA_CORE{e});
my $D=new Math::BigInt($RSA_CORE{d});
print "N=$N D=$D E=$E\n";
sub RSA_ENCRYPT
{
my $r_mess = shift @_;
my ($c,$i,$M,$C,$cmess);
for($i=0;$i < length($$r_mess);$i++)
{
$c=ord(substr($$r_mess,$i,1));
$M=Math::BigInt->new($c);
$C=$M->copy(); $C->bmodpow($D,$N);
$c=sprintf "%03X",$C;
$cmess.=$c;
}
return \$cmess;
}
sub RSA_DECRYPT
{
my $r_mess = shift @_;
my ($c,$i,$M,$C,$dmess);
for($i=0;$i < length($$r_mess);$i+=3)
{
$c=substr($$r_mess,$i,3);
$c=hex($c);
$M=Math::BigInt->new($c);
$C=$M->copy(); $C->bmodpow($E,$N);
$c=chr($C);
$dmess.=$c;
}
return \$dmess;
}
my $mess="RSA 娃哈哈哈~~~";
$mess=$ARGV[0] if @ARGV >= 1;
print "原始串:",$mess,"\n";
my $r_cmess = RSA_ENCRYPT(\$mess);
print "加密串:",$$r_cmess,"\n";
my $r_dmess = RSA_DECRYPT($r_cmess);
print "解密串:",$$r_dmess,"\n";
#EOF
測試壹下:
C:\Temp>perl rsa-test.pl
N=2773 D=847 E=63
原始串:RSA 娃哈哈哈~~~
加密串:5CB6CD6BC58A7709470AA74A0AA74A0AA74A6C70A46C70A46C70A4
解密串:RSA 娃哈哈哈~~~
C:\Temp>perl rsa-test.pl 安全焦點(xfocus)
N=2773 D=847 E=63
原始串:安全焦點(xfocus)
加密串:3393EC12F0A466E0AA9510D025D7BA0712DC3379F47D51C325D67B
解密串:安全焦點(xfocus)
<四>提高
前面已經提到,rsa的安全來源於n足夠大,我們測試中使用的n是非常小的,根本不能保障安全性,
我們可以通過RSAKit、RSATool之類的工具獲得足夠大的N 及D E。
通過工具,我們獲得1024位的N及D E來測試壹下:
n=0x328C74784DF31119C526D18098EBEBB943B0032B599CEE13CC2BCE7B5FCD15F90B66EC3A85F5005D
BDCDED9BDFCB3C4C265AF164AD55884D8278F791C7A6BFDAD55EDBC4F017F9CCF1538D4C2013433B383B
47D80EC74B51276CA05B5D6346B9EE5AD2D7BE7ABFB36E37108DD60438941D2ED173CCA50E114705D7E2
BC511951
d=0x10001
e=0xE760A3804ACDE1E8E3D7DC0197F9CEF6282EF552E8CEBBB7434B01CB19A9D87A3106DD28C523C2995
4C5D86B36E943080E4919CA8CE08718C3B0930867A98F635EB9EA9200B25906D91B80A47B77324E66AFF2
C4D70D8B1C69C50A9D8B4B7A3C9EE05FFF3A16AFC023731D80634763DA1DCABE9861A4789BD782A592D2B
1965
設原始信息
M=0x11111111111122222222222233333333333
完成這麽大數字的計算依賴於大數運算庫,用perl來運算非常簡單:
A) 用d對M進行加密如下:
c=M**d%n :
C:\Temp>perl -Mbigint -e " $x=Math::BigInt->bmodpow(0x11111111111122222222222233
333333333, 0x10001, 0x328C74784DF31119C526D18098EBEBB943B0032B599CEE13CC2BCE7B5F
CD15F90B66EC3A85F5005DBDCDED9BDFCB3C4C265AF164AD55884D8278F791C7A6BFDAD55EDBC4F0
17F9CCF1538D4C2013433B383B47D80EC74B51276CA05B5D6346B9EE5AD2D7BE7ABFB36E37108DD6
0438941D2ED173CCA50E114705D7E2BC511951);print $x->as_hex"
0x17b287be418c69ecd7c39227ab681ac422fcc84bb35d8a632543b304de288a8d4434b73d2576bd
45692b007f3a2f7c5f5aa1d99ef3866af26a8e876712ed1d4cc4b293e26bc0a1dc67e247715caa6b
3028f9461a3b1533ec0cb476441465f10d8ad47452a12db0601c5e8beda686dd96d2acd59ea89b91
f1834580c3f6d90898
即用d對M加密後信息為:
c=0x17b287be418c69ecd7c39227ab681ac422fcc84bb35d8a632543b304de288a8d4434b73d2576bd
45692b007f3a2f7c5f5aa1d99ef3866af26a8e876712ed1d4cc4b293e26bc0a1dc67e247715caa6b
