如何在没有“*”运算符的情况下执行乘法？

Question

I was just going through some basic stuff as I am learning C. I came upon a question to multiply a number by 7 without using the * operator. Basically it's like this

      (x << 3) - x;

Now I know about basic bit manipulation operations, but I can't get how do you multiply a number by any other odd number without using the * operator? Is there a general algorithm for this?

Answer 1

Think about how you multiply in decimal using pencil and paper:

  12
x 26
----
  72
 24
----
 312

What does multiplication look like in binary?

   0111
x  0101
-------
   0111
  0000
 0111
-------
 100011

Notice anything? Unlike multiplication in decimal, where you need to memorize the "times table," when multiplying in binary, you are always multiplying one of the terms by either 0 or 1 before writing it down in the list addends. There's no times table needed. If the digit of the second term is 1, you add in the first term. If it's 0, you don't. Also note how the addends are progressively shifted over to the left.

If you're unsure of this, do a few binary multiplications on paper. When you're done, convert the result back to decimal and see if it's correct. After you've done a few, I think you'll get the idea how binary multiplication can be implemented using shifts and adds.

Answer 2

Everyone is overlooking the obvious. No multiplication is involved:

10^(log10(A) + log10(B))

Answer 3

The question says:

multiply a number by 7 without using * operator

This doesn't use * :

 number / (1 / 7) 

Edit:
This compiles and works fine in C:

 int number,result;
 number = 8;
 result = number / (1. / 7);
 printf("result is %d\n",result);

Answer 4

An integer left shift is multiplying by 2, provided it doesn't overflow. Just add or subtract as appropriate once you get close.

Answer 5

int multiply(int multiplicand, int factor)
{
    if (factor == 0) return 0;

    int product = multiplicand;
    for (int ii = 1; ii < abs(factor); ++ii) {
        product += multiplicand;
    }

    return factor >= 0 ? product : -product;
}

You wanted multiplication without * , you got it, pal!

Answer 6

It's easy to avoid the '*' operator:

mov eax, 1234h
mov edx, 5678h
imul edx

No '*' in sight. Of course, if you wanted to get into the spirit of it, you could also use the trusty old shift and add algorithm:

mult proc
; Multiplies eax by ebx and places result in edx:ecx
    xor ecx, ecx
    xor edx, edx
mul1:
    test ebx, 1
    jz  mul2
    add ecx, eax
    adc edx, 0
mul2:
    shr ebx, 1
    shl eax, 1
    test ebx, ebx
    jnz  mul1
done:
    ret
mult endp

Of course, with modern processors, all (?) have multiplication instructions, but back when the PDP-11 was shiny and new, code like this saw real use.

Answer 7

Mathematically speaking, multiplication distributes over addition. Essentially, this means:

x * (a + b + c ...) = (x * a) + (x * b) + (x * c) ...

Any real number (in your case 7 ), can be presented as a series of additions (such as 8 + (-1) , since subtraction is really just addition going the wrong way). This allows you to represent any single multiplication statement as an equivalent series of multiplication statements, which will come up with the same result:

x * 7
= x * (8 + (-1))
= (x * 8) + (x * (-1))
= (x * 8) - (x * 1)
= (x * 8) - x

The bitwise shift operator essentially just multiplies or divides a number by a power of 2. So long as your equation is only dealing with such values, bit shifting can be used to replace all occurrence of the multiplication operator.

(x * 8) - x = (x * 2 ³ ) - x = (x << 3) - x

A similar strategy can be used on any other integer, and it makes no difference whether it's odd or even.

Answer 8

等同于x*8-x = x*(8-1) = x*7

Answer 9

Any number, odd or even, can be expressed as a sum of powers of two. For example,

     1   2   4   8
------------------
 1 = 1
 2 = 0 + 2
 3 = 1 + 2
 4 = 0 + 0 + 4
 5 = 1 + 0 + 4
 6 = 0 + 2 + 4
 7 = 1 + 2 + 4
 8 = 0 + 0 + 0 + 8
11 = 1 + 2 + 0 + 8

So, you can multiply x by any number by performing the right set of shifts and adds.

 1x = x
 2x = 0 + x<<1
 3x = x + x<<1
 4x = 0 +  0   + x<<2
 5x = x +  0   + x<<2
11x = x + x<<1 +   0  + x<<3

Answer 10

When it comes down to it, multiplication by a positive integer can be done like this:

int multiply(int a, int b) {
  int ret = 0;
  for (int i=0; i<b; i++) {
    ret += b;
  }
  return ret;
}

Efficient? Hardly. But it's correct (factoring in limits on ints and so forth).

So using a left-shift is just a shortcut for multiplying by 2. But once you get to the highest power-of-2 under b you just add a the necessary number of times, so:

int multiply(int a, int b) {
  int ret = a;
  int mult = 1;
  while (mult <= b) {
    ret <<= 1;
    mult <<= 1;
  }
  while (mult < b) {
    ret += a;
  }
  return ret;
}

or something close to that.

