"""
Unit tests for PerfectPlasticity material model.

Tests cover:
1. Construction and validation
2. State initialization
3. Elastic loading (f < 0)
4. Plastic loading (f > 0)
5. Radial return mapping
6. Hardening behavior
7. Cyclic loading (Bauschinger effect)
8. Consistency (yield surface constraint)
9. Zero allocation (after compilation)
10. Type stability
"""

using Test
using Tensors
using LinearAlgebra

# Load implementation
include("../src/materials/perfect_plasticity.jl")

@testset "Perfect Plasticity Material" begin

    @testset "Material Construction" begin
        # Valid construction
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
        @test steel.E == 200e9
        @test steel.ν == 0.3
        @test steel.σ_y == 250e6
        @test steel.H == 1e9
        @test steel.μ ≈ 200e9 / (2 * (1 + 0.3))
        @test steel.λ ≈ 200e9 * 0.3 / ((1 + 0.3) * (1 - 2 * 0.3))

        # Perfect plasticity (H=0)
        perfect = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=0.0)
        @test perfect.H == 0.0

        # Invalid inputs
        @test_throws ArgumentError PerfectPlasticity(E=-200e9, ν=0.3, σ_y=250e6, H=1e9)  # Negative E
        @test_throws ArgumentError PerfectPlasticity(E=200e9, ν=0.6, σ_y=250e6, H=1e9)   # ν too large
        @test_throws ArgumentError PerfectPlasticity(E=200e9, ν=0.3, σ_y=-250e6, H=1e9)  # Negative σ_y
        @test_throws ArgumentError PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=-1e9)  # Negative H
    end

    @testset "State Construction" begin
        # Default state (zero)
        state0 = PlasticityState()
        @test state0.ε_p == zero(SymmetricTensor{2,3})
        @test state0.α == zero(SymmetricTensor{2,3})
        @test state0.κ == 0.0

        # Custom state
        ε_p = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))
        α = SymmetricTensor{2,3}((1e8, 0.0, 0.0, 0.0, 0.0, 0.0))
        state = PlasticityState(ε_p, α, 0.01)
        @test state.ε_p == ε_p
        @test state.α == α
        @test state.κ == 0.01

        # Invalid state (negative κ)
        @test_throws ArgumentError PlasticityState(ε_p, α, -0.01)
    end

    @testset "Elastic Loading (Small Strain)" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)

        # Small strain (well below yield)
        ε_small = SymmetricTensor{2,3}((1e-5, 0.0, 0.0, 0.0, 0.0, 0.0))

        σ, 𝔻, state_new = compute_stress(steel, ε_small, nothing, 0.0)

        # Should remain elastic
        @test state_new.ε_p == zero(SymmetricTensor{2,3})  # No plastic strain
        @test state_new.α == zero(SymmetricTensor{2,3})    # No backstress
        @test state_new.κ == 0.0                            # No plastic work

        # Stress should be elastic
        μ = steel.μ
        λ = steel.λ
        I = one(ε_small)
        σ_elastic = λ * tr(ε_small) * I + 2μ * ε_small
        @test σ ≈ σ_elastic rtol = 1e-12

        # Tangent should be elastic
        @test 𝔻 isa SymmetricTensor{4,3}
    end

    @testset "Plastic Loading (Yield)" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)

        # Strain beyond yield (uniaxial tension)
        # Yield strain: ε_y = σ_y / E ≈ 0.00125
        ε_plastic = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))

        σ, 𝔻, state_new = compute_stress(steel, ε_plastic, nothing, 0.0)

        # Should have plastic strain
        @test norm(state_new.ε_p) > 0.0
        @test state_new.κ > 0.0

        # Check yield criterion (should be satisfied)
        s = dev(σ - state_new.α)
        von_mises = √(3 / 2) * √(s ⊡ s)
        @test von_mises ≈ steel.σ_y rtol = 1e-6  # On yield surface

        # Plastic strain should be deviatoric
        @test abs(tr(state_new.ε_p)) < 1e-12
    end

    @testset "Radial Return Mapping" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)

        # Large strain (far beyond yield)
        ε_large = SymmetricTensor{2,3}((0.01, 0.0, 0.0, 0.0, 0.0, 0.0))

        σ, 𝔻, state_new = compute_stress(steel, ε_large, nothing, 0.0)

        # Check yield criterion (must be satisfied)
        s = dev(σ - state_new.α)
        von_mises = √(3 / 2) * √(s ⊡ s)
        @test von_mises ≈ steel.σ_y rtol = 1e-6

        # Stress should be less than elastic prediction
        μ = steel.μ
        λ = steel.λ
        I = one(ε_large)
        σ_elastic = λ * tr(ε_large) * I + 2μ * ε_large
        @test norm(σ) < norm(σ_elastic)

        # Plastic strain should be significant
        @test norm(state_new.ε_p) > 1e-4
    end

    @testset "Hardening Behavior" begin
        # Compare hardening vs perfect plasticity
        steel_hard = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
        steel_perf = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=0.0)

        ε_test = SymmetricTensor{2,3}((0.005, 0.0, 0.0, 0.0, 0.0, 0.0))

        σ_hard, _, state_hard = compute_stress(steel_hard, ε_test, nothing, 0.0)
        σ_perf, _, state_perf = compute_stress(steel_perf, ε_test, nothing, 0.0)

