upload result plot

.gitignore

@@ -1 +1,5 @@
 .vscode
+.DS_Store
+__pycache__
+*.pyc
+*Zone.Identifier

homework10/Homework10_Q1.py

@@ -0,0 +1,249 @@
+import numpy as np
+import matplotlib.pyplot as plt
+import sys
+
+# Configuration for plotting
+plt.style.use('seaborn-v0_8')
+plt.rcParams['figure.figsize'] = (18, 6)
+
+def newton_method(f, df, x0, tolerance=1e-12, max_iter=20):
+    """
+    Implements Newton's Method with overflow protection.
+    Returns: list of iteration history [(iter, x, error)]
+    """
+    history = []
+    x = x0
+    
+    for i in range(max_iter):
+        try:
+            # Check for divergence first
+            if abs(x) > 1e100:
+                break
+
+            fx = f(x)
+            dfx = df(x)
+            
+            if abs(dfx) < 1e-15: # Avoid division by zero
+                break
+                
+            history.append(x)
+            
+            x_new = x - fx / dfx
+            
+            # Check convergence
+            if abs(x_new - x) < tolerance:
+                x = x_new
+                history.append(x)
+                break
+                
+            x = x_new
+        except OverflowError:
+            # Catch errors if numbers become too large
+            break
+        
+    return history
+
+def chord_method(f, df_x0, x0, tolerance=1e-12, max_iter=50):
+    """
+    Implements Chord Method using fixed slope m = f'(x0) with overflow protection.
+    Returns: list of iteration history
+    """
+    history = []
+    x = x0
+    m = df_x0 # Fixed slope
+    
+    for i in range(max_iter):
+        try:
+            if abs(x) > 1e100:
+                break
+                
+            fx = f(x)
+            history.append(x)
+            
+            if abs(m) < 1e-15:
+                break
+                
+            x_new = x - fx / m
+            
+            if abs(x_new - x) < tolerance:
+                x = x_new
+                history.append(x)
+                break
+            x = x_new
+        except OverflowError:
+            break
+            
+    return history
+
+def secant_method(f, x0, x_minus_1, tolerance=1e-12, max_iter=20):
+    """
+    Implements Secant Method with overflow protection.
+    Returns: list of iteration history
+    """
+    history = []
+    x_curr = x0
+    x_prev = x_minus_1
+    
+    for i in range(max_iter):
+        try:
+            if abs(x_curr) > 1e100:
+                break
+
+            history.append(x_curr)
+            
+            fx_curr = f(x_curr)
+            fx_prev = f(x_prev)
+            
+            if abs(fx_curr - fx_prev) < 1e-15:
+                break
+                
+            # Secant update
+            x_new = x_curr - fx_curr * (x_curr - x_prev) / (fx_curr - fx_prev)
+            
+            if abs(x_new - x_curr) < tolerance:
+                x_curr = x_new
+                history.append(x_curr)
+                break
+                
+            x_prev = x_curr
+            x_curr = x_new
+        except OverflowError:
+            break
+        
+    return history
+
+def solve_and_save():
+    # Definition of problems
+    problems = [
+        {
+            'id': 'a',
+            'title': r'$f(x) = \cos(x) - x, x_0 = 0.5$',
+            'func_name': 'cos(x) - x',
+            'f': lambda x: np.cos(x) - x,
+            'df': lambda x: -np.sin(x) - 1,
+            'x0': 0.5,
+            'true_root': 0.73908513321516064165531208767387340401341175890075746496568063577328465488354759 # From wolframalpha 
+        },
+        {
+            'id': 'b',
+            'title': r'$f(x) = \sin(x), x_0 = 3$',
+            'func_name': 'sin(x)',
+            'f': lambda x: np.sin(x),
+            'df': lambda x: np.cos(x),
+            'x0': 3.0,
+            'true_root': np.pi
+        },
+        {
+            'id': 'c',
+            'title': r'$f(x) = x^2 + 1, x_0 = 0.5$',
+            'func_name': 'x^2 + 1',
+            'f': lambda x: x**2 + 1,
+            'df': lambda x: 2*x,
+            'x0': 0.5,
+            'true_root': None # No real root
+        }
+    ]
+    
+    # Open file for writing text results
+    output_filename = 'numerical_results.txt'
+    image_filename = 'convergence_results.png'
+    
+    with open(output_filename, 'w', encoding='utf-8') as txt_file:
+        
+        fig, axes = plt.subplots(1, 3)
+        
+        for idx, prob in enumerate(problems):
+            header_str = f"\n{'='*60}\nProblem ({prob['id']}): {prob['func_name']}, x0={prob['x0']}\n{'='*60}\n"
+            print(header_str) # Print to console to show progress
+            txt_file.write(header_str)
+            
+            x0 = prob['x0']
+            f = prob['f']
+            df = prob['df']
+            
+            # 1. Run Newton
+            hist_newton = newton_method(f, df, x0)
+            
+            # 2. Run Chord (m = f'(x0))
+            hist_chord = chord_method(f, df(x0), x0)
+            
+            # 3. Run Secant (x_-1 = 0.99 * x0)
+            x_minus_1 = 0.99 * x0
+            hist_secant = secant_method(f, x0, x_minus_1)
+            
+            methods_data = {
+                "Newton": hist_newton,
+                "Chord": hist_chord,
+                "Secant": hist_secant
+            }
+            
+            ax = axes[idx]
+            
+            for m_name, history in methods_data.items():
+                iters = list(range(len(history)))
+                errors = []
+                
+                # Calculate errors for plotting
+                # Note: Handling overflow for plotting large errors in problem c
+                for x in history:
+                    try:
+                        if prob['true_root'] is not None:
+                            val = abs(x - prob['true_root'])
+                        else:
+                            val = abs(f(x)) # Residual for problem c
+                        errors.append(val)
+                    except OverflowError:
+                        errors.append(float('inf'))
+
+                # Write to TXT file
+                txt_file.write(f"\nMethod: {m_name}\n")
+                txt_file.write(f"{'Iter':<5} | {'Value (x_k)':<25} | {'Error/Resid':<25}\n")
+                txt_file.write("-" * 60 + "\n")
+                
+                for i, val in enumerate(history):
+                    # Only write full history if short, or head/tail if long
+                    if len(history) < 20 or i < 5 or i > len(history) - 3:
+                        if prob['true_root'] is not None:
+                            err_val = abs(val - prob['true_root'])
+                        else:
+                            try:
+                                err_val = abs(f(val))
+                            except OverflowError:
+                                err_val = float('inf')
+                                
+                        txt_file.write(f"{i:<5} | {val:<25.10g} | {err_val:<25.6e}\n")
+                    elif i == 5:
+                        txt_file.write("...\n")
+
+                # Add to plot
+                if prob['true_root'] is not None:
+                    ylabel_text = "Error |x_k - x*|"
+                    log_scale = True
+                    ax.plot(iters, errors, marker='o', label=m_name)
+                else:
+                    ylabel_text = "Residual |f(x_k)|"
+                    log_scale = False
+                    # For problem C, data explodes, so we limit plotting range to avoid ruining the graph
+                    valid_errors = [e for e in errors if e < 1e10]
+                    valid_iters = iters[:len(valid_errors)]
+                    ax.plot(valid_iters, valid_errors, marker='x', label=m_name)
+
+            ax.set_title(f"Problem ({prob['id']})\n{prob['title']}")
+            ax.set_xlabel("Iteration")
+            ax.set_ylabel(ylabel_text)
+            if log_scale:
+                ax.set_yscale('log')
+            ax.legend()
+            if prob['id'] == 'c':
+                ax.set_ylim(0, 20)
+                ax.set_xticks(range(0, len(iters), 1))
+            ax.grid(True)
+
+    print(f"\nText results saved to: {output_filename}")
+    
+    plt.tight_layout()
+    plt.savefig(image_filename)
+    print(f"Chart saved to: {image_filename}")
+
+if __name__ == "__main__":
+    solve_and_save()

homework10/comment.txt

@@ -0,0 +1 @@
+For problems (a) and (b), which have real roots, all methods converged; Newton's method was the fastest (quadratic), followed by the Secant method (superlinear), with the Chord method being the slowest (linear). However, for problem (c) (x^2 + 1), which has no real roots, all methods failed: the Chord method rapidly diverged to infinity, while Secant and Newton's method oscillated.

