# Finite-element solution of the 2D Poisson equation  -Laplacian(u) = f  on the
# unit square with u = 0 on the boundary, using linear (P1) triangular elements.
#
# Manufactured solution u(x,y) = x(1-x) y(1-y), so f = 2[x(1-x) + y(1-y)] and
# u = 0 on the boundary automatically. Validation: the RMS nodal error falls ~4x
# per mesh halving (O(h^2), the theoretical P1 rate), and u at the center -> 0.0625.
#
# Data layout is struct-of-arrays (node_x/node_y coordinates, elem connectivity) --
# the same paradigm as the MD and QC examples. This baseline establishes whether
# FEM needs any object/record support beyond what Viv already has.

# --- dense linear solver: A x = b by Gaussian elimination w/ partial pivoting ---
def solve_linear(A, b, x, n) {
    for k to n {
        piv = k
        big = A[k, k]
        if big < 0.0 { big = -big }
        for i to n {
            if i > k {
                ai = A[i, k]
                if ai < 0.0 { ai = -ai }
                if ai > big {
                    big = ai
                    piv = i
                }
            }
        }
        if piv > k {
            for j to n {
                tmp = A[k, j]
                A[k, j] = A[piv, j]
                A[piv, j] = tmp
            }
            tb = b[k]
            b[k] = b[piv]
            b[piv] = tb
        }
        for i to n {
            if i > k {
                f = A[i, k] / A[k, k]
                for j to n {
                    A[i, j] = A[i, j] - f * A[k, j]
                }
                b[i] = b[i] - f * b[k]
            }
        }
    }
    for kk to n {
        i = n - 1.0 - kk
        s = b[i]
        for j to n {
            if j > i {
                s = s - A[i, j] * x[j]
            }
        }
        x[i] = s / A[i, i]
    }
}

# --- mesh: unit square split into 2*N*N right triangles ---
N = 8
nn = (N + 1.0) * (N + 1.0)
ne = 2.0 * N * N
node_x = zeros(nn)
node_y = zeros(nn)
for i to N + 1.0 {
    for j to N + 1.0 {
        idx = i * (N + 1.0) + j
        node_x[idx] = i / N
        node_y[idx] = j / N
    }
}
elem = zeros(ne, 3)
ecount = 0.0
for i to N {
    for j to N {
        n00 = i * (N + 1.0) + j
        n10 = (i + 1.0) * (N + 1.0) + j
        n01 = i * (N + 1.0) + (j + 1.0)
        n11 = (i + 1.0) * (N + 1.0) + (j + 1.0)
        elem[ecount, 0] = n00
        elem[ecount, 1] = n10
        elem[ecount, 2] = n11
        ecount = ecount + 1.0
        elem[ecount, 0] = n00
        elem[ecount, 1] = n11
        elem[ecount, 2] = n01
        ecount = ecount + 1.0
    }
}

# --- assemble global stiffness K and load vector rhs ---
K = zeros(nn, nn)
rhs = zeros(nn)
nl = zeros(3)
bv = zeros(3)
cv = zeros(3)
for el to ne {
    n0 = elem[el, 0]
    n1 = elem[el, 1]
    n2 = elem[el, 2]
    x0 = node_x[n0]
    y0 = node_y[n0]
    x1 = node_x[n1]
    y1 = node_y[n1]
    x2 = node_x[n2]
    y2 = node_y[n2]
    detJ = (x1 - x0) * (y2 - y0) - (x2 - x0) * (y1 - y0)
    area = 0.5 * detJ
    if area < 0.0 { area = -area }
    bv[0] = y1 - y2
    bv[1] = y2 - y0
    bv[2] = y0 - y1
    cv[0] = x2 - x1
    cv[1] = x0 - x2
    cv[2] = x1 - x0
    nl[0] = n0
    nl[1] = n1
    nl[2] = n2
    xc = (x0 + x1 + x2) / 3.0
    yc = (y0 + y1 + y2) / 3.0
    fval = 2.0 * (xc * (1.0 - xc) + yc * (1.0 - yc))
    for a to 3 {
        gI = nl[a]
        for c to 3 {
            gJ = nl[c]
            ke = (bv[a] * bv[c] + cv[a] * cv[c]) / (4.0 * area)
            K[gI, gJ] = K[gI, gJ] + ke
        }
        rhs[gI] = rhs[gI] + fval * area / 3.0
    }
}

# --- Dirichlet boundary condition u = 0 on the square's edges (row replacement) ---
for gI to nn {
    xx = node_x[gI]
    yy = node_y[gI]
    onb = xx < 1.0e-9 or xx > 1.0 - 1.0e-9 or yy < 1.0e-9 or yy > 1.0 - 1.0e-9
    if onb {
        for gJ to nn {
            K[gI, gJ] = 0.0
        }
        K[gI, gI] = 1.0
        rhs[gI] = 0.0
    }
}

# --- solve and measure the error against the exact solution ---
u = zeros(nn)
solve_linear(K, rhs, u, nn)
err2 = 0.0
for gI to nn {
    ue = node_x[gI] * (1.0 - node_x[gI]) * node_y[gI] * (1.0 - node_y[gI])
    d = u[gI] - ue
    err2 = err2 + d * d
}
rms = sqrt(err2 / nn)
center = N / 2.0 * (N + 1.0) + N / 2.0
u_center = u[center]
show N
show rms
show u_center

# SPDX-License-Identifier: MIT OR Apache-2.0
# SPDX-FileCopyrightText: 2026 The Viv authors