        # Hardening material should have backstress
        @test norm(state_hard.α) > 0.0
        @test norm(state_perf.α) == 0.0

        # Hardening material should have higher stress
        @test norm(σ_hard) > norm(σ_perf)
    end

    @testset "Incremental Loading" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)

        # Load in increments
        n_steps = 10
        ε_max = 0.005

        state = PlasticityState()
        stresses = []
        plastic_strains = []

        for i in 1:n_steps
            ε = SymmetricTensor{2,3}((i * ε_max / n_steps, 0.0, 0.0, 0.0, 0.0, 0.0))
            σ, _, state = compute_stress(steel, ε, state, 0.0)
            push!(stresses, σ[1, 1])
            push!(plastic_strains, state.κ)
        end

        # Stress should increase monotonically (hardening)
        @test all(diff(stresses) .≥ 0)

        # Plastic strain should increase monotonically
        @test all(diff(plastic_strains) .≥ 0)

        # Final plastic strain should be positive
        @test plastic_strains[end] > 0.0
    end

    @testset "Bauschinger Effect (Cyclic Loading)" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=10e9)  # High H for visibility

        # Step 1: Tension to plastic regime
        ε_tension = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))
        σ_t, _, state_t = compute_stress(steel, ε_tension, nothing, 0.0)

        # Step 2: Reverse to compression
        ε_compression = SymmetricTensor{2,3}((-0.002, 0.0, 0.0, 0.0, 0.0, 0.0))
        σ_c, _, state_c = compute_stress(steel, ε_compression, state_t, 0.0)

        # Should yield in compression earlier (Bauschinger effect from backstress)
        @test state_c.κ > state_t.κ  # Additional plastic strain
        @test norm(state_c.α) > 0.0   # Backstress present
    end

    @testset "Pure Shear" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)

        # Pure shear strain
        γ = 0.005
        ε_shear = SymmetricTensor{2,3}((0.0, γ / 2, 0.0, 0.0, 0.0, 0.0))

        σ, _, state = compute_stress(steel, ε_shear, nothing, 0.0)

        # Check shear stress
        @test abs(σ[1, 2]) > 0.0

        # Check yield in shear
        # For pure shear: τ_yield = σ_y / √3
        s = dev(σ - state.α)
        von_mises = √(3 / 2) * √(s ⊡ s)

        if von_mises > steel.σ_y - 1e-3  # Plastic
            @test von_mises ≈ steel.σ_y rtol = 1e-6
        end
    end

    @testset "Consistency Check" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)

        # Multiple strain levels
        strain_levels = [0.001, 0.002, 0.005, 0.01, 0.02]

        for ε_mag in strain_levels
            ε = SymmetricTensor{2,3}((ε_mag, 0.0, 0.0, 0.0, 0.0, 0.0))
            σ, _, state = compute_stress(steel, ε, nothing, 0.0)

            # Check yield criterion
            s = dev(σ - state.α)
            von_mises = √(3 / 2) * √(s ⊡ s)

            # Must satisfy: f = von_mises - σ_y ≤ 0
            f = von_mises - steel.σ_y
            @test f ≤ 1e-6  # On or inside yield surface
        end
    end

    @testset "Simplified Interface" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
        ε = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))

        # Test with and without explicit state/Δt
        σ1, 𝔻1, state1 = compute_stress(steel, ε)
        σ2, 𝔻2, state2 = compute_stress(steel, ε, nothing, 0.0)

        @test σ1 ≈ σ2
        @test 𝔻1 ≈ 𝔻2
        @test state1.κ ≈ state2.κ
    end

    @testset "Zero Allocation" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
        state = PlasticityState()

        # Test elastic path (no state change)
        ε_elastic = SymmetricTensor{2,3}((1e-5, 0.0, 0.0, 0.0, 0.0, 0.0))

        # First call to compile
        compute_stress(steel, ε_elastic, state, 0.0)

        # Check allocations on elastic path
        allocs_elastic = @allocated compute_stress(steel, ε_elastic, state, 0.0)
        @test allocs_elastic == 0  # Elastic path should have zero allocations

        # Test plastic path (state changes)
        ε_plastic = SymmetricTensor{2,3}((0.003, 0.0, 0.0, 0.0, 0.0, 0.0))

        # First call to compile
        compute_stress(steel, ε_plastic, state, 0.0)

        # Check allocations on plastic path
        allocs_plastic = @allocated compute_stress(steel, ε_plastic, state, 0.0)
        # Note: Plastic path allocates ~128 bytes for PlasticityState struct
        # This is acceptable for stateful materials
        @test allocs_plastic ≤ 256  # Allow some allocation for state
    end

    @testset "Type Stability" begin
        steel = PerfectPlasticity(E=200e9, ν=0.3, σ_y=250e6, H=1e9)
        ε = SymmetricTensor{2,3}((0.002, 0.0, 0.0, 0.0, 0.0, 0.0))
        state = PlasticityState()

        # Infer return types
        result = @inferred compute_stress(steel, ε, state, 0.0)

        @test result isa Tuple{SymmetricTensor{2,3,Float64},
            SymmetricTensor{4,3,Float64},
            PlasticityState}
    end

end