Creation date: 2020-01-19
This is kind of part 16 of the series about: How to build a Constraint Programming solver? Even if you are not deep into constraint programming this post tries to show you what kind of problems occur when you deal with floating point numbers. If you are new to this blog and are interested in constraint programming, there are several posts already waiting for you:
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At first I thought: Oh yeah I just have to change the type of some structs to support it. Well I forgot the problems using floating point numbers...
You might have seen it before\[ 0.2+0.1 \neq 0.3 \]
Yes we are living in a new decade but we still need to deal with the problems that computers can't calculate even though it's kind of the thing everyone thinks they are much better than we humans...
There are way better posts and explanations out there why computers don't sum up \(0.2+0.1\) to \(0.3\) but instead to something like \(0.30000000000000004\).
It might be obvious how this applies to the problem in this constraint solver project but maybe a short look back on how these equality constraints are tackled is helpful:\[ 0.2x+0.1y = 0.3z \]
We have \(1 \leq x,y,z \leq 3\) and as always they are integers. First we write that as:\[ 0.2x+0.1y-0.3z = 0 \]
Let's assume we have set \(x = y = 1\) during our backtracking process. Then we would like to get \(z=1\) but instead we get: "The problem is infeasible". Because\[ 0.2+0.1-0.3 = 5.551115123125783e-17 \neq 0 \]
If you want to read the full post on how the whole constraint works: Part 6: Sum constraint first part
Everything works without any problems for integers as computers are actually very good at adding discrete numbers. Wow good job computer!
Now there are different solutions to solve the problem above. We could either say:
Well we just don't allow real coefficients. Which I did before :D
We can convert everything to integers and be happy.
Okay the result is close so it's probably just the case that I miscalculated the result -> Maybe if it doesn't add up to 0 it's still okay.
The first one is not really a solution so we should have a look at option 2. The above example can be easily converted to:\[ 2x+1y = 3z \]
If I have \(0.0000002x+0.0000001y=0.0000001z\) we can convert to the same thing but how about: \(0.0000002x+0.1000001y=0.1000001z\)? Clearly the result stays \(x=y=z=1\) but that is not obvious to a computer. We can multiply everything by 10 million to get: \(2x+1000001 = 3000001z\) and the numbers might get bigger and we sum over it which brings us eventually to deal with overflow errors or we use something bigger than
Int64 for example
BigInt. Which makes it slower so this is an interesting thing to do but it might be not the best choice and is certainly not the one used the most. Which brings us to option 3:
Let us just check whether it is close enough or not. Julia is of course prepared as I'm not the first guy who encountered this problem. Therefore there are functions like
>? isapprox help?> isapprox search: isapprox isapprox(x, y; rtol::Real=atol>0 ? 0 : √eps, atol::Real=0, nans::Bool=false, norm::Function) Inexact equality comparison: true if norm(x-y) <= max(atol, rtol*max(norm(x), norm(y))). The default atol is zero and the default rtol depends on the types of x and y. The keyword argument nans determines whether or not NaN values are considered equal (defaults to false). For real or complex floating-point values, if an atol > 0 is not specified, rtol defaults to the square root of eps of the type of x or y, whichever is bigger (least precise). This corresponds to requiring equality of about half of the significand digits. Otherwise, e.g. for integer arguments or if an atol > 0 is supplied, rtol defaults to zero. ... Examples ≡≡≡≡≡≡≡≡≡≡ julia> 0.1 ≈ (0.1 - 1e-10) true
The example also shows an interesting unicode feature. You can use the mathematical approx sign \(\approx\) using
\approx in your editor for such comparisons. Which is a short notation for the default values used by
isapprox. Normally solvers give it as an option to define
rtol so these are options for the constraint solver now.
Let's have a look how this can be applied to our problem.
We normally remove values from the search space in the following way. Our slightly changed example:\[ 0.2x+0.1y - 0.15z = 0 \]
and we still have \(1 \leq x,y \leq 3\) but \(1 \leq z \leq 2\). We check which value is the highest we can reach so use the maximum value for positive coefficients and the lowest for negative coefficients:\[ 0.2\cdot 3 + 0.1 \cdot 3 - 0.15 \cdot 1 = 0.7500000000000001 \]
I use Julia now to compute this which shows that we have some floating point issues.
We would have a maximum value of \(0.7500000000000001\) and a minimum value of \(5.551115123125783e-17\) which means of course that we can't achieve our goal of \(0.0\).
Normally this gets removed immediately as there is no way to make it feasible. Therefore we need to change the point where we
return false in these cases a bit by using our absolute tolerance
# if the maximum is smaller than 0 (and not even near zero) # or if the minimum is bigger than 0 (and not even near zero) # the equation can't sum to 0 => infeasible if full_max < -com.options.atol || full_min > +com.options.atol com.bt_infeasible[indices] .+= 1 return false end
Okay what are the next steps? We remove one value at a time from the equation and compute it again and gain more information about the bound of the current value. This is explained in more detail in previous posts.
If we want to change the bounds of a variable we need to be careful again. We had this before: See on GitHub
if maxs[i] < pre_maxs[i] if constraint.coeffs[i] > 0 still_feasible = remove_above!(com, search_space[idx], fld(maxs[i], constraint.coeffs[i])) else still_feasible = remove_below!(com, search_space[idx], fld(maxs[i], constraint.coeffs[i])) end if !still_feasible return false end end
or more specifically:
remove_above!(com, search_space[idx], fld(maxs[i], constraint.coeffs[i])) which means we found a new bound (including the coefficient) and divide by the coefficient and round down. This can result in something like:
julia> 0.3/0.1 2.9999999999999996
Which means we would have
fld(0.3, 0.1) = 2.0 and would remove the value 3 but actually the value
3 is a possible value. Therefore we need to compute a safe upper bound which means if the result is close to the above integer we probably don't want to round down but instead round up.
This needs to be done on several steps throughout the constraint. I hope the idea behind this is clear.
That's all for this a bit smaller post. There are two PRs currently in the reviewing process with different things which will get their own post as well:
Better dispatch on functions:
Currently each constraint is not perfectly typed as the constraint function is of type
::Function which might lead to worse performance. Additionally there exists a smoother way by using what is used for the MathOptInterface interface anyway.
Support for linear objectives: Currently we only support
@objective(m, Min, x) but not
@objective(m, Min, 2x+4.1y).
After all changes and reviews I'll blog about those two and make a final post on the Kaggle Santa Challenge.
The next post on linear objectives is published: Have a look: Support Linear Objectives
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