3028f9461a3b1533ec0cb476441465f10d8ad47452a12db0601c5e8beda686dd96d2acd59ea89b91
f1834580c3f6d90898
B) 用e對c進行解密如下:
m=c**e%n :
C:\Temp>perl -Mbigint -e " $x=Math::BigInt->bmodpow(0x17b287be418c69ecd7c39227ab
681ac422fcc84bb35d8a632543b304de288a8d4434b73d2576bd45692b007f3a2f7c5f5aa1d99ef3
866af26a8e876712ed1d4cc4b293e26bc0a1dc67e247715caa6b3028f9461a3b1533ec0cb4764414
65f10d8ad47452a12db0601c5e8beda686dd96d2acd59ea89b91f1834580c3f6d90898, 0xE760A
3804ACDE1E8E3D7DC0197F9CEF6282EF552E8CEBBB7434B01CB19A9D87A3106DD28C523C29954C5D
86B36E943080E4919CA8CE08718C3B0930867A98F635EB9EA9200B25906D91B80A47B77324E66AFF
2C4D70D8B1C69C50A9D8B4B7A3C9EE05FFF3A16AFC023731D80634763DA1DCABE9861A4789BD782A
592D2B1965, 0x328C74784DF31119C526D18098EBEBB943B0032B599CEE13CC2BCE7B5FCD15F90
B66EC3A85F5005DBDCDED9BDFCB3C4C265AF164AD55884D8278F791C7A6BFDAD55EDBC4F017F9CCF
1538D4C2013433B383B47D80EC74B51276CA05B5D6346B9EE5AD2D7BE7ABFB36E37108DD60438941
D2ED173CCA50E114705D7E2BC511951);print $x->as_hex"
0x11111111111122222222222233333333333
(我的P4 1.6G的機器上計算了約5秒鐘)
得到用e解密後的m=0x11111111111122222222222233333333333 == M
C) RSA通常的實現
RSA簡潔幽雅,但計算速度比較慢,通常加密中並不是直接使用RSA 來對所有的信息進行加密,
最常見的情況是隨機產生壹個對稱加密的密鑰,然後使用對稱加密算法對信息加密,之後用
RSA對剛才的加密密鑰進行加密。
最後需要說明的是,當前小於1024位的N已經被證明是不安全的
自己使用中不要使用小於1024位的RSA,最好使用2048位的。
----------------------------------------------------------
壹個簡單的RSA算法實現JAVA源代碼:
filename:RSA.java
/*
* Created on Mar 3, 2005
*
* TODO To change the template for this generated file go to
* Window - Preferences - Java - Code Style - Code Templates
*/
import java.math.BigInteger;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.FileWriter;
import java.io.FileReader;
import java.io.BufferedReader;
import java.util.StringTokenizer;
/**
* @author Steve
*
* TODO To change the template for this generated type comment go to
* Window - Preferences - Java - Code Style - Code Templates
*/
public class RSA {
/**
* BigInteger.ZERO
*/
private static final BigInteger ZERO = BigInteger.ZERO;
/**
* BigInteger.ONE
*/
private static final BigInteger ONE = BigInteger.ONE;
/**
* Pseudo BigInteger.TWO
*/
private static final BigInteger TWO = new BigInteger("2");
private BigInteger myKey;
private BigInteger myMod;
private int blockSize;
public RSA (BigInteger key, BigInteger n, int b) {
myKey = key;
myMod = n;
blockSize = b;
}
public void encodeFile (String filename) {
byte[] bytes = new byte[blockSize / 8 + 1];
byte[] temp;
int tempLen;
InputStream is = null;
FileWriter writer = null;
try {
is = new FileInputStream(filename);
writer = new FileWriter(filename + ".enc");
}
catch (FileNotFoundException e1){
System.out.println("File not found: " + filename);
}
catch (IOException e1){
System.out.println("File not found: " + filename + ".enc");
