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- Introduce the notion of
*polynomial-time reductions*as a way to relate the complexity of problems to one another. - See several examples of such reductions.
- 3SAT as a basic starting point for reductions.

Let us consider several of the problems we have encountered before:

- Finding the longest path in a graph
- Finding the maximum cut in a graph
- The 3SAT problem: deciding whether a given 3CNF formula has a satisfying assignment.
- Solving quadratic equations over \(n\) variables \(x_0,\ldots,x_{n-1} \in \R\).

All of these have the following properties:

- These are important problems, and people have spent significant effort on trying to find better algorithms for them.
- Each one of these problems is a
*search*problem, whereby we search for a solution that is “good” in some easy to define sense (e.g., a long path, a satisfying assignment, etc..). - Each of these problems has trivial exponential time algorithms that involve enumerating all possible solutions.
- At the moment, for all these problems the best known algorithms are not much better than the trivial one in the worst case.

In this chapter and in cooklevinchap we will see that, despite
their apparent differences, we can relate these problems by their
complexity. In fact, it turns out that all these problems are
*computationally equivalent*, in the sense that solving one of them
immediately implies solving the others. This phenomenon, known as
*\(\mathbf{NP}\) completeness*, is one of the surprising discoveries of
theoretical computer science, and we will see that it has far-reaching
ramifications.

For reasons of technical conditions rather than anything substantial, we
will concern ourselves with *decision problems* (i.e., Yes/No questions)
or in other words *Boolean* (i.e., one-bit output) functions. Thus, we
will model all the problems as functions mapping \(\{0,1\}^*\) to
\(\{0,1\}\):

- The
*3SAT*problem can be phrased as the function \(3SAT:\{0,1\}^* \rightarrow \{0,1\}\) that maps a 3CNF formula \(\varphi\) to \(1\) if there exists some assignment \(x\) that satisfies it, and to \(0\) otherwise.We assume some representation of formulas as strings, and define the function to output \(0\) if its input is not a valid representation. We will use the same convention for all the other functions below. - The
*quadratic equations*problem corresponds to the function \(QUADEQ:\{0,1\}^* \rightarrow \{0,1\}\) that maps a set of quadratic equations \(E\) to \(1\) if there is an assignment \(x\) that satisfies all equations, and to \(0\) otherwise. - The
*longest path*problem corresponds to the function \(LONGPATH:\{0,1\}^* \rightarrow \{0,1\}\) that maps a graph \(G\) and a number \(k\) to \(1\) if there is a simpleRecall that a path in \(G\) of length at least \(k\), and maps \((G,k)\) to \(0\) otherwise. The longest path problem is a generalization of the well-known Hamiltonian Path Problem of determining whether a path of length \(n\) exists in a given \(n\) vertex graph.*simple*path in a graph is one that does not visit any vertex more than once. For the*shortest path problem*we can assume that a path is simple without loss of generality since removing a loop (a portion of the path that starts from the same vertex and returns to it) only makes the path shorter. For the*longest path problem*we need to make this restriction to avoid “degenerate” paths such as paths that repeat endlessly the same loop. - The
*maximum cut*problem corresponds to the function \(MAXCUT:\{0,1\}^* \rightarrow \{0,1\}\) that maps a graph \(G\) and a number \(k\) to \(1\) if there is a cut in \(G\) that cuts at least \(k\) edges, and maps \((G,k)\) to \(0\) otherwise.

Suppose that \(F,G:\{0,1\}^* \rightarrow \{0,1\}\) are two functions. How
can we show that they are “computationally equivalent”? The idea is that
we show that an efficient algorithm for \(F\) would imply an efficient
algorithm for \(G\) and vice versa. The key to this is the notion of a
*reduction*. Roughly speaking, we will say that *\(F\) reduces to \(G\)*
(denoted as \(F \leq_p G\)) if \(F\) is “no harder” than \(G\), in the sense
that a polynomial-time algorithm for \(G\) implies a polynomial-time
algorithm for \(F\). The formal definition is as follows:*mapping reduction*, *many to one reduction* or a *Karp reduction*.

