test: update discrete system tests

Author: AayushSabharwal

test/discrete_system.jl

@@ -40,7 +40,6 @@ u = ModelingToolkit.varmap_to_vars(
     Dict([S(k - 1) => 1, I(k - 1) => 2, R(k - 1) => 3]), unknowns(syss))
 p = MTKParameters(syss, [c, nsteps, δt, β, γ] .=> collect(1:5))
 df.f(du, u, p, 0)
-@test_broken getu(syss, [S, I, R])
 reorderer = getu(syss, [S(k - 1), I(k - 1), R(k - 1)])
 @test reorderer(du) ≈ [0.01831563888873422, 0.9816849729159067, 4.999999388195359]
 
@@ -58,7 +57,7 @@ prob_map = DiscreteProblem(syss, [u0; p], tspan)
 # Solution
 using OrdinaryDiffEq
 sol_map = solve(prob_map, FunctionMap());
-@test_broken sol_map[S] isa Vector
+@test sol_map[S] isa Vector
 @test sol_map[S(k - 1)] isa Vector
 
 # Using defaults constructor
@@ -86,8 +85,8 @@ sol_map2 = solve(prob_map2, FunctionMap());
 for p in parameters(sys)
     @test sol_map.prob.ps[p] ≈ sol_map2.prob.ps[p]
 end
-@test sol_map2[R2][begin:(end - 1)] == sol_map2[R(k - 1)][(begin + 1):end]
-@test_broken sol_map2[R2(k + 1)][begin:(end - 1)] == sol_map2[R][(begin + 1):end]
+@test sol_map2[R2][begin:(end - 1)] == sol_map2[R(k - 1)][(begin + 1):end] ==
+      sol_map2[R][begin:(end - 1)]
 # Direct Implementation
 
 function sir_map!(u_diff, u, p, t)
@@ -103,14 +102,12 @@ function sir_map!(u_diff, u, p, t)
     end
     nothing
 end;
-@test_broken prob_map2[[S, I, R]]
-u0 = prob_map2[[S(k - 1), I(k - 1), R(k - 1)]];
+u0 = prob_map2[[S, I, R]];
 p = [0.05, 10.0, 0.25, 0.1];
 prob_map = DiscreteProblem(sir_map!, u0, tspan, p);
 sol_map2 = solve(prob_map, FunctionMap());
 
-@test_broken reduce(hcat, sol_map[[S, I, R]]) ≈ Array(sol_map2)
-@test reduce(hcat, sol_map[[S(k - 1), I(k - 1), R(k - 1)]]) ≈ Array(sol_map2)
+@test reduce(hcat, sol_map[[S, I, R]]) ≈ Array(sol_map2)
 
 # Delayed difference equation
 # @variables x(..) y(..) z(t)
@@ -260,70 +257,6 @@ end
     end
 end
 
-@testset "Initialization" begin
-    @test_broken begin
-        # test that default values apply to the entire history
-        @variables x(t) = 1.0
-        @mtkcompile de = System([x ~ x(k - 1) + x(k - 2)], t)
-        prob = DiscreteProblem(de, [], (0, 10))
-        @test prob[x] == 2.0
-        @test prob[x(k - 1)] == 1.0
-
-        # must provide initial conditions for history
-        @test_throws ErrorException DiscreteProblem(de, [x => 2.0], (0, 10))
-        @test_throws ErrorException DiscreteProblem(de, [x(k + 1) => 2.0], (0, 10))
-
-        # initial values only affect _that timestep_, not the entire history
-        prob = DiscreteProblem(de, [x(k - 1) => 2.0], (0, 10))
-        @test prob[x] == 3.0
-        @test prob[x(k - 1)] == 2.0
-        @variables xₜ₋₁(t)
-        @test prob[xₜ₋₁] == 2.0
-
-        # Test initial assignment with lowered variable
-        prob = DiscreteProblem(de, [xₜ₋₁(k - 1) => 4.0], (0, 10))
-        @test prob[x(k - 1)] == prob[xₜ₋₁] == 1.0
-        @test prob[x] == 5.0
-
-        # Test missing initial throws error
-        @variables x(t)
-        @mtkcompile de = System([x ~ x(k - 1) + x(k - 2) * x(k - 3)], t)
-        @test_throws ErrorException prob=DiscreteProblem(de, [x(k - 3) => 2.0], (0, 10))
-        @test_throws ErrorException prob=DiscreteProblem(
-            de, [x(k - 3) => 2.0, x(k - 1) => 3.0], (0, 10))
-
-        # Test non-assigned initials are given default value
-        @variables x(t) = 2.0
-        @mtkcompile de = System([x ~ x(k - 1) + x(k - 2) * x(k - 3)], t)
-        prob = DiscreteProblem(de, [x(k - 3) => 12.0], (0, 10))
-        @test prob[x] == 26.0
-        @test prob[x(k - 1)] == 2.0
-        @test prob[x(k - 2)] == 2.0
-
-        # Elaborate test
-        @variables xₜ₋₂(t) zₜ₋₁(t) z(t)
-        eqs = [x ~ x(k - 1) + z(k - 2),
-            z ~ x(k - 2) * x(k - 3) - z(k - 1)^2]
-        @mtkcompile de = System(eqs, t)
-        u0 = [x(k - 1) => 3,
-            xₜ₋₂(k - 1) => 4,
-            x(k - 2) => 1,
-            z(k - 1) => 5,
-            zₜ₋₁(k - 1) => 12]
-        prob = DiscreteProblem(de, u0, (0, 10))
-        @test prob[x] == 15
-        @test prob[z] == -21
-
-        import ModelingToolkit: shift2term
-        # unknowns(de) = xₜ₋₁, x, zₜ₋₁, xₜ₋₂, z
-        vars = sort(ModelingToolkit.value.(unknowns(de)); by = string)
-        @test isequal(shift2term(Shift(t, 1)(vars[2])), vars[1])
-        @test isequal(shift2term(Shift(t, 1)(vars[3])), vars[2])
-        @test isequal(shift2term(Shift(t, -1)(vars[4])), vars[5])
-        @test isequal(shift2term(Shift(t, -2)(vars[1])), vars[3])
-    end
-end
-
 @testset "Shifted array variables" begin
     @variables x(t)[1:2] y(t)[1:2]
     k = ShiftIndex(t)
@@ -339,6 +272,5 @@ end
         (0, 4))
     @test all(isone, prob.u0)
     sol = solve(prob, FunctionMap())
-    @test_broken sol[[x..., y...], end]
-    @test sol[[x(k - 1)..., y(k - 1)...], end] == 8ones(4)
+    @test sol[[x..., y...], end] == 8ones(4)
 end

test/implicit_discrete_system.jl

@@ -1,4 +1,4 @@
-using ModelingToolkit, Test
+using ModelingToolkit, SymbolicIndexingInterface, Test
 using ModelingToolkit: t_nounits as t
 using StableRNGs
 
@@ -45,6 +45,8 @@ end
             1 - (u_next[1] + u_next[2])^2 - u_next[3]^2]
     end
 
+    reorderer = getu(sys, [x(k - 2), x(k - 1), y])
+
     for _ in 1:10
         u_next = rand(rng, 3)
         u = rand(rng, 3)
@@ -73,6 +75,6 @@ end
         y(k) ~ x(k - 1) + x(k - 2),
         z(k) * x(k) ~ 3]
     @mtkcompile sys = System(eqs, t)
-    @test occursin("var\"Shift(t, 1)(z(t))\"",
+    @test occursin("var\"Shift(t, 1)(x(t))\"",
         string(ImplicitDiscreteFunction(sys; expression = Val{true})))
 end