给定两个表示两个整数值的二进制字符串，找到两个字符串的乘积。例如，如果第一位字符串是“ 1100”，第二位字符串是“ 1010”，则输出应为120。
为简单起见，让两个字符串的长度相同且为n。
天真的方法是遵循我们在学校学习的过程。逐个取第二个数字的所有位，然后将其与第一个数字的所有位相乘。最后添加所有乘法。该算法花费O(n ^ 2)时间。
使用Divide和Conquer ，我们可以将两个整数相乘，从而降低了时间复杂度。我们将给定的数字分为两半。让给定的数字为X和Y。
为简单起见，我们假设n为偶数
X = Xl*2n/2 + Xr [Xl and Xr contain leftmost and rightmost n/2 bits of X]
Y = Yl*2n/2 + Yr [Yl and Yr contain leftmost and rightmost n/2 bits of Y]乘积XY可以写成如下形式。
XY = (Xl*2n/2 + Xr)(Yl*2n/2 + Yr)
= 2n XlYl + 2n/2(XlYr + XrYl) + XrYr如果我们看上面的公式，有四个大小为n / 2的乘法，所以我们基本上将大小为n / 2的问题分为四个大小为n / 2的子问题。但这无济于事，因为递归T(n)= 4T(n / 2)+ O(n)的解是O(n ^ 2)。该算法的棘手部分是将中间的两个项更改为其他形式，以便仅执行一次额外的乘法就足够了。以下是中间两个术语的棘手表达式。
XlYr + XrYl = (Xl + Xr)(Yl + Yr) - XlYl- XrYr因此XY的最终值变为
XY = 2n XlYl + 2n/2 * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr有了以上技巧，递归变为T(n)= 3T(n / 2)+ O(n)，并且此递归的解为O(n 1.59 )。
如果输入字符串的长度不同并且不相等怎么办?为了处理不同长度的情况，我们在开头添加0。为了处理奇数长度，我们将floor(n / 2)位放在左半部分，将ceil(n / 2)位放在右半部分。因此，XY的表达式变为以下表达式。
XY = 22ceil(n/2) XlYl + 2ceil(n/2) * [(Xl + Xr)(Yl + Yr) - XlYl - XrYr] + XrYr上面的算法称为Karatsuba算法，它可以用于任何基础。
以下是上述算法的C++实现。
// C++ implementation of Karatsuba algorithm for bit string multiplication.
#include
#include
using namespace std;
// FOLLOWING TWO FUNCTIONS ARE COPIED FROM http://goo.gl/q0OhZ
// Helper method: given two unequal sized bit strings, converts them to
// same length by adding leading 0s in the smaller string. Returns the
// the new length
int makeEqualLength(string &str1, string &str2)
{
int len1 = str1.size();
int len2 = str2.size();
if (len1 < len2)
{
for (int i = 0 ; i < len2 - len1 ; i++)
str1 = '0' + str1;
return len2;
}
else if (len1 > len2)
{
for (int i = 0 ; i < len1 - len2 ; i++)
str2 = '0' + str2;
}
return len1; // If len1 >= len2
}
// The main function that adds two bit sequences and returns the addition
string addBitStrings( string first, string second )
{
string result; // To store the sum bits
// make the lengths same before adding
int length = makeEqualLength(first, second);
int carry = 0; // Initialize carry
// Add all bits one by one
for (int i = length-1 ; i >= 0 ; i--)
{
int firstBit = first.at(i) - '0';
int secondBit = second.at(i) - '0';
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry)+'0';
result = (char)sum + result;
// boolean expression for 3-bit addition
carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);
}
// if overflow, then add a leading 1
if (carry) result = '1' + result;
return result;
}
// A utility function to multiply single bits of strings a and b
int multiplyiSingleBit(string a, string b)
{ return (a[0] - '0')*(b[0] - '0'); }
// The main function that multiplies two bit strings X and Y and returns
// result as long integer
long int multiply(string X, string Y)
{
// Find the maximum of lengths of x and Y and make length
// of smaller string same as that of larger string
int n = makeEqualLength(X, Y);
// Base cases
if (n == 0) return 0;
if (n == 1) return multiplyiSingleBit(X, Y);
int fh = n/2; // First half of string, floor(n/2)
int sh = (n-fh); // Second half of string, ceil(n/2)
// Find the first half and second half of first string.
