116 lines
3.2 KiB
Plaintext
116 lines
3.2 KiB
Plaintext
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{
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"cells": [
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"cell_type": "code",
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"execution_count": null,
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"id": "acb06966-8cfa-4b28-9fa9-30286c33d20c",
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"metadata": {},
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"outputs": [],
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"source": []
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},
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{
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"cell_type": "markdown",
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"id": "203b80b6-7559-4861-ab9c-538885b8d223",
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"metadata": {
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"tags": []
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},
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"source": [
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"## Inverse?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 16,
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"id": "66031921-f6b4-4c32-bc8b-97a92ea935df",
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"metadata": {
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"jupyter": {
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"source_hidden": true
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},
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"tags": []
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},
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"outputs": [],
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"source": [
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"# Q: If you compute the inverse of one of these matrixes and multiply it to the right of the position\n",
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"# where it was originally multiplied it should \"delete\" that element as if the element was never added.\n",
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"# I'm uncertain of the possibility and frequency of such inverse matrices over finite fields.\n",
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"# If this scenario is possible to construct, I think we should \"define\" this operation as deletion.\n",
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"\n",
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"# See: https://en.wikipedia.org/wiki/Determinant#Properties_of_the_determinant\n",
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"\n",
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"def det2(mat, m=256):\n",
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" ((a, b),\n",
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" (c, c)) = mat\n",
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" return (a*c - b*c) % m\n",
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"\n",
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"def det3(mat, m=256):\n",
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" ((a, b, c),\n",
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" (d, e, f),\n",
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" (g, h, i)) = mat\n",
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" return (a*e*i + b*f*g + c*d*h - c*e*g - b*d*i - a*f*h) % m\n",
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"\n",
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"def det(mat, m=256):\n",
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" # I dislike having to round here, but it returns a float\n",
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" # I wish it returned a native \"B\" type\n",
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" return int(round(np.linalg.det(mat),0)) % m\n",
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"\n",
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"a = ((-2,2,-3),(-1,1,3),(2,0,-1))\n",
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"assert_equal(det(a), 18)\n",
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"assert_equal(det(a), det3(a))\n",
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"\n",
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"# ?\n",
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"# assert that they didn't change since the last time I ran it\n",
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"assert_equal(det(f1), 128)\n",
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"assert_equal(det(f2), 160)\n",
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"assert_equal(det(f3), 128)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "568ad4ed-7894-465b-9bf3-f19f2269c0f3",
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"metadata": {
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"tags": []
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},
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"source": [
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"[This](https://math.stackexchange.com/questions/273054/how-to-find-matrix-inverse-over-finite-field) suggests to find the inverse matrix by first finding the adjugate then applying\n",
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"\n",
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"> (1/det(A))adj(A)=inv(A)\n",
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"\n",
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"Under mod p\n",
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"\n",
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"---\n",
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"\n",
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"The [adjugate](https://en.wikipedia.org/wiki/Adjugate_matrix) is the inverse of the cofactor matrix.\n",
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"\n",
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"---\n",
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"\n",
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"The cofactor matrix is made by multiplying the matrix minor of each (i,j) of the original matrix times a sign factor.\n",
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"\n",
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"---\n",
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"\n",
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"The (i,j)-minor of matrix A is the determinant of the matrix A with row i & column j removed."
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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"pygments_lexer": "ipython3",
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"version": "3.9.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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