- Commit
- 9c7392a31b59f45a69505c0415b9b7e6387205d5
- Author
- Pablo <pablo-pie@riseup.net>
- Date
Initial commit
Educational material on the SciPy implementation of numerical linear algebra algorithms
Initial commit
5 files changed, 1060 insertions, 0 deletions
Status | File Name | N° Changes | Insertions | Deletions |
Added | .gitignore | 5 | 5 | 0 |
Added | LICENSE | 675 | 675 | 0 |
Added | getrf.py | 106 | 106 | 0 |
Added | linear-solvers.ipynb | 196 | 196 | 0 |
Added | test_getrf.py | 78 | 78 | 0 |
diff --git a/.gitignore b/.gitignore @@ -0,0 +1,5 @@ +testlogs.json +lapack + +.ipynb_checkpoints +__pycache__
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diff --git a/getrf.py b/getrf.py @@ -0,0 +1,106 @@ +# "SciPy-transpiled" version of LAPACK's GETRF family of subroutines! +import numpy as np +import scipy.linalg as la + +def getrf(A: np.ndarray) -> (np.ndarray, np.ndarray, np.ndarray): + """Returns the P, L, U + + * A is m by n + * P is m by m + * L is m by n if m >= n and m by m if m <= n + * U is n by n if m >= n and m by n if m <= n + """ + m, n = A.shape + + # A is a row + if m == 1: + return np.eye(1), np.eye(1), A + + # A is a column + elif n == 1: + i0 = 0 + + for i in range(m): + if abs(A[i, 0]) > abs(A[i0, 0]): i0 = i + + # P permutes the 0-th and i0-th basis vectors + P = np.eye(m) + P[0,0], P[i0,i0] = 0, 0 + P[i0,0], P[0,i0] = 1, 1 + + if A[i0, 0] != 0: + L = P@A / A[i0, 0] + U = A[i0, 0] * np.eye(1) + else: + L = A + U = np.zeros((1, 1)) + + return P, L, U + else: + n1 = min(m, n)//2 + n2 = n - n1 + + # Write + # + # A = [[A11, A12], + # [A21, A22]], + # + # A1 = [[A11, + # A21]], + # + # A2 = [[A12, + # A22]] + # + # where A11 is n1 by n1 and A22 is n2 by n2 + A11, A12 = A[:n1,:n1], A[:n1,n1:] + A21, A22 = A[n1:,:n1], A[n1:,n1:] + A1, A2 = A[:,:n1], A[:,n1:] + + # Solve the A1 block + P1, L1, U11 = getrf(A1) + + # Apply pivots + # A2 is m by n2 + A2 = la.inv(P1) @ A2 + A12, A22 = A2[:n1,:], A2[n1:,:] + + # Solve A12 + L11, L21 = L1[:n1,:], L1[n1:,:] + A12 = la.inv(L11) @ A12 + + # Update A22 + A22 = -L21@A12 + A22 + + # Solve the A22 block + P2, L22, U22 = getrf(A22) + + # Take P = P1 @ P2_ for + # + # P2_ = [[1, 0, + # 0, P2]] + P2_ = np.eye(m) + P2_[n1:,n1:] = P2 + P = P1 @ P2_ + + # Apply interchanges to L21 + L21 = la.inv(P2) @ L21 + + # Take + # + # L = [[L11, 0], + # [L21, L22]], + # + # U = [[U11, A12], + # [0, U22]] + if m >= n: + L, U = np.zeros((m, n)), np.zeros((n, n)) + else: + L, U = np.zeros((m, m)), np.zeros((m, n)) + L[:n1,:n1] = L11 + L[n1:,:n1] = L21 + L[n1:,n1:] = L22 + U[:n1,:n1] = U11 + U[:n1,n1:] = A12 + U[n1:,n1:] = U22 + + return P, L, U
diff --git a/linear-solvers.ipynb b/linear-solvers.ipynb @@ -0,0 +1,196 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "id": "e1791697-e733-4721-9e10-fcdcfbda9064", + "metadata": {}, + "source": [ + "# Solving Linear systems in SciPy\n", + "\n", + "A linear system is a system of equations of the form\n", + "$$ \\left\\{ \\begin{aligned} a_{1 1} x_1 + a_{1 2} x_2 + \\cdots + a_{1 n} x_n &= b_1 \\\\ a_{2 1} x_1 + a_{2 2} x_2 + \\cdots + a_{2 n} x_n &= b_2 \\\\ & \\vdots \\\\ a_{n 1} x_1 + a_{n 2} x_2 + \\cdots + a_{n n} x_n &= b_n\\end{aligned} \\right.$$\n", + "on the variables $x_1, \\ldots, x_n$.\n", + "\n", + "Solving this system is equivalent to solving the equation $A x = b$ on $x$ where $A = (a_{ij})_{ij}$ is a $n\\times n$ matrix and $x = (x_1, \\ldots, x_n) \\; \\& \\; b = (b_1, \\ldots, b_n)$ are vectors, which can always be done provided $A$ is invertible." + ] + }, + { + "cell_type": "markdown", + "id": "26bc6a30-7853-4c78-adfb-320f0a65dd10", + "metadata": {}, + "source": [ + "In SciPy, we can solve linear systems using the `la.solve` function." + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "id": "b1ced47f-6783-48ed-a4e4-4f1f4e2b8835", + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "array([[-0.12672176],\n", + " [ 0.1046832 ],\n", + " [ 1.19008264]])" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + "import numpy as np\n", + "import scipy.linalg as la\n", + "\n", + "A, b = np.array([[6,15,1],[8,7,12],[2,7,8]]), np.array([[2], [14], [10]])\n", + "la.solve(A, b)" + ] + }, + { + "cell_type": "markdown", + "id": "bfdacdbf-150f-4b4d-8072-ee18758d3b60", + "metadata": {}, + "source": [ + "## But how does `la.solve` work???\n", + "\n", + "Internally, the `la.solve` function uses the the [LAPACK library](https://netlib.org/lapack), a Fortran package for numerical linear algebra. The LAPACK generic linear solver algorithm goes something like the following:\n", + "\n", + "1. Decompose $A$ as $A = PL U$, where $P$ is permutation matrix, $L$ is a lower triangular matrix with $1$ in the diagonal and $U$ is an upper triangular matrix.\n", + "3. Solve $P b' = b$ for $b'$, i.e. compute $b' = P^{-1} b$. Since $P$ is a permutation matrix, this operation is $O(n)$.\n", + "4. Solve $L b'' = b'$ for $b''$, i.e. compute $b'' = L^{-1} P^{-1} b$. Since $L$ is known to be lower triangular, this operation is $O(n^2)$.\n", + "3. Solve $U x = b''$ for $x$, i.e. compute $x = U^{-1} L^{-1} P^{-1} b = A^{-1} b$. Since $U$ is known to be upper triangular, this operation is $O(n^2)$.\n", + "\n", + "This is implemented in the `GETRS` family of subroutines." + ] + }, + { + "cell_type": "markdown", + "id": "5b7ad45f-ea81-4d46-9097-735cc159cf1a", + "metadata": {}, + "source": [ + "As for the decomposition of $A$ in the first step, LAPACK uses a method called [_partial pivoting_](https://en.wikipedia.org/wiki/LU_decomposition#LU_factorization_with_partial_pivoting). A simple simple recurssive algorithm using such method might look something like the following:\n", + "\n", + "1. If $A = a_{11}$ is $1 \\times 1$ then take $P = L = 1$ and $U = a_{11}$.\n", + "2. If $A$ is $n \\times n$ for $n > 1$, choose $i_0$ that maximizes $|a_{i_0, 1}|$ and consider the $n \\times n$ permutation matrix $S_{i_0}$ that swaps the first and $i_0$-th basis vectors.\n", + "3. Write\n", + " $$S_{i_0} A = \\left( \\begin{array}{c|c} a_{i_0} & A_{12}' \\\\ \\hline A_{21}' & A_{22}' \\end{array} \\right), $$\n", + " where $A_{22}'$ is $(n - 1) \\times (n - 1)$. Since $A$ is invertible, $a_{i_0} \\ne 0$.\n", + "4. We want to solve the equation\n", + " $$S_{i_0} A = \\left( \\begin{array}{c|c} 1 & 0 \\\\ \\hline 0 & P_{22} \\end{array} \\right) \\left( \\begin{array}{c|c} 1 & 0 \\\\ \\hline L_{21} & L_{22} \\end{array} \\right) \\cdot \\left( \\begin{array}{c|c} u_{11} & U_{12} \\\\ \\hline 0 & U_{22} \\end{array} \\right),$$\n", + " where $P_{22}$ is a permutation matrix, $L_{22}$ is lower triangular with $1$ in the diagonal entries and $U_{22}$ is upper triangular. In other words, we want to solve the equations\n", + " $$\n", + " \\begin{aligned}\n", + " a_{i_0} &= u_{11} & A_{12}' &= U_{12} \\\\\n", + " A_{21}' &= u_{11} P_{22} L_{21} & A_{22}' &= P_{22} L_{21} U_{12} + P_{22} L_{22} U_{22}.\n", + " \\end{aligned}\n", + " $$\n", + " We must take $u_{11} = a_{i_0}$, $U_{12} = A_{12}'$ and $L_{21} = a_{i_0}^{-1} P_{22}^{-1} A_{12}$, so it remains to solve the bottom-right equation.\n", + "5. Write $(A_{22}' - a_{i_0}^{-1} A_{21}' A_{12}') = P_{22} L_{22} U_{22}$, where $P_{22}$ is a permutation matrix, $L_{22}$ is lower triangular with $1$ in the diagonals and $U_{22}$ is upper triangular.\n", + "6. Take\n", + " $$\n", + " \\begin{aligned}\n", + " L &= \\begin{pmatrix} 1 & 0 \\\\ L_{21} & L_{22} \\end{pmatrix} &\n", + " U &= \\begin{pmatrix} u_{11} & U_{12} \\\\ 0 & U_{22} \\end{pmatrix}\n", + " \\end{aligned}\n", + " $$\n", + " for $L_{21}, L_{22}, u_{11}, U_{12}, U_{22}$ as above, so that\n", + " $$\n", + " S_{i_0} A = \\begin{pmatrix} 1 & 0 \\\\ 0 & P_{22} \\end{pmatrix} L U.\n", + " $$\n", + "7. Hence by taking\n", + " $$\n", + " P = S_{i_0} \\begin{pmatrix} 1 & 0 \\\\ 0 & P_{22} \\end{pmatrix}\n", + " $$\n", + " we get $A = P L U$ as desired!" + ] + }, + { + "cell_type": "markdown", + "id": "0e1e9c6b-5e65-4cf1-aa53-a7202abff788", + "metadata": { + "jp-MarkdownHeadingCollapsed": true + }, + "source": [ + "## Scrap" + ] + }, + { + "cell_type": "markdown", + "id": "4dde0b0b-08cc-4bca-9dd5-c2bced871cef", + "metadata": {}, + "source": [ + "As for the decomposition of $A$ in the first step, given some $m \\times n$ matrix $A$, LAPACK uses a method called [_partial pivoting_](https://en.wikipedia.org/wiki/LU_decomposition#LU_factorization_with_partial_pivoting) to write $A = P L U$, where $P$ is a $m \\times m$ permutation matrix, $L$ is a $m \\times n$ lower trapezoidal matrix with $1$ in the diagona entries and $U$ is a $n \\times n$ upper triangular matrix. Their algorithm goes something like the following:\n", + "\n", + "1. If $A = 0$ then take $P = 1$, $L$ the $m \\times n$ matrix with $1$ in the diagonals and zero elsewhere and $U = 0$.\n", + "2. If $m = 1$, so that $A = \\begin{pmatrix} a_1 & \\cdots & a_n \\end{pmatrix}$, *DIE*\n", + "3. If $n = 1$, so that\n", + " $$ A = \\begin{pmatrix} a_1 \\\\ \\vdots \\\\ a_m \\end{pmatrix} $$\n", + " choose $i_0$ that maximizes $|a_{i_0}|$. Since $A \\ne 0$, $a_{i_0} \\ne 0$. Return $P = \\sigma_{1, i_0}$, $L = a_{i_0}^{-1} P^{-1} A$ and $U = 1$.\n", + "4. Otherwise do the following series of steps.\n", + " 1. Take $n_1 = \\left\\lfloor \\frac{\\min \\{n, m\\}}{2} \\right \\rfloor$ and $n_2 = n - n1$.\n", + " 2. Write\n", + " $$A = \\left( \\begin{array}{c|c} A_{11} & A_{12} \\\\ \\hline A_{21} & A_{22} \\end{array} \\right), $$\n", + " where $A_{11}$ is $n_1 \\times n_2$ and $A_{22}$ is $n_2 \\times n_2$.\n", + " 3. Write\n", + " $$\\begin{pmatrix} A_{11} \\\\ A_{21} \\end{pmatrix} = P_1 L_1 U_1 \\quad L_1 = \\begin{pmatrix} L_1' \\\\ L_1'' \\end{pmatrix} ,$$\n", + " where $P_1$ is a $m \\times m$ permutation matrix, $L_1$ is a $m \\times n_1$ lower trapezoidal matrix, $L_1'$ is $n_1 \\times n_1$ and $U_1$ is a $n_1 \\times n_1$ upper triangular matrix.