Let \(F,G:\{0,1\}^* \rightarrow \{0,1\}^*\). We say that *\(F\) reduces to
\(G\)*, denoted by \(F \leq_p G\) if there is a polynomial-time computable
\(R:\{0,1\}^* \rightarrow \{0,1\}^*\) such that for every
\(x\in \{0,1\}^*\), \[
F(x) = G(R(x)) \;. \label{eq:reduction}
\] We say that \(F\) and \(G\) have *equivalent complexity* if \(F \leq_p G\)
and \(G \leq_p F\).

Prove that if \(F \leq_p G\) and \(G \in \mathbf{P}\) then \(F\in \mathbf{P}\).

As usual, solving this exercise on your own is an excellent way to make sure you understand reduction-def. This exercise justifies the informal description of \(F \leq_p G\) as saying that “\(F\) is no harder than \(G\).”

Suppose there was an algorithm \(B\) that compute \(F\) in time \(p(n)\) where \(p\) is its input size. Then, \eqref{eq:reduction} directly gives an algorithm \(A\) to compute \(F\) (see reductionsfig). Indeed, on input \(x\in \{0,1\}^*\), Algorithm \(A\) will run the polynomial-time reduction \(R\) to obtain \(y=R(x)\) and then return \(B(y)\). By \eqref{eq:reduction}, \(G(R(x)) = F(x)\) and hence Algorithm \(A\) will indeed compute \(F\).

We now show that \(A\) runs in polynomial time. By assumption, \(R\) can be computed in time \(q(n)\) for some polynomial \(q\). In particular, this means that \(|y| \leq q(|x|)\) (as just writing down \(y\) takes \(|y|\) steps). This, computing \(B(y)\) will take at most \(p(|y|) \leq p(q(|x|))\) steps. Thus the total running time of \(A\) on inputs of length \(n\) is at most the time to compute \(y\), which is bounded by \(q(n)\), and the time to compute \(B(y)\), which is bounded by \(p(q(n))\), and since the composition of two polynomials is a polynomial, \(A\) runs in polynomial time.

Since we think of \(F \leq_p G\) as saying that (as far as polynomial-time computation is concerned) \(F\) is “easier or equal in difficulty to” \(G\), we would expect that if \(F \leq_p G\) and \(G \leq_p H\), then it would hold that \(F \leq_p H\). Indeed this is the case:

For every \(F,G,H :\{0,1\}^* \rightarrow \{0,1\}\), if \(F \leq_p G\) and \(G \leq_p H\) then \(F \leq_p H\).

We leave the proof of transitivitylem as transitivity-reductions-ex. Pausing now and doing this exercise is an excellent way to verify that you understood the definition of reductions.

We have seen reductions before in the context of proving the
uncomputability of problems such as \(HALTONZERO\) and others. The most
crucial difference between the notion in reduction-def and
previously occuring notions is that in the context of relating the time
complexity of problems, we need the reduction to be computable in
*polynomial time*, as opposed to merely computable.
reduction-def also restricts reductions to have a very
specific format. That is, to show that \(F \leq_p G\), rather than
allowing a general algorithm for \(F\) that uses a “magic box” that
computes \(G\), we only allow an algorithm that computes \(F(x)\) by
outputting \(G(R(x))\). This restricted form is convenient for us, but
people have defined and used more general reductions as well.

Since both \(F\) and \(G\) are Boolean functions, the condition
\(F(x)=G(R(x))\) in \eqref{eq:reduction} is equivalent to the
following two implications: **(i)** if \(F(x)=1\) then \(G(R(x))=1\), and
**(ii)** if \(G(R(x))=1\) then \(F(x)=1\). Traditionally, condition **(i)**
is known as *completness* and condition **(ii)** is known as
*soundness*. We can think of this as saying that the reduction \(R\) is
*complete* if every \(1\)-input of \(F\) (i.e. \(x\) such that \(F(x)=1\)) is
mapped by \(R\) to a \(1\)-input of \(G\), and that it is *sound* if no
\(0\)-input of \(F\) will ever be mapped to a \(1\)-input of \(G\). As we will
see below, it is often the case that establishing **(ii)** is the more
challenging part.

