Skip to content

Instantly share code, notes, and snippets.

Embed
What would you like to do?
Root of a number - error range - June 26, 2017 - debugging code, bug fix on line 74 (int)(value * 1000), forget () to wrap value * 1000.
using System;
class Solution
{
// 0 , 0.001, 3=> 0.1, x = 7, n = 3
static double root(double x, uint n)
{
// your code goes here
if( x < 0)
{
return -1;
}
if( Math.Abs(x) < 0.001)
{
return 0; // 0
}
// assume that x is nonnagative
var search = binarySearch(x, n); // 0.001, 3
return search;
}
/// 0.001, 3 => 0.1 from 0 to 1, 1000 numbers,
private static double binarySearch(double x, uint n)
{
double start = 0;
double end = Math.Max(1, x); // Math.Max(1, x), x < 1, x > 1, x , [0 7]
int index = 0;
while(start <= end) // 0, 0.001
{
index++;
if(index < 20)
{
Console.WriteLine("start=" + start);
Console.WriteLine("end=" + end);
Console.WriteLine(index);
}
else
{
break;
}
var middle = keep3Decimal(start + (end - start)/ 2.0); // keep to 3 decimal, 0, middle = 7/2.0 = 3.500, 1.7
var middleValue = Math.Pow(middle, n); // 3.5^3 > 7
bool isEqual = Math.Abs(middleValue - x) < 0.001;
bool isLess = x - middleValue > 0.001;
if(isEqual)
{
return middle; //
}
else if(isLess)
{
start = middle + 0.001; // 1.701 - 1.7
}
else
{
end = middle - 0.001; // 3.5 - 0.001 = 3.499
}
}
return -1; // not found
}
///
private static double keep3Decimal(double value)
{
int integerValue = (int)(value * 1000);
return integerValue/ 1000.0; // double
}
static void Main(string[] args)
{
Console.WriteLine(root(7, 3));
}
}
// x = 7, n = 3, 1.913
// x > 1, Math.Abs(1.913 ^ 3 - 7) <= 0.001
// x > 1, > 1, lower bound 1, upper bound <= 7
// binary search - error range 0.001, 7000, log(n) = 10 * log7 , 30 times, find the number
// x = 9 , n = 2, 3.0 = 9, 2.999 -3 = 0.001, 3.000, 2.999
// x < 1, upper bound 1, lower bound - nonnegative number
// min(0, 1), max ( 1, x)
// < 0.001, they are equal
// double -> convert to 3 decimal
// double d , d * 1000 -> integer - type conversion - double -> integer -> 1000.0
// 0.1001 -> 100 -> 100/ 1000.0 = 0.1 00 -> fourth decimal 0.0005 -> 0.001
@jianminchen

This comment has been minimized.

Copy link
Owner Author

commented Jun 27, 2017

Julia did white boarding test on her own code, and then she found a bug on line 29 on test case 0.001, n = 3, search value 0.1. Learn to test your own code very carefully, and then find a bug on line 29. But Julia missed the bug on line 74. The peer asked the question to point out the bug on line 75, return integerValue/ 1000.0; not integerValue/ 1.0.

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
You can’t perform that action at this time.