কিউবিক স্প্লাইন ইন্টারপোলেশন

def runge(x):
    return 1.0 / (1.0 + 25.0 * x**2)

def equally_spaced(n, a=-1.0, b=1.0):
    return [a + i*(b-a)/(n-1) for i in range(n)]

def natural_cubic_spline_coeffs(xs, ys):
    n = len(xs) - 1
    h = [xs[i+1] - xs[i] for i in range(n)]
    size = n - 1
    a = [0.0]*size
    b = [0.0]*size
    c = [0.0]*size
    d = [0.0]*size
    for i in range(1, n):
        idx = i - 1
        b[idx] = 2*(h[i-1] + h[i])
        if idx - 1 >= 0:
            a[idx] = h[i-1]
        if idx + 1 < size:
            c[idx] = h[i]
        d[idx] = 6*((ys[i+1]-ys[i])/h[i] - (ys[i]-ys[i-1])/h[i-1])

    # Thomas অ্যালগরিদম (M3/L15-এর মতো)
    cp = [0.0]*size
    dp = [0.0]*size
    cp[0] = c[0]/b[0]
    dp[0] = d[0]/b[0]
    for i in range(1, size):
        m = b[i] - a[i]*cp[i-1]
        cp[i] = c[i]/m if i < size-1 else 0.0
        dp[i] = (d[i] - a[i]*dp[i-1])/m

    Msol = [0.0]*size
    Msol[-1] = dp[-1]
    for i in range(size-2, -1, -1):
        Msol[i] = dp[i] - cp[i]*Msol[i+1]

    M = [0.0]*(n+1)
    for i in range(size):
        M[i+1] = Msol[i]
    M[0] = 0.0
    M[n] = 0.0
    return xs, ys, h, M

def spline_eval(xs, ys, h, M, x):
    n = len(xs) - 1
    i = 0
    for k in range(n):
        if xs[k] <= x <= xs[k+1]:
            i = k
            break
    hi = h[i]
    A = (xs[i+1] - x) / hi
    B = (x - xs[i]) / hi
    term1 = A*ys[i] + B*ys[i+1]
    term2 = ((A**3 - A)*M[i] + (B**3 - B)*M[i+1]) * (hi**2) / 6.0
    return term1 + term2

n = 11
xs = equally_spaced(n)
ys = [runge(x) for x in xs]
xs2, ys2, h, M = natural_cubic_spline_coeffs(xs, ys)

print(f"ন্যাচারাল কিউবিক স্প্লাইন, n = {n} নোড (L18-এর একই ডেটাসেট):")
print()

poly_errors_from_L18 = {0.95: 1.881191, 0.98: 1.190333, 0.0: 0.000000}
for xt in [0.95, 0.98, 0.0]:
    s = spline_eval(xs, ys, h, M, xt)
    true_v = runge(xt)
    err = abs(s - true_v)
    print(f"x={xt}: স্প্লাইন S(x)={s:.6f}  true={true_v:.6f}  স্প্লাইন_এরর={err:.6f}   (ডিগ্রি-১০ পলিনোমিয়ালের এরর ছিল {poly_errors_from_L18[xt]:.6f})")

# (আগের সেলের xs, ys, h, M ব্যবহার করে — একই Pyodide সেশনে চালাতে হবে)
print("নোডের মাঝামাঝি বিন্দুতে স্প্লাইন মূল্যায়ন:")
for xt in [0.1, 0.5, -0.7]:
    s = spline_eval(xs, ys, h, M, xt)
    true_v = runge(xt)
    print(f"  x={xt}: S(x)={s:.6f}  true={true_v:.6f}  এরর={abs(s-true_v):.6f}")