We will now use reductions to relate the computational complexity of the
problems mentioned above \(-\) 3SAT, Quadratic Equations, Maximum Cut, and
Longest Path. We start by reducing 3SAT to the latter three problems,
demonstrating that solving any one of them will solve 3SAT. Along the
way we will introduce one more problem: the *independent set* problem.
Like the others, it shares the characteristics that it is an important
and well-motivated computational problem, and that the best known
algorithm for it takes exponential time. In cooklevinchap we
will show the other direction: reducing each one of these problems to
3SAT in one fell swoop.

Let us now see our first example of a reduction. Recall that in the
*quadratic equation* problem, the input is a list of \(n\)-variate
polynomials \(p_0,\ldots,p_{m-1}:\R^n \rightarrow \R\) that are all of
degree at most
two (i.e., they are *quadratic*) and with integer coefficients.

For example, the following is a set of quadratic equations over the variables \(x_0,x_1,x_2\): \[ \begin{aligned} x_0^2 - x_0 &= 0 \\ x_1^2 - x_1 &= 0 \\ x_2^2 - x_2 &= 0 \\ 1-x_0-x_1+x_0x_1 &= 0 \end{aligned} \] You can verify that \(x \in \R^3\) satisfies this set of equations if and only if \(x \in \{0,1\}^3\) and \(x_0 \vee x_1 = 1\).

We will show how to reduce 3SAT to the problem of Quadratic Equations.

\[3SAT \leq_p QUADEQ\] where \(3SAT\) is the function that maps a 3SAT formula \(\varphi\) to \(1\) if it is satisfiable and to \(0\) otherwise, and \(QUADEQ\) is the function that maps a set \(E\) of quadratic equations over \(\{0,1\}^n\) to \(1\) it has a solution and to \(0\) otherwise.

At the end of the day, a 3SAT formula can be thought of as a list of
equations on some variables \(x_0,\ldots,x_{n-1}\). Namely, the equations
are that each of the \(x_i\)’s should be equal to either \(0\) or \(1\), and
that the variables should satisfy some set of constraints which
corresponds to the OR of three variables or their negation. To show that
\(3SAT \leq_p QUADEQ\) we need to give a polynomial-time reduction that
maps a 3SAT formula \(\varphi\) into a set of quadratic equations \(E\) such
that \(E\) has a solution if and only if \(\varphi\) is satisfiable. The
idea is that we can transform a 3SAT formula \(\varphi\) first to a set of
*cubic* equations by mapping every constraint of the form
\((x_{12} \vee \overline{x}_{15} \vee x_{24})\) into an equation of the
form \((1-x_{12})x_{15}(1-x_{24})=0\). We can then turn this into a
*quadratic equation* by mapping any cubic equation of the form
\(x_ix_jx_k =0\) into the two quadratic equations \(y_{i,j}=x_ix_j\) and
\(y_{i,j}x_k=0\).

To prove quadeq-thm we need to give a polynomial-time transformation of every 3SAT formula \(\varphi\) into a set of quadratic equations \(E\), and prove that \(3SAT(\varphi)=QUADEQ(E)\).

We now describe the transformation of a formula \(\varphi\) to equations
\(E\) and show the completeness and soundness conditions. Recall that a
*3SAT formula* \(\varphi\) is a formula such as
\((x_{17} \vee \overline{x}_{101} \vee x_{57}) \wedge ( x_{18} \vee \overline{x}_{19} \vee \overline{x}_{101}) \wedge \cdots\).
That is, \(\varphi\) is composed of the AND of \(m\) *3SAT clauses* where a
3SAT clause is the OR of three variables or their negation. A *quadratic
equations* instance \(E\) is composed of a list of equations, each of
involving a sum of variables or their products, such as
\(x_{19}x_{52} - x_{12} + 2x_{33} = 2\), etc.. We will include the
constraints \(x_i^2-x_i=0\) for every \(i\in [n]\) in our equations, which
means that we can restrict attention to assignments where
\(x_i \in \{0,1\}\) for every \(i\).