homework10/numerical_results.txt

@@ -0,0 +1,112 @@
+
+============================================================
+Problem (a): cos(x) - x, x0=0.5
+============================================================
+
+Method: Newton
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 0.5                       | 2.390851e-01             
+1     | 0.7552224171              | 1.613728e-02             
+2     | 0.7391416661              | 5.653293e-05             
+3     | 0.7390851339              | 7.056461e-10             
+4     | 0.7390851332              | 0.000000e+00             
+5     | 0.7390851332              | 0.000000e+00             
+
+Method: Chord
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 0.5                       | 2.390851e-01             
+1     | 0.7552224171              | 1.613728e-02             
+2     | 0.7369022576              | 2.182876e-03             
+3     | 0.7393704622              | 2.853290e-04             
+4     | 0.7390476612              | 3.747205e-05             
+...          
+13    | 0.7390851332              | 4.333200e-13             
+14    | 0.7390851332              | 5.695444e-14             
+
+Method: Secant
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 0.5                       | 2.390851e-01             
+1     | 0.7556018135              | 1.651668e-02             
+2     | 0.738106944               | 9.781892e-04             
+3     | 0.7390815948              | 3.538433e-06             
+4     | 0.739085134               | 7.646579e-10             
+5     | 0.7390851332              | 5.551115e-16             
+6     | 0.7390851332              | 0.000000e+00             
+
+============================================================
+Problem (b): sin(x), x0=3.0
+============================================================
+
+Method: Newton
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 3                         | 1.415927e-01             
+1     | 3.142546543               | 9.538895e-04             
+2     | 3.141592653               | 2.893161e-10             
+3     | 3.141592654               | 0.000000e+00             
+4     | 3.141592654               | 0.000000e+00             
+
+Method: Chord
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 3                         | 1.415927e-01             
+1     | 3.142546543               | 9.538895e-04             
+2     | 3.141583011               | 9.642404e-06             
+3     | 3.141592751               | 9.747184e-08             
+4     | 3.141592653               | 9.853101e-10             
+5     | 3.141592654               | 9.960477e-12             
+6     | 3.141592654               | 1.003642e-13             
+7     | 3.141592654               | 1.332268e-15             
+
+Method: Secant
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 3                         | 1.415927e-01             
+1     | 3.142873442               | 1.280788e-03             
+2     | 3.141588403               | 4.250744e-06             
+3     | 3.141592654               | 1.158629e-12             
+4     | 3.141592654               | 0.000000e+00             
+5     | 3.141592654               | 0.000000e+00             
+
+============================================================
+Problem (c): x^2 + 1, x0=0.5
+============================================================
+
+Method: Newton
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 0.5                       | 1.250000e+00             
+1     | -0.75                     | 1.562500e+00             
+2     | 0.2916666667              | 1.085069e+00             
+3     | -1.568452381              | 3.460043e+00             
+4     | -0.4654406117             | 1.216635e+00             
+...
+18    | -1.820817243              | 4.315375e+00             
+19    | -0.6358066524             | 1.404250e+00             
+
+Method: Chord
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 0.5                       | 1.250000e+00             
+1     | -0.75                     | 1.562500e+00             
+2     | -2.3125                   | 6.347656e+00             
+3     | -8.66015625               | 7.599831e+01             
+4     | -84.65846252              | 7.168055e+03             
+...  
+8     | -7.660265913e+30          | 5.867967e+61             
+9     | -5.867967385e+61          | 3.443304e+123            
+
+Method: Secant
+Iter  | Value (x_k)               | Error/Resid              
+------------------------------------------------------------
+0     | 0.5                       | 1.250000e+00             
+1     | -0.756281407              | 1.571962e+00             
+2     | 5.37745098                | 2.991698e+01             
+3     | -1.096446714              | 2.202195e+00             
+4     | -1.610857647              | 3.594862e+00             
+...
+18    | 0.3478146096              | 1.120975e+00             
+19    | -2.615940634              | 7.843145e+00             

homework11/Homework11_Q5.py

@@ -0,0 +1,159 @@
+import numpy as np
+import matplotlib.pyplot as plt
+
+# --- 1. Define Function, Gradient, and Hessian ---
+# Objective function: Rosenbrock function
+def objective_function(x):
+    x1, x2 = x[0], x[1]
+    return 100 * (x2 - x1**2)**2 + (1 - x1)**2
+
+# Gradient vector calculation
+def gradient(x):
+    x1, x2 = x[0], x[1]
+    df_dx1 = -400 * x1 * (x2 - x1**2) - 2 * (1 - x1)
+    df_dx2 = 200 * (x2 - x1**2)
+    return np.array([df_dx1, df_dx2])
+
+# Hessian matrix calculation
+def hessian(x):
+    x1, x2 = x[0], x[1]
+    d2f_dx1dx1 = -400 * (x2 - 3 * x1**2) + 2
+    d2f_dx1dx2 = -400 * x1
+    d2f_dx2dx1 = -400 * x1
+    d2f_dx2dx2 = 200
+    return np.array([[d2f_dx1dx1, d2f_dx1dx2], [d2f_dx2dx1, d2f_dx2dx2]])
+
+# --- 2. Backtracking Line Search ---
+def backtracking_line_search(x, p, grad, alpha_bar=1.0, rho=0.5, c1=1e-3):
+    """
+    Finds a step size alpha that satisfies the 1st Wolfe condition (Armijo condition).
+    Condition: f(x + alpha*p) <= f(x) + c1 * alpha * grad^T * p
+    """
+    alpha = alpha_bar
+    f_val = objective_function(x)
+    grad_dot_p = np.dot(grad, p)
+    
+    # Loop until condition is met
+    while objective_function(x + alpha * p) > f_val + c1 * alpha * grad_dot_p:
+        alpha *= rho
+        # Safety break to prevent infinite loops if alpha becomes too small
+        if alpha < 1e-16: 
+            break
+    return alpha
+
+# --- 3. Optimization Algorithms ---
+def run_optimization(method, x0, tol_ratio=1e-12, max_iter=20000):
+    """
+    Runs the specified optimization method.
+    Returns: list of alphas used, total iterations
+    """
+    x = x0.copy()
+    grad_0_norm = np.linalg.norm(gradient(x0))
+    
+    alphas = []
+    
+    print(f"[{method}] Starting from {x0}...")
+    
+    for i in range(max_iter):
+        grad = gradient(x)
+        grad_norm = np.linalg.norm(grad)
+        
+        # Convergence Check: ||grad f(xk)|| < 10^-12 * ||grad f(x0)||
+        if grad_norm < tol_ratio * grad_0_norm:
+            print(f"   -> Converged at iteration {i}. Final Loss: {objective_function(x):.6e}")
+            return alphas, i
+            
+        # Determine Search Direction
+        if method == "Steepest_Descent":
+            direction = -grad
+        elif method == "Newton":
+            hess = hessian(x)
+            try:
+                # Solve H * d = -g for d
+                direction = np.linalg.solve(hess, -grad) 
+            except np.linalg.LinAlgError:
+                # Fallback if Hessian is singular
+                direction = -grad
+
+        # Perform Backtracking Line Search
+        alpha = backtracking_line_search(x, direction, grad, alpha_bar=1.0, rho=0.5, c1=1e-3)
+        
+        # Update Position
+        x = x + alpha * direction
+        
+        alphas.append(alpha)
+        
+    print(f"   -> Did not converge within {max_iter} iterations.")
+    print(f"   -> Final Gradient Norm: {grad_norm:.6e}")
+    return alphas, max_iter
+
+# --- 4. Main Execution and Plotting ---
+
+# Configuration
+initial_points = [np.array([1.2, 1.2]), np.array([-1.2, 1.0])]
+methods = ["Steepest_Descent", "Newton"]
+
+# Dictionary to hold simulation results
+results = {}
+
+# Run simulations
+for x0 in initial_points:
+    for method in methods:
+        # Construct a unique key for storage
+        key = f"{method}_x0_{x0}"
+        alphas, num_iters = run_optimization(method, x0)
+        results[key] = {'alphas': alphas, 'iters': num_iters, 'x0': x0}
+
+# Plot configuration
+fig, axes = plt.subplots(2, 2, figsize=(14, 10))
+plt.subplots_adjust(hspace=0.4, wspace=0.3)
+
+# Mapping results to subplots
+plot_mapping = {
+    ("Steepest_Descent", str(initial_points[0])): (0, 0),
+    ("Newton", str(initial_points[0])): (0, 1),
+    ("Steepest_Descent", str(initial_points[1])): (1, 0),
+    ("Newton", str(initial_points[1])): (1, 1),
+}
+
+for key, data in results.items():
+    method_name = key.split("_x0_")[0]
+    x0_str = str(data['x0'])
+    
+    ax_idx = plot_mapping[(method_name, x0_str)]
+    ax = axes[ax_idx]
+    
+    y_data = data['alphas']
+    x_data = range(len(y_data))
+    
+    # --- PLOTTING LOGIC MODIFICATION ---
+    # For Steepest Descent, show only first {limit} steps
+    if method_name == "Steepest_Descent":
+        limit = 30
+        if len(y_data) > limit:
+            y_data = y_data[:limit]
+            x_data = range(limit)
+            title_suffix = f" (First {limit} Steps Shown)"
+        else:
+            title_suffix = ""
+    else:
+        title_suffix = ""
+        
+    ax.plot(x_data, y_data, marker='o', linestyle='-', markersize=4, linewidth=1.5)
+    
+    # Titles and Labels
+    ax.set_title(f"{method_name}\nStart: {data['x0']}\nTotal Iters: {data['iters']}{title_suffix}")
+    ax.set_xlabel("Iteration")
+    ax.set_ylabel("Step Size (Alpha)")
+    ax.grid(True, which='both', linestyle='--', alpha=0.7)
+    
+    # Add annotation for total iterations
+    ax.text(0.95, 0.95, f"Total Iters: {data['iters']}", 
+            transform=ax.transAxes, 
+            verticalalignment='top', horizontalalignment='right',
+            bbox=dict(boxstyle='round', facecolor='white', alpha=0.8))
+
+# Save and Show
+#plt.suptitle("Step Size vs Iteration (Rosenbrock Function)", fontsize=16)
+plt.savefig("optimization_results.png")
+#plt.show()

homework8/Homework8_Q5.py

@@ -0,0 +1,81 @@
+import numpy as np
+import matplotlib.pyplot as plt
+
+# --- FUNCTION DEFINITION CORRECTED HERE ---
+def g(t):
+    """
+    Calculates g(t) = t^4 - 3t^3 + 2t^2 in a vectorized way.
+    It handles both single float inputs and NumPy array inputs.
+    If an element's absolute value is too large, it returns infinity
+    to prevent overflow and handle plotting of diverging paths.
+    """
+    # Use np.where for a vectorized if-else statement.
+    # This correctly handles both scalar and array inputs.
+    return np.where(np.abs(t) > 1e10,  # Condition
+                    np.inf,             # Value if condition is True
+                    t**4 - 3*t**3 + 2*t**2) # Value if condition is False
+
+# Define the gradient (derivative) of g(t)
+def grad_g(t):
+    # This function is only called with scalars in our loop,
+    # but it's good practice for it to be vectorization-safe as well.
+    # Standard arithmetic operations are already vectorized in NumPy.
+    return 4*t**3 - 9*t**2 + 4*t
+
+# Implement the gradient descent algorithm
+def gradient_descent(x0, alpha, iterations=100):
+    history = [x0]
+    x = x0
+    for _ in range(iterations):
+        grad = grad_g(x)
+        
+        if np.isinf(x) or np.isnan(x) or np.abs(x) > 1e9:
+            print(f"Divergence detected for alpha = {alpha}. Stopping early.")
+            break
+
+        x = x - alpha * grad
+        history.append(x)
+    if not np.isinf(x) and not np.isnan(x) and np.abs(x) <= 1e9:
+        print(f"Alpha: {alpha:<.2f}, Current x: {x:<.2f}, g(x): {g(x):<.2f}")
+    return history
+
+# --- Main part of the experiment ---
+
+# Set an initial point
+initial_xs = [-1.0, 0.5, 0.65, 2.5]
+
+# Define a list of different alpha values to test
+alphas_to_test = [0.05, 0.15, 0.25, 0.4, 0.5, 0.6]
+
+# Run the experiment for each alpha value and plot the results
+for initial_x in initial_xs:
+    print(f"Running gradient descent from initial x = {initial_x}")
+    # Create a plot
+    plt.figure(figsize=(14, 8))
+
+    # Plot the function g(t) for context
+    t_plot = np.linspace(-1.5, 3.5, 400)
+    plt.plot(t_plot, g(t_plot), 'k-', label='g(t) = t^4 - 3t^3 + 2t^2', linewidth=2)
+    for alpha in alphas_to_test:
+        history = gradient_descent(initial_x, alpha)
+        history_np = np.array(history)
+        plt.plot(history_np, g(history_np), 'o-', label=f'alpha = {alpha}', markersize=4, alpha=0.8)
+
+    # Add stationary points
+    minima1 = 0.0
+    minima2 = (9 + np.sqrt(17)) / 8
+    maxima = (9 - np.sqrt(17)) / 8
+    plt.plot(minima1, g(minima1), 'g*', markersize=15, label=f'Local Minimum at t={minima1:.2f}')
+    plt.plot(minima2, g(minima2), 'g*', markersize=15, label=f'Local Minimum at t≈{minima2:.2f}')
+    plt.plot(maxima, g(maxima), 'rX', markersize=10, label=f'Local Maximum at t≈{maxima:.2f}')
+
+    # Final plot formatting
+    plt.title('Gradient Descent Convergence for Different Step Sizes (alpha, initial x={})'.format(initial_x))
+    plt.xlabel('t')
+    plt.ylabel('g(t)')
+    plt.legend()
+    plt.grid(True)
+    plt.ylim(-1, 5)
+    plt.xlim(-1.5, 3.5)
+    plt.savefig(f'Homework8_Q5_Gradient_Descent_Convergence_x={initial_x}.png')
+

homework8/Homework8_Q5.txt

@@ -0,0 +1,28 @@
+Running gradient descent from initial x = -1.0
+Alpha: 0.05, Current x: -0.00, g(x): 0.00
+Alpha: 0.15, Current x: 1.64, g(x): -0.62
+Divergence detected for alpha = 0.25. Stopping early.
+Divergence detected for alpha = 0.4. Stopping early.
+Divergence detected for alpha = 0.5. Stopping early.
+Divergence detected for alpha = 0.6. Stopping early.
+Running gradient descent from initial x = 0.5
+Alpha: 0.05, Current x: 0.00, g(x): 0.00
+Alpha: 0.15, Current x: 0.00, g(x): 0.00
+Alpha: 0.25, Current x: 0.00, g(x): 0.00
+Alpha: 0.40, Current x: -0.00, g(x): 0.00
+Alpha: 0.50, Current x: 0.02, g(x): 0.00
+Alpha: 0.60, Current x: 0.18, g(x): 0.05
+Running gradient descent from initial x = 0.65
+Alpha: 0.05, Current x: 1.64, g(x): -0.62
+Alpha: 0.15, Current x: 1.64, g(x): -0.62
+Alpha: 0.25, Current x: 1.64, g(x): -0.62
+Alpha: 0.40, Current x: 1.51, g(x): -0.57
+Alpha: 0.50, Current x: 0.01, g(x): 0.00
+Divergence detected for alpha = 0.6. Stopping early.
+Running gradient descent from initial x = 2.5
+Alpha: 0.05, Current x: 1.64, g(x): -0.62
+Alpha: 0.15, Current x: 0.00, g(x): 0.00
+Divergence detected for alpha = 0.25. Stopping early.
+Divergence detected for alpha = 0.4. Stopping early.
+Divergence detected for alpha = 0.5. Stopping early.
+Divergence detected for alpha = 0.6. Stopping early.

project1/Project1.md

@@ -16,15 +16,15 @@ As $\epsilon \rightarrow 0$, the normal equation method numerically unstable. A
 Results:
 | $\epsilon$ | Normal Equation Method `proj(A, b)` | SVD Method `proj_SVD(A, b)` | Difference (Normal Eq - SVD) |
 |:---:|:---|:---|:---|
-| **1.0** | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[ 1.78e-15 -2.22e-15 -8.88e-16 8.88e-16 0.00e+00]` |
+| **1.0** | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[ 2.22e-16 1.55e-15 -4.44e-16 2.22e-16 -8.88e-16]` |
-| **0.1** | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[ 7.28e-14 -4.44e-16 -2.66e-14 -1.62e-14 -5.28e-14]` |
+| **0.1** | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[ 8.08e-14 5.87e-14 -8.88e-16 4.82e-14 -1.38e-14]` |
-| **0.01** | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[-5.45e-12 -7.28e-12 -3.45e-13 -4.24e-12 -4.86e-12]` |
+| **0.01** | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[-4.77e-12 -1.84e-14 5.54e-13 -4.00e-12 1.50e-12]` |
-| **1e-4** | `[1.85714297 1.00000012 3.14285716 1.28571442 3.85714302]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[1.11e-07 1.19e-07 1.92e-08 1.36e-07 1.67e-07]` |
+| **1e-4** | `[1.85714282 1. 3.14285716 1.28571427 3.85714291]` | `[1.85714286 1. 3.14285714 1.28571429 3.85714286]` | `[-3.60e-08 9.11e-13 1.94e-08 -1.20e-08 4.80e-08]` |
-| **1e-8** | **Error:** `LinAlgError: Singular matrix` | `[1.85714286 1. 3.14285714 1.28571428 3.85714286]` | `Could not compute difference due to previous error.` |
+| **1e-8** | `[-1.87500007 0.99999993 -3.12499997 -2.62500007 3.49999996]` | `[1.85714286 1. 3.14285714 1.28571427 3.85714286]` | `[-3.73e+00 -7.45e-08 -6.27e+00 -3.91e+00 -3.57e-01]` |
-| **1e-16**| **Error:** `ValueError: Matrix A must be full rank` | `[1.81820151 1. 3.18179849 2.29149804 2.89030045]` | `Could not compute difference due to previous error.` |
+| **1e-16**| **Error:** `ValueError: Matrix A must be full rank` | `[3.4 1. 1.6 1.8 1.8]` | `Could not compute difference due to previous error.` |
-| **1e-32**| **Error:** `ValueError: Matrix A must be full rank` | `[2. 1. 3. 2.5 2.5]` | `Could not compute difference due to previous error.` |
+| **1e-32**| **Error:** `ValueError: Matrix A must be full rank` | `[3.4 1. 1.6 1.8 1.8]` | `Could not compute difference due to previous error.` |
 
-Numerical experiments show that as $\epsilon$ becomes small, the difference between the two methods increases. When $\epsilon$ becomes very small (e.g., 1e-8), the normal equation method fails while the SVD method continues to provide a stable solution.
+Numerical experiments show that as $\epsilon$ becomes small, the difference between the two methods increases. When $\epsilon$ becomes very small (e.g., 1e-8), the normal equation method can't give a valid result and fails when $\epsilon$ continues to decrease (e.g., 1e-16), while the SVD method continues to provide a stable solution.
 
 ***
 

project1/Project1_A1.txt

@@ -4,8 +4,8 @@ Projection of b onto the column space of A(1): (Using proj function)
 Projection of b onto the column space of A(1): (Using proj_SVD function)
 [1.85714286 1.         3.14285714 1.28571429 3.85714286]
 Difference between the two methods:
-[ 1.77635684e-15 -2.22044605e-15 -8.88178420e-16  8.88178420e-16
+[ 2.22044605e-16  1.55431223e-15 -4.44089210e-16  2.22044605e-16
-  0.00000000e+00]
+ -8.88178420e-16]
 
 Question 1(b):
 
@@ -15,8 +15,8 @@ Projection of b onto the column space of A(1.0):
 Projection of b onto the column space of A(1.0) using SVD:
 [1.85714286 1.         3.14285714 1.28571429 3.85714286]
 Difference between the two methods:
-[ 1.77635684e-15 -2.22044605e-15 -8.88178420e-16  8.88178420e-16
+[ 2.22044605e-16  1.55431223e-15 -4.44089210e-16  2.22044605e-16
-  0.00000000e+00]
+ -8.88178420e-16]
 
 For ε = 0.1:
 Projection of b onto the column space of A(0.1):
@@ -24,8 +24,8 @@ Projection of b onto the column space of A(0.1):
 Projection of b onto the column space of A(0.1) using SVD:
 [1.85714286 1.         3.14285714 1.28571429 3.85714286]
 Difference between the two methods:
-[ 7.28306304e-14 -4.44089210e-16 -2.66453526e-14 -1.62092562e-14
+[ 8.08242362e-14  5.87307980e-14 -8.88178420e-16  4.81836793e-14
- -5.28466160e-14]
+ -1.37667655e-14]
 
 For ε = 0.01:
 Projection of b onto the column space of A(0.01):
@@ -33,35 +33,37 @@ Projection of b onto the column space of A(0.01):
 Projection of b onto the column space of A(0.01) using SVD:
 [1.85714286 1.         3.14285714 1.28571429 3.85714286]
 Difference between the two methods:
-[-5.45297141e-12 -7.28239691e-12 -3.44613227e-13 -4.24371649e-12
+[-4.76907402e-12 -1.84297022e-14  5.53779245e-13 -4.00324218e-12
- -4.86499729e-12]
+  1.49613655e-12]
 
 For ε = 0.0001:
 Projection of b onto the column space of A(0.0001):
-[1.85714297 1.00000012 3.14285716 1.28571442 3.85714302]
+[1.85714282 1.         3.14285716 1.28571427 3.85714291]
 Projection of b onto the column space of A(0.0001) using SVD:
 [1.85714286 1.         3.14285714 1.28571429 3.85714286]
 Difference between the two methods:
-[1.11406275e-07 1.19209290e-07 1.91642426e-08 1.35516231e-07
+[-3.59740875e-08  9.10937992e-13  1.94238092e-08 -1.19938544e-08
- 1.67459071e-07]
+  4.79721098e-08]
 
 For ε = 1e-08:
-LinAlgError for eps=1e-08: Singular matrix
+Projection of b onto the column space of A(1e-08):
+[-1.87500007  0.99999993 -3.12499997 -2.62500007  3.49999996]
 Projection of b onto the column space of A(1e-08) using SVD:
-[1.85714286 1.         3.14285714 1.28571428 3.85714286]
+[1.85714286 1.         3.14285714 1.28571427 3.85714286]
 Difference between the two methods:
-Could not compute difference due to previous error.
+[-3.73214294e+00 -7.45058057e-08 -6.26785711e+00 -3.91071435e+00
+ -3.57142909e-01]
 
 For ε = 1e-16:
 ValueError for eps=1e-16: Matrix A must be full rank.
 Projection of b onto the column space of A(1e-16) using SVD:
-[1.81820151 1.         3.18179849 2.29149804 2.89030045]
+[3.4 1.  1.6 1.8 1.8]
 Difference between the two methods:
 Could not compute difference due to previous error.
 
