//! Guided Derivation Studio for Special Functions
//!
//! This example provides step-by-step mathematical derivations with:
//! - Interactive proof construction
//! - Hints and explanations at each step
//! - Validation of mathematical reasoning
//! - Multiple derivation paths for the same result
//! - Historical context and alternative approaches
//! - Connection to physical and mathematical applications
//!
//! Run with: cargo run --example guided_derivation_studio

use std::io::{self, Write};

#[derive(Debug, Clone)]
struct DerivationStep {
    description: String,
    mathematical_content: String,
    hints: Vec<String>,
    alternative_approaches: Vec<String>,
    validation_questions: Vec<ValidationQuestion>,
    #[allow(dead_code)]
    difficulty_level: u32,
}

#[derive(Debug, Clone)]
struct ValidationQuestion {
    question: String,
    options: Vec<String>,
    correct_answer: usize,
    explanation: String,
}

#[derive(Debug, Clone)]
struct DerivationSession {
    title: String,
    steps: Vec<DerivationStep>,
    current_step: usize,
    completed_steps: Vec<bool>,
    hints_used: u32,
    start_time: std::time::Instant,
    difficulty_level: u32,
}

impl DerivationSession {
    fn new(title: String, steps: Vec<DerivationStep>, difficulty: u32) -> Self {
        let num_steps = steps.len();
        Self {
            title,
            steps,
            current_step: 0,
            completed_steps: vec![false; num_steps],
            hints_used: 0,
            start_time: std::time::Instant::now(),
            difficulty_level: difficulty,
        }
    }

    fn progress_percentage(&self) -> f64 {
        let completed = self.completed_steps.iter().filter(|&&x| x).count();
        (completed as f64 / self.steps.len() as f64) * 100.0
    }

    fn is_complete(&self) -> bool {
        self.completed_steps.iter().all(|&x| x)
    }
}

#[allow(dead_code)]
fn main() -> Result<(), Box<dyn std::error::Error>> {
    println!("📚 Guided Derivation Studio for Special Functions");
    println!("================================================\n");

    println!("🎯 Master mathematical derivations through guided exploration!");
    println!("Each derivation is broken down into digestible steps with hints and validation.\n");

    loop {
        display_derivation_menu();
        let choice = get_user_input("Choose a derivation (1-8, or 'q' to quit): ")?;

        if choice.to_lowercase() == "q" {
            println!("👋 Thank you for using the Guided Derivation Studio!");
            println!("Keep exploring the beautiful mathematics of special functions!");
            break;
        }

        match choice.parse::<u32>() {
            Ok(1) => gamma_half_derivation()?,
            Ok(2) => stirling_approximation_derivation()?,
            Ok(3) => bessel_generating_function_derivation()?,
            Ok(4) => error_function_series_derivation()?,
            Ok(5) => legendre_orthogonality_derivation()?,
            Ok(6) => hypergeometric_integral_derivation()?,
            Ok(7) => spherical_harmonics_derivation()?,
            Ok(8) => wright_function_asymptotic_derivation()?,
            _ => println!("❌ Invalid choice. Please try again.\n"),
        }
    }

    Ok(())
}

#[allow(dead_code)]
fn display_derivation_menu() {
    println!("📖 Available Derivations:");
    println!("1. 🎲 Γ(1/2) = √π (Beginner)");
    println!("2. 📈 Stirling's Approximation (Intermediate)");
    println!("3. 🌊 Bessel Function Generating Function (Intermediate)");
    println!("4. 📊 Error Function Series Expansion (Beginner)");
    println!("5. 📐 Legendre Polynomial Orthogonality (Advanced)");
    println!("6. 🔢 Hypergeometric Integral Representation (Advanced)");
    println!("7. 🌍 Spherical Harmonics from Laplace Equation (Expert)");
    println!("8. 🧮 Wright Function Asymptotic Behavior (Expert)");
    println!();
}

#[allow(dead_code)]
fn gamma_half_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n🎲 Derivation: Γ(1/2) = √π");
    println!("===========================\n");

    let steps = vec![
        DerivationStep {
            description: "Start with the gamma function definition".to_string(),
            mathematical_content: "Γ(1/2) = ∫₀^∞ t^(1/2-1) e^(-t) dt = ∫₀^∞ t^(-1/2) e^(-t) dt".to_string(),
            hints: vec![
                "The gamma function is defined as Γ(z) = ∫₀^∞ t^(z-1) e^(-t) dt".to_string(),
                "For z = 1/2, we have z-1 = 1/2-1 = -1/2".to_string(),
            ],
            alternative_approaches: vec![
                "Could also use the beta function relationship".to_string(),
                "Or the duplication formula for gamma functions".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the exponent of t in the integrand?".to_string(),
                    options: vec!["1/2".to_string(), "-1/2".to_string(), "0".to_string(), "1".to_string()],
                    correct_answer: 1,
                    explanation: "Since Γ(z) uses t^(z-1), for z=1/2 we get t^(-1/2)".to_string(),
                }
            ],
            difficulty_level: 1,
        },

        DerivationStep {
            description: "Apply the substitution t = u²".to_string(),
            mathematical_content: "Let t = u², then dt = 2u du\nΓ(1/2) = ∫₀^∞ (u²)^(-1/2) e^(-u²) · 2u du = ∫₀^∞ u^(-1) e^(-u²) · 2u du = 2∫₀^∞ e^(-u²) du".to_string(),
            hints: vec![
                "When t = u², we have t^(-1/2) = (u²)^(-1/2) = u^(-1)".to_string(),
                "Don't forget the Jacobian: dt = 2u du".to_string(),
                "The u^(-1) and u terms cancel to give just the exponential".to_string(),
            ],
            alternative_approaches: vec![
                "Could use trigonometric substitution t = tan²θ".to_string(),
                "Or relate to the beta function B(1/2, 1/2)".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "After substitution, what cancels in the integrand?".to_string(),
                    options: vec!["e^(-u²) terms".to_string(), "u^(-1) and u terms".to_string(), "The constant 2".to_string(), "Nothing cancels".to_string()],
                    correct_answer: 1,
                    explanation: "u^(-1) · 2u = 2, leaving just 2e^(-u²)".to_string(),
                }
            ],
            difficulty_level: 2,
        },