There is a natural way to map a 3SAT instance into a set of *cubic*
equations \(E'\), and that is to map a clause such as
\((x_{17} \vee \overline{x}_{101} \vee x_{57})\) (which is equivalent to
the negation of
\(\overline{x}_{17} \wedge x_{101} \wedge \overline{x}_{57}\)) to the
equation \((1-x_{17})x_{101}(1-x_{57})=0\). Therefore, we can map a
formula \(\varphi\) with \(n\) variables \(m\) clauses into a set \(E'\) of
\(m+n\) cubic equations on \(n\) variables (that is, one equation per each
clause, plus one equation of the form \(x_i^2-x_i=0\) for each variable to
ensure that its value is in \(\{0,1\}\)) such that every assignment
\(a\in \{0,1\}^n\) to the \(n\) variables satisfies the original formula if
and only if it satisfies the equations of \(E'\).

To make the equations *quadratic* we introduce for every two distinct
\(i,j \in [n]\) a variable \(y_{i,j}\) and include the constraint
\(y_{i,j}-x_ix_j=0\) in the equations. This is a quadratic equation that
ensures that \(y_{i,j}=x_ix_j\) for every such \(i,j\in [n]\). Now we can
turn any cubic equation in the \(x\)’s into a quadratic equation in the
\(x\) and \(y\) variables. For example, we can “open up the parentheses” of
an equation such as \((1-x_{17})x_{101}(1-x_{57})=0\) to
\(x_{101} -x_{17}x_{101}-x_{101}x_{57}+x_{17}x_{101}x_{57}=0\). We can now
replace the cubic term \(x_{17}x_{101}x_{57}\) with the quadratic term
\(y_{17,101}x_{57}\). This can be done for every cubic equation in the
same way, replacing any cubic term \(x_ix_jx_k\) with the term
\(y_{i,j}x_k\). The end result will be a set of \(m+n+\binom{n}{2}\)
equations (one equation per clause, one equation per variable to ensure
\(x_i^2-x_i=0\), and one equation per pair \(i,j\) to ensure
\(y_{i,j}=x_ix_j=0\)) on the \(n + \binom{n}{2}\) variables
\(x_0,\ldots,x_{n-1}\) and \(y_{i,j}\) for all pairs of distinct variables
\(i,j\).

To complete the proof we need to show that if we transform \(\varphi\) to \(E\) in this way then the 3SAT formula \(\varphi\) is satisfiable if and only if the equations \(E\) have a solution. This is essentially immediate from the construction, but as this is our first reduction, we spell this out fully:

**Completeness:**We claim that if \(\varphi\) is satisfiable then the equations \(E\) have a solution. To prove this we need to show how to transform a satisfying assignment \(a\in \{0,1\}^n\) to the variables of \(\varphi\) (that is, \(a_i\) is the value assigned to \(x_i\)) to a solution to the variables of \(E\). Specifically, if \(a\in \{0,1\}^n\) is such an assignment then by design \(a\) satisfies all the*cubic*equations \(E'\) that we constructed above. But then, if we assign to the \(n+\binom{n}{2}\) variables the values \(a_0,\ldots,a_{n-1}\) and \(\{ a_ia_j \}\) for all \(\{i,j\} \subseteq [n]\) then by construction this will satisfy all the quadratic equations of \(E\) as well.**Soundness:**We claim that if the equations \(E\) have a solution then \(\varphi\) is satisfiable. Indeed, suppose that \(z \in \R^{n + \binom{n}{2}}\) is a solution to the equations \(E\). A priori \(z\) could be any vector of \(n+ \binom{n}{2}\) numbers, but the fact that \(E\) contains the equations \(x_i^2 - x_i =0\) and \(y_{i,j} - x_ix_j = 0\) means that if \(z\) satisfies these equations then the values it assigns to \(x_i\) must be in \(\{0,1\}\) for every \(i\), and the value it assigns to \(y_{i,j}\) must be \(x_ix_j\) for every \(\{i,j\} \subseteq [n]\). Therefore by the way we constructed our equations, the value assigned \(x\) must be a solution of the original cubic equations \(E'\) and hence also of the original formula \(\varphi\), which in particular implies \(\varphi\) is satisfiable.