To put it another way, to multiply by 7.

• Left shift by 2 (times 4). Left shift 3 is 8 which is >7;
• Add b 3 times.

Answer 11

One evening, I found that I was extremely bored, and cooked this up:

#include <iostream>

typedef unsigned int uint32;

uint32 add(uint32 a, uint32 b) {
    do {
        uint32 s = a ^ b;
        uint32 c = a & b;
        a = s;
        b = c << 1;
    } while (a & b)
    return (a | b)
}

uint32 mul(uint32 a, uint32 b) {
    uint32 total = 0;
    do {
        uint32 s1 = a & (-(b & 1))
        b >>= 1; a <<= 1;
        total = add(s1, total)
    } while (b)
    return total;
}

int main(void) {
    using namespace std;
    uint32 a, b;

    cout << "Enter two numbers to be multiplied: ";
    cin >> a >> b;

    cout << "Total: " << mul(a,b) << endl;
    return 0;
}

The code above should be quite self-explanatory, as I tried to keep it as simple as possible. It should work, more or less, the way a CPU might perform these operations. The only bug I'm aware of is that a is not permitted to be greater than 32,767 and b is not permitted to be large enough to overflow a (that is, multiply overflow is not handled, so 64-bit results are not possible). It should even work with negative numbers, provided the inputs are appropriately reinterpret_cast<> .

Answer 12

O(log(b)) method

public int multiply_optimal(int a, int b) {

    if (a == 0 || b == 0)
        return 0;
    if (b == 1)
        return a;
    if ((b & 1) == 0)
        return multiply_optimal(a + a, b >> 1);
    else
        return a + multiply_optimal(a + a, b >> 1);

}

The resursive code works as follows:
Base case:
if either of the number is 0 ,product is 0.
if b=1, product =a.

If b is even:
ab can be written as 2a(b/2)
2a(b/2)=(a+a) (b/2)=(a+a) (b>>1) where'>>' arithematic right shift operator in java.

If b is odd:
ab can be written as a+a(b-1)
a+a(b-1)=a+2a(b-1)/2=a+(a+a)(b-1)/2=a+(a+a)((b-1)>>1)
Since b is odd (b-1)/2=b/2=b>>1
So ab=a+(2a*(b>>1))
NOTE:each recursive call b is halved => O(log(b))

Answer 13

unsigned int Multiply(unsigned int m1, unsigned int m2)
{
    unsigned int numBits = sizeof(unsigned int) * 8; // Not part of the core algorithm
    unsigned int product = 0;
    unsigned int mask = 1;
    for(int i =0; i < numBits; ++i, mask = mask << 1)
    {
        if(m1 & mask)
        {
            product += (m2 << i);
        }
    }
    return product;
}

Answer 14

@Wang, that's a good generalization. But here is a slightly faster version. But it assumes no overflow and a is non-negative.

int mult(int a, int b){
    int p=1;
    int rv=0;
    for(int i=0; a >= p && i < 31; i++){
        if(a & p){
            rv += b;
        }
        p = p << 1;
        b = b << 1;
    }

    return rv;
}

It will loop at most 1+log_2(a) times. Could be faster if you swap a and b when a > b.

Answer 15

unsigned int Multiply( unsigned int a, unsigned int b )
{
    int ret = 0;
    // For each bit in b
    for (int i=0; i<32; i++) {

        // If that bit is not equal to zero
        if (( b & (1 << i)) != 0) {

            // Add it to our return value
            ret += a << i;
        }
    }
    return ret;
}

I avoided the sign bit, because it's kind of not the subject of the post. This is an implementation of what Wayne Conrad said basically.Here is another problem is you want to try more low level math operations. Project Euler is cool!

Answer 16

Shift and add doesn't work (even with sign extension) when the multiplicand is negative. Signed multiplication has to be done using Booth encoding :

Starting from the LSB , a change from 0 to 1 is -1; a change from 1 to 0 is 1, otherwise 0. There is also an implicit extra bit 0 below the LSB.

For example, the number 5 (0101) will be encoded as: (1)(-1)(1)(-1). You can verify this is correct:

5 = 2^3 - 2^2 + 2 -1

This algorithm also works with negative numbers in 2's complement form:

-1 in 4-bit 2's complement is 1111. Using the Booth algorithm: (1)(0)(0)(0)(-1), where there is no space for the leftmost bit 1 so we get: (0)(0)(0)(-1) which is -1.

