Rounding algorithm for integer programs

The next step is to apply the method of conditional probabilities to the existence proof. Although the concept is simple, implementing it requires a careful calculation bounding the conditional probability of failure as the scheme proceeds.

The original existence proof implicitly bounds the unconditioned probability of failure by the expectation of. The form of Q mirrors the probabilistic proof in a systematic way.

It contributes at least 1 to Q if the cost is too high. It contributes at least 1 to Q if the coverage fails. The algorithm will choose each set to keep the conditional expectation of Q , given the choices made so far, below 1. Consider the state after the t th choice. The conditional expectation of Q , given this state, is. This dependence is not surprising, but it is not inevitable.

You can read more about that here. I was curious to know how I can round a number to the nearest whole number. For instance, if I had:. In response to a comment "Why is this actually working? You get a rounded result if you add half of the denominator to the numerator before dividing, but only if numerator and denominator have the same sign. If the signs differ, you must subtract half of the denominator before dividing.

Putting all that together:. Edit: This solution will only work in the simplest of cases. A more robust solution would be:. Edited Rounding integers with floating point is the easiest solution to this problem; however, depending on the problem set is may be possible.

For example, in embedded systems the floating point solution may be too costly. Doing this using integer math turns out to be kind of hard and a little unintuitive. The first posted solution worked okay for the the problem I had used it for but after characterizing the results over the range of integers it turned out to be very bad in general. Looking through several books on bit twiddling and embedded math return few results.

A couple of notes. First, I only tested for positive integers, my work does not involve negative numerators or denominators. Second, and exhaustive test of 32 bit integers is computational prohibitive so I started with 8 bit integers and then mades sure that I got similar results with 16 bit integers. My thought was that the first version would overflow with big numbers and the second underflow with small numbers.

I did not take 2 things into consideration. In the first case you often overflow and then underflow, the two canceling each other out. Here is an error plot of the two incorrect algorithms:. Unexpectedly it actually handles large numerators better than the 2nd version. Here is a plot of the 2nd algorithm:. As expected it fails for small numerators but also fails for more large numerators than the 1st version. It turns out that larger denominators only have problems with larger numerators.

For 8 bit signed number the problem points are. This is not exact but close. I am sure there is an exact formula for determining the largest N that will work for a particular D and number of bits but I don't have any more time to work on this problem I seem to be missing this graph at the moment, I will edit and add later.

This is a graph of the 8 bit version with the special cases noted above:! As written, you're performing integer arithmetic, which automatically just truncates any decimal results. To perform floating point arithmetic, either change the constants to be floating point values:. Either way, you need to do the final rounding with the round function in the math.

Checking if there is a remainder allows you to manually roundup the quotient of integer division. The code in this answer below is for positive integers only.

Some of these answers are crazy looking! Codeface nailed it though! I really like the type-free macro or gcc statement expression form over the function form, however, so, I wrote this answer with a detailed explanation of what I'm doing ie: why this mathematically works and put it into 2 forms:.

See a little more on gcc statement expressions here. Here's my solution. I like it because I find it more readable and because it has no branching neither ifs nor ternaries. The following correctly rounds the quotient to the nearest integer for both positive and negative operands WITHOUT floating point or conditional branches see assembly output below.

Assumes N-bit 2's complement integers. If you're dividing positive integers you can shift it up, do the division and then check the bit to the right of the real b0. In this paper, we consider solving the integer linear systems, i. R m , and a positive integer d , to compute an integer vector x? D n such that A x?

View 3 excerpts, cites methods. Sparse Linear Complementarity Problems. Polyhedral sets having a least element. Related Papers. Abstract 34 Citations 2 References Related Papers. By clicking accept or continuing to use the site, you agree to the terms outlined in our Privacy Policy , Terms of Service , and Dataset License.

Hassin, Refael ; Segev, Danny. In: Operations Research Letters. In: Operations Research Letters , Vol. Hassin R , Segev D. Operations Research Letters.