        DerivationStep {
            description: "Recognize the Gaussian integral".to_string(),
            mathematical_content: "We know that ∫_{-∞}^∞ e^(-u²) du = √π (the famous Gaussian integral)\nBy symmetry: ∫₀^∞ e^(-u²) du = (1/2)√π".to_string(),
            hints: vec![
                "The Gaussian integral is a fundamental result in analysis".to_string(),
                "It can be proven using polar coordinates and the identity ∫∫ e^(-(x²+y²)) dx dy = π".to_string(),
                "The integral from 0 to ∞ is exactly half of the integral from -∞ to ∞".to_string(),
            ],
            alternative_approaches: vec![
                "Could derive the Gaussian integral using polar coordinates".to_string(),
                "Or use complex analysis and the residue theorem".to_string(),
                "Or employ Feynman's trick with parameter differentiation".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why is ∫₀^∞ e^(-u²) du = (1/2)∫_{-∞}^∞ e^(-u²) du?".to_string(),
                    options: vec!["By substitution".to_string(), "By symmetry of e^(-u²)".to_string(), "By integration by parts".to_string(), "It's not true".to_string()],
                    correct_answer: 1,
                    explanation: "e^(-u²) is an even function, so the integral from -∞ to 0 equals the integral from 0 to ∞".to_string(),
                }
            ],
            difficulty_level: 2,
        },

        DerivationStep {
            description: "Complete the calculation".to_string(),
            mathematical_content: "Therefore: Γ(1/2) = 2 · (1/2)√π = √π ✓\n\nThis beautiful result connects the gamma function to π!".to_string(),
            hints: vec![
                "Just substitute the known value of the Gaussian integral".to_string(),
                "2 · (1/2) = 1, so we get exactly √π".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify numerically: Γ(0.5) ≈ 1.7725 and √π ≈ 1.7725".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the numerical value of √π to 4 decimal places?".to_string(),
                    options: vec!["1.4142".to_string(), "1.7725".to_string(), "2.7183".to_string(), "3.1416".to_string()],
                    correct_answer: 1,
                    explanation: "√π ≈ 1.7725, which matches the numerical value of Γ(1/2)".to_string(),
                }
            ],
            difficulty_level: 1,
        },
    ];

    let session = DerivationSession::new("Γ(1/2) = √π".to_string(), steps, 1);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn stirling_approximation_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n📈 Derivation: Stirling's Approximation");
    println!("=======================================\n");

    let steps = vec![
        DerivationStep {
            description: "Express ln Γ(z) using the integral representation".to_string(),
            mathematical_content: "ln Γ(z) = ln[∫₀^∞ t^(z-1) e^(-t) dt] = ∫₀^∞ e^((z-1)ln t - t) dt\n\nWe want to find the asymptotic behavior for large z.".to_string(),
            hints: vec![
                "Taking logarithm of the integral allows us to work with the exponent".to_string(),
                "The integrand becomes e^(f(t)) where f(t) = (z-1)ln t - t".to_string(),
                "For large z, this integral will be dominated by the maximum of f(t)".to_string(),
            ],
            alternative_approaches: vec![
                "Could start with the factorial formula n! and use ln(n!)".to_string(),
                "Or use the Euler-Maclaurin formula".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is f(t) in the exponent e^(f(t))?".to_string(),
                    options: vec!["(z-1)ln t".to_string(), "-t".to_string(), "(z-1)ln t - t".to_string(), "z ln t".to_string()],
                    correct_answer: 2,
                    explanation: "f(t) = (z-1)ln t - t comes from t^(z-1) e^(-t) = e^((z-1)ln t - t)".to_string(),
                }
            ],
            difficulty_level: 3,
        },

        DerivationStep {
            description: "Find the saddle point (maximum of the exponent)".to_string(),
            mathematical_content: "To find the maximum of f(t) = (z-1)ln t - t:\nf'(t) = (z-1)/t - 1 = 0\n\nSolving: (z-1)/t = 1 ⟹ t₀ = z-1\n\nSecond derivative: f''(t₀) = -(z-1)/t₀² = -1/(z-1)".to_string(),
            hints: vec![
                "The maximum occurs where the derivative is zero".to_string(),
                "At the maximum, the integrand contributes most to the integral".to_string(),
                "The second derivative tells us about the curvature at the maximum".to_string(),
            ],
            alternative_approaches: vec![
                "Could use Lagrange multipliers to find the constrained maximum".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the saddle point t₀?".to_string(),
                    options: vec!["z".to_string(), "z-1".to_string(), "1/(z-1)".to_string(), "ln(z-1)".to_string()],
                    correct_answer: 1,
                    explanation: "Setting f'(t) = 0 gives (z-1)/t = 1, so t₀ = z-1".to_string(),
                }
            ],
            difficulty_level: 3,
        },

        DerivationStep {
            description: "Apply the method of steepest descent".to_string(),
            mathematical_content: "Expand around t₀: f(t) ≈ f(t₀) + (1/2)f''(t₀)(t-t₀)²\n\nf(t₀) = (z-1)ln(z-1) - (z-1)\nf''(t₀) = -1/(z-1)\n\nThe integral becomes: ∫ e^(f(t₀)) e^(-1/2 · 1/(z-1) · (t-t₀)²) dt".to_string(),
            hints: vec![
                "This is a second-order Taylor expansion around the maximum".to_string(),
                "The first-order term vanishes because we're at a critical point".to_string(),
                "The integral now looks like a Gaussian integral".to_string(),
            ],
            alternative_approaches: vec![
                "Could use the WKB approximation from quantum mechanics".to_string(),
                "Or the stationary phase method from complex analysis".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why does the first-order term (t-t₀) disappear?".to_string(),
                    options: vec!["It's too small".to_string(), "f'(t₀) = 0 at the critical point".to_string(), "It cancels with another term".to_string(), "By symmetry".to_string()],
                    correct_answer: 1,
                    explanation: "At the critical point, f'(t₀) = 0, so the linear term vanishes".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Evaluate the Gaussian integral and derive Stirling's formula".to_string(),
            mathematical_content: "∫_{-∞}^∞ e^(-1/2 · 1/(z-1) · u²) du = √(2π(z-1))\n\nTherefore: Γ(z) ≈ e^((z-1)ln(z-1) - (z-1)) √(2π(z-1))\n= √(2π(z-1)) ((z-1)/e)^(z-1)\n\nFor large z: Γ(z) ≈ √(2π/z) (z/e)^z".to_string(),
            hints: vec![
                "The Gaussian integral ∫ e^(-au²) du = √(π/a)".to_string(),
                "For large z, (z-1) ≈ z in the leading behavior".to_string(),
                "This gives the famous Stirling's approximation".to_string(),
            ],
            alternative_approaches: vec![
                "Could include higher-order corrections for better accuracy".to_string(),
                "Or derive the complete asymptotic series".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the leading behavior of ln Γ(z) for large z?".to_string(),
                    options: vec!["z ln z".to_string(), "z ln z - z".to_string(), "(z-1/2) ln z - z".to_string(), "z²".to_string()],
                    correct_answer: 2,
                    explanation: "ln Γ(z) ≈ (z-1/2) ln z - z + (1/2) ln(2π) for large z".to_string(),
                }
            ],
            difficulty_level: 4,
        },
    ];