 For ε = 1e-32:
 ValueError for eps=1e-32: Matrix A must be full rank.
 Projection of b onto the column space of A(1e-32) using SVD:
-[2.  1.  3.  2.5 2.5]
+[3.4 1.  1.6 1.8 1.8]
 Difference between the two methods:
 Could not compute difference due to previous error.

project1/Project1_Q1.py

@@ -38,9 +38,13 @@ def proj_SVD(A:np.ndarray, b:np.ndarray) -> np.ndarray:
     = U S V^* (V S^2 V^*)^(-1) V S U^* b
     = U S V^* V S^(-2) V^* V S U^* b
     = U U^* b
+    If A = U S V^*, then the projection onto the column space of A is:
+        proj_A(b) = U_r U_r^* b
+    where U_r are the left singular vectors corresponding to nonzero singular values.
     """
+    U_r = U[:, :np.linalg.matrix_rank(A)]  # Take only the relevant columns
     # Compute the projection using the SVD components
-    projection = U @ U.conj().T @ b
+    projection = U_r @ U_r.conj().T @ b
     return projection
 
 def build_A(eps: float) -> np.ndarray:

project1/Project1_Q3.py

@@ -73,18 +73,18 @@ def f1(t: float) -> float:
     """
     return 1.0 / (1.0 + t**2)
 
-def build_A(N: int, n: int, f: callable) -> tuple[np.ndarray, np.ndarray]:
+def build_A(N: int, n: int, f: callable, f_range: tuple) -> tuple[np.ndarray, np.ndarray]:
     """
     Build the matrix A for the least squares problem.
     
     Parameters:
     n + 1 (int): The number of columns in the matrix A.
-    
+    f_range (tuple): The range of the function f.
     Returns:
     np.ndarray: The constructed matrix A.
     """
     """
-    {t_j}j=0->N, t_0=-5, t_N=5, t_j divides [-5, 5] equally
+    {t_j}j=0->N, t_0=f_range[0], t_N=f_range[1], t_j divides [f_range[0], f_range[1]] equally
     A = [[1, t_0, t_0^2, t_0^3, ..., t_0^n],
         [1, t_1, t_1^2, t_1^3, ..., t_1^n],
         ...
@@ -95,7 +95,7 @@ def build_A(N: int, n: int, f: callable) -> tuple[np.ndarray, np.ndarray]:
     Ax = b, x is the coeffs of the polynomial
     """
     A = np.zeros((N + 1, n + 1), dtype=float)
-    t = np.linspace(-5, 5, N + 1)
+    t = np.linspace(f_range[0], f_range[1], N + 1)
     for i in range(N + 1):
         A[i, :] = [t[i]**j for j in range(n + 1)]
     b = np.array([f(t_i) for t_i in t])
@@ -113,7 +113,7 @@ def question_3c():
         N = Ns[i]
         n = ns[i]
         A = np.zeros((N + 1, n + 1), dtype=float)
-        A, b = build_A(N, n, f1)
+        A, b = build_A(N, n, f1, (-5, 5))
         x = householder_lstsq(A, b)
         p_t = sum([x[j] * t**j for j in range(n + 1)])
         plt.subplot(1, 3, i + 1)
@@ -135,6 +135,32 @@ def f2(n:int, t: float) -> float:
     """
     return sum([t**i for i in range(n + 1)])
 
+def question_3d():
+    Ns = [25, 50, 100]
+    ns = Ns
+    t = np.linspace(-5, 5, 1000)
+    f_t = f1(t)
+    plt.figure(figsize=(15, 10))
+    for i in range(len(Ns)):
+        N = Ns[i]
+        n = ns[i]
+        A = np.zeros((N + 1, n + 1), dtype=float)
+        A, b = build_A(N, n, f1, (-5, 5))
+        x = householder_lstsq(A, b)
+        p_t = sum([x[j] * t**j for j in range(n + 1)])
+        plt.subplot(1, 3, i + 1)
+        plt.plot(t, f_t, label='f(t)', color='blue')
+        plt.plot(t, p_t, label='p(t)', color='red', linestyle='--')
+        plt.title(f'N={N}, n={n}')
+        plt.xlabel('t')
+        plt.ylabel('y')
+        plt.legend()
+        plt.grid(True, alpha=0.3)
+    plt.suptitle('Least Squares Polynomial Approximation using Householder QR')
+    plt.tight_layout(rect=[0, 0.03, 1, 0.95])
+    plt.savefig('Project1_A3d_least_squares_approximation.png')
+    #plt.show()
+
 def question_3e():
     # n = 4, 6, 8, 10, ..., 20
     ns = list(range(4, 21, 2))
@@ -143,7 +169,7 @@ def question_3e():
     losses = []
     for n in ns:
         N = 2 * n
-        A, b = build_A(N, n, lambda t: f2(n, t))
+        A, b = build_A(N, n, lambda t: f2(n, t), (0, 1))
         x = householder_lstsq(A, b)
         true_coeffs = np.array([1.0 for _ in range(n + 1)])
         loss = np.linalg.norm(x - true_coeffs)
@@ -161,6 +187,7 @@ def question_3e():
 
 if __name__ == "__main__":
     question_3c()
+    question_3d()
     question_3e()
 
 

project1/README.md

@@ -6,7 +6,7 @@ conda activate math6601_env
 pip install -r requirements.txt
 ```
 
-### Question 1
+### Run programs
 