This reduction can be easily implemented in about a dozen lines of Python or any other programming language, see sattoqefig.

For a graph \(G=(V,E)\), an independent
set (also
known as a *stable set*) is a subset \(S \subseteq V\) such that there are
no edges with both endpoints in \(S\) (in other words,
\(E(S,S)=\emptyset\)). Every “singleton” (set consisting of a single
vertex) is trivially an independent set, but finding larger independent
sets can be challenging. The *maximum independent set* problem
(henceforth simply “independent set”) is the task of finding the largest
independent set in the graph.*maximal* (as opposed to *maximum*) independent set, which is the
task of finding a “local maximum” of an independent set: an
independent set \(S\) such that one cannot add a vertex to it without
losing the independence property (such a set is known as a *vertex
cover*). Finding a maximal independent set can be done efficiently
by a greedy algorithm, but this local maximum can be much smaller
than the global maximum.*scheduling problems*: if we put an edge between
two conflicting tasks, then an independent set corresponds to a set of
tasks that can all be scheduled together without conflicts. But it also
arises in very different settings, including trying to find structure in
protein-protein interaction
graphs.

To phrase independent set as a decision problem, we think of it as a function \(ISET:\{0,1\}^* \rightarrow \{0,1\}\) that on input a graph \(G\) and a number \(k\) outputs \(1\) if and only if the graph \(G\) contains an independent set of size at least \(k\). We will now reduce 3SAT to Independent set.

\(3SAT \leq_p ISET\).

The idea is that finding a satisfying assignment to a 3SAT formula corresponds to satisfying many local constraints without creating any conflicts. One can think of “\(x_{17}=0\)” and “\(x_{17}=1\)” as two conflicting events, and of the constraints \(x_{17} \vee \overline{x}_5 \vee x_9\) as creating a conflict between the events “\(x_{17}=0\)”, “\(x_5=1\)” and “\(x_9=0\)”, saying that these three cannot simultaneosly co-occur. Using these ideas, we can we can think of solving a 3SAT problem as trying to schedule non conflicting events, though the devil is, as usual, in the details.

Given a 3SAT formula \(\varphi\) on \(n\) variables and with \(m\) clauses, we will create a graph \(G\) with \(3m\) vertices as follows: (see threesattoisfig for an example)

- A clause \(C\) in \(\varphi\) has the form \(C = y \vee y' \vee y''\)
where \(y,y',y''\) are
*literals*(variables or their negation). For each such clause \(C\), we will add three vertices to \(G\), and label them \((C,y)\), \((C,y')\), and \((C,y'')\) respectively. We will also add the three edges between all pairs of these vertices, so they form a*triangle*. Since there are \(m\) clauses in \(\varphi\), the graph \(G\) will have \(3m\) vertices. - In addition to the above edges, we also add an edge between every
pair vertices of the form \((C,y)\) and \((C',y')\) where \(y\) and \(y'\)
are
*conflicting*literals. That is, we add an edge between \((C,y)\) and \((C,y')\) if there is an \(i\) such that \(y=x_i\) and \(y' = \overline{x}_i\) or vice versa.

The above construction of \(G\) based on \(\varphi\) can clearly be carried out in polynomial time. Hence to prove the theorem we need to show that \(\varphi\) is satisfiable if and only if \(G\) contains an independent set of \(m\) vertices. We now show both directions of this equivalence:

**Part 1: Completeness.** The “completeness” direction is to show that
if \(\varphi\) has a satisfying assignment \(x^*\), then \(G\) has an
independent set \(S^*\) of \(m\) vertices. Let us now show this.