    let session = DerivationSession::new("Stirling's Approximation".to_string(), steps, 3);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn bessel_generating_function_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n🌊 Derivation: Bessel Function Generating Function");
    println!("=================================================\n");

    let steps = vec![
        DerivationStep {
            description: "Start with the exponential generating function".to_string(),
            mathematical_content: "Consider: G(x,t) = exp(x/2 · (t - 1/t))\n\nWe will show that: G(x,t) = Σ_{n=-∞}^∞ J_n(x) t^n".to_string(),
            hints: vec![
                "This is known as the generating function for Bessel functions".to_string(),
                "The argument x/2(t - 1/t) has a special structure".to_string(),
                "We'll expand this exponential and identify coefficients".to_string(),
            ],
            alternative_approaches: vec![
                "Could derive from the integral representation of Bessel functions".to_string(),
                "Or start from the differential equation".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What symmetry does the exponent x/2(t - 1/t) have?".to_string(),
                    options: vec!["Even in t".to_string(), "Odd in t".to_string(), "Anti-symmetric under t → 1/t".to_string(), "No special symmetry".to_string()],
                    correct_answer: 2,
                    explanation: "If we replace t with 1/t, we get x/2(1/t - t) = -x/2(t - 1/t)".to_string(),
                }
            ],
            difficulty_level: 3,
        },

        DerivationStep {
            description: "Expand the exponential function".to_string(),
            mathematical_content: "G(x,t) = exp(x/2 · (t - 1/t)) = exp(xt/2) · exp(-x/(2t))\n\n= [Σ_{m=0}^∞ (xt/2)^m/m!] · [Σ_{k=0}^∞ (-x/(2t))^k/k!]\n\n= [Σ_{m=0}^∞ (x/2)^m t^m/m!] · [Σ_{k=0}^∞ (-1)^k (x/2)^k t^(-k)/k!]".to_string(),
            hints: vec![
                "Use the fact that e^(A+B) = e^A · e^B when A and B commute".to_string(),
                "Each exponential can be expanded as a power series".to_string(),
                "Collect powers of t to find coefficients".to_string(),
            ],
            alternative_approaches: vec![
                "Could use the binomial theorem for complex arguments".to_string(),
                "Or employ generating function techniques directly".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "In the product of series, what determines the coefficient of t^n?".to_string(),
                    options: vec!["Only terms with m = n".to_string(), "Terms where m - k = n".to_string(), "All terms".to_string(), "Terms with m + k = n".to_string()],
                    correct_answer: 1,
                    explanation: "We need t^m · t^(-k) = t^n, so m - k = n".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Collect coefficients of t^n".to_string(),
            mathematical_content: "Coefficient of t^n:\nΣ_{m} (x/2)^m/m! · (-1)^(m-n) (x/2)^(m-n)/(m-n)!\n\nwhere the sum is over m ≥ max(0, n) and m-n ≥ 0.\n\nThis gives: J_n(x) = (x/2)^n Σ_{k=0}^∞ (-1)^k (x/2)^(2k)/(k!(n+k)!)".to_string(),
            hints: vec![
                "Change the summation index: let k = m - n".to_string(),
                "The constraints become k ≥ 0 and m = n + k ≥ 0".to_string(),
                "This leads to the series representation of J_n(x)".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify by substituting back into the differential equation".to_string(),
                "Or check specific cases like J_0(x) and J_1(x)".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the leading term in J_n(x) for small x?".to_string(),
                    options: vec!["x^n".to_string(), "(x/2)^n/n!".to_string(), "x^(2n)".to_string(), "1".to_string()],
                    correct_answer: 1,
                    explanation: "The k=0 term gives (x/2)^n/n!, which dominates for small x".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Verify the result and explore consequences".to_string(),
            mathematical_content: "The generating function gives us:\nexp(x/2(t - 1/t)) = Σ_{n=-∞}^∞ J_n(x) t^n\n\nConsequences:\n• J_{-n}(x) = (-1)^n J_n(x) (from t → 1/t symmetry)\n• Addition formula for Bessel functions\n• Recurrence relations".to_string(),
            hints: vec![
                "The symmetry property comes from the anti-symmetry of the exponent".to_string(),
                "Setting t = e^(iθ) gives the integral representation".to_string(),
                "This generating function is fundamental to Bessel function theory".to_string(),
            ],
            alternative_approaches: vec![
                "Could derive recurrence relations by differentiating".to_string(),
                "Or explore connections to physics applications".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What happens when we set t = 1 in the generating function?".to_string(),
                    options: vec!["We get 0".to_string(), "We get 1".to_string(), "We get J_0(x)".to_string(), "We get e^x".to_string()],
                    correct_answer: 2,
                    explanation: "When t = 1, only the n = 0 term survives, giving J_0(x)".to_string(),
                }
            ],
            difficulty_level: 3,
        },
    ];

    let session = DerivationSession::new("Bessel Generating Function".to_string(), steps, 3);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn error_function_series_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n📊 Derivation: Error Function Series Expansion");
    println!("===============================================\n");

    let steps = vec![
        DerivationStep {
            description: "Start with the definition of the error function".to_string(),
            mathematical_content: "erf(x) = (2/√π) ∫₀ˣ e^(-t²) dt\n\nOur goal: Find the power series expansion of erf(x).".to_string(),
            hints: vec![
                "The error function is defined as an integral".to_string(),
                "To find a series, we need to expand the integrand".to_string(),
                "Then integrate term by term".to_string(),
            ],
            alternative_approaches: vec![
                "Could use the Taylor series of erf(x) directly".to_string(),
                "Or derive from the differential equation erf'(x) = (2/√π)e^(-x²)".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is erf(0)?".to_string(),
                    options: vec!["0".to_string(), "1".to_string(), "1/2".to_string(), "2/√π".to_string()],
                    correct_answer: 0,
                    explanation: "erf(0) = (2/√π) ∫₀⁰ e^(-t²) dt = 0".to_string(),
                }
            ],
            difficulty_level: 1,
        },

