remove nu in R2N & R2DH

Author: MaxenceGollier

src/R2DH.jl

@@ -121,9 +121,9 @@ or
 - `max_iter::Int = 10000`: maximum number of iterations;
 - `verbose::Int = 0`: if > 0, display iteration details every `verbose` iteration;
 - `σmin::T = eps(T)`: minimum value of the regularization parameter;
+- `σk::T = eps(T)^(1 / 5)`: initial value of the regularization parameter;
 - `η1::T = √√eps(T)`: very successful iteration threshold;
 - `η2::T = T(0.9)`: successful iteration threshold;
-- `ν::T = eps(T)^(1 / 5)`: inverse of the initial regularization parameter: ν = 1/σ;
 - `γ::T = T(3)`: regularization parameter multiplier, σ := σ/γ when the iteration is very successful and σ := σγ when the iteration is unsuccessful.
 - `θ::T = 1/(1 + eps(T)^(1 / 5))`: is the model decrease fraction with respect to the decrease of the Cauchy model. 
 - `m_monotone::Int = 6`: monotoneness parameter. By default, R2DH is non-monotone but the monotone variant can be used with `m_monotone = 1`
@@ -173,7 +173,6 @@ function R2DH(
     σmin = options.σmin,
     η1 = options.η1,
     η2 = options.η2,
-    ν = options.ν,
     γ = options.γ,
     θ = options.θ,
     kwargs_dict...,
@@ -206,7 +205,6 @@ function R2DH(
     σmin = options.σmin,
     η1 = options.η1,
     η2 = options.η2,
-    ν = options.ν,
     γ = options.γ,
     θ = options.θ,
     kwargs...,
@@ -242,7 +240,6 @@ function SolverCore.solve!(
   σmin::T = eps(T),
   η1::T = √√eps(T),
   η2::T = T(0.9),
-  ν::T = eps(T)^(1 / 5),
   γ::T = T(3),
   θ::T = 1/(1 + eps(T)^(1 / 5)),
 ) where {T, V}

src/R2N.jl

@@ -128,7 +128,6 @@ For advanced usage, first define a solver "R2NSolver" to preallocate the memory
 - `σk::T = eps(T)^(1 / 5)`: initial value of the regularization parameter;
 - `η1::T = √√eps(T)`: successful iteration threshold;
 - `η2::T = T(0.9)`: very successful iteration threshold;
-- `ν::T = eps(T)^(1 / 5)`: inverse of the initial regularization parameter: ν = 1/σ;
 - `γ::T = T(3)`: regularization parameter multiplier, σ := σ/γ when the iteration is very successful and σ := σγ when the iteration is unsuccessful;
 - `θ::T = 1/(1 + eps(T)^(1 / 5))`: is the model decrease fraction with respect to the decrease of the Cauchy model;
 - `m_monotone::Int = 1`: monotonicity parameter. By default, R2N is monotone but the non-monotone variant will be used if `m_monotone > 1`;
@@ -180,7 +179,6 @@ function R2N(
     σk = options.σk,
     η1 = options.η1,
     η2 = options.η2,
-    ν = options.ν,
     γ = options.γ,
     sub_kwargs = sub_kwargs;
     kwargs_dict...,
@@ -214,7 +212,6 @@ function SolverCore.solve!(
   σmin::T = eps(T),
   η1::T = √√eps(T),
   η2::T = T(0.9),
-  ν::T = eps(T)^(1 / 5),
   γ::T = T(3),
   β::T = 1 / eps(T),
   θ::T = 1/(1 + eps(T)^(1 / 5)),
@@ -355,16 +352,12 @@ function SolverCore.solve!(
 
     solver.subpb.model.σ = σk
     isa(solver.subsolver, R2DHSolver) && (solver.subsolver.D.d[1] = 1/ν₁)
-    sub_ν = isa(solver.subsolver, R2DHSolver) ? 1/σk : ν₁
-    solve!(
-      solver.subsolver,
-      solver.subpb,
-      solver.substats;
-      x = s1,
-      atol = sub_atol,
-      ν = sub_ν,
-      sub_kwargs...,
-    )
+    if isa(solver.subsolver, R2Solver) #FIXME
+      sub_kwargs[:ν] = ν₁
+    else
+      sub_kwargs[:σk] = σk
+    end
+    solve!(solver.subsolver, solver.subpb, solver.substats; x = s1, atol = sub_atol, sub_kwargs...)
 
     s .= solver.substats.solution
 

test/test_allocs.jl

@@ -48,7 +48,12 @@ end
         reg_nlp = RegularizedNLPModel(LBFGSModel(bpdn), h)
         solver = eval(solver)(reg_nlp)
         stats = RegularizedExecutionStats(reg_nlp)
-        @test @wrappedallocs(solve!(solver, reg_nlp, stats, ν = 1.0, atol = 1e-6, rtol = 1e-6)) == 0
+        solver_name == "R2" &&
+          @test @wrappedallocs(solve!(solver, reg_nlp, stats, ν = 1.0, atol = 1e-6, rtol = 1e-6)) ==
+                0
+        solver_name == "R2DH" && @test @wrappedallocs(
+          solve!(solver, reg_nlp, stats, σk = 1.0, atol = 1e-6, rtol = 1e-6)
+        ) == 0
         @test stats.status == :first_order
       end
     end