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Add mod #145

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Add mod #145

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lbenet
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@lbenet lbenet commented May 3, 2018

This PR addresses #129.

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coveralls commented May 3, 2018

Coverage Status

Coverage increased (+0.2%) to 92.463% when pulling e0eaa54 on lb/mod into e6e0d0d on master.

@lbenet lbenet mentioned this pull request May 3, 2018
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codecov-io commented May 4, 2018

Codecov Report

Merging #145 into master will increase coverage by 0.17%.
The diff coverage is 100%.

Impacted file tree graph

@@            Coverage Diff             @@
##           master     #145      +/-   ##
==========================================
+ Coverage   92.29%   92.46%   +0.17%     
==========================================
  Files          23       23              
  Lines        1064     1088      +24     
==========================================
+ Hits          982     1006      +24     
  Misses         82       82
Impacted Files Coverage Δ
src/IntervalArithmetic.jl 100% <ø> (ø) ⬆️
src/intervals/functions.jl 97.65% <100%> (+0.54%) ⬆️

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@dpsanders
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Does this work with e.g. mod(X, pi_interval(Float64))?
Maybe we could add some tests for cases like this.

@lbenet
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lbenet commented Jun 4, 2018

No, it doesn't, and actually yields a segmentation fault. Let me take a look on this...

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lbenet commented Jun 4, 2018

The problem seems to be that mod(x::Interval, y::Real) assumes y different from an Interval. When y is an interval, fld (used internally) makes a promotion which ends calling (recursively) mod which yields the segmentation fault.

Any idea about how to restrict y to avoid being an Interval?

@dpsanders
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You can do

mod(x::Interval, y::AbstractFloat)

Or use / instead of fld?

@lbenet
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lbenet commented Jun 7, 2018

Using / instead of fld would not guarantee rounding it down. In any case, I'll take your suggestion into account.

@dpsanders
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Maybe the following?

julia> x = 7.3..7.3
[7.29999, 7.30001]

julia> y = 5.5..5.5
[5.5, 5.5]

julia> floor(x / y)
[1, 1]

@dpsanders
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dpsanders commented Jun 7, 2018

julia> y = pi..pi
Interval(3.141592653589793, 3.1415926535897936)

julia> x / y
Interval(2.3236621691416715, 2.3236621691416723)

julia> floor(x / y)
Interval(2.0, 2.0)

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lbenet commented Jun 7, 2018

I'll certainly test it, since it is simply faster:

julia> @btime floor(7.3/5.5)
  1.895 ns (0 allocations: 0 bytes)
1.0

julia> @btime fld(7.3, 5.5)
  10.660 ns (0 allocations: 0 bytes)
1.0

@dpsanders
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Any update here, @lbenet ?

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lbenet commented Jun 29, 2018

Let me rebase to current master and check if everything runs smoothly.

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lbenet commented Jun 29, 2018

Rebased to current master and implemented this suggestion.

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lbenet commented Jun 29, 2018

Locally, tests pass.

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Lgtm, thanks! Could you add some tests with mod respect to pi, please?

@lbenet
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lbenet commented Jun 29, 2018

Thanks for the suggestion. I'll try to add more tests.

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lbenet commented Jun 30, 2018

I just pushed few tests involving pi, as you suggested.

Yet, I just noticed that the implementation allows to compute mod(a, b) with b::Interval. The result is weird (an Interval of Intervals). Should it throw an error (and which one) in that case? Do you have another suggestion?

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lbenet commented Jun 30, 2018

Tests pass locally.

@dpsanders
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But any mod with pi should use the interval version of pi, no?
We certainly want to allow mod with (narrow -- probably atomic?) intervals, for this reason, and the answer should be an interval.

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Isn't this basically the same as the issue with quadrants for sin and cos? And the solution should be similar, I guess.

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dpsanders commented Jul 1, 2018

I think it should be something like

julia> X = 10..11
[10, 11]

julia> p = pi_interval(Float64)
[3.14159, 3.1416]

julia> division = X / p
[3.18309, 3.50141]

julia> division_floor = floor(division)
[3, 3]

julia> result = (division - division_floor) * p
[0.575222, 1.57523]

But sometimes we will have:

julia> X = 2p
[6.28318, 6.28319]

julia> X / p
[1.99999, 2.00001]

If the lo and hi of the floor are different then the interval spans two "quadrants", so the mod should be the whole interval 0..(p.hi).

For extended_mod, we can then do whatever the correct thing is.

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lbenet commented Jul 2, 2018

You are right. Let me think it over how to implement it.

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4 participants