        DerivationStep {
            description: "Expand e^(-t²) as a power series".to_string(),
            mathematical_content: "e^(-t²) = Σ_{n=0}^∞ (-t²)^n/n! = Σ_{n=0}^∞ (-1)^n t^(2n)/n!\n\nThis is valid for all t ∈ ℂ (entire function).".to_string(),
            hints: vec![
                "This is the standard exponential series with argument -t²".to_string(),
                "Each term involves even powers of t".to_string(),
                "The series converges everywhere".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify by differentiating the series".to_string(),
                "Or use the complex exponential e^(iz) = cos(z) + i sin(z)".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What type of powers of t appear in the expansion?".to_string(),
                    options: vec!["All powers".to_string(), "Only even powers".to_string(), "Only odd powers".to_string(), "Only integer powers".to_string()],
                    correct_answer: 1,
                    explanation: "Since we have (-t²)^n = (-1)^n t^(2n), only even powers appear".to_string(),
                }
            ],
            difficulty_level: 2,
        },

        DerivationStep {
            description: "Integrate term by term".to_string(),
            mathematical_content: "erf(x) = (2/√π) ∫₀ˣ [Σ_{n=0}^∞ (-1)^n t^(2n)/n!] dt\n\n= (2/√π) Σ_{n=0}^∞ (-1)^n/n! ∫₀ˣ t^(2n) dt\n\n= (2/√π) Σ_{n=0}^∞ (-1)^n/n! · x^(2n+1)/(2n+1)".to_string(),
            hints: vec![
                "Integration and summation can be exchanged for uniform convergence".to_string(),
                "∫₀ˣ t^(2n) dt = x^(2n+1)/(2n+1)".to_string(),
                "This gives the final series form".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify convergence using the ratio test".to_string(),
                "Or check by differentiating the series".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the power of x in the nth term of erf(x)?".to_string(),
                    options: vec!["2n".to_string(), "2n+1".to_string(), "n".to_string(), "n+1".to_string()],
                    correct_answer: 1,
                    explanation: "After integrating t^(2n), we get x^(2n+1)/(2n+1)".to_string(),
                }
            ],
            difficulty_level: 2,
        },

        DerivationStep {
            description: "Write the final series and verify".to_string(),
            mathematical_content: "erf(x) = (2/√π) Σ_{n=0}^∞ (-1)^n x^(2n+1)/(n!(2n+1))\n\nExpanded:\nerf(x) = (2/√π)[x - x³/3 + x⁵/(5·2!) - x⁷/(7·3!) + ...]\n\nVerification: erf'(x) = (2/√π)e^(-x²) ✓".to_string(),
            hints: vec![
                "The first few terms give good approximation for small x".to_string(),
                "Differentiating the series should give (2/√π)e^(-x²)".to_string(),
                "This series converges for all x".to_string(),
            ],
            alternative_approaches: vec![
                "Could compare with numerical values".to_string(),
                "Or explore asymptotic behavior for large x".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the coefficient of x³ in erf(x)?".to_string(),
                    options: vec!["2/√π".to_string(), "-2/(3√π)".to_string(), "1/3".to_string(), "-1/3".to_string()],
                    correct_answer: 1,
                    explanation: "The x³ term comes from n=1: (2/√π)(-1)¹x³/(1!·3) = -2/(3√π)".to_string(),
                }
            ],
            difficulty_level: 2,
        },
    ];

    let session = DerivationSession::new("Error Function Series".to_string(), steps, 1);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn legendre_orthogonality_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n📐 Derivation: Legendre Polynomial Orthogonality");
    println!("================================================\n");

    let steps = vec![
        DerivationStep {
            description: "Start with Legendre's differential equation".to_string(),
            mathematical_content: "Legendre polynomials P_n(x), satisfy:\n\n(1-x²)P_n''(x) - 2xP_n'(x) + n(n+1)P_n(x) = 0\n\nThis can be written as:\nd/dx[(1-x²)P_n'(x)] + n(n+1)P_n(x) = 0".to_string(),
            hints: vec![
                "This is Sturm-Liouville form with weight function w(x) = 1".to_string(),
                "The equation is self-adjoint".to_string(),
                "Different eigenvalues lead to orthogonal eigenfunctions".to_string(),
            ],
            alternative_approaches: vec![
                "Could start with Rodrigues' formula".to_string(),
                "Or use the generating function approach".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the eigenvalue in this Sturm-Liouville problem?".to_string(),
                    options: vec!["n".to_string(), "n+1".to_string(), "n(n+1)".to_string(), "n²".to_string()],
                    correct_answer: 2,
                    explanation: "The eigenvalue λ_n = n(n+1) appears in the equation".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Apply the orthogonality theorem for Sturm-Liouville problems".to_string(),
            mathematical_content: "For distinct eigenvalues λ_m ≠ λ_n, the eigenfunctions satisfy:\n\n∫_{-1}^1 P_m(x) P_n(x) w(x) dx = 0\n\nwhere w(x) = 1 for Legendre polynomials.\n\nSince λ_m = m(m+1) ≠ n(n+1) = λ_n for m ≠ n, we have orthogonality.".to_string(),
            hints: vec![
                "This is a general theorem for self-adjoint operators".to_string(),
                "The weight function w(x) = 1 on the interval [-1,1]".to_string(),
                "We need to find the normalization constant".to_string(),
            ],
            alternative_approaches: vec![
                "Could prove directly using integration by parts".to_string(),
                "Or use the generating function method".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why are λ_m and λ_n distinct for m ≠ n?".to_string(),
                    options: vec!["By definition".to_string(), "m(m+1) ≠ n(n+1) for m ≠ n".to_string(), "They're not always distinct".to_string(), "By orthogonality".to_string()],
                    correct_answer: 1,
                    explanation: "The function f(k) = k(k+1) is strictly increasing for k ≥ 0".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Calculate the normalization integral".to_string(),
            mathematical_content: "For the diagonal case m = n:\n\n∫_{-1}^1 [P_n(x)]² dx = ?\n\nUsing Rodrigues' formula: P_n(x) = (1/2ⁿn!) dⁿ/dxⁿ[(x²-1)ⁿ]\n\nBy repeated integration by parts:\n∫_{-1}^1 [P_n(x)]² dx = 2/(2n+1)".to_string(),
            hints: vec![
                "Rodrigues' formula relates P_n to derivatives of (x²-1)ⁿ".to_string(),
                "Integration by parts eliminates boundary terms (they vanish)".to_string(),
                "The calculation involves factorials and combinatorics".to_string(),
            ],
            alternative_approaches: vec![
                "Could use the generating function and Parseval's theorem".to_string(),
                "Or derive using recurrence relations".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why do boundary terms vanish in integration by parts?".to_string(),
                    options: vec!["P_n(±1) = 0".to_string(), "(x²-1)ⁿ and its derivatives vanish at x = ±1".to_string(), "By symmetry".to_string(), "They don't vanish".to_string()],
                    correct_answer: 1,
                    explanation: "(x²-1)ⁿ = 0 at x = ±1, and its first n-1 derivatives also vanish there".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "State the complete orthogonality relation".to_string(),
            mathematical_content: "The complete orthogonality relation for Legendre polynomials is:\n\n∫_{-1}^1 P_m(x) P_n(x) dx = (2/(2n+1)) δ_{mn}\n\nwhere δ_{mn} is the Kronecker delta.\n\nThis makes {P_n(x)} an orthogonal basis for L²[-1,1].".to_string(),
            hints: vec![
                "This relation is fundamental for expanding functions in Legendre series".to_string(),
                "The normalization factor 2/(2n+1) comes from the detailed calculation".to_string(),
                "This forms a complete orthogonal system".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify numerically for specific values".to_string(),
                "Or explore applications to boundary value problems".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is ∫_{-1}^1 P_0(x) P_1(x) dx?".to_string(),
                    options: vec!["0".to_string(), "1".to_string(), "2".to_string(), "2/3".to_string()],
                    correct_answer: 0,
                    explanation: "Since 0 ≠ 1, the orthogonality relation gives 0".to_string(),
                }
            ],
            difficulty_level: 3,
        },
    ];

