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| a /= a' = error "Incomparable"
| otherwise = b <= b'
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/* This file is part of ASCIIVN.
*
* Copyright (C) 2018 Adrian Parvin D. Ouano
*
* 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, version 3 of the License.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
/* This file is part of ASCIIVN.
*
* Copyright (C) 2018 Adrian Parvin D. Ouano
*
* 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, version 3 of the License.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
struct imagebuffer *
new_imagebuffer(size_t width, size_t height)
{
size_t size = width * height * color_type_to_bytes(DEFAULT_COLOR_TYPE);
struct imagebuffer *imagebuffer = malloc(sizeof(struct imagebuffer) + size);
*imagebuffer = (struct imagebuffer)
{
.width = width,
.height = height,

Let G be a group and S a subset of G. We shall say that S generates G, or that S is a set of generators for G, if every element of G can be expressed as a product of elements of S or inverses of elements of S, i.e. as a product x_1 \cdots x_n, where each x_i or x_i-1 is in S. It is clear that the set of all such products is a subgroup of G, and is the smallest subgroup of G containing S. Thus S generates G if and only if the smallest subgroup of G containing S is G itself. If G is generated by S, then we write G = \langle S \rangle

\documentclass{article}
\usepackage[margin=0in]{geometry}
\usepackage{array}
\setlength\parindent{0pt}
\begin{document}
\begin{tabular}{@{\hspace{0pt}}p{0.5\textwidth}@{\hspace{0pt}}>{\raggedleft}p{0.5\textwidth}@{\hspace{0pt}}}
The quick brown fox & The quick brown fox.
\end{tabular}
\end{document}
\documentclass{article}
\usepackage[margin=0in]{geometry}
\usepackage{array}
\setlength\parindent{0pt}
\begin{document}
\begin{tabular}{@{\hspace{0pt}}m{0.5\textwidth}@{\hspace{0pt}}>{\raggedleft}m{0.5\textwidth}@{\hspace{0pt}}}
The quick & {The \\
quick \\
brown \\}
\end{tabular}

A homomorphism whose kernel is trivial is injective. Proof: Given \(f(x) = f(y)\), then \(f(xy-1) = f(x)f(y-1) = e’\) Therefore \(xy-1 = e\) and \(x = y\)

Notes for page 26

A kernel of f is the set \[\{ x ∈ G | f(x) = e’ \}\] An injective homomorphism \(f:G → G’\) is called an embedding.

A homomorphism whose kernel is trivial is injective.

Proof: