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clogarithm

C: Safe to Calculate Number of Digits in Integer (N) of Base (b) Using Logarithmic Functions


Given the formula:

NUM_DIGITS_IN_N_FOR_BASE_B = 1 + floor(ln(abs(N)) / ln(b))

such that b is a base between 2 and 36 and N is of type int.

According to this post post 1 it seems okay to assume that the value returned for the range of all integers [INT_MIN, INT_MAX] would work with no round-off or overflow error. I am a bit skeptical about this. My skepticism comes from a recent post of mine @post 2. If using the mathematical definition is not "safe" in a computer program, is there another trick to use to compute the number of digits in a number N given a base b?


Solution

  • Is it safe to calculate the number of digits in an integer N in base b using logarithmic functions?

    No absolutely not, due to rounding errors. Specifically, the risk is that log(N)/log(b) will be slightly less than the exact value. This is most likely to occur when N is an exact multiple of b.

    How does one count the number of digits in an integer N in base b?

    Divide N by b in a loop until N is zero. See the countDigits function in the code below. Handling values of N <= 0 is left as an exercise for the reader.


    For example, consider the following code

    int countDigits( int N, int b )
    {
        int count;
        for ( count = 0; N; count++ )
            N /= b;
        return count;
    }
    
    int main( void )
    {
        int N, b;
        if ( scanf( "%d %d", &N, &b ) != 2 )
            return 1;
    
        printf( "log(%3d) = %.50lf\n", N, log(N) );
        printf( "log(%3d) = %.50lf\n", b, log(b) );
        printf( "ratio    = %.50lf\n", log(N)/log(b) );
        printf( "expected = %d\n", countDigits(N, b) );
        double digits = 1 + floor(log(N) / log(b));
        printf( "computed = %lf\n", digits );
    }
    

    If the user enters 243 for N, and 3 for b, the output is

    log(243) = 5.49306144334054824440727315959520637989044189453125
    log(  3) = 1.09861228866810978210821758693782612681388854980469
    ratio    = 4.99999999999999911182158029987476766109466552734375
    expected = 6
    computed = 5.000000
    

    The expected number of digits is 6 since 24310 = 1000003. The problem with the logarithm method is that either log(243) is slightly too small or log(3) is slightly too large, resulting in a ratio that is just below 5 when it should be exactly 5.