    let session = DerivationSession::new("Legendre Orthogonality".to_string(), steps, 4);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn hypergeometric_integral_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n🔢 Derivation: Hypergeometric Integral Representation");
    println!("====================================================\n");

    let steps = vec![
        DerivationStep {
            description: "Start with the beta function integral".to_string(),
            mathematical_content: "Recall the beta function:\nB(p,q) = ∫₀¹ t^(p-1)(1-t)^(q-1) dt = Γ(p)Γ(q)/Γ(p+q)\n\nWe want to derive:\n₂F₁(a,b;c;z) = (Γ(c)/[Γ(b)Γ(c-b)]) ∫₀¹ t^(b-1)(1-t)^(c-b-1)(1-zt)^(-a) dt".to_string(),
            hints: vec![
                "The hypergeometric function is defined by its series".to_string(),
                "We need to connect the series to this integral".to_string(),
                "The key is expanding (1-zt)^(-a)".to_string(),
            ],
            alternative_approaches: vec![
                "Could start from the differential equation".to_string(),
                "Or use complex contour integration".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What condition ensures convergence of the integral?".to_string(),
                    options: vec!["Re(b) > 0, Re(c-b) > 0".to_string(), "Re(c) > Re(b) > 0".to_string(), "Both are correct".to_string(), "No conditions needed".to_string()],
                    correct_answer: 2,
                    explanation: "Both conditions ensure the integrand is well-behaved".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Expand (1-zt)^(-a) using the binomial series".to_string(),
            mathematical_content: "(1-zt)^(-a) = Σ_{n=0}^∞ (a)_n (zt)^n/n!\n\nwhere (a)_n = a(a+1)...(a+n-1) is the Pochhammer symbol.\n\nSubstituting:\n∫₀¹ t^(b-1)(1-t)^(c-b-1)(1-zt)^(-a) dt = Σ_{n=0}^∞ (a)_n z^n/n! ∫₀¹ t^(b+n-1)(1-t)^(c-b-1) dt".to_string(),
            hints: vec![
                "The binomial series (1-w)^(-a) = Σ (a)_n w^n/n! for |w| < 1".to_string(),
                "Here w = zt and we need |zt| < 1 in the integration region".to_string(),
                "We can exchange sum and integral for appropriate convergence".to_string(),
            ],
            alternative_approaches: vec![
                "Could use complex analysis to extend the domain".to_string(),
                "Or derive using generating functions".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is (a)₀ in the Pochhammer symbol?".to_string(),
                    options: vec!["0".to_string(), "1".to_string(), "a".to_string(), "undefined".to_string()],
                    correct_answer: 1,
                    explanation: "By convention, (a)₀ = 1 (empty product)".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Evaluate the beta function integrals".to_string(),
            mathematical_content: "Each integral is a beta function:\n∫₀¹ t^(b+n-1)(1-t)^(c-b-1) dt = B(b+n, c-b) = Γ(b+n)Γ(c-b)/Γ(c+n)\n\nUsing the identity Γ(z+n) = (z)_n Γ(z):\nΓ(b+n) = (b)_n Γ(b)\nΓ(c+n) = (c)_n Γ(c)\n\nTherefore: B(b+n, c-b) = (b)_n Γ(b) Γ(c-b)/[(c)_n Γ(c)]".to_string(),
            hints: vec![
                "The rising factorial (Pochhammer symbol) relates to the gamma function".to_string(),
                "Use the identity Γ(z+n) = z(z+1)...(z+n-1)Γ(z)".to_string(),
                "This simplifies the beta function expression".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify using the duplication formula".to_string(),
                "Or use contour integration for complex parameters".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What is the relationship between (b)_n and Γ(b+n)?".to_string(),
                    options: vec!["(b)_n = Γ(b+n)".to_string(), "Γ(b+n) = (b)_n Γ(b)".to_string(), "No simple relationship".to_string(), "(b)_n = n! Γ(b)".to_string()],
                    correct_answer: 1,
                    explanation: "Γ(b+n) = b(b+1)...(b+n-1)Γ(b) = (b)_n Γ(b)".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Complete the derivation".to_string(),
            mathematical_content: "Putting everything together:\n\n₂F₁(a,b;c;z) = Σ_{n=0}^∞ (a)_n (b)_n z^n/[(c)_n n!]\n\n= (Γ(c)/[Γ(b)Γ(c-b)]) Σ_{n=0}^∞ (a)_n z^n/n! · (b)_n Γ(b) Γ(c-b)/[(c)_n Γ(c)]\n\n= (Γ(c)/[Γ(b)Γ(c-b)]) ∫₀¹ t^(b-1)(1-t)^(c-b-1)(1-zt)^(-a) dt ✓\n\nThis is Euler's integral representation!".to_string(),
            hints: vec![
                "The series definition matches exactly with our integral".to_string(),
                "This representation is valid for Re(c) > Re(b) > 0".to_string(),
                "This connects series and integral representations beautifully".to_string(),
            ],
            alternative_approaches: vec![
                "Could extend to other hypergeometric functions".to_string(),
                "Or explore special cases like complete elliptic integrals".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What happens when z = 0 in the integral representation?".to_string(),
                    options: vec!["We get 0".to_string(), "We get B(b, c-b)".to_string(), "We get 1".to_string(), "The integral diverges".to_string()],
                    correct_answer: 2,
                    explanation: "When z = 0, (1-zt)^(-a) = 1, so we get B(b, c-b) = Γ(b)Γ(c-b)/Γ(c), and the hypergeometric function ₂F₁(a,b;c;0) = 1".to_string(),
                }
            ],
            difficulty_level: 4,
        },
    ];

