## 187 – Selected solutions to problems from Chapter 1

September 19, 2011

Professor Warren Esty, one of the authors of our main textbook, has made available a list of solutions to some of the problems from Chapter 1. They are most of the odd numbered problems. Please let him (or me) know if you find errors or typos, or if you have suggestions for alternative solutions or different approaches.

Solutions (Chapter 1)

## 187 – The resolution method

September 19, 2011

This note is based on lecture notes for the Caltech course Math 6c, prepared with A. Kechris and M. Shulman.

We would like to have a mechanical procedure (algorithm) for checking whether a given set of formulas logically implies another, that is, given ${A_1, \dots, A_n, A}$, whether

$\displaystyle (A_1\land \dots \land A_n)\Rightarrow A$

is a tautology (i.e., it is true under all truth-value assignments.)

This happens iff

$\displaystyle A_1\wedge\dots\wedge A_n\wedge\neg A\text{ is unsatisfiable.}$

So it suffices to have an algorithm to check the (un)satisfiability of a single propositional formula. The method of truth tables gives one such algorithm. We will now develop another method which is often (with various improvements) more efficient in practice.

It will be also an example of a formal calculus. By that we mean a set of rules for generating a sequence of strings in a language. Formal calculi usually start with a certain string or strings as given, and then allow the application of one or more “rules of production” to generate other strings.

A formula ${A}$ is in conjunctive normal form iff it has the form

$\displaystyle A_1\land A_2\land\dots\land A_n$

where each ${A_i}$ has the form

$\displaystyle B_1\lor B_2\lor\dots\lor B_k$

and each ${B_i}$ is either a propositional variable, or its negation. So ${A}$ is in conjunctive normal form iff it is a conjunction of disjunctions of variables and negated variables. The common terminology is to call a propositional variable or its negation a literal.

Suppose ${A}$ is a propositional statement which we want to test for satisfiability. First we note (without proof) that although there is no known efficient algorithm for finding ${A'}$ in cnf (conjunctive normal form) equivalent to ${A}$, it is not hard to show that there is an efficient algorithm for finding ${A^*}$ in cnf such that:

${A}$ is satisfiable iff ${A^*}$ is satisfiable.

(But, in general, ${A^*}$ has more variables than ${A}$.)

So from now on we will only consider ${A}$s in cnf, and the Resolution Method applies to such formulas only. Say

${A=(\ell_{1,1}\vee\dots\vee\ell_{1,n_1})\wedge\dots \wedge (\ell_{k,1}\vee\dots\vee\ell_{k,n_k})}$

with ${\ell_{i,j}}$ literals. Since order and repetition in each conjunct (1):

$\displaystyle \ell_{i,1}\vee\dots\vee\ell_{i,n_i}$

are irrelevant, we can replace (1) by the set of literals

$\displaystyle c_i=\{\ell_{i,1},\ell_{i,2},\dots ,\ell_{i,n_i}\}.$

Such a set of literals is called a clause. It corresponds to the formula (1). So the formula ${A}$ above can be simply written as a set of clauses (again since the order of the conjunctions is irrelevant): ${C=\{c_1,\dots ,c_k\}}$

$=\{\{\ell_{i,1},\dots \ell_{i,n_1}\},\dots ,\{\ell_{k,1},\dots , \ell_{k,n_k}\}\}$

Satisfiability of ${A}$ means then simultaneous satisfiability of all of its clauses ${c_1,\dots ,c_k}$, i.e., finding a valuation ${\nu}$ which makes ${c_i}$ true for each ${i}$, i.e., which for each ${i}$ makes some ${\ell_{i,j}}$ true.

Example 1 $A=(p_1\vee\neg p_2)\wedge (p_3\vee p_3)$

$c_1=\{p_1,\neg p_2\}$

$c_2=\{p_3\}$

$C=\{\{p_1,\neg p_2\},\{p_3\}\}.$

From now on we will deal only with a set of clauses ${C=\{c_1, c_2,\dots,c_n \}}$. It is possible to consider also infinite sets ${C}$, but we will not do that here.

Satisfying ${C}$ means (again) that there is a valuation which satisfies all ${c_1,c_2,\dots}$, i.e. if ${c_i=\ell_{i,1}\vee\dots \vee\ell_{i,n_i}}$, then for all ${i}$ there is ${j}$ so that it makes ${\ell_{i,j}}$ true.

