Bruteforce integer multiplication in Java

Question

Intro

This time, I have attempted to prove the following bruteforce multiplication formula: $$ (a_n \cdots a_0) \cdot (b_m \cdots b_0) = \sum_{i = 0}^n \sum_{j = 0}^m a_i b_j \cdot 10^{i + j}, $$ where \$a_n, \ldots, a_0, b_m, \ldots, b_0 \in \{ 0, 1, \ldots, 9\}\$ are decimal digits.

Code

com.github.coderodde.research.multiply.BruteForceMultiply.java:

package com.github.coderodde.research.multiply;

import java.util.Scanner;

/**
 * This class provides the method for doing brute-force integer multiplication.
 */
public final class BruteForceMultiply {

    /**
     * This method performs the brute-force integer operation for non-negative
     * long integers.
     * 
     * @param number1 the first number.
     * @param number2 the second number.
     * @return the product of {@code number1} and {@code number2}.
     */
    public static long multiply(final long number1, final long number2) {
        checkNotNegative(number1);
        checkNotNegative(number2);
        
        final String number1String = Long.toString(number1);
        final String number2String = Long.toString(number2);
        
        final long[] a = convertLongStringToDecimalDigitArray(number1String);
        final long[] b = convertLongStringToDecimalDigitArray(number2String);
        
        long product = 0L;
        
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < b.length; j++) {
                product += a[i] * b[j] * myPow(10, i + j);
            }
        }
        
        return product;
    }
    
    public static void main(String[] args) {
        final Scanner scanner = new Scanner(System.in);
        
        while (true) {
            System.out.print(">>> ");
            final long a = scanner.nextLong();
            final long b = scanner.nextLong();
            System.out.println(multiply(a, b));
        }
    }
    
    private static void checkNotNegative(final long number) {
        if (number < 0) {
            final String exceptionMessage = 
                    String.format("number(%d) < 0", number);
            
            throw new IllegalArgumentException(exceptionMessage);
        }
    }
    
    private static long[] convertLongStringToDecimalDigitArray(
            final String numberString) {
        final long[] digitArray = new long[numberString.length()];
        
        for (int i = 0; i < digitArray.length; i++) {
            digitArray[i] = numberString.charAt(digitArray.length - 1 - i)
                          - '0';
        }
        
        return digitArray;
    }
    
    private static long myPow(final long base, final long exponent) {
        long p = 1L;
        
        for (int i = 0; i < exponent; i++) {
            p *= base;
        }
        
        return p;
    }
}

com.github.coderodde.research.multiply.BruteForceMultiplyTest.java:

package com.github.coderodde.research.multiply;

import org.junit.Test;
import static org.junit.Assert.*;

public final class BruteForceMultiplyTest {
    
    @Test
    public void bruteforceTest() {
        for (long a = 0L; a < 1000L; a++) {
            for (long b = 0L; b < 1000L; b++) {
                assertEquals(a * b, 
                             BruteForceMultiply.multiply(a, b));
            }
        }
    }
}

Critique request

Please, as always, give me any critique regarding any aspect of my code.

Answer 1

I get a feeling that you're doing this string conversion just to get the number of digits in the input:

    final String number1String = Long.toString(number1);
    final long[] a = convertLongStringToDecimalDigitArray(number1String);

It seems a bit confusing. You can get the length of a number using (int) (Math.log10(number) + 1) as long as you add a special case for 0. But then again you could replace the whole conversion part with series of divisions and modulo 10s.

public static long multiply(final long number1, final long number2) {
    long product = 0L;

    int i = 0;
    long a = number1;
    while (a > 0) {
        int j = 0;
        long b = number2;
        while (b > 0) {
            product += (a % 10) * (b % 10) * myPow(10, i + j);
            b /= 10;
            j++;
        }
        a /= 10;
        i++;
    }

    return product;
}

Pardon me, I did not test that at all.

Personally I am not a fan of having a this many single character variables but I suppose the context in which they are used is so short that you don't lose track of what each of them means.

This class JavaDoc seems a bit forced, given the class name. You probably should explain what the brute force means in this case. Starting the comments with "This class" and "This method" is unnecessary.

/**
 * This class provides the method for doing brute-force integer multiplication.
 */
public final class BruteForceMultiply {

Answer 2

You can avoid the repeated calls to myPow if you store the result in an array.

final long[] c = new int[a.length + b.length - 1];

        c[i+j] = a[i] * b[j];

long multiplier = 1;
for (long multiplicand : c) {
    product += multiplicand * multiplier;
    multiplier *= 10;
}

This does one more multiplication than the greatest call to myPow but replaces all the calls.

It's trickier, but you could probably do the same thing without the array.