    let session = DerivationSession::new("Hypergeometric Integral".to_string(), steps, 4);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn spherical_harmonics_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n🌍 Expert Derivation: Spherical Harmonics from Laplace Equation");
    println!("===============================================================\n");

    println!("⚠️ Expert Level: This derivation requires advanced knowledge of PDEs and complex analysis.");

    let steps = vec![
        DerivationStep {
            description: "Start with Laplace's equation in spherical coordinates".to_string(),
            mathematical_content: "∇²Ψ = 0 in spherical coordinates (r,θ,φ):\n\n(1/r²)[∂/∂r(r²∂Ψ/∂r) + (1/sin θ)∂/∂θ(sin θ ∂Ψ/∂θ) + (1/sin²θ)∂²Ψ/∂φ²] = 0\n\nSeek solutions of the form: Ψ(r,θ,φ) = R(r)Y(θ,φ)".to_string(),
            hints: vec![
                "Separation of variables works because of the coordinate system symmetry".to_string(),
                "The angular part Y(θ,φ) will give us spherical harmonics".to_string(),
                "The radial part R(r) involves powers of r".to_string(),
            ],
            alternative_approaches: vec![
                "Could start from the hydrogen atom Schrödinger equation".to_string(),
                "Or use group theory and rotation symmetry".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why can we separate r and angular variables?".to_string(),
                    options: vec!["Lucky guess".to_string(), "Coordinate system symmetry".to_string(), "Mathematical convenience".to_string(), "It doesn't always work".to_string()],
                    correct_answer: 1,
                    explanation: "Spherical coordinates respect the rotational symmetry of the problem".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Separate variables and derive the angular equation".to_string(),
            mathematical_content: "Substituting Ψ = R(r)Y(θ,φ) and dividing by RY:\n\n(1/R)d/dr[r²dR/dr] = -(1/Y)[(1/sin θ)∂/∂θ(sin θ ∂Y/∂θ) + (1/sin²θ)∂²Y/∂φ²]\n\nSince LHS depends only on r and RHS only on (θ,φ), both equal a constant ℓ(ℓ+1):\n\n(1/sin θ)∂/∂θ(sin θ ∂Y/∂θ) + (1/sin²θ)∂²Y/∂φ² + ℓ(ℓ+1)Y = 0".to_string(),
            hints: vec![
                "The separation constant ℓ(ℓ+1) is chosen for convenience".to_string(),
                "This form will lead to Legendre polynomials".to_string(),
                "The angular equation is an eigenvalue problem".to_string(),
            ],
            alternative_approaches: vec![
                "Could use the angular momentum operator L²".to_string(),
                "Or start with the generating function method".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why is the separation constant written as ℓ(ℓ+1)?".to_string(),
                    options: vec!["Mathematical tradition".to_string(), "It simplifies later calculations".to_string(), "It gives integer eigenvalues".to_string(), "All of the above".to_string()],
                    correct_answer: 3,
                    explanation: "This choice leads to integer ℓ values and simplifies the Legendre equation".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Further separate θ and φ variables".to_string(),
            mathematical_content: "Let Y(θ,φ) = Θ(θ)Φ(φ). The φ equation gives:\n\nd²Φ/dφ² = -m²Φ\n\nSolution: Φ(φ) = e^(imφ)\n\nPeriodicity requires Φ(φ+2π) = Φ(φ), so m ∈ ℤ.\n\nThe θ equation becomes:\n(1/sin θ)d/dθ[sin θ dΘ/dθ] + [ℓ(ℓ+1) - m²/sin²θ]Θ = 0".to_string(),
            hints: vec![
                "The φ equation is simple harmonic oscillator type".to_string(),
                "Periodicity in φ quantizes m to integers".to_string(),
                "The θ equation is the associated Legendre equation".to_string(),
            ],
            alternative_approaches: vec![
                "Could solve using power series methods".to_string(),
                "Or transform to standard Legendre form".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What values can m take?".to_string(),
                    options: vec!["Any real number".to_string(), "Integers: ..., -2, -1, 0, 1, 2, ...".to_string(), "Only positive integers".to_string(), "Only 0, ±1".to_string()],
                    correct_answer: 1,
                    explanation: "Periodicity in φ requires m to be an integer".to_string(),
                }
            ],
            difficulty_level: 4,
        },

        DerivationStep {
            description: "Solve the associated Legendre equation".to_string(),
            mathematical_content: "Substituting x = cos θ transforms the θ equation to:\n\n(1-x²)d²Θ/dx² - 2x dΘ/dx + [ℓ(ℓ+1) - m²/(1-x²)]Θ = 0\n\nFor solutions finite at x = ±1, we need:\n• ℓ ∈ {0,1,2,...} (non-negative integers)\n• |m| ≤ ℓ\n\nSolutions: Θ(θ) = P_ℓ^m(cos θ) (associated Legendre polynomials)".to_string(),
            hints: vec![
                "The substitution x = cos θ is standard for this type of equation".to_string(),
                "Boundary conditions at θ = 0, π require finite solutions".to_string(),
                "This quantizes both ℓ and m".to_string(),
            ],
            alternative_approaches: vec![
                "Could use Frobenius method to find series solutions".to_string(),
                "Or derive from Rodrigues' formula".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "For ℓ = 2, what values can m take?".to_string(),
                    options: vec!["0, 1, 2".to_string(), "-2, -1, 0, 1, 2".to_string(), "Any integer".to_string(), "Only 0".to_string()],
                    correct_answer: 1,
                    explanation: "For given ℓ, m can range from -ℓ to +ℓ".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Construct the spherical harmonics and normalize".to_string(),
            mathematical_content: "The spherical harmonics are:\n\nY_ℓ^m(θ,φ) = N_ℓ^m P_ℓ^m(cos θ) e^(imφ)\n\nNormalization: ∫₀^(2π) ∫₀^π |Y_ℓ^m|² sin θ dθ dφ = 1\n\nThis gives: N_ℓ^m = √[(2ℓ+1)/(4π) · (ℓ-m)!/(ℓ+m)!]\n\nFinal result:\nY_ℓ^m(θ,φ) = √[(2ℓ+1)/(4π) · (ℓ-m)!/(ℓ+m)!] P_ℓ^m(cos θ) e^(imφ)".to_string(),
            hints: vec![
                "Normalization ensures orthonormality on the sphere".to_string(),
                "The factorial ratio comes from properties of associated Legendre polynomials".to_string(),
                "These form a complete orthonormal basis on the sphere".to_string(),
            ],
            alternative_approaches: vec![
                "Could derive normalization using generating functions".to_string(),
                "Or use the connection to angular momentum operators".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "How many spherical harmonics are there for a given ℓ?".to_string(),
                    options: vec!["ℓ".to_string(), "ℓ+1".to_string(), "2ℓ+1".to_string(), "ℓ²".to_string()],
                    correct_answer: 2,
                    explanation: "For each ℓ, m ranges from -ℓ to +ℓ, giving 2ℓ+1 values".to_string(),
                }
            ],
            difficulty_level: 4,
        },
    ];

    let session = DerivationSession::new("Spherical Harmonics".to_string(), steps, 5);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn wright_function_asymptotic_derivation() -> Result<(), Box<dyn std::error::Error>> {
    println!("\n🧮 Expert Derivation: Wright Function Asymptotic Behavior");
    println!("=========================================================\n");

    println!("⚠️ Expert Level: Advanced complex analysis and asymptotic methods required.");

    let steps = vec![
        DerivationStep {
            description: "Start with the Mellin transform representation".to_string(),
            mathematical_content: "The Wright function has the integral representation:\n\nΦ(α,β;z) = (1/2πi) ∫_C Γ(-s) Γ(β+αs) (-z)^s ds\n\nwhere C is a suitable contour in the complex s-plane.\n\nFor large |z|, we use the saddle-point method.".to_string(),
            hints: vec![
                "This Mellin transform representation is key for asymptotic analysis".to_string(),
                "The saddle-point method finds the dominant contribution for large z".to_string(),
                "We need to locate where the exponent is stationary".to_string(),
            ],
            alternative_approaches: vec![
                "Could use the series representation with Stirling's formula".to_string(),
                "Or employ Watson's lemma for asymptotic expansions".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What makes this suitable for saddle-point analysis?".to_string(),
                    options: vec!["Large parameter z".to_string(), "Exponential factor (-z)^s".to_string(), "Both are important".to_string(), "Neither helps".to_string()],
                    correct_answer: 2,
                    explanation: "Large z and the exponential dependence (-z)^s make saddle-point methods applicable".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Find the saddle point".to_string(),
            mathematical_content: "The exponent in the integrand is:\nf(s) = ln Γ(-s) + ln Γ(β+αs) + s ln(-z)\n\nSaddle point condition: f'(s₀) = 0\n-ψ(-s₀) + α ψ(β+αs₀) + ln(-z) = 0\n\nwhere ψ(z) = Γ'(z)/Γ(z) is the digamma function.\n\nFor large |z|: s₀ ≈ (z/α)^(1/α)/α (leading approximation)".to_string(),
            hints: vec![
                "The digamma function ψ(z) is the logarithmic derivative of Γ(z)".to_string(),
                "For large z, ψ(z) ≈ ln z - 1/(2z) + O(z^(-2))".to_string(),
                "The saddle point equation is transcendental but solvable asymptotically".to_string(),
            ],
            alternative_approaches: vec![
                "Could use iterative methods to solve the saddle point equation".to_string(),
                "Or employ perturbative expansion around known solutions".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "Why does the saddle point depend on z^(1/α)?".to_string(),
                    options: vec!["Mathematical coincidence".to_string(), "From the balance of terms in ψ functions".to_string(), "Dimensional analysis".to_string(), "It doesn't".to_string()],
                    correct_answer: 1,
                    explanation: "The balance between -ψ(-s) and α ψ(β+αs) for large z gives this scaling".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Apply the saddle-point approximation".to_string(),
            mathematical_content: "Near the saddle point s₀:\nf(s) ≈ f(s₀) + (1/2)f''(s₀)(s-s₀)²\n\nThe integral becomes:\nΦ(α,β;z) ≈ (1/2πi) e^(f(s₀)) ∫ e^((1/2)f''(s₀)(s-s₀)²) ds\n\nEvaluating the Gaussian integral:\nΦ(α,β;z) ≈ e^(f(s₀)) / √(2π|f''(s₀)|)".to_string(),
            hints: vec![
                "This is the standard saddle-point approximation".to_string(),
                "The Gaussian integral gives √(2π/|f''(s₀)|)".to_string(),
                "The main contribution comes from the exponential factor e^(f(s₀))".to_string(),
            ],
            alternative_approaches: vec![
                "Could include higher-order corrections in the expansion".to_string(),
                "Or use steepest descent contour deformation".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "What determines the exponential growth rate?".to_string(),
                    options: vec!["f''(s₀)".to_string(), "f(s₀)".to_string(), "The parameter α".to_string(), "The variable z".to_string()],
                    correct_answer: 1,
                    explanation: "The exponential factor e^(f(s₀)) dominates the asymptotic behavior".to_string(),
                }
            ],
            difficulty_level: 5,
        },

        DerivationStep {
            description: "Derive the final asymptotic formula".to_string(),
            mathematical_content: "Computing f(s₀) with s₀ ≈ (z/α)^(1/α)/α:\n\nf(s₀) ≈ (1/α)(z/α)^(1/α) + lower order terms\n\nThis gives the asymptotic formula:\n\nΦ(α,β;z) ∼ (1/√(2πα)) z^((β-1)/(2α)) exp((1/α)(z/α)^(1/α))\n\nfor large |z| and α > 0.\n\nThis shows exponential growth faster than any polynomial!".to_string(),
            hints: vec![
                "The fractional power (z/α)^(1/α) creates super-exponential growth".to_string(),
                "The pre-exponential factor comes from the saddle-point curvature".to_string(),
                "This behavior is characteristic of entire functions of exponential type".to_string(),
            ],
            alternative_approaches: vec![
                "Could verify using known special cases".to_string(),
                "Or compare with numerical computations".to_string(),
            ],
            validation_questions: vec![
                ValidationQuestion {
                    question: "How does this compare to exponential growth e^z?".to_string(),
                    options: vec!["Same growth".to_string(), "Slower growth".to_string(), "Faster (super-exponential) growth".to_string(), "Depends on α".to_string()],
                    correct_answer: 2,
                    explanation: "The factor (z/α)^(1/α) grows faster than z for α < 1, giving super-exponential behavior".to_string(),
                }
            ],
            difficulty_level: 5,
        },
    ];

    let session = DerivationSession::new("Wright Function Asymptotics".to_string(), steps, 5);
    run_derivation_session(session)
}