/* Multiply two signed integers using the Booth algorithm */
int booth(int x, int y)
{
    int prev_bit = 0;
    int result = 0;

    while (x != 0) {
        int current_bit = x & 0x1;
        if (prev_bit & ~current_bit) {
            result += y;
        } else if (~prev_bit & current_bit) {
            result -= y;
        }

        prev_bit = current_bit;

        x = static_cast<unsigned>(x) >> 1;
        y <<= 1;
    }

    if (prev_bit)
        result += y;

    return result;
}

The above code does not check for overflow. Below is a slightly modified version that multiplies two 16 bit numbers and returns a 32 bit number so it never overflows:

/* Multiply two 16-bit signed integers using the Booth algorithm */
/* Returns a 32-bit signed integer */
int32_t booth(int16_t x, int16_t y)
{
    int16_t prev_bit = 0;
    int16_t sign_bit = (x >> 16) & 0x1;
    int32_t result = 0;
    int32_t y1 = static_cast<int32_t>(y);

    while (x != 0) {
        int16_t current_bit = x & 0x1;
        if (prev_bit & ~current_bit) {
            result += y1;
        } else if (~prev_bit & current_bit) {
            result -= y1;
        }

        prev_bit = current_bit;

        x = static_cast<uint16_t>(x) >> 1;
        y1 <<= 1;
    }

    if (prev_bit & ~sign_bit)
        result += y1;

    return result;
}

Answer 17

import java.math.BigInteger;

public class MultiplyTest {
    public static void main(String[] args) {
        BigInteger bigInt1 = new BigInteger("5");
        BigInteger bigInt2 = new BigInteger("8");
        System.out.println(bigInt1.multiply(bigInt2));
    }
}

Answer 18

In C#:

private static string Multi(int a, int b)
{
    if (a == 0 || b == 0)
        return "0";

    bool isnegative = false;

    if (a < 0 || b < 0)
    {
        isnegative = true;

        a = Math.Abs(a);

        b = Math.Abs(b);
    }

    int sum = 0;

    if (a > b)
    {
        for (int i = 1; i <= b; i++)
        {
            sum += a;
        }
    }
    else
    {
        for (int i = 1; i <= a; i++)
        {
            sum += b;
        }
    }

    if (isnegative == true)
        return "-" + sum.ToString();
    else
        return sum.ToString();
}

Answer 19

JAVA:
Considering the fact, that every number can be splitted into powers of two:

1 = 2 ^ 0
2 = 2 ^ 1
3 = 2 ^ 1 + 2 ^ 0
...

We want to get x where:
x = n * m

So we can achieve that by doing following steps:

1.   while m is greater or equal to 2^pow:
     1.1  get the biggest number pow, such as 2^pow is lower or equal to m
     1.2  multiply n*2^pow and decrease m to m-2^pow
2.   sum the results

Sample implementation using recursion:

long multiply(int n, int m) {
    int pow = 0;
    while (m >= (1 << ++pow)) ;
    pow--;
    if (m == 1 << pow) return (n << pow);
    return (n << pow) + multiply(n, m - (1 << pow));
}

I got this question in last job interview and this answer was accepted.

EDIT: solution for positive numbers

Answer 20

这是最简单的正数 C99/C11 解决方案：

unsigned multiply(unsigned x, unsigned y) { return sizeof(char[x][y]); }

Answer 21

If you can use the log function:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    //Find the 2^b which is larger than "a" which turns out to be the 
    //ceiling of (Log base 2 of b) == numbers of digits to shift
    double logBase2 = Math.log(absB) / Math.log(2);
    long bits = (long)Math.ceil(logBase2);

    //Get the value of 2^bits
    long biggerInteger = (int)Math.pow(2, bits);

    //Find the difference of the bigger integer and "b"
    long difference = biggerInteger - absB;

    //Shift "bits" places to the left
    long result = absA<<bits;

    //Subtract the "difference" "a" times
    int diffLoop = Math.abs(a);
    while (diffLoop>0) {
        result -= difference;
        diffLoop--;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

If you cannot use the log function:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    //Get the number of bits for a 2^(b+1) larger number
    int bits = 0;
    int bitInteger = absB;
    while (bitInteger>0) {
        bitInteger /= 2;
        bits++;
    }

    //Get the value of 2^bit
    long biggerInteger = (int)Math.pow(2, bits);

    //Find the difference of the bigger integer and "b"
    long difference = biggerInteger - absB;

    //Shift "bits" places to the left
    long result = absA<<bits;

    //Subtract the "difference" "a" times
    int diffLoop = absA;
    while (diffLoop>0) {
        result -= difference;
        diffLoop--;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

I found this to be more efficient:

public static final long multiplyUsingShift(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);

    long result = 0L;
    while (absA>0) {
        if ((absA&1)>0) result += absB; //Is odd
        absA >>= 1;
        absB <<= 1;
    }

    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

and yet another way.