Indeed, suppose that \(\varphi\) has a satisfying assignment
\(x^* \in \{0,1\}^n\). Then for every clause \(C = y \vee y' \vee y''\) of
\(\varphi\), one of the literals \(y,y',y''\) must evaluate to *true* under
the assignment \(x^*\) (as otherwise it would not satisfy \(\varphi\)). We
let \(S\) be a set of \(m\) vertices that is obtained by choosing for every
clause \(C\) one vertex of the form \((C,y)\) such that \(y\) evaluates to
true under \(x^*\). (If there is more than one such vertex for the same
\(C\), we arbitrarily choose one of them.)

We claim that \(S\) is an independent set. Indeed, suppose otherwise that
there was a pair of vertices \((C,y)\) and \((C',y')\) in \(S\) that have an
edge between them. Since we picked one vertex out of each triangle
corresponding to a clause, it must be that \(C \neq C'\). Hence the only
way that there is an edge between \((C,y)\) and \((C,y')\) is if \(y\) and
\(y'\) are conflicting literals (i.e. \(y=x_i\) and \(y'=\overline{x}_i\) for
some \(i\)). But that would that they can’t both evaluate to *true* under
the assignment \(x^*\), which contradicts the way we constructed the set
\(S\). This completes the proof of the completeness condition.

**Part 2: Soundness.** The “soundness” direction is to show that if \(G\)
has an independent set \(S^*\) of \(m\) vertices, then \(\varphi\) has a
satisfying assignment \(x^* \in \{0,1\}^n\). Let us now show this.

Indeed, suppose that \(G\) has an independent set \(S*\) with \(m\) vertices. We will define an assignment \(x^* \in \{0,1\}^n\) for the variables of \(\varphi\) as follows. For every \(i\in [n]\), we set \(x^*_i\) according to the following rules:

- If \(S^*\) contains a vertex of the form \((C,x_i)\) then we set \(x^*_i=1\).
- If \(S^*\) contains a vertex of the form \((C,\overline{x_i})\) then we set \(x^*_i=0\).
- If \(S^*\) does not contain a vertex of either of these forms, then it does not matter which value we give to \(x^*_i\), but for concreteness we’ll set \(x^*_i=0\).

The first observation is that \(x^*\) is indeed well defined, in the sense
that the rules above do not conflict with one another, and ask to set
\(x^*_i\) to be both \(0\) and \(1\). This follows from the fact that \(S^*\) is
an *independent set* and hence if it contains a vertex of the form
\((C,x_i)\) then it cannot contain a vertex of the form
\((C',\overline{x_i})\).

We now claim that \(x^*\) is a satisfying assignment for \(\varphi\).
Indeed, since \(S^*\) is an independent set, it cannot have more than one
vertex inside each one of the \(m\) triangles \((C,y),(C,y'),(C,y'')\)
corresponding to a clause of \(\varphi\). Hence since \(|S^*|=m\), it must
have exactly one vertex in each such triangle. For every clause \(C\) of
\(\varphi\), if \((C,y)\) is the vertex in \(S^*\) in the triangle
corresponding to \(C\), then by the way we defined \(x^*\), the literal \(y\)
must evaluate to *true*, which means that \(x^*\) satisfies this clause.
Therefore \(x^*\) satisfies all clauses of \(\varphi\), which is the
definition of a satisfying assignment.

This completes the proof of isetnpc

\(ISET \leq_p MAXCUT\)

We will map a graph \(G\) into a graph \(H\) such that a large independent set in \(G\) becomes a partition cutting many edges in \(H\). We can think of a cut in \(H\) as coloring each vertex either “blue” or “red”. We will add a special “source” vertex \(s^*\), connect it to all other vertices, and assume without loss of generality that it is colored blue. Hence the more vertices we color red, the more edges from \(s^*\) we cut. Now, for every edge \(u,v\) in the original graph \(G\) we will add a special “gadget” which will be a small subgraph that involves \(u\),\(v\), the source \(s^*\), and two other additional vertices. We design the gadget in a way so that if the red vertices are not an independent set in \(G\) then the corresponding cut in \(H\) will be “penalized” in the sense that it would not cut as many edges. Once we set for ourselves this objective, it is not hard to find a gadget that achieves it\(-\) see the proof below.

