\documentclass[12pt]{article} \usepackage{tikz} \usetikzlibrary{automata,positioning,arrows} \newcommand{\lf}{\left\lfloor} \newcommand{\rf}{\right\rfloor} \newcommand{\lc}{\left\lceil} \newcommand{\rc}{\right\rceil} \newcommand{\Ceil}[1]{\left\lceil {#1}\right\rceil} \newcommand{\ceil}[1]{\left\lceil {#1}\right\rceil} \newcommand{\floor}[1]{\left\lfloor{#1}\right\rfloor} \newcommand{\N}{{\sf N}} \newcommand{\goes}{\rightarrow} \newcommand{\st}{\mathrel{:}} \newcommand{\nth}{n^{th}} \newcommand{\es}{\emptyset} \begin{document} \newif{\ifshowsoln} \showsolntrue % comment out to NOT show solution inside \ifshowsoln and \fi blocks. \centerline{\bf Homework 7 DUE April 21 at 11:00 AM (Dead Cat April 23 at 11:00 AM)} \begin{enumerate} \item (25 points) Find a function $T(n)$ such that $$NP \subseteq DTIME(T(n)).$$ \item (25 points) Construct a decidable set $A$ that is NOT in NP. You have to construct $A$. You cannot just invoke some theorem to show $A$ exists. (Hint: Use Problem 1.) \item (25 points) Let $A,B,C$ be sets. Show that if $A\le B$ and $B\le C$ then $A\le C$. (Recall that $X\le Y$ means that there is a polynomial time function $f$ such that $$x\in X \hbox{ iff }f(x)\in Y.$$ ) \item (25 points) Let $$COL_c = \{ G \st \hbox{$G$ is $c$-colorable} \}.$$ \begin{enumerate} \item (10 points) Show $COL_3 \le COL_4$. (That is, there exists a function $f$ that takes a graph $G$ and produces a graph $G'$ such that $G\in COL_3$ iff $G'\in COL_4$.) \item (15 points) Show that $COL_3 \le COL_5$. \item (0 points) Think about: is $COL_4 \le COL_3$? \end{enumerate} \end{enumerate} \end{document}