}
/**
* Write encoded message to 'filename'.enc
*/
try {
while ((tempLen = is.read(bytes, 1, blockSize / 8)) > 0) {
for (int i = tempLen + 1; i < bytes.length; ++i) {
bytes[i] = 0;
}
writer.write(encodeDecode(new BigInteger(bytes)) + " ");
}
}
catch (IOException e1) {
System.out.println("error writing to file");
}
/**
* Close input stream and file writer
*/
try {
is.close();
writer.close();
}
catch (IOException e1) {
System.out.println("Error closing file.");
}
}
public void decodeFile (String filename) {
FileReader reader = null;
OutputStream os = null;
try {
reader = new FileReader(filename);
os = new FileOutputStream(filename.replaceAll(".enc", ".dec"));
}
catch (FileNotFoundException e1) {
if (reader == null)
System.out.println("File not found: " + filename);
else
System.out.println("File not found: " + filename.replaceAll(".enc", "dec"));
}
BufferedReader br = new BufferedReader(reader);
int offset;
byte[] temp, toFile;
StringTokenizer st = null;
try {
while (br.ready()) {
st = new StringTokenizer(br.readLine());
while (st.hasMoreTokens()){
toFile = encodeDecode(new BigInteger(st.nextToken())).toByteArray();
System.out.println(toFile.length + " x " + (blockSize / 8));
if (toFile[0] == 0 && toFile.length != (blockSize / 8)) {
temp = new byte[blockSize / 8];
offset = temp.length - toFile.length;
for (int i = toFile.length - 1; (i <= 0) && ((i + offset) <= 0); --i) {
temp[i + offset] = toFile[i];
}
toFile = temp;
}
/*if (toFile.length != ((blockSize / 8) + 1)){
temp = new byte[(blockSize / 8) + 1];
System.out.println(toFile.length + " x " + temp.length);
for (int i = 1; i < temp.length; i++) {
temp[i] = toFile[i - 1];
}
toFile = temp;
}
else
System.out.println(toFile.length + " " + ((blockSize / 8) + 1));*/
os.write(toFile);
}
}
}
catch (IOException e1) {
System.out.println("Something went wrong");
}
/**
* close data streams
*/
try {
os.close();
reader.close();
}
catch (IOException e1) {
System.out.println("Error closing file.");
}
}
/**
* Performs <tt>base</tt>^<sup><tt>pow</tt></sup> within the modular
* domain of <tt>mod</tt>.
*
* @param base the base to be raised
* @param pow the power to which the base will be raisded
* @param mod the modular domain over which to perform this operation
* @return <tt>base</tt>^<sup><tt>pow</tt></sup> within the modular
* domain of <tt>mod</tt>.
*/
public BigInteger encodeDecode(BigInteger base) {
BigInteger a = ONE;
BigInteger s = base;
BigInteger n = myKey;
while (!n.equals(ZERO)) {
if(!n.mod(TWO).equals(ZERO))
a = a.multiply(s).mod(myMod);
s = s.pow(2).mod(myMod);
n = n.divide(TWO);
}
return a;
}
}
在這裏提供兩個版本的RSA算法JAVA實現的代碼下載:
1. 來自於 /code.aspx?ID=27020 的RSA算法實現源代碼包:
/rsa/