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simplify internal_itp #1127

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Mar 27, 2025
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simplify internal_itp #1127

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simplify internal_itp
Mar 19, 2025
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Mar 26, 2025
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Mar 27, 2025
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95 changes: 37 additions & 58 deletions src/internal_itp.jl
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Original file line number Diff line number Diff line change
Expand Up @@ -20,12 +20,11 @@ end
simpler dependencies.
"""
struct InternalITP
k1::Float64
k2::Float64
scaled_k1::Float64
n0::Int
end

InternalITP() = InternalITP(0.007, 1.5, 10)
InternalITP() = InternalITP(0.2, 10)

function SciMLBase.solve(prob::IntervalNonlinearProblem{IP, Tuple{T, T}}, alg::InternalITP,
args...;
Expand All @@ -36,78 +35,58 @@ function SciMLBase.solve(prob::IntervalNonlinearProblem{IP, Tuple{T, T}}, alg::I
ϵ = eps(T)
if iszero(fl)
return SciMLBase.build_solution(prob, alg, left, fl;
retcode = ReturnCode.ExactSolutionLeft, left = left,
right = right)
retcode = ReturnCode.ExactSolutionLeft, left, right)
elseif iszero(fr)
return SciMLBase.build_solution(prob, alg, right, fr;
retcode = ReturnCode.ExactSolutionRight, left = left,
right = right)
retcode = ReturnCode.ExactSolutionRight, left, right)
end
#defining variables/cache
k1 = T(alg.k1)
k2 = T(alg.k2)
span = abs(right - left)
k1 = T(alg.scaled_k1)/span
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do you know if the lack of scaling based on span was intentional here? The NonlinearSolveBase verison already had it.

n0 = T(alg.n0)
n_h = ceil(log2(abs(right - left) / (2 * ϵ)))
mid = (left + right) / 2
x_f = (fr * left - fl * right) / (fr - fl)
xt = left
xp = left
r = zero(left) #minmax radius
δ = zero(left) # truncation error
σ = 1.0
ϵ_s = ϵ * 2^(n_h + n0)
i = 0 #iteration
while i <= maxiters
#mid = (left + right) / 2
n_h = exponent(span / (2 * ϵ))
ϵ_s = ϵ * exp2(n_h + n0)
T0 = zero(fl)

i = 1
while i ≤ maxiters
stats.nsteps += 1
span = abs(right - left)
mid = (left + right) / 2
r = ϵ_s - (span / 2)
δ = k1 * (span^k2)

## Interpolation step ##
x_f = left + (right - left) * (fl / (fl - fr))
x_f = left + span * (fl / (fl - fr)) # Interpolation Step

## Truncation step ##
σ = sign(mid - x_f)
if δ <= abs(mid - x_f)
xt = x_f + (σ * δ)
else
xt = mid
end
δ = max(k1 * span^2, eps(x_f))
diff = mid - x_f

## Projection step ##
if abs(xt - mid) <= r
xp = xt
else
xp = mid - (σ * r)
end
xt = ifelse(δ ≤ abs(diff), x_f + copysign(δ, diff), mid) # Truncation Step

## Update ##
tmin, tmax = minmax(left, right)
xp >= tmax && (xp = prevfloat(tmax))
xp <= tmin && (xp = nextfloat(tmin))
xp = ifelse(abs(xt - mid) ≤ r, xt, mid - copysign(r, diff)) # Projection Step
if span < 2ϵ
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Is this correct? ϵ here is the machine epsilon eps(T) (not eps(left)), i.e. a relative difference, compared to an absolute value span. If left and right are >> 1 in absolute value and different this condition cannot be true, while if they are << 1 in absolute value this condition can be met even if the interval is not tight.

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good catch. This was based on SciML/NonlinearSolve.jl#548 where we have an abstol. I think this branch can just be deleted.

return SciMLBase.build_solution(
prob, alg, xt, f(xt); retcode = ReturnCode.Success, left, right
)
end
yp = f(xp)
stats.nf += 1
yps = yp * sign(fr)
if yps > 0
right = xp
fr = yp
elseif yps < 0
left = xp
fl = yp
if yps > T0
right, fr = xp, yp
elseif yps < T0
left, fl = xp, yp
else
left = prevfloat_tdir(xp, prob.tspan...)
right = xp
return SciMLBase.build_solution(prob, alg, left, f(left);
retcode = ReturnCode.Success, left = left,
right = right)
return SciMLBase.build_solution(
prob, alg, xp, yps; retcode = ReturnCode.Success, left, right
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Another question: here we're no longer contracting the left - right interval for the solution object, does this change behavior in any case? I'm investigating a small outcome variation in a test in some internal code of ours that opts into rootfind=SciMLBase.RightRootFind for a callback and it seemingly appears between DiffEqBase v6.167.0 and v6.167.2 (though this is just a tentative assessment as of now).

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It's definitely possible that this changes behavior slightly. It's very unclear why the previous code thought that this was trying to do. Specifically, this case only is hit if the function hits exactly 0 (i.e. it will not hit this even if left==right which seems likely unintended).

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@DaniGlez DaniGlez Apr 1, 2025 •
edited
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Yes, that makes sense; my concern is that while the actual crossing (xp) is provided as the "main" value of the solution sol.u, the way this solution is used ( https://github.com/SciML/DiffEqBase.jl/blob/master/src/callbacks.jl#L365 ) this value is apparently discarded; instead, the left and right fields of the solution seem to be the relevant ones (depending on the user-defined rootfinding side). However, in this file, the solution is built without updating the left and right variables, which could be arbitrarily far from xp (which is I guess the value we want) if the iteration hitting this branch is particularly lucky. The previous code basically used xp (the value producing the exact zero) to update the right value and its prevfloat as the left value, thus producing a tighter interval around the crossing.

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I guess another connected question (that is probably not very relevant practically except for sorting simultaneous events) is what should happen if there are multiple zeros in a row (e.g. for a sequence of adjacent floats for t the condition produces something like -a, 0, 0, 0, a where a is small). It's not as important, because for this to happen you need to have small absolute values of t (so that the expected difference between consecutive values is comparable to numerical error) and/or a flat-ish f, but maybe it's worth thinking about what's the desirable behavior in that case (it should be cheap to seek the first or last zero with a few refinement iterations, probably with straightforward bisection for smaller overhead).

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Upon further testing it appears that setting left and right to xp here reverts my test results to their pre-6.167.1 values, I will make a PR once I double-check.

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Upon further testing it appears that setting left and right to xp here reverts my test results to their pre-6.167.1 values, I will make a PR once I double-check.

)
end

i += 1
mid = (left + right) / 2
ϵ_s /= 2

if nextfloat_tdir(left, prob.tspan...) == right
return SciMLBase.build_solution(prob, alg, left, fl;
retcode = ReturnCode.FloatingPointLimit, left = left,
right = right)
if nextfloat_tdir(left, left, right) == right
return SciMLBase.build_solution(
prob, alg, right, fr; retcode = ReturnCode.FloatingPointLimit, left, right
)
end
end
return SciMLBase.build_solution(prob, alg, left, fl; retcode = ReturnCode.MaxIters,
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