 ```bash
 python Project1_Q1.py > Project1_A1.txt

project1/main.tex

@@ -0,0 +1,162 @@
+\documentclass{article}
+
+% Language setting
+\usepackage[english]{babel}
+
+% Set page size and margins
+\usepackage[letterpaper,top=2cm,bottom=2cm,left=3cm,right=3cm,marginparwidth=1.75cm]{geometry}
+
+% Useful packages
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{amsfonts}
+\usepackage[colorlinks=true, allcolors=blue]{hyperref}
+\usepackage{array} % For better table column definitions
+
+\title{MATH 6601 - Project 1 Report}
+\author{Zhe Yuan (yuan.1435)}
+\date{Sep 2025}
+
+\begin{document}
+\maketitle
+
+\section{Projections}
+
+\subsection*{(a)}
+The projection of vector $b$ onto the column space of a full-rank matrix $A$ is given by $p=A(A^*A)^{-1}A^*b$. A function \texttt{proj(A, b)} was implemented based on this formula (see \texttt{Project1\_Q1.py}). For $\epsilon = 1$, the projection is:
+\begin{verbatim}
+p = [1.85714286, 1.0, 3.14285714, 1.28571429, 3.85714286]
+\end{verbatim}
+
+\subsection*{(b)}
+As $\epsilon \rightarrow 0$, the normal equation method becomes numerically unstable. A more robust SVD-based method, \texttt{proj\_SVD(A, b)}, was implemented to handle this (see \texttt{Project1\_Q1.py}). The results are summarized in Table~\ref{tab:proj_results}.
+
+\begin{table}[h!]
+\centering
+\small % Use a smaller font for the table
+\begin{tabular}{|c|>{\ttfamily}p{3cm}|>{\ttfamily}p{3cm}|p{3.5cm}|}
+\hline
+\textbf{$\epsilon$} & \textbf{Normal Equation Method \texttt{proj(A,b)}} & \textbf{SVD Method \texttt{proj\_SVD(A,b)}} & \textbf{Difference (Normal Eq - SVD)} \\
+\hline
+\textbf{1.0} & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [ 2.22e-16  1.55e-15 -4.44e-16  2.22e-16 -8.88e-16] \\
+\hline
+\textbf{0.1} & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [ 8.08e-14  5.87e-14 -8.88e-16  4.82e-14 -1.38e-14] \\
+\hline
+\textbf{0.01} & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [-4.77e-12 -1.84e-14  5.54e-13 -4.00e-12  1.50e-12] \\
+\hline
+\textbf{1e-4} & [1.85714282 1. 3.14285716 1.28571427 3.85714291] & [1.85714286 1. 3.14285714 1.28571429 3.85714286] & [-3.60e-08  9.11e-13  1.94e-08 -1.20e-08  4.80e-08] \\
+\hline
+\textbf{1e-8} & [-1.87500007 0.99999993 -3.12499997 -2.62500007 3.49999996] & [1.85714286 1. 3.14285714 1.28571427 3.85714286] & [-3.73e+00 -7.45e-08 -6.27e+00 -3.91e+00 -3.57e-01] \\
+\hline
+\textbf{1e-16} & \textbf{Error:} \texttt{ValueError: Matrix A must be full rank} & [3.4 1. 1.6 1.8 1.8] & Could not compute difference due to previous error. \\
+\hline
+\textbf{1e-32} & \textbf{Error:} \texttt{ValueError: Matrix A must be full rank} & [3.4 1. 1.6 1.8 1.8] & Could not compute difference due to previous error. \\
+\hline
+\end{tabular}
+\caption{Comparison of results from the normal equation and SVD-based methods.}
+\label{tab:proj_results}
+\end{table}
+
+\noindent Numerical experiments show that as $\epsilon$ becomes small, the difference between the two methods increases. When $\epsilon$ becomes very small (e.g., 1e-8), the normal equation method can't give a valid result and fails when $\epsilon$ continues to decrease (e.g., 1e-16), while the SVD method continues to provide a stable solution.
+
+\section{QR Factorisation}
+
+\subsection*{(a, b, c)}
+Four QR factorization algorithms were implemented to find an orthonormal basis $Q$ for the column space of the $n \times n$ Hilbert matrix (for $n=2$ to $20$):
+\begin{enumerate}
+    \item Classical Gram-Schmidt (CGS, \texttt{cgs\_q})
+    \item Modified Gram-Schmidt (MGS, \texttt{mgs\_q})
+    \item Householder QR (\texttt{householder\_q})
+    \item Classical Gram-Schmidt Twice (CGS-Twice, \texttt{cgs\_twice\_q})
+\end{enumerate}
+For implementation details, please see the \texttt{Project1\_Q2.py} file.
+
+\subsection*{(d)}
+The quality of the computed $Q$ from each method was assessed by plotting the orthogonality loss, $\log_{10}(\|I - Q^T Q\|_F)$, versus the matrix size $n$, as shown in Figure~\ref{fig:ortho_loss}.
+
+\begin{figure}[h!]
+    \centering
+    \includegraphics[width=0.8\textwidth]{Project1_A2d_orthogonality_loss.png}
+    \caption{Orthogonality Loss vs. Matrix Size for QR Methods.}
+    \label{fig:ortho_loss}
+\end{figure}
+
+A summary of the loss and computation time for $n = 4, 9, 11$ is provided in Table~\ref{tab:qr_results}.
+
+\begin{table}[h!]
+    \centering
+    \begin{tabular}{|c|l|r|r|}
+        \hline
+        \textbf{n} & \textbf{Method} & \textbf{Loss ($log_{10}$)} & \textbf{Time (s)} \\
+        \hline
+        4 & Classical Gram-Schmidt & -10.311054 & 4.28e-05 \\
+        4 & Modified Gram-Schmidt & -12.374458 & 5.89e-05 \\
+        4 & Householder & -15.351739 & 1.42e-04 \\
+        4 & Classical Gram-Schmidt Twice & -15.570756 & 8.99e-05 \\
+        \hline
+        9 & Classical Gram-Schmidt & 0.389425 & 2.89e-04 \\
+        9 & Modified Gram-Schmidt & -5.257679 & 3.50e-04 \\
+        9 & Householder & -14.715202 & 6.19e-04 \\
+        9 & Classical Gram-Schmidt Twice & -8.398044 & 6.07e-04 \\
+        \hline
+        11 & Classical Gram-Schmidt & 0.609491 & 5.23e-04 \\
+        11 & Modified Gram-Schmidt & -2.005950 & 5.77e-04 \\
+        11 & Householder & -14.697055 & 8.68e-04 \\
+        11 & Classical Gram-Schmidt Twice & -1.745087 & 1.07e-03 \\
+        \hline
+    \end{tabular}
+    \caption{Loss and computation time for selected matrix sizes.}
+    \label{tab:qr_results}
+\end{table}
+
+\subsection*{(e)}
+The speed and accuracy of the four methods were compared based on the plots in Figure~\ref{fig:ortho_loss} and Figure~\ref{fig:time_taken}.
+
+\begin{figure}[h!]
+    \centering
+    \includegraphics[width=0.8\textwidth]{Project1_A2d_time_taken.png}
+    \caption{Computation Time vs. Matrix Size for QR Methods.}
+    \label{fig:time_taken}
+\end{figure}
+
+\begin{itemize}
+    \item \textbf{Accuracy:} Householder is the most accurate and stable. MGS is significantly better than CGS. CGS-Twice improves on CGS, achieving accuracy comparable to MGS. CGS is highly unstable and loses orthogonality quickly.
+    \item \textbf{Speed:} CGS and MGS are the fastest. Householder is slightly slower, and CGS-Twice is the slowest.
+\end{itemize}
+
+\section{Least Square Problem}
+
+\subsection*{(a)}
+The problem is formulated as $Ax = b$, where $A$ is an $(N+1) \times (n+1)$ Vandermonde matrix, $x$ is the $(n+1) \times 1$ vector of unknown polynomial coefficients $[a_0, \dots, a_n]^T$, and $b$ is the $(N+1) \times 1$ vector of known function values $f(t_j)$.
+
+\subsection*{(b)}
+The least squares problem was solved using the implemented Householder QR factorization (\texttt{householder\_lstsq} in \texttt{Project1\_Q3.py}, the same function in Question 2) to find the coefficient vector $x$.
+
+\subsection*{(c)}
+The function $f(t) = 1 / (1 + t^2)$ and its polynomial approximation $p(t)$ were plotted for $N=30$ and $n=5, 15, 30$, as shown in Figure~\ref{fig:ls_approx}.
+
+\begin{figure}[h!]
+    \centering
+    \includegraphics[width=0.8\textwidth]{Project1_A3c_least_squares_approximation.png}
+    \caption{Least squares approximation of $f(t)=1/(1+t^2)$ for varying polynomial degrees $n$.}
+    \label{fig:ls_approx}
+\end{figure}
+
+When $n=15$, the approximation is excellent. When $n=30$, the polynomial interpolates the points but exhibits wild oscillations between them.
+
+\subsection*{(d)}
+No, $p(t)$ will not converge to $f(t)$ if $N = n$ tends to infinity. Polynomial interpolation on equally spaced nodes diverges for this function, as demonstrated by the severe oscillations in the $n=30$ case shown in Figure~\ref{fig:ls_approx}.
+
+\subsection*{(e)}
+The error between the computed and true coefficients was plotted against the polynomial degree $n$, as shown in Figure~\ref{fig:coeff_error}.
+
+\begin{figure}[h!]
+    \centering
+    \includegraphics[width=0.8\textwidth]{Project1_A3e_coefficient_error.png}
+    \caption{Error in computed polynomial coefficients vs. polynomial degree $n$.}
+    \label{fig:coeff_error}
+\end{figure}
+
+The error in the computed coefficients grows significantly as $n$ increases.
+
+\end{document}

project2/iterative_solvers.py

@@ -0,0 +1,154 @@
+import numpy as np
+from typing import Callable, Any, List, Tuple
+
+# Tolerance and maximum iterations
+TOL = 1e-10
+MAX_ITER = 20000
+
+def stationary_method(
+        b: np.ndarray,
+        method: str,
+        get_diag_inv: Callable[[int, Any], float],
+        get_off_diag_sum: Callable[[np.ndarray, int, Any], float],
+        mat_vec_prod: Callable[[np.ndarray, Any], np.ndarray],
+        omega: float = 1.0,
+        **kwargs
+    ) -> Tuple[np.ndarray, int, List[float]]:
+    """
+    Solves Ax=b using stationary methods (Jacobi, Gauss-Seidel, SOR).
+    """
+    n = len(b)
+    x = np.zeros(n)
+    r0 = b.copy()
+    r0_norm = np.linalg.norm(r0)
+    if r0_norm == 0:
+        return x, 0, [0.0]
+        
+    residual_history = []
+
+    for k in range(MAX_ITER):
+        # 1. Calculate current residual and its norm
+        r = b - mat_vec_prod(x, **kwargs)
+        res_norm = np.linalg.norm(r)
+        
+        # 2. Append to history
+        residual_history.append(res_norm)
+        
+        # 3. Check for convergence
+        if res_norm / r0_norm < TOL:
+            return x, k, residual_history
+        
+        # 4. If not converged, perform the next iteration update
+        if method == 'jacobi':
+            x_old = x.copy()
+            for i in range(n):
+                off_diag_val = get_off_diag_sum(x_old, i, **kwargs)
+                diag_inv_val = get_diag_inv(i, **kwargs)
+                x[i] = diag_inv_val * (b[i] - off_diag_val)
+        else: # Gauss-Seidel and SOR
+            x_k_iter = x.copy() # Needed for SOR
+            for i in range(n):
+                off_diag_val = get_off_diag_sum(x, i, **kwargs)
+                diag_inv_val = get_diag_inv(i, **kwargs)
+                x_gs = diag_inv_val * (b[i] - off_diag_val)
+                
+                if method == 'gauss_seidel':
+                    x[i] = x_gs
+                elif method == 'sor':
+                    x[i] = (1 - omega) * x_k_iter[i] + omega * x_gs
+    
+    # If loop finishes (MAX_ITER reached), calculate and append final residual
+    r_final = b - mat_vec_prod(x, **kwargs)
+    residual_history.append(np.linalg.norm(r_final))
+    return x, MAX_ITER, residual_history
+
+
+def steepest_descent(
+        mat_vec_prod: Callable[[np.ndarray, Any], np.ndarray],
+        b: np.ndarray,
+        **kwargs
+    ) -> Tuple[np.ndarray, int, List[float]]:
+    """
+    Solves Ax=b using Steepest Descent, with corrected residual logging.
+    """
+    x = np.zeros_like(b)
+    r = b.copy()
+    
+    r0_norm = np.linalg.norm(r)
+    if r0_norm == 0:
+        return x, 0, [0.0]
+        
+    residual_history = []
+    
+    for k in range(MAX_ITER):
+        # 1. Current residual norm is from previous step (or initial)
+        res_norm = np.linalg.norm(r)
+        
+        # 2. Append to history
+        residual_history.append(res_norm)
+        
+        # 3. Check for convergence BEFORE the update
+        if res_norm / r0_norm < TOL:
+            return x, k, residual_history
+            
+        # 4. If not converged, compute update
+        Ar = mat_vec_prod(r, **kwargs)
+        r_dot_Ar = np.dot(r, Ar)
+        if r_dot_Ar == 0: # Should not happen for SPD if r!=0
+            break
+        alpha = res_norm**2 / r_dot_Ar
+        
+        x += alpha * r
+        r -= alpha * Ar
+        
+    # If loop finishes, append final residual
+    residual_history.append(np.linalg.norm(r))
+    return x, MAX_ITER, residual_history
+
+
+def conjugate_gradient(
+        mat_vec_prod: Callable[[np.ndarray, Any], np.ndarray],
+        b: np.ndarray,
+        **kwargs
+    ) -> Tuple[np.ndarray, int, List[float]]:
+    """
+    Solves Ax=b using Conjugate Gradient, with corrected residual logging.
+    """
+    x = np.zeros_like(b)
+    r = b.copy()
+    p = r.copy()
+    
+    r0_norm = np.linalg.norm(r)
+    if r0_norm == 0:
+        return x, 0, [0.0]
+        
+    res_norm_old = r0_norm
+    residual_history = []
+    
+    for k in range(MAX_ITER):
+        # 1. Current residual norm is from previous step
+        # 2. Append to history
+        residual_history.append(res_norm_old)
+        
+        # 3. Check for convergence
+        if res_norm_old / r0_norm < TOL:
+            return x, k, residual_history
+            
+        # 4. If not converged, compute update
+        Ap = mat_vec_prod(p, **kwargs)
+        p_dot_Ap = np.dot(p, Ap)
+        if p_dot_Ap == 0:
+            break
+        alpha = res_norm_old**2 / p_dot_Ap
+        
+        x += alpha * p
+        r -= alpha * Ap
+        
+        res_norm_new = np.linalg.norm(r)
+        beta = res_norm_new**2 / res_norm_old**2
+        
+        p = r + beta * p
+        res_norm_old = res_norm_new
+    # If loop finishes, append final residual
+    residual_history.append(res_norm_old)
+    return x, MAX_ITER, residual_history

