numeric-linalg

Educational material on the SciPy implementation of numerical linear algebra algorithms

git clone: git://git.pablopie.xyz/numeric-linalg
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Commit: 9c7392a31b59f45a69505c0415b9b7e6387205d5
Author: Pablo <pablo-pie@riseup.net>
Date: Mon, 25 Nov 2024 23:09:32 +0000

Initial commit

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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
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+testlogs.json
+lapack
+
+.ipynb_checkpoints
+__pycache__

diff --git a/LICENSE b/LICENSE
@@ -0,0 +1,675 @@
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+author or copyright holder as a result of your choosing to follow a
+later version.
+
+  15. Disclaimer of Warranty.
+
+  THERE IS NO WARRANTY FOR THE PROGRAM, TO THE EXTENT PERMITTED BY
+APPLICABLE LAW.  EXCEPT WHEN OTHERWISE STATED IN WRITING THE COPYRIGHT
+HOLDERS AND/OR OTHER PARTIES PROVIDE THE PROGRAM "AS IS" WITHOUT WARRANTY
+OF ANY KIND, EITHER EXPRESSED OR IMPLIED, INCLUDING, BUT NOT LIMITED TO,
+THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
+PURPOSE.  THE ENTIRE RISK AS TO THE QUALITY AND PERFORMANCE OF THE PROGRAM
+IS WITH YOU.  SHOULD THE PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF
+ALL NECESSARY SERVICING, REPAIR OR CORRECTION.
+
+  16. Limitation of Liability.
+
+  IN NO EVENT UNLESS REQUIRED BY APPLICABLE LAW OR AGREED TO IN WRITING
+WILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MODIFIES AND/OR CONVEYS
+THE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES, INCLUDING ANY
+GENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING OUT OF THE
+USE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED TO LOSS OF
+DATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY YOU OR THIRD
+PARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER PROGRAMS),
+EVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE POSSIBILITY OF
+SUCH DAMAGES.
+
+  17. Interpretation of Sections 15 and 16.
+
+  If the disclaimer of warranty and limitation of liability provided
+above cannot be given local legal effect according to their terms,
+reviewing courts shall apply local law that most closely approximates
+an absolute waiver of all civil liability in connection with the
+Program, unless a warranty or assumption of liability accompanies a
+copy of the Program in return for a fee.
+
+                     END OF TERMS AND CONDITIONS
+
+            How to Apply These Terms to Your New Programs
+
+  If you develop a new program, and you want it to be of the greatest
+possible use to the public, the best way to achieve this is to make it
+free software which everyone can redistribute and change under these terms.
+
+  To do so, attach the following notices to the program.  It is safest
+to attach them to the start of each source file to most effectively
+state the exclusion of warranty; and each file should have at least
+the "copyright" line and a pointer to where the full notice is found.
+
+    <one line to give the program's name and a brief idea of what it does.>
+    Copyright (C) <year>  <name of author>
+
+    This program is free software: you can redistribute it and/or modify
+    it under the terms of the GNU General Public License as published by
+    the Free Software Foundation, either version 3 of the License, or
+    (at your option) any later version.
+
+    This program is distributed in the hope that it will be useful,
+    but WITHOUT ANY WARRANTY; without even the implied warranty of
+    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+    GNU General Public License for more details.
+
+    You should have received a copy of the GNU General Public License
+    along with this program.  If not, see <https://www.gnu.org/licenses/>.
+
+Also add information on how to contact you by electronic and paper mail.
+
+  If the program does terminal interaction, make it output a short
+notice like this when it starts in an interactive mode:
+
+    <program>  Copyright (C) <year>  <name of author>
+    This program comes with ABSOLUTELY NO WARRANTY; for details type `show w'.
+    This is free software, and you are welcome to redistribute it
+    under certain conditions; type `show c' for details.
+
+The hypothetical commands `show w' and `show c' should show the appropriate
+parts of the General Public License.  Of course, your program's commands
+might be different; for a GUI interface, you would use an "about box".
+
+  You should also get your employer (if you work as a programmer) or school,
+if any, to sign a "copyright disclaimer" for the program, if necessary.
+For more information on this, and how to apply and follow the GNU GPL, see
+<https://www.gnu.org/licenses/>.
+
+  The GNU General Public License does not permit incorporating your program
+into proprietary programs.  If your program is a subroutine library, you
+may consider it more useful to permit linking proprietary applications with
+the library.  If this is what you want to do, use the GNU Lesser General
+Public License instead of this License.  But first, please read
+<https://www.gnu.org/licenses/why-not-lgpl.html>.
+

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)
+