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

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merged 12 commits into from
Jul 3, 2023
Merged

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

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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
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1 change: 1 addition & 0 deletions src/IntervalArithmetic.jl
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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,
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18 changes: 18 additions & 0 deletions src/intervals/functions.jl
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Expand Up @@ -373,3 +373,21 @@ 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` where y != zero(y) and y is not inteval`.
"""
function mod(x::Interval, y::Real)
@assert y != zero(y) """mod(x::Interval, y::Real)
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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)
∅

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@petvana petvana May 27, 2022 •
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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)

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

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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.

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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

mod(x::Interval, y::Interval) = throw(ArgumentError("mod not defined for interval as divisor `y`"))
mod(x::Real, y::Interval) = throw(ArgumentError("mod not defined for interval as divisor `y`"))
41 changes: 41 additions & 0 deletions test/interval_tests/numeric.jl
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Expand Up @@ -434,3 +434,44 @@ 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
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@testset "`mod`" begin
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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 ArgumentError mod(1..2, 1.4..1.5)
@test_throws ArgumentError mod(1.0, 1.4..1.5)
end