project2/main.tex

@@ -0,0 +1,187 @@
+\documentclass{article}
+
+% Language setting
+\usepackage[english]{babel}
+
+% Set page size and margins
+\usepackage[letterpaper,top=2cm,bottom=2cm,left=3cm,right=3cm,marginparwidth=1.75cm]{geometry}
+
+% Useful packages
+\usepackage{amsmath}
+\usepackage{graphicx}
+\usepackage{amsfonts}
+\usepackage{float} % For better table and figure placement
+\usepackage[colorlinks=true, allcolors=blue]{hyperref}
+\usepackage{array} % For better table column definitions
+
+\title{MATH 6601 - Project 2 Report}
+\author{Zhe Yuan (yuan.1435)}
+\date{Nov 2025}
+
+\begin{document}
+\maketitle
+
+\section*{Common Iterative Methods for Both Problems}
+This project implements five iterative methods to solve large, sparse, symmetric positive-definite (SPD) linear systems of the form $Ax=b$. The implemented methods are:
+\begin{itemize}
+    \item Jacobi method
+    \item Gauss-Seidel method
+    \item Successive Over-Relaxation (SOR) method
+    \item Steepest Descent method
+    \item Conjugate Gradient method
+\end{itemize}
+All five methods are implemented in Python. The core logic for each solver is contained in the `iterative\_solvers.py' file. The specific problems are set up and solved in `problem1.py' and `problem2.py', respectively.
+
+The initial guess was $x^{(0)}=0$, and the stopping condition was $\|r^{(k)}\|_2 / \|r^{(0)}\|_2 < 10^{-10}$. All $A$ were not explicitly constructed.
+
+\section{Problem 1: Ordinary Differential Equations}
+This problem solves the boundary value problem $-u'' = t$ on $(0,1)$ with $u(0)=u(1)=0$. To solve the problem numerically, the interval $(0,1)$ is discretized into $n+1$ subintervals of width $h=1/(n+1)$. The second derivative $u''(t_i)$ is approximated using a central difference scheme:
+\[ u''(t_i) = -t_i \approx \frac{u(t_{i-1}) - 2u(t_i) + u(t_{i+1})}{h^2} \quad \implies \quad -u_{i-1} + 2u_i - u_{i+1} = h^2 t_i \]
+for the internal grid points $i=1, \dots, n$. Combined with the boundary conditions $u_0=0$ and $u_{n+1}=0$, this forms a linear system $Au=b$, where $u = [u_1, \dots, u_n]^T$. The matrix $A \in \mathbb{R}^{n \times n}$ is a tridiagonal matrix, and the vector $b \in \mathbb{R}^n$ is given by:
+\[
+A = \begin{pmatrix}
+    2 & -1 & & & \\
+    -1 & 2 & -1 & & \\
+    & \ddots & \ddots & \ddots & \\
+    & & -1 & 2 & -1 \\
+    & & & -1 & 2
+\end{pmatrix}, \quad
+b = h^2 \begin{pmatrix} t_1 \\ t_2 \\ \vdots \\ t_n \end{pmatrix}
+\]
+
+\subsection*{(a) Exact Solution}
+The boundary value problem is given by $-u'' = t$ on the interval $(0, 1)$ with boundary conditions $u(0) = u(1) = 0$.
+Integrating twice gives the general solution:
+\[ u(t) = -\frac{t^3}{6} - C_1 t - C_2 \]
+Applying the boundary conditions $u(0)=0$ and $u(1)=0$, we find the constants $C_2=0$ and $C_1 = -1/6$. Thus, the exact solution is:
+\[ u(t) = \frac{t - t^3}{6} \]
+The plot of this exact solution is shown in Figure~\ref{fig:p1_exact_solution}.
+\begin{figure}[H]
+    \centering
+    \includegraphics[width=0.8\textwidth]{problem1_exact_solution.png}
+    \caption{Plot of the exact solution $u(t) = (t-t^3)/6$.}
+    \label{fig:p1_exact_solution}
+\end{figure}
+
+\subsection*{(b, c) Iteration Solution}
+According to the requirments from 1(b), The linear system was solved for $n=20$ and $n=40$. For SOR, the optimal relaxation parameter $\omega = \frac{2}{1+\sin(\pi h)}$ was used. The number of iterations required for each method to converge is summarized in Table~\ref{tab:p1_iterations}.
+
+\begin{table}[H]
+\centering
+\begin{tabular}{|l|r|r|}
+\hline
+\textbf{Method} & \textbf{Iterations (n=20)} & \textbf{Iterations (n=40)} \\
+\hline
+Jacobi & 2031 & 7758 \\
+Gauss-Seidel & 1018 & 3882 \\
+SOR & 94 & 183 \\
+Steepest Descent & 2044 & 7842\\
+Conjugate Gradient & 20 & 40 \\
+\hline
+\end{tabular}
+\caption{Number of iterations for each method for Problem 1.}
+\label{tab:p1_iterations}
+\end{table}
+
+\noindent \textbf{Observations:}
+\begin{enumerate}
+    \item The Conjugate Gradient method is the most efficient, requiring the fewest iterations. SOR is the second-best with a significant speedup over Gauss-Seidel. Gauss-Seidel converges about twice as fast as Jacobi. The Steepest Descent method performs poorly, rivaling Jacobi in its slow convergence.
+    \item As $n$ doubles, the number of iterations increases for all methods. The iteration count for Jacobi, Gauss-Seidel, and Steepest Descent increases by a factor of roughly 4, while for SOR and CG, the increase is closer to a factor of 2.
+\end{enumerate}
+
+\subsection*{(d) Convergence History}
+The convergence history is shown in Figures~\ref{fig:p1_conv_n20} and \ref{fig:p1_conv_n40}. The stationary methods and SD show linear convergence, while CG shows superlinear convergence.
+
+\begin{figure}[H]
+    \centering
+    \includegraphics[width=0.8\textwidth]{problem1_convergence_n20.png}
+    \caption{Convergence history for Problem 1 with n=20.}
+    \label{fig:p1_conv_n20}
+\end{figure}
+
+\begin{figure}[H]
+    \centering
+    \includegraphics[width=0.8\textwidth]{problem1_convergence_n40.png}
+    \caption{Convergence history for Problem 1 with n=40.}
+    \label{fig:p1_conv_n40}
+\end{figure}
+
+\newpage
+\section{Problem 2: Partial Differential Equation}
+This problem involves solving the PDE $-\nabla^2 u + u = 1$ (this should be $-\nabla^2 u + u = 1$ instead of $-\nabla u(\text{vector}) + u(\text{scalar}) = 1$ in question) on the unit square $[0,1]^2$ with zero boundary conditions. To solve the problem numerically, a finite difference approximation on an $n \times n$ grid of interior points leads to the discrete equations:
+\[ (4+h^2)u_{i,j} - u_{i-1,j} - u_{i,j-1} - u_{i+1,j} - u_{i,j+1} = h^2, \quad i,j=1,\dots,n \]
+To formulate this as a standard matrix system $Ax=b$, we reshape the 2D array of unknowns $u_{i,j}$ into a single 1D vector $x$ of length $N=n^2$. After flattening, $u_{i,j}$ maps to the index $k = (i-1)n + j$ in the vector $x$, the matrix $A \in \mathbb{R}^{N \times N}$ becomes a large, sparse, block tridiagonal matrix:
+\[
+A = \begin{pmatrix}
+    T & -I & & & \\
+    -I & T & -I & & \\
+    & \ddots & \ddots & \ddots & \\
+    & & -I & T & -I \\
+    & & & -I & T
+\end{pmatrix}_{n^2 \times n^2}
+\]
+where $I$ is the $n \times n$ identity matrix, and $T$ is an $n \times n$ tridiagonal matrix given by:
+\[
+T = \begin{pmatrix}
+    4+h^2 & -1 & & & \\
+    -1 & 4+h^2 & -1 & & \\
+    & \ddots & \ddots & \ddots & \\
+    & & -1 & 4+h^2 & -1 \\
+    & & & -1 & 4+h^2
+\end{pmatrix}_{n \times n}
+\]
+The right-hand side vector $b \in \mathbb{R}^{N}$ is a constant vector where every element is $h^2$. So:
+
+\[
+(A_{i,i}=T)_{j,j}x_{(i-1)n + j} = (4+h^2)u_{i,j}, (A_{i-1,i}=I)_{j,j}x_{(i-2)n + j} = u_{i-1,j},\dots
+\]
+
+\subsection*{(a) Iterative Solution}
+The system was solved for $n=16$ ($N=256$) and $n=32$ ($N=1024$) using the same five iterative methods. The number of iterations to reach convergence is reported in Table~\ref{tab:p2_iterations}.
+
+\begin{table}[H]
+\centering
+\begin{tabular}{|l|r|r|}
+\hline
+\textbf{Method} & \textbf{Iterations (n=16)} & \textbf{Iterations (n=32)} \\
+\hline
+Jacobi & 1268 & 4792 \\
+Gauss-Seidel & 635 & 2397 \\
+SOR & 71 & 138 \\
+Steepest Descent & 1247 & 4807 \\
+Conjugate Gradient & 31 & 66 \\
+\hline
+\end{tabular}
+\caption{Number of iterations for each method for Problem 2.}
+\label{tab:p2_iterations}
+\end{table}
+
+\noindent \textbf{Observations:}
+The relative performance of the methods remains consistent with Problem 1: CG is the fastest, followed by SOR, Gauss-Seidel, Jacobi, and Steepest Descent. Doubling $n$ from 16 to 32 quadruples the number of unknowns. Again, the iteration counts for slower methods increase by a factor of approximately 4, while for CG and SOR, the increase is only about a factor of 2. This reinforces the conclusion that CG and SOR scale much better with problem size.
+
+\subsection*{(b) Convergence History}
+The convergence history plots are shown in Figures~\ref{fig:p2_conv_n16} and \ref{fig:p2_conv_n32}.
+
+\begin{figure}[H]
+    \centering
+    \includegraphics[width=0.8\textwidth]{problem2_convergence_n16.png}
+    \caption{Convergence history for Problem 2 with n=16.}
+    \label{fig:p2_conv_n16}
+\end{figure}
+
+\begin{figure}[H]
+    \centering
+    \includegraphics[width=0.8\textwidth]{problem2_convergence_n32.png}
+    \caption{Convergence history for Problem 2 with n=32.}
+    \label{fig:p2_conv_n32}
+\end{figure}
+
+\subsection*{(c) Advantages of Iterative Methods}
+For large, sparse linear systems arising from PDE discretizations, iterative methods offer significant advantages over direct methods:
+\begin{enumerate}
+    \item Iterative methods typically only require storing a few vectors of size $N$, leading to a memory complexity of $O(N)$. In contrast, direct methods require denser helper matrices like $Q$ and $R$. 
+    \item The cost per iteration for these problems is dominated by the sparse matrix-vector product, which is $O(N)$. If the method converges in $K$ iterations, the total cost is $O(KN)$. If a good iterative method is used, $K$ can be much smaller than $N$ (like C.G.). Direct sparse solvers often have a higher complexity. When $N$ is very large, iterative methods can be much faster.
+    \item Implementing iterative methods is often simpler than implementing a direct solver for sparse matrix.
+\end{enumerate}
+
+\end{document}