We will transform a graph \(G\) of \(n\) vertices and \(m\) edges into a graph
\(H\) of \(n+1+2m\) vertices and \(n+5m\) edges in the following way: the
graph \(H\) will contain all vertices of \(G\) (though not the edges between
them!) and in addition to that will contain:

* A special vertex \(s^*\) that is connected to all the vertices of \(G\)

* For every edge \(e=\{u,v\} \in E(G)\), two vertices \(e_0,e_1\) such that
\(e_0\) is connected to \(u\) and \(e_1\) is connected to \(v\), and moreover we
add the edges \(\{e_0,e_1 \},\{ e_0,s^* \},\{e_1,s^*\}\) to \(H\).

isettomaxcut will follow by showing that \(G\) contains an independent set of size at least \(k\) if and only if \(H\) has a cut cutting at least \(k+4m\) edges. We now prove both directions of this equivalence:

**Part 1: Completeness.** If \(I\) is an independent \(k\)-sized set in \(G\),
then we can define \(S\) to be a cut in \(H\) of the following form: we let
\(S\) contain all the vertices of \(I\) and for every edge
\(e=\{u,v \} \in E(G)\), if \(u\in I\) and \(v\not\in I\) then we add \(e_1\) to
\(S\); if \(u\not\in I\) and \(v\in I\) then we add \(e_0\) to \(S\); and if
\(u\not\in I\) and \(v\not\in I\) then we add both \(e_0\) and \(e_1\) to \(S\).
(We don’t need to worry about the case that both \(u\) and \(v\) are in \(I\)
since it is an independent set.) We can verify that in all cases the
number of edges from \(S\) to its complement in the gadget corresponding
to \(e\) will be four (see ISETtoMAXCUTfig). Since \(s^*\) is not
in \(S\), we also have \(k\) edges from \(s^*\) to \(I\), for a total of \(k+4m\)
edges.

**Part 2: Soundness.** Suppose that \(S\) is a cut in \(H\) that cuts at
least \(C=k+4m\) edges. We can assume that \(s^*\) is not in \(S\) (otherwise
we can “flip” \(S\) to its complement \(\overline{S}\), since this does not
change the size of the cut). Now let \(I\) be the set of vertices in \(S\)
that correspond to the original vertices of \(G\). If \(I\) was an
independent set of size \(k\) then would be done. This might not always be
the case but we will see that if \(I\) is not an independent set then its
also larger than \(k\). Specifically, we define \(m_{in}=|E(I,I)|\) be the
set of edges in \(G\) that are contained in \(I\) and let \(m_{out}=m-m_{in}\)
(i.e., if \(I\) is an independent set then \(m_{in}=0\) and \(m_{out}=m\)). By
the properties of our gadget we know that for every edge \(\{u,v\}\) of
\(G\), we can cut at most three edges when both \(u\) and \(v\) are in \(S\),
and at most four edges otherwise. Hence the number \(C\) of edges cut by
\(S\) satisfies
\(C \leq |I| + 3m_{in}+4m_{out} = |I|+ 3m_{in} + 4(m-m_{in})=|I|+4m-m_{in}\).
Since \(C = k +4m\) we get that \(|I|-m_{in} \geq k\). Now we can transform
\(I\) into an independent set \(I'\) by going over every one of the \(m_{in}\)
edges that are inside \(I\) and removing one of the endpoints of the edge
from it. The resulting set \(I'\) is an independent set in the graph \(G\)
of size \(|I|-m_{in} \geq k\) and so this concludes the proof of the
soundness condition.