// Refer http://goo.gl/lLmgn for substr method
string Xl = X.substr(0, fh);
string Xr = X.substr(fh, sh);
// Find the first half and second half of second string
string Yl = Y.substr(0, fh);
string Yr = Y.substr(fh, sh);
// Recursively calculate the three products of inputs of size n/2
long int P1 = multiply(Xl, Yl);
long int P2 = multiply(Xr, Yr);
long int P3 = multiply(addBitStrings(Xl, Xr), addBitStrings(Yl, Yr));
// Combine the three products to get the final result.
return P1*(1<<(2*sh)) + (P3 - P1 - P2)*(1< 输出：
120
60
30
10
0
49
9时间复杂度：上述解决方案的时间复杂度为O(n log 2 3 )= O(n 1.59 )。
使用另一种分而治之算法，快速傅里叶变换，可以进一步提高乘法的时间复杂度。我们很快将在单独的文章中讨论快速傅立叶变换。
锻炼
上面的程序返回一个long int值，不适用于大字符串。扩展上述程序以返回字符串而不是long int值。
解决方案
大量的乘法过程是计算机科学中的一个重要问题。给定的方法使用分而治之方法。
运行代码以查看常规二进制乘法和Karatsuba算法的时间复杂度比较。
您可以在此存储库中看到完整的代码
例子：
Fist Binary Input : 101001010101010010101001010100101010010101010010101
Second Binary Input : 101001010101010010101001010100101010010101010010101
Decimal Output : Not Representable
Output : 2.1148846e+30
Fist Binary Input : 1011
Second Binary Input : 1000
Decimal Output : 88
Output : 5e-05
#include
#include
#include
#include
#include
#include
using namespace std;
// classical method class
class BinaryMultiplier
{
public:
string MakeMultiplication(string,string);
string MakeShifting(string,int);
string addBinary(string,string);
void BinaryStringToDecimal(string);
};
// karatsuba method class
class Karatsuba
{
public:
int lengthController(string &,string &);
string addStrings(string,string);
string multiply(string,string);
string DecimalToBinary(long long int);
string Subtraction(string,string);
string MakeShifting(string,int);
};
// this function get strings and go over str2 bit bit
// if it sees 1 it calculates the shifted version according to position bit
// Makes add operation for binary strings
// returns result string
string BinaryMultiplier::MakeMultiplication(string str1, string str2)
{
string allSum = "";
for (int j = 0 ; j= 0 || j >= 0 || s == 1)
{
s += ((i >= 0)? a[i] - '0': 0);
s += ((j >= 0)? b[j] - '0': 0);
result = char(s % 2 + '0') + result;
s /= 2;
i--;
j--;
}
return result;
}
// this function shifts the given string according to given number
// returns shifted version
string BinaryMultiplier::MakeShifting(string str, int stepnum)
{
string shifted = str;
for (int i = 0 ; i < stepnum ; i++)
shifted = shifted + '0';
return shifted;
}
// this function converts Binary String Number to Decimal Number
// After 32 bits it gives 0 because it overflows the size of int
void BinaryMultiplier::BinaryStringToDecimal(string result)
{
cout<<"Binary Result : "<= 0; i--)
{
if (result[i] == '1')
{
val += pow(2,(result.length()-1)-i);
}
}
cout<<"Decimal Result (Not proper for Large Binary Numbers):" < len2)
{
for (int i = 0 ; i < len1 - len2 ; i++)
str2 = '0' + str2;
}
return len1;
}
// this function add strings with carry
// uses one by one bit addition methodology
// returns result string
string Karatsuba::addStrings(string first, string second)
{
string result; // To store the sum bits
// make the lengths same before adding
int length = lengthController(first, second);