project2/problem1.py

@@ -0,0 +1,160 @@
+import numpy as np
+import matplotlib.pyplot as plt
+from iterative_solvers import *
+
+# (a): Exact Solution
+def exact_solution(t: np.ndarray) -> np.ndarray:
+    """Computes the exact solution u(t) = t/6 * (1 - t^2)."""
+    return t / 6.0 * (1 - t**2)
+
+def plot_exact_solution():
+    """Plots the exact solution of the BVP."""
+    t_fine = np.linspace(0, 1, 500)
+    u_exact_fine = exact_solution(t_fine)
+    
+    plt.figure(figsize=(10, 6))
+    plt.plot(t_fine, u_exact_fine, 'k-', label='Exact Solution $u(t) = t/6 \cdot (1 - t^2)$')
+    plt.title('Exact Solution of $-u\'\'=t$ with $u(0)=u(1)=0$')
+    plt.xlabel('$t$')
+    plt.ylabel('$u(t)$')
+    plt.grid(True)
+    plt.legend()
+    plt.savefig('problem1_exact_solution.png')
+    # plt.show()
+    print("Exact solution plot saved to 'problem1_exact_solution.png'")
+
+
+# (b): Helper functions for iterative methods
+def mat_vec_prod_p1(x: np.ndarray, n: int) -> np.ndarray:
+    """Computes the matrix-vector product Ax for Problem 1."""
+    y = np.zeros(n)
+    # A is tridiagonal with [ -1, 2, -1 ]
+    # y_i = -x_{i-1} + 2x_i - x_{i+1}
+    for i in range(n):
+        term_prev = -x[i-1] if i > 0 else 0
+        term_next = -x[i+1] if i < n - 1 else 0
+        y[i] = term_prev + 2 * x[i] + term_next
+    return y
+
+def get_diag_inv_p1(i: int, n: int) -> float:
+    """Returns 1/A_ii for Problem 1. Diagonal elements are all 2, so inverse is 1/2."""
+    return 1.0 / 2.0
+
+def get_off_diag_sum_p1(x: np.ndarray, i: int, n: int) -> float:
+    """Computes sum_{j!=i} A_ij * x_j. For Problem 1, A_ij = -1 for j=i-1, i+1, else 0."""
+    sum_val = 0.0
+    if i > 0:
+        sum_val += -x[i-1]
+    if i < n - 1:
+        sum_val += -x[i+1]
+    return sum_val
+
+def solve_and_report(n: int):
+    """
+    Solves the linear system for a given n and reports results.
+    """
+    print(f"\n--- Solving for n = {n} ---")
+    h = 1.0 / (n + 1)
+    
+    # Setup the right-hand side vector b
+    t = np.linspace(h, 1.0 - h, n)
+    b = (h**2) * t
+    
+    # Calculate SOR parameter omega
+    omega = 2.0 / (1.0 + np.sin(np.pi * h))
+    
+    # Store results
+    iteration_counts = []
+    all_residuals = {}
+    all_solutions = {}
+    # Define methods and run solvers
+    methods_to_run = {
+        'Jacobi': ('jacobi', {}),
+        'Gauss-Seidel': ('gauss_seidel', {}),
+        'SOR': ('sor', {'omega': omega}),
+        'Steepest Descent': (steepest_descent, {}),
+        'Conjugate Gradient': (conjugate_gradient, {})
+    }
+    
+    print("\nStarting iterative solvers...")
+    for name, (method_func, params) in methods_to_run.items():
+        print(f"Running {name}...")
+        common_args = {'b': b, 'n': n}
+        if isinstance(method_func, str):
+            solver_args = { 'method': method_func, 'get_diag_inv': get_diag_inv_p1,
+                            'get_off_diag_sum': get_off_diag_sum_p1, 'mat_vec_prod': mat_vec_prod_p1,
+                            **common_args, **params }
+            solution, iters, res_hist = stationary_method(**solver_args)
+        else:
+            solver_args = {'mat_vec_prod': mat_vec_prod_p1, **common_args}
+            solution, iters, res_hist = method_func(**solver_args)
+        iteration_counts.append((name, iters))
+        all_residuals[name] = res_hist
+        all_solutions[name] = solution if iters < MAX_ITER else None
+    # (c): Report iteration counts
+    print("\n--- Iteration Counts ---")
+    for name, iters in iteration_counts:
+        status = "converged" if iters < MAX_ITER else "did NOT converge"
+        print(f"{name:<20}: {iters} iterations ({status}, last residual = {all_residuals[name][-1]:.2e})")
+    # (d): Plot residue history
+    plt.figure(figsize=(12, 8))
+    for name, res_hist in all_residuals.items():
+        res_log = np.log10(np.array(res_hist) + 1e-20)
+        plt.plot(res_log, label=name, linestyle='-')
+    
+    plt.title(f'Convergence History for n = {n}')
+    plt.xlabel('Iteration Step (m)')
+    plt.ylabel('log10(||r^(m)||_2)')
+    plt.legend()
+    plt.grid(True)
+    
+    # Set the y-axis limit to make the plot readable
+    current_ylim = plt.ylim()
+    plt.ylim(bottom=max(current_ylim[0], -13), top=current_ylim[1])
+    plt.savefig(f'problem1_convergence_n{n}.png')
+    # plt.show()
+    print(f"\nConvergence plot saved to 'problem1_convergence_n{n}.png'")
+    
+    if all_solutions:
+        plt.figure(figsize=(12, 8))
+        
+        # Plot exact solution on a fine grid
+        t_fine = np.linspace(0, 1, 500)
+        u_exact_fine = exact_solution(t_fine)
+        plt.plot(t_fine, u_exact_fine, 'k-', label='Exact Solution', linewidth=1, zorder=10) # Black, thick, on top
+        
+        # Define some styles for the different numerical methods
+        styles = {
+            'Jacobi':          {'color': 'red',   'linestyle': '--'},
+            'Gauss-Seidel':    {'color': 'blue',  'linestyle': '-.'},
+            'SOR':             {'color': 'green', 'linestyle': ':'},
+            'Steepest Descent':{'color': 'purple','linestyle': '--'},
+            'Conjugate Gradient':{'color': 'orange','linestyle': '-.'},
+        }
+        
+        # Plot each numerical solution
+        for name, solution in all_solutions.items():
+            # Add boundary points for a complete plot
+            t_numerical_with_bounds = np.concatenate(([0], t, [1]))
+            u_numerical_with_bounds = np.concatenate(([0], solution, [0]))
+            
+            style = styles.get(name, {'color': 'gray', 'linestyle': '-'})
+            plt.plot(t_numerical_with_bounds, u_numerical_with_bounds, 
+                     label=f'Numerical ({name})', **style, linewidth=2)
+        
+        plt.title(f'Comparison of Exact and Numerical Solutions for n = {n}')
+        plt.xlabel('$t$')
+        plt.ylabel('$u(t)$')
+        plt.legend()
+        plt.grid(True)
+        plt.savefig(f'problem1_all_solutions_comparison_n{n}.png')
+        # plt.show()
+        print(f"All solutions comparison plot saved to 'problem1_all_solutions_comparison_n{n}.png'")
+    
+if __name__ == "__main__":
+    # Part (a)
+    plot_exact_solution()
+    
+    # Part (b), (c), (d)
+    solve_and_report(n=20)
+    solve_and_report(n=40)

