Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Jump to bottom

Simplified version of mod(x::Interval, y::Real) #525

Merged
merged 12 commits into from
Jul 3, 2023
Merged

Simplified version of mod(x::Interval, y::Real) #525

Show file tree
Hide file tree
Changes from 8 commits
Commits
Show all changes
12 commits
Select commit Hold shift + click to select a range
fea1b34
Introduce simplified version of mod
petvana May 24, 2022
153d749
Improve testing
petvana May 24, 2022
d23df89
Update src/intervals/functions.jl
petvana May 24, 2022
b56f4be
Use ⊇ operator
petvana May 25, 2022
ea00b23
Imrpove test coverage + use zero()
petvana May 25, 2022
ff47910
Add todo for mod with between two intervals
petvana May 26, 2022
45abc46
Implementation for strictly negative divisors for mod
petvana May 26, 2022
d2603d6
Update docs
petvana May 26, 2022
439723f
Disable divisor for mod to be an interval
petvana May 27, 2022
6fdc809
Merge branch 'simplified-mod' of github.com:petvana/IntervalArithmeti…
petvana May 27, 2022
f9f4733
Throw ArgumentError for Interval divisor for mod
petvana May 27, 2022
7bef23e
Cleanup
petvana May 27, 2022
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
1 change: 1 addition & 0 deletions src/IntervalArithmetic.jl
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -24,6 +24,7 @@ import Base:
in, zero, one, eps, typemin, typemax, abs, abs2, real, min, max,
sqrt, exp, log, sin, cos, tan, cot, inv, cbrt, csc, hypot, sec,
exp2, exp10, log2, log10,
mod,
asin, acos, atan,
sinh, cosh, tanh, coth, csch, sech, asinh, acosh, atanh, sinpi, cospi,
union, intersect, isempty,
Expand Down
15 changes: 15 additions & 0 deletions src/intervals/functions.jl
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -373,3 +373,18 @@ function nthroot(a::Interval{T}, n::Integer) where T
b = nthroot(bigequiv(a), n)
return convert(Interval{T}, b)
end

"""
Calculate `x::Interval mod y::Real`, limited by `y != 0`.
"""
function mod(x::Interval, y::Real)
@assert y != zero(y) """mod(x::Interval, y::Real)
Copy link
Member

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

an alternative would be to return an empty interval, I think this would be more conformant with the IEEE standard behavior, e.g.

julia> sqrt(-1.. -1)
∅

julia> log(-2.. -1)
∅

Copy link
Contributor Author

@petvana petvana May 27, 2022 •
edited
Loading

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Or we can define mod(a..b, 0) == 0..0 ⊇ ∅ as Wolfram says
https://www.wolframalpha.com/input?i=limit+x-%3E0%2B+mod%281%2C+x%29

Consequently, for 0 ∈ y, we can define smth like mod(x,y) = union(mod(x,(y.lo..y.lo/3)), mod(x, (y.hi/3)..(y.hi))) because the maximum value needs to by in (y.hi/2)..(y.hi)

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

I propose to solve this in some future PR, and move the discussion to #129, to keep track once this PR merged.

Copy link
Member

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Or we can define mod(a..b, 0) == 0..0 ⊇ ∅ as Wolfram says https://www.wolframalpha.com/input?i=limit+x-%3E0%2B+mod%281%2C+x%29

Julia (and Wolfram) return NaN (undefined) for mod(3.2, 0.0), which makes sense due to the division in the algorithm.

Consequently, for 0 ∈ y, we can define smth like mod(x,y) = union(mod(x,(y.lo..y.lo/3)), mod(x, (y.hi/3)..(y.hi))) because the maximum value needs to by in (y.hi/2)..(y.hi)

This idea is interesting, and reminds me a little bit what we have for extended_div. That certainly should be addressed in another PR. Yet, I think for (mathematical) reals, mod(x,y) is a number in the interval [0,y] for y>0, or [y, 0] if y<0, isn't it?

From the range of mod(x,y) for real y, then for a strictly positive interval y, we should return [0, sup(y)], or if it is strictly negative [inf(y), 0]. If 0 ∈ y, we should return the hull of these intervals.

Copy link
Member

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

I think my last comment corresponds to these results... Or not?

is currently implemented only for a strictly positive or negative divisor y."""
division = x / y
fl = floor(division)
if !isthin(fl)
return y > zero(y) ? Interval(zero(y), y) : Interval(y, zero(y))
else
return y * (division - fl)
end
end
40 changes: 40 additions & 0 deletions test/interval_tests/numeric.jl
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -434,3 +434,43 @@ end
@test nthroot(Interval{BigFloat}(-81, -16), -4) == ∅
@test nthroot(Interval{BigFloat}(-81, -16), 1) == Interval{BigFloat}(-81, -16)
end

# approximation used for testing (not to rely on ≈ for intervals)
# ⪆(x, y) = (x ≈ y) && (x ⊇ y)
⪆(x::Interval, y::Interval) = x.lo ≈ y.lo && x.hi ≈ y.hi && x ⊇ y
petvana marked this conversation as resolved.
Show resolved Hide resolved

@testset "`mod`" begin
petvana marked this conversation as resolved.
Show resolved Hide resolved
r = 0.0625
x = r..(1+r)
@test mod(x, 1) == mod(x, 1.0) == 0..1
@test mod(x, 2) == mod(x, 2.0) ⪆ x
@test mod(x, 2.5) ⪆ x
@test mod(x, 0.5) == 0..0.5
@test mod(x, -1) == mod(x, -1.0) == -1..0
@test mod(x, -2) == mod(x, -2.0) ⪆ -2+x
@test mod(x, -2.5) ⪆ -2.5+x
@test mod(x, -0.5) == -0.5..0

x = (-1+r) .. -r
@test mod(x, 1) == mod(x, 1.0) ⪆ 1+x
@test mod(x, 2) == mod(x, 2.0) ⪆ 2+x
@test mod(x, 2.5) ⪆ 2.5+x
@test mod(x, 0.5) == 0..0.5
@test mod(x, -1) == mod(x, -1.0) ⪆ x
@test mod(x, -2) == mod(x, -2.0) ⪆ x
@test mod(x, -2.5) ⪆ x
@test mod(x, -0.5) == -0.5..0

x = -r .. 1-r
@test mod(x, 1) == mod(x, 1.0) == 0..1
@test mod(x, 2) == mod(x, 2.0) == 0..2
@test mod(x, 2.5) == 0..2.5
@test mod(x, 0.5) == 0..0.5
@test mod(x, -1) == mod(x, -1.0) == -1..0
@test mod(x, -2) == mod(x, -2.0) == -2..0
@test mod(x, -2.5) == -2.5..0
@test mod(x, -0.5) == -0.5..0

# TODO - implement mod for two intervals
@test_throws TypeError mod(1..2, 1.4..1.5)
petvana marked this conversation as resolved.
Show resolved Hide resolved
end