Notice that if the set of clauses ${C_A}$ is associated as above to ${A}$ (in cnf) and ${C_B}$ to ${B}$, then

$\displaystyle A\wedge B\text{ is satisfiable iff }C_A\cup C_B\text{ is satisfiable.}$

By convention we also have the empty clause ${\Box}$, which contains no literals. The empty clause is (by definition) unsatisfiable, since for a clause to be satisfied by a valuation, there has to be some literal in the clause which it makes true, but this is impossible for the empty clause, which has no literals. For a literal ${u}$, let ${\bar u}$ denote its “conjugate”, i.e. $\bar u=\neg p,$ if $u=p,$ and $\bar u=p$ if $u=\neg p.$

Definition 1 Suppose now ${c_1,c_2,c}$ are three clauses. We say that ${c}$ is a resolvent of ${c_1,c_2}$ if there is a ${u}$ such that ${u\in c_1}$, ${\bar u\in c_2}$ and

$\displaystyle c=(c_1\setminus\{u\})\cup (c_2\setminus\{\bar u\}).$

We allow here the case ${c=\Box}$, i.e. ${c_1=\{u\},\ c_2= \{\bar u\}}$.

## 187 – The P vs NP problem

September 19, 2011

This note is based on lecture notes for the Caltech course Math 6c, prepared with A. Kechris and M. Shulman.

1. Decision problems

Consider a finite alphabet ${A}$, and “words” on that alphabet (the “alphabet” may consist of digits, of abstract symbols, of actual letters, etc).

We use the notation ${A^*}$ to indicate the set of all “words” from the alphabet ${A}$. Here, a word is simply a finite sequence of symbols from ${A}$. For example, if ${A}$ is the usual alphabet, then

$\displaystyle awwweeeedddfDDkH$

would be a word.

We are also given a set ${V}$ of words, and we say that the words in ${V}$ are valid. (${V}$ may be infinite.)

In the decision problem associated to ${V}$, we are given as input a word in this alphabet. As output we say yes or no, depending on whether the word is in ${V}$ or not (i.e., whether it is “valid”).

We are interested in whether there is an algorithm that allows us to decide the right answer.

## 414/514 – Metric spaces

September 19, 2011

This is homework 2, due Monday September 26 at the beginning of lecture.

Let $(X,d)$ be a metric space.

• Show that if $d_1:X\times X\to{\mathbb R}$ is defined by $\displaystyle d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$, then $d_1$ is also a metric on $X$.
• Show that if $U$ is open in $(X,d)$ then it is open in $(X,d_1)$, and viceversa.

Recall that $U$ is open iff it is a union of open balls. Use this to explain why it suffices to show that if $U$ is open in $(X,d)$ then for any $x\in U$ there is an $\epsilon>0$ such that

$B_\epsilon^{d_1}(x):=\{y\mid d_1(x,y)<\epsilon\}\subseteq U$,

and similarly, if $V$ is open in $(X,d_1)$ then for any $z\in V$ there is a $\delta>0$ such that

$B_\delta^d(z):=\{w\mid d(z,w)<\delta\}\subseteq V$.

In turn, explain why to show this it suffices to prove that for any $x\in X$ and any $\eta>0$ there is a $\rho>0$ such that

$B^{d_1}_\eta(x)\supseteq B^d_\rho(x)$

and, similarly, for any $\tau>0$ there is a $\mu>0$ such that

$B_\tau^d(x)\supseteq B^{d_1}_\mu(x)$.

Finally, prove this by showing that we can take $\rho=\epsilon$ (no matter what $x$ is) and similarly, find an appropriate $\mu$ that works for $\tau$ (again, independently of $x$).

• Illustrate the above in ${\mathbb R}^2$ as accurately as possible.
• Suppose that a sequence $(x_n)_{n\in{\mathbb N}}$ converges to $x$ in $(X,d)$ and to $x'$ in $(X,d_1)$. Show that $x=x'$.
• Is it true that a sequence $(x_n)_{n\in{\mathbb N}}$ is Cauchy in $(X,d)$ iff it is Cauchy in $(X,d_1)$? (Give a proof or else exhibit a counterexample, with a proof that it is indeed a counterexample.)
• $(*)$ Show that any dense $G_\delta$ subset of ${\mathbb R}$ has the same size as ${\mathbb R}$.