project2/problem1_output.txt

@@ -0,0 +1,39 @@
+Exact solution plot saved to 'problem1_exact_solution.png'
+
+--- Solving for n = 20 ---
+
+Starting iterative solvers...
+Running Jacobi...
+Running Gauss-Seidel...
+Running SOR...
+Running Steepest Descent...
+Running Conjugate Gradient...
+
+--- Iteration Counts ---
+Jacobi              : 2031 iterations (converged, last residual = 5.78e-13)
+Gauss-Seidel        : 1018 iterations (converged, last residual = 5.72e-13)
+SOR                 : 94 iterations (converged, last residual = 4.52e-13)
+Steepest Descent    : 2044 iterations (converged, last residual = 5.74e-13)
+Conjugate Gradient  : 20 iterations (converged, last residual = 1.39e-18)
+
+Convergence plot saved to 'problem1_convergence_n20.png'
+All solutions comparison plot saved to 'problem1_all_solutions_comparison_n20.png'
+
+--- Solving for n = 40 ---
+
+Starting iterative solvers...
+Running Jacobi...
+Running Gauss-Seidel...
+Running SOR...
+Running Steepest Descent...
+Running Conjugate Gradient...
+
+--- Iteration Counts ---
+Jacobi              : 7758 iterations (converged, last residual = 2.15e-13)
+Gauss-Seidel        : 3882 iterations (converged, last residual = 2.16e-13)
+SOR                 : 183 iterations (converged, last residual = 2.04e-13)
+Steepest Descent    : 7842 iterations (converged, last residual = 2.16e-13)
+Conjugate Gradient  : 40 iterations (converged, last residual = 5.40e-19)
+
+Convergence plot saved to 'problem1_convergence_n40.png'
+All solutions comparison plot saved to 'problem1_all_solutions_comparison_n40.png'

project2/problem2.py

@@ -0,0 +1,147 @@
+import numpy as np
+import matplotlib.pyplot as plt
+from iterative_solvers import *
+
+def mat_vec_prod_p2(x_flat: np.ndarray, n: int, h_sq: float) -> np.ndarray:
+    """
+    Computes the matrix-vector product y = Ax for the 2D PDE problem.
+    """
+    x_2d = x_flat.reshape((n, n))
+    x_padded = np.zeros((n + 2, n + 2))
+    x_padded[1:-1, 1:-1] = x_2d
+    
+    y_2d = np.zeros((n, n))
+    
+    # Apply the finite difference stencil: (4+h^2)u_ij - u_{i-1,j} - u_{i+1,j} - u_{i,j-1} - u_{i,j+1}
+    for i in range(n):
+        for j in range(n):
+            y_2d[i, j] = (4.0 + h_sq) * x_padded[i + 1, j + 1] \
+                       - x_padded[i, j + 1] \
+                       - x_padded[i + 2, j + 1] \
+                       - x_padded[i + 1, j] \
+                       - x_padded[i + 1, j + 2]
+                       
+    return y_2d.flatten()
+
+def get_diag_inv_p2(i: int, n: int, h_sq: float) -> float:
+    """Returns 1/A_ii for Problem 2. Diagonal is constant: 4 + h^2."""
+    return 1.0 / (4.0 + h_sq)
+
+def get_off_diag_sum_p2(x_flat: np.ndarray, i_flat: int, n: int, **kwargs) -> float:
+    """Computes the off-diagonal sum for a given row `i_flat` in Problem 2."""
+    row = i_flat // n
+    col = i_flat % n
+    sum_val = 0.0
+    
+    if row > 0: sum_val += -x_flat[i_flat - n]
+    if row < n - 1: sum_val += -x_flat[i_flat + n]
+    if col > 0: sum_val += -x_flat[i_flat - 1]
+    if col < n - 1: sum_val += -x_flat[i_flat + 1]
+        
+    return sum_val
+
+def solve_and_report_p2(n: int):
+    """Solves the 2D PDE linear system for a given n and reports results."""
+    print(f"\n--- Solving for n = {n} (System size: {n*n} x {n*n}) ---")
+    h = 1.0 / (n + 1)
+    h_sq = h**2
+    N = n * n
+    
+    b = np.full(N, h_sq)
+    omega = 2.0 / (1.0 + np.sin(np.pi * h))
+    
+    iteration_counts = []
+    all_residuals = {}
+    all_solutions = {}
+
+    methods_to_run = {
+        'Jacobi': ('jacobi', {}),
+        'Gauss-Seidel': ('gauss_seidel', {}),
+        'SOR': ('sor', {'omega': omega}),
+        'Steepest Descent': (steepest_descent, {}),
+        'Conjugate Gradient': (conjugate_gradient, {})
+    }
+    
+    print("\nStarting iterative solvers...")
+    for name, (method_func, params) in methods_to_run.items():
+        print(f"Running {name}...")
+        
+        solver_kwargs = {'n': n, 'h_sq': h_sq}
+        
+        if isinstance(method_func, str):
+            solver_args = {'b': b, 'method': method_func, 'get_diag_inv': get_diag_inv_p2, 
+                           'get_off_diag_sum': get_off_diag_sum_p2, 'mat_vec_prod': mat_vec_prod_p2,
+                           **solver_kwargs, **params}
+            solution, iters, res_hist = stationary_method(**solver_args)
+        else:
+            solver_args = {'b': b, 'mat_vec_prod': mat_vec_prod_p2, **solver_kwargs}
+            solution, iters, res_hist = method_func(**solver_args)
+        
+        iteration_counts.append((name, iters))
+        all_residuals[name] = res_hist
+        all_solutions[name] = solution if iters < MAX_ITER else None
+
+    # Report iteration counts
+    print("\n--- Iteration Counts ---")
+    for name, iters in iteration_counts:
+        status = "converged" if iters < MAX_ITER else "did NOT converge"
+        print(f"{name:<20}: {iters} iterations ({status})")
+
+    # Plot residue history
+    plt.figure(figsize=(12, 8))
+    for name, res_hist in all_residuals.items():
+        res_log = np.log10(np.array(res_hist) + 1e-20)
+        plt.plot(res_log, label=name, linestyle='-')
+    
+    plt.title(f'Convergence History for PDE Problem, n = {n}')
+    plt.xlabel('Iteration Step (m)')
+    plt.ylabel('log10(||r^(m)||_2)')
+    plt.legend()
+    plt.grid(True)
+    
+    current_ylim = plt.ylim()
+    plt.ylim(bottom=max(current_ylim[0], -13), top=current_ylim[1])
+    
+    plt.savefig(f'problem2_convergence_n{n}.png')
+    print(f"\nConvergence plot saved to 'problem2_convergence_n{n}.png'")
+
+    if all_solutions:
+        print("Generating 3D surface plots of the solutions...")
+        
+        # Create the X, Y grid for the surface plot, including boundaries
+        plot_grid = np.linspace(0, 1, n + 2)
+        X, Y = np.meshgrid(plot_grid, plot_grid)
+        
+        # Create a figure for the subplots
+        fig = plt.figure(figsize=(18, 10))
+        fig.suptitle(f'3D Surface Plots of Numerical Solutions for n = {n}', fontsize=16)
+        # Loop through the stored solutions and create a subplot for each
+        for i, (name, solution_flat) in enumerate(all_solutions.items()):
+            # Reshape the 1D solution vector into a 2D grid
+            solution_2d = solution_flat.reshape((n, n))
+            
+            # Pad the solution with zero boundaries for plotting
+            u_padded = np.zeros((n + 2, n + 2))
+            u_padded[1:-1, 1:-1] = solution_2d
+            
+            # Add a new 3D subplot
+            ax = fig.add_subplot(2, 3, i + 1, projection='3d')
+            
+            # Plot the surface
+            ax.plot_surface(X, Y, u_padded, cmap='viridis', edgecolor='none')
+            
+            # Set titles and labels for each subplot
+            ax.set_title(name)
+            ax.set_xlabel('x')
+            ax.set_ylabel('y')
+            ax.set_zlabel('u(x,y)')
+            ax.view_init(elev=30, azim=-60) # Set a nice viewing angle
+        plt.tight_layout(rect=[0, 0.03, 1, 0.95]) # Adjust layout to make room for suptitle
+        plt.savefig(f'problem2_3d_solutions_n{n}.png')
+        print(f"3D surface comparison plot saved to 'problem2_3d_solutions_n{n}.png'")
+
+if __name__ == "__main__":
+    # Solve Problem 2 for n=16 and n=32
+    solve_and_report_p2(n=16)
+    solve_and_report_p2(n=32)
+

project2/problem2_output.txt

@@ -0,0 +1,40 @@
+
+--- Solving for n = 16 (System size: 256 x 256) ---
+
+Starting iterative solvers...
+Running Jacobi...
+Running Gauss-Seidel...
+Running SOR...
+Running Steepest Descent...
+Running Conjugate Gradient...
+
+--- Iteration Counts ---
+Jacobi              : 1268 iterations (converged)
+Gauss-Seidel        : 635 iterations (converged)
+SOR                 : 71 iterations (converged)
+Steepest Descent    : 1247 iterations (converged)
+Conjugate Gradient  : 31 iterations (converged)
+
+Convergence plot saved to 'problem2_convergence_n16.png'
+Generating 3D surface plots of the solutions...
+3D surface comparison plot saved to 'problem2_3d_solutions_n16.png'
+
+--- Solving for n = 32 (System size: 1024 x 1024) ---
+
+Starting iterative solvers...
+Running Jacobi...
+Running Gauss-Seidel...
+Running SOR...
+Running Steepest Descent...
+Running Conjugate Gradient...
+
+--- Iteration Counts ---
+Jacobi              : 4792 iterations (converged)
+Gauss-Seidel        : 2397 iterations (converged)
+SOR                 : 138 iterations (converged)
+Steepest Descent    : 4807 iterations (converged)
+Conjugate Gradient  : 66 iterations (converged)
+
+Convergence plot saved to 'problem2_convergence_n32.png'
+Generating 3D surface plots of the solutions...
+3D surface comparison plot saved to 'problem2_3d_solutions_n32.png'