public static final long multiplyUsingLogs(int a, int b) {
    int absA = Math.abs(a);
    int absB = Math.abs(b);
    long result = Math.round(Math.pow(10, (Math.log10(absA)+Math.log10(absB))));
    return (a>0&&b>0 || a<0&&b<0)?result:-result;
}

Answer 22

package com.amit.string;

// Here I am passing two values, 7 and 3 and method getResult() will
// return 21 without use of any operator except the increment operator, ++.
//
public class MultiplyTwoNumber {

    public static void main(String[] args) {
        int a = 7;
        int b = 3;
        System.out.println(new MultiplyTwoNumber().getResult(a, b));
    }

    public int getResult(int i, int j) {
        int result = 0;

        // Check for loop logic it is key thing it will go 21 times
        for (int k = 0; k < i; k++) {
            for (int p = 0; p < j; p++) {
                result++;
            }
        }
        return result;
    }
}

Answer 23

public static int multiply(int a, int b) 
{
    int temp = 0;
    if (b == 0) return 0;
    for (int ii = 0; ii < abs(b); ++ii) {
        temp = temp + a;
    }

    return b >= 0 ? temp : -temp;
}

public static int abs(int val) {

    return val>=0 ? val : -val;
}

Answer 24

Another thinking-outside-the-box answer:

BigDecimal a = new BigDecimal(123);
BigDecimal b = new BigDecimal(2);
BigDecimal result = a.multiply(b);
System.out.println(result.intValue());

Answer 25

Loop it. Run a loop seven times and iterate by the number you are multiplying with seven.

Pseudocode:

total = 0
multiply = 34

loop while i < 7

    total = total + multiply

endloop

Answer 26

public static void main(String[] args) {
    System.out.print("Enter value of A -> ");
    Scanner s=new Scanner(System.in);
    double j=s.nextInt();
    System.out.print("Enter value of B -> ");
    Scanner p=new Scanner(System.in);
    double k=p.nextInt();
    double m=(1/k);
    double l=(j/m);
    System.out.print("Multiplication of A & B=> "+l);
}

Answer 27

Let N be the number that we want to multiply with 7.

N x 7 = N + N + N + N + N + N + N
N x 7 = N + N + N + N + N + N + N + (N - N)
N x 7 = (N + N + N + N + N + N + N + N) - N
N x 7 = 8xN - N

As we know that, left shifting any number by one bit multiply it by 2. Hence, multiplying any number with 8 is equivalent to right shifting it by 3 bits

N x 7 = (N << 3) - N 

I found this description here http://www.techcrashcourse.com/2016/02/c-program-to-multiply-number-by-7-bitwise-operator.html

Hope it helps.

Answer 28

By making use of recursion, we can multiply two integers with the given constraints.

To multiply x and y, recursively add xy times.

     #include<stdio.h>
     /* function to multiply two numbers x and y*/
     int multiply(int x, int y)
     {
        /* multiplied with anything gives */
        if(y == 0)
        return 0;

        /* Add x one by one */
        if(y > 0 )
        return (x + multiply(x, y-1));

        /* the case where y is negative */
        if(y < 0 )
        return -multiply(x, -y);
     }

     int main()
     {
       printf("\n %d", multiply(5, -11));
       getchar();
       return 0;
     }

Answer 29

A JavaScript approach for positive numbers

function recursiveMultiply(num1, num2){
    const bigger = num1 > num2 ? num1 : num2; 
    const smaller = num1 <= num2 ? num1 : num2; 
    const indexIncrement = 1;
    const resultIncrement = bigger;

    return recursiveMultiplyHelper(bigger, smaller, 0, indexIncrement, resultIncrement)
}

function recursiveMultiplyHelper(num1, num2, index, indexIncrement, resultIncrement){
    let result = 0;
    if (index === num2){
        return result;
    } 

    if ((index+indexIncrement+indexIncrement) >= num2){
        indexIncrement = 1;
        resultIncrement = num1;
    } else{
        indexIncrement += indexIncrement;
        resultIncrement += resultIncrement;
    }

    result = recursiveMultiplyHelper(num1, num2, (index+indexIncrement), indexIncrement, resultIncrement);
    result += resultIncrement;
    console.log(num1, num2, index, result);

    return result;
}

Answer 30

Think about the normal multiplication method we use

         1101 x        =>13
         0101          =>5
---------------------
         1101
        0000
       1101
      0000
===================        
      1000001 .        => 65

Writing the same above in the code

#include<stdio.h>

int multiply(int a, int b){
    int res = 0,count =0;
    while(b>0) {
        if(b & 0x1)
            res = res + (a << count);
        b = b>>1;
        count++;
    }
    return res;
}
int main() {
    printf("Sum of x+y = %d", multiply(5,10));
    return 0;
}

Answer 31

非常简单，朋友...每次当你离开一个数字时，这意味着你将数字乘以2，这意味着答案是（x << 3）-x。