\n", + " 4. Write\n", + " $$\\begin{pmatrix} A_{12}' \\\\ A_{22}' \\end{pmatrix} = P_1^{-1} \\begin{pmatrix} A_{12} \\\\ A_{22} \\end{pmatrix}$$\n", + " 5. Take\n", + " $$A_{12}'' = L_1'^{-1} A_{12}' \\quad A_{22}'' = -L_1'1A_{12}'' + A_{22}'$$\n", + " 6. Write $A_{12}'' = P_2 L_2 U_2$, where $P_2$ is a $n_1 \\times n_1$ permutation matrix, $L_2$ is a $n_1 \\times n_2$ lower trapezoidal matrix and $U_2$ is a $n_2 \\times n_2$ upper triangular matrix.\n", + " 7. Take\n", + " $$\n", + " \\begin{aligned}\n", + " P &= \\begin{pmatrix} P_1 & 0 \\\\ 0 & P_2 \\end{pmatrix} &\n", + " L &= \\begin{pmatrix} L_1 & 0 \\\\ A_{21}'' & L_2 \\end{pmatrix} &\n", + " U &= \\begin{pmatrix} U_1 & A_{12}'' \\\\ 0 & U_2 \\end{pmatrix},\n", + " \\end{aligned}\n", + " $$\n", + " where $A_{21}'' = P_2^{-1} L_1'1$." + ] + }, + { + "cell_type": "markdown", + "id": "93106465-de78-418d-8c7b-9774b97f6b07", + "metadata": {}, + "source": [ + "Althought similar to the naive method of computing $A^{-1}$ and then applying it to $b$, this algorithm has better performance characteristics since it only computes $A^{-1} b$ (without calculating the product $A^{-1} = U^{-1} L^{-1} P^{-1}$). The `la.solve` function also uses optimized linear solvers when $A$ is known to be symmetric or Hermitian.\n", + "\n", + "For more details please refer to the implementations of the `GETRS`, `SYSV` and `HESV` families of subroutines in LAPACK." + ] + }, + { + "cell_type": "code", + "execution_count": null, + "id": "6d77288f-56a8-4037-965d-56f2c443a73c", + "metadata": {}, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.12.6" + } + }, + "nbformat": 4, + "nbformat_minor": 5 +}
diff --git a/test_getrf.py b/test_getrf.py @@ -0,0 +1,78 @@ +import numpy as np +import scipy.linalg as la +from getrf import getrf +import json + +N_TESTS = 1000 +EPS = 1e-12 + +def matrix_eq(A: np.array, B: np.array) -> bool: + return A.shape == B.shape and np.all(np.abs(A - B) < EPS) + +def test(M: int, N: int, n_tests = N_TESTS): + print(f"Comparing getrf and la.lu (running {n_tests} tests)") + print("=" * 20) + + fails = [] + + for i in range(n_tests): + m = np.random.randint(1, M+1) + n = np.random.randint(1, N+1) + + A = np.random.rand(m, n) + P, L, U = la.lu(A) + + print(f"{i:05d}: Testing a {m:03d}x{n:03d} matrix: ", end="") + + try: + my_P, my_L, my_U = getrf(A) + except Exception as e: + d = {"A": A.tolist(), "runtime error": str(e)} + fails.append(d) + + print(f"\033[91mfailed!\033[0m") + print(" > getrf raised an error!") + continue + + P_test = matrix_eq(P, my_P) + L_test = matrix_eq(L, my_L) + U_test = matrix_eq(U, my_U) + + if (not P_test) or (not L_test) or (not U_test): + d = {"A": A.tolist(),} + + if not P_test: + d["expected P"] = P.tolist() + d["actual P"] = my_P.tolist() + + if not L_test: + d["expected L"] = L.tolist() + d["actual L"] = my_L.tolist() + + if not U_test: + d["expected U"] = U.tolist() + d["actual U"] = my_U.tolist() + + fails.append(d) + + print(f"\033[91mfailed!\033[0m") + print(" > getrf returned wrong values") + continue + + print(f"\033[32mpassed!\033[0m") + + print("\n" + "=" * 20) + if len(fails) == 0: + print(f"All {n_tests} tests passed!") + else: + testlogs = "./testlogs.json" + with open(testlogs, "w") as f: + json.dump(fails, f) + + print(f"{n_tests-len(fails)} tests passed ({len(fails)} fails)") + print(f"Check {testlogs} for the failing inputs") + +m, n = 200, 200 + +test(m, n, 10000) +