One of the most basic algorithms in Computer Science is Dijkstra’s
algorithm to find the *shortest path* between two vertices. We now show
that in contrast, an efficient algorithm for the *longest path* problem
would imply a polynomial-time algorithm for 3SAT.

\[3SAT \leq_p LONGPATH\]

To prove longpaththm need to show how to transform a 3CNF formula \(\varphi\) into a graph \(G\) and two vertices \(s,t\) such that \(G\) has a path of length at least \(k\) if and only if \(\varphi\) is satisfiable. The idea of the reduction is sketched in longpathfig and longpathfigtwo. We will construct a graph that contains a potentially long “snaking path” that corresponds to all variables in the formula. We will add a “gadget” corresponding to each clause of \(\varphi\) in a way that we would only be able to use the gadgets if we have a satisfying assignment.

We build a graph \(G\) that “snakes” from \(s\) to \(t\) as follows. After \(s\) we add a sequence of \(n\) long loops. Each loop has an “upper path” and a “lower path”. A simple path cannot take both the upper path and the lower path, and so it will need to take exactly one of them to reach \(s\) from \(t\).

Our intention is that a path in the graph will correspond to an assignment \(x\in \{0,1\}^n\) in the sense that taking the upper path in the \(i^{th}\) loop corresponds to assigning \(x_i=1\) and taking the lower path corresponds to assigning \(x_i=0\). When we are done snaking through all the \(n\) loops corresponding to the variables to reach \(t\) we need to pass through \(m\) “obstacles”: for each clause \(j\) we will have a small gadget consisting of a pair of vertices \(s_j,t_j\) that have three paths between them. For example, if the \(j^{th}\) clause had the form \(x_{17} \vee \overline{x}_{55} \vee x_{72}\) then one path would go through a vertex in the lower loop corresponding to \(x_{17}\), one path would go through a vertex in the upper loop corresponding to \(x_{55}\) and the third would go through the lower loop corresponding to \(x_{72}\). We see that if we went in the first stage according to a satisfying assignment then we will be able to find a free vertex to travel from \(s_j\) to \(t_j\). We link \(t_1\) to \(s_2\), \(t_2\) to \(s_3\), etc and link \(t_m\) to \(t\). Thus a satisfying assignment would correspond to a path from \(s\) to \(t\) that goes through one path in each loop corresponding to the variables, and one path in each loop corresponding to the clauses. We can make the loop corresponding to the variables long enough so that we must take the entire path in each loop in order to have a fighting chance of getting a path as long as the one corresponds to a satisfying assignment. But if we do that, then the only way if we are able to reach \(t\) is if the paths we took corresponded to a satisfying assignment, since otherwise we will have one clause \(j\) where we cannot reach \(t_j\) from \(s_j\) without using a vertex we already used before.

- The computational complexity of many seemingly unrelated
computational problems can be related to one another through the use
of
*reductions*. - If \(F \leq_p G\) then a polynomial-time algorithm for \(G\) can be transformed into a polynomial-time algorithm for \(F\).
- Equivalently, if \(F \leq_p G\) and \(F\) does
*not*have a polynomial-time algorithm then neither does \(G\). - We’ve developed many techniques to show that \(3SAT \leq_p F\) for
interesting functions \(F\). Sometimes we can do so by using
*transitivity*of reductions: if \(3SAT \leq_p G\) and \(G \leq_p F\) then \(3SAT \leq_p F\).

Most of the exercises have been written in the summer of 2018 and haven’t yet been fully debugged. While I would prefer people do not post online solutions to the exercises, I would greatly appreciate if you let me know of any bugs. You can do so by posting a GitHub issue about the exercise, and optionally complement this with an email to me with more details about the attempted solution.

Prove product-lem

Prove that if \(F \leq_p G\) and \(G \leq_p H\) then \(F \leq_p H\).

Reduction of independent set to max cut taken from these notes.

Some topics related to this chapter that might be accessible to advanced students include: (to be completed)

Copyright 2018, Boaz Barak.

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