In which segment a given number lies in? [duplicate]

Answer 1

how would you solve the problem "In which segment a given number lie in?"

You should divide the number by the segment length, then truncate the fractional part away. Like this:

int segmentId = (int) floor(x/l);

It seems that you have already figured this out.

Of course, due to finite arithmetic, this does not work well.

If the result of 6.6 / 1.1 happens to be5.9999999999999991118215802998747676610946655273438, then 5 is in fact the correct segment for the result.

If you would like 6.6 / 1.1 to be exactly 6, then your problem is with finite precision division, which doesn't do what you want and with finite precision representation of floating point numbers that has no exact representation for all numbers. The segmentation itself worked perfectly.

I really don't know how to proceed further

Either don't use finite precision floating point (use fixed or arbitrary precision), or don't require the results of calculations to be exact.

Answer 2

Your problem is that neither 1.1, nor 6.6 are representable exactly in binary floating point. So when you type

double l = 1.1;
double x = 6.6;

you get 2 numbers stored in l and in x, which are slightly different than 1.1 and 6.6. After that, int segmentId = (int) floor(x/l); determines the correct segment for those slightly different numbers, but not for the original numbers.

You can solve this problem by using a decimal floating point data type instead of binary. You can check C++ decimal data types and Exact decimal datatype for C++? for the libraries, or implement the decimal data type yourself.

But still the problem will remain for numbers, which are not representable in finite decimal floating point, such as 1/3 (circulating fraction), sqrt(2) (irrational), pi (transcendental), etc.

Answer 3

What precision does your solution require? There can always be a problem with marginal values for given segment, cause they are most likely unrepresentable. I think adding a very small epsilon in this case could help. However it may fail in other case.

Answer 4

Just in case u don't specifically want an O(1) answer you can go for the O(logn) answer by just doing a binary search on the segments.

Answer 5

Check the segments again after the division.

bool inSegment(double x, double l, double segment)
{
    return (x >= l*(segment-1)) && (x < l*segment);
}

int segmentId;
double segment = floor(x/l);

if (inSegment(x, l, segment-1))
    segmentId = segment - 1;
else if (inSegment(x, l, segment))
    segmentId = segment;
else if (inSegment(x, l, segment+1))
    segmentId = segment + 1;
else
    printf("Something wrong happened\n");

Or use an epsilon and round the value up if the value is close enough to an integer above.