int carry = 0; // Initialize carry
// Add all bits one by one
for (int i = length-1 ; i >= 0 ; i--)
{
int firstBit = first.at(i) - '0';
int secondBit = second.at(i) - '0';
// boolean expression for sum of 3 bits
int sum = (firstBit ^ secondBit ^ carry)+'0';
result = (char)sum + result;
// Boolean expression for 3-bit addition
carry = (firstBit&secondBit) | (secondBit&carry) | (firstBit&carry);
}
// if overflow, then add a leading 1
if (carry)
{
result = '1' + result;
}
return result;
}
// this function converts decimal number to binary string
string Karatsuba::DecimalToBinary(long long int number)
{
string result = "";
if (number <= 0)
{
return "0";
}
else
{
int i = 0;
while (number > 0)
{
long long int num= number % 2;
stringstream ss;
ss<= 0; i--)
{
diff = (lhs[i]-'0') - (rhs[i]-'0');
if (diff >= 0)
{
result = DecimalToBinary(diff) + result;
}
else
{
for (int j = i-1; j>=0; j--)
{
lhs[j] = ((lhs[j]-'0') - 1) % 10 + '0';
if (lhs[j] != '1')
{
break;
}
}
result = DecimalToBinary(diff+2) + result;
}
}
return result;
}
// this function makes shifting
string Karatsuba::MakeShifting(string str, int stepnum)
{
string shifted = str;
for (int i = 0 ; i < stepnum ; i++)
shifted = shifted + '0';
return shifted;
}
// this function is the core of the Karatsuba
// divides problem into 4 subproblems
// recursively multiplies them
// returns the result string
string Karatsuba::multiply(string X, string Y)
{
int n = lengthController(X, Y);
if (n == 1) return ((Y[0]-'0' == 1) && (X[0]-'0' == 1)) ? "1" : "0";
int fh = n/2; // First half of string, floor(n/2)
int sh = (n-fh); // Second half of string, ceil(n/2)
// Find the first half and second half of first string.
string Xl = X.substr(0, fh);
string Xr = X.substr(fh, sh);
// Find the first half and second half of second string
string Yl = Y.substr(0, fh);
string Yr = Y.substr(fh, sh);
// Recursively calculate the three products of inputs of size n/2
string P1 = multiply(Xl, Yl);
string P2 = multiply(Xr, Yr);
string P3 = multiply(addStrings(Xl, Xr), addStrings(Yl, Yr));
// return added string version
return addStrings(addStrings(MakeShifting(P1, 2*(n-n/2)),P2),MakeShifting(Subtraction(P3,addStrings(P1,P2)), n-(n/2)));
}
int main(int argc, const char * argv[])
{
// get the binary numbers as strings
string firstNumber,secondNumber;
cout<<"Please give the First Binary number : ";
cin>>firstNumber;
cout<>secondNumber;
cout << endl;
// make the initial lengths equal by adding zeros
int len1 = firstNumber.size();
int len2 = secondNumber.size();
int general_len = firstNumber.size();
if (len1 < len2)
{
for (int i = 0 ; i < len2 - len1 ; i++)
firstNumber = '0' + firstNumber;
general_len = firstNumber.size();
}
else if (len1 > len2)
{
for (int i = 0 ; i < len1 - len2 ; i++)
secondNumber = '0' + secondNumber;
general_len = secondNumber.size();
}
// In classical methodology Binary String Multiplication
cout<<"Classical Algorithm : "< 相关文章：
大数表示为字符串
参考：
Karatsuba算法的维基百科页面
算法第一版，作者：Sanjoy Dasgupta，Christos Papadimitriou和Umesh Vazirani
http://courses.csail.mit.edu/6.006/spring11/exams/notes3-karatsuba
http://www.cc.gatech.edu/~ninamf/Algos11/lectures/lect0131.pdf