#[allow(dead_code)]
fn run_derivation_session(
    mut session: DerivationSession,
) -> Result<(), Box<dyn std::error::Error>> {
    println!("🎯 Starting derivation: {}", session.title);
    println!("Difficulty level: {}/5", session.difficulty_level);
    println!("Total steps: {}\n", session.steps.len());

    while !session.is_complete() {
        display_session_status(&session);

        if session.current_step < session.steps.len() {
            let step = &session.steps[session.current_step].clone();

            println!(
                "📖 Step {} of {}: {}",
                session.current_step + 1,
                session.steps.len(),
                step.description
            );
            println!("\n📝 Mathematical Content:");
            println!("{}", step.mathematical_content);

            // Interactive component
            println!("\n🎯 Choose an action:");
            println!("1. 💡 Get a hint");
            println!("2. 🔄 See alternative approaches");
            println!("3. ❓ Answer validation questions");
            println!("4. ✅ Mark step as understood and continue");
            println!("5. 📊 Show progress and summary");

            let choice = get_user_input("Your choice (1-5): ")?;

            match choice.parse::<u32>() {
                Ok(1) => show_hints(step, &mut session),
                Ok(2) => show_alternatives(step),
                Ok(3) => run_validation_questions(step)?,
                Ok(4) => {
                    session.completed_steps[session.current_step] = true;
                    session.current_step += 1;
                    println!("✅ Step completed! Moving to next step...\n");
                }
                Ok(5) => show_session_progress(&session),
                _ => println!("❌ Invalid choice. Please try again."),
            }
        } else {
            break;
        }
    }

    // Session completion
    if session.is_complete() {
        println!(
            "\n🎉 Congratulations! You've completed the derivation: {}",
            session.title
        );
        let duration = session.start_time.elapsed();
        println!(
            "⏱️ Time taken: {:.1} minutes",
            duration.as_secs_f64() / 60.0
        );
        println!("💡 Hints used: {}", session.hints_used);

        // Final summary
        println!("\n📚 What you've learned:");
        println!("• Mathematical rigor in step-by-step derivations");
        println!("• Connection between different mathematical concepts");
        println!("• Alternative approaches to the same problem");
        println!("• Applications and significance of the result");

        let difficulty_bonus = session.difficulty_level * 10;
        let time_bonus = if duration.as_secs() < 300 { 20 } else { 0 }; // 5-minute bonus
        let hint_penalty = session.hints_used * 5;
        let total_score = (100 + difficulty_bonus + time_bonus).saturating_sub(hint_penalty);

        println!("\n🏆 Session Score: {} points", total_score);
        println!("  Base score: 100");
        println!("  Difficulty bonus: +{}", difficulty_bonus);
        if time_bonus > 0 {
            println!("  Speed bonus: +{}", time_bonus);
        }
        if hint_penalty > 0 {
            println!("  Hint penalty: -{}", hint_penalty);
        }
    }

    Ok(())
}

#[allow(dead_code)]
fn display_session_status(session: &DerivationSession) {
    println!(
        "📊 Progress: {:.1}% ({}/{} steps completed)",
        session.progress_percentage(),
        session.completed_steps.iter().filter(|&&x| x).count(),
        session.steps.len()
    );

    // Progress bar
    let completed = session.completed_steps.iter().filter(|&&x| x).count();
    let total = session.steps.len();
    let bar_length = 20;
    let filled = (completed * bar_length) / total;

    print!("Progress: [");
    for i in 0..bar_length {
        if i < filled {
            print!("█");
        } else if i == filled && session.current_step < total {
            print!("▶");
        } else {
            print!("░");
        }
    }
    println!("]\n");
}

#[allow(dead_code)]
fn show_hints(step: &DerivationStep, session: &mut DerivationSession) {
    println!("\n💡 Hints for this _step:");
    for (i, hint) in step.hints.iter().enumerate() {
        println!("{}. {}", i + 1, hint);
    }
    session.hints_used += 1;
    println!();
}

#[allow(dead_code)]
fn show_alternatives(step: &DerivationStep) {
    println!("\n🔄 Alternative approaches:");
    for (i, approach) in step.alternative_approaches.iter().enumerate() {
        println!("{}. {}", i + 1, approach);
    }
    println!();
}

#[allow(dead_code)]
fn run_validation_questions(step: &DerivationStep) -> Result<(), Box<dyn std::error::Error>> {
    if step.validation_questions.is_empty() {
        println!("ℹ️ No validation questions for this step.");
        return Ok(());
    }

    println!("\n❓ Validation Questions:");

    for (i, question) in step.validation_questions.iter().enumerate() {
        println!("\nQuestion {}: {}", i + 1, question.question);

        for (j, option) in question.options.iter().enumerate() {
            println!("  {}. {}", (b'a' + j as u8) as char, option);
        }

        let answer = get_user_input("Your answer: ")?;
        let answer_index = answer.to_lowercase().chars().next().and_then(|c: char| {
            if c.is_ascii_lowercase() {
                Some((c as u8 - b'a') as usize)
            } else {
                None
            }
        });

        if let Some(idx) = answer_index {
            if idx == question.correct_answer {
                println!("✅ Correct! {}", question.explanation);
            } else {
                println!("❌ Incorrect. {}", question.explanation);
            }
        } else {
            println!("❌ Invalid answer format. {}", question.explanation);
        }
    }

    Ok(())
}

#[allow(dead_code)]
fn show_session_progress(session: &DerivationSession) {
    println!("\n📊 Session Progress Report");
    println!("=========================");
    println!("Derivation: {}", session.title);
    println!("Difficulty: {}/5", session.difficulty_level);
    println!(
        "Time elapsed: {:.1} minutes",
        session.start_time.elapsed().as_secs_f64() / 60.0
    );
    println!("Hints used: {}", session.hints_used);
    println!("Progress: {:.1}%", session.progress_percentage());

    println!("\nStep Status:");
    for (i, completed) in session.completed_steps.iter().enumerate() {
        let status = if *completed {
            "✅"
        } else if i == session.current_step {
            "▶️"
        } else {
            "⏳"
        };
        println!(
            "  Step {}: {} {}",
            i + 1,
            status,
            session.steps[i].description
        );
    }

    if !session.is_complete() {
        let remaining =
            session.steps.len() - session.completed_steps.iter().filter(|&&x| x).count();
        println!("\n{} steps remaining.", remaining);
    }

    println!();
}

#[allow(dead_code)]
fn get_user_input(prompt: &str) -> Result<String, Box<dyn std::error::Error>> {
    print!("{}", prompt);
    io::stdout().flush()?;
    let mut input = String::new();
    io::stdin().read_line(&mut input)?;
    Ok(input.trim().to_string())
}
