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- Introduce the class \(\mathbf{NP}\) capturing a great many important
computational problems
- \(\mathbf{NP}\)-completeness: evidence that a problem might be
intractable.
- The \(\mathbf{P}\) vs \(\mathbf{NP}\) problem.

“In this paper we give theorems that suggest, but do not imply, that these problems, as well as many others, will remain intractable perpetually”, Richard Karp, 1972

“Sad to say, but it will be many more years, if ever before we really understand the Mystical Power of Twoness… 2-SAT is easy, 3-SAT is hard, 2-dimensional matching is easy, 3-dimensional matching is hard. Why? oh, Why?”Eugene Lawler

So far we have shown that 3SAT is no harder than Quadratic Equations,
Independent Set, Maximum Cut, and Longest Path. But to show that these
problems are *computationally equivalent* we need to give reductions in
the other direction, reducing each one of these problems to 3SAT as
well. It turns out we can reduce all three problems to 3SAT in one fell
swoop.

In fact, this result extends far beyond these particular problems. All
of the problems we discussed in reductionchap, and a great
many other problems, share the same commonality: they are all *search*
problems, where the goal is to decide, given an instance \(x\), whether
there exists a *solution* \(y\) that satisfies some condition that can be
verified in polynomial time. For example, in 3SAT, the instance is a
formula and the solution is an assignment to the variable; in Max-Cut
the instance is a graph and the solution is a cut in the graph; and so
on and so forth. It turns out that *every* such search problem can be
reduced to 3SAT.

To make this precise, we make the following mathematical definition: we
define the class \(\mathbf{NP}\) to contain all Boolean functions that
correspond to a *search problem* of the form above\(-\) that is, functions
that output \(1\) on \(x\) if and only if there exists a solution \(w\) such
that the pair \((x,w)\) satisfies some polynomial-time checkable
condition. Formally, \(\mathbf{NP}\) is defined as follows:

We say that \(F:\{0,1\}^* \rightarrow \{0,1\}\) is in \(\mathbf{NP}\) if there exists some constants \(a,b \in \N\) and \(V:\{0,1\}^* \rightarrow \{0,1\}\) such that \(V\in \mathbf{P}\) and for every \(x\in \{0,1\}^n\), \[ F(x)=1 \Leftrightarrow \exists_{w \in \{0,1\}^{an^b}} \text{ s.t. } V(xw)=1 \;. \label{NP:eq} \]

In other words, for \(F\) to be in \(\mathbf{NP}\), there needs to exist
some polynomial-time computable verification function \(V\), such that if
\(F(x)=1\) then there must exist \(w\) (of length polynomial in \(|x|\)) such
that \(V(xw)=1\), and if \(F(x)=0\) then for *every* such \(w\), \(V(xw)=0\).
Since the existence of this string \(w\) certifies that \(F(x)=1\), \(w\) is
often referred to as a *certificate*, *witness*, or *proof* that
\(F(x)=1\).

See also NPdeffigfig for an illustration of NP-def.
The name \(\mathbf{NP}\) stands for “nondeterministic polynomial time” and
is used for historical reasons; see the bibiographical notes. The string
\(w\) in \eqref{{NP:eq}} is sometimes known as a *solution*,
*certificate*, or *witness* for the instance \(x\).

The definition of \(\mathbf{NP}\) means that for every \(F\in \mathbf{NP}\)
and string \(x\in \{0,1\}^*\), \(F(x)=1\) if and only if there is a *short
and efficiently verifiable proof* of this fact. That is, we can think of
the function \(V\) in NP-def as a *verifier* algorithm, similar
to what we’ve seen in godelproofdef. The verifier checks
whether a given string \(w\in \{0,1\}^*\) is a valid proof for the
statement “\(F(x)=1\)”. Essentially all proof systems considered in
mathematics involve line-by-line checks that can be carried out in
polynomial time. Thus the heart of \(\mathbf{NP}\) is asking for
statements that have *short* (i.e., polynomial in the size of the
statements) proof. Indeed, as we will see in #chappvsnp, Kurt
Gödel phrased the question of whether \(\mathbf{NP}=\mathbf{P}\) as asking
whether “the mental work of a mathematician [in proving theorems]
could be completely replaced by a machine”.

NP-def is *assymetric* in the sense that there is a difference
between an output of \(1\) and an output of \(0\). You should make sure you
understand why this definition does *not* guarantee that if
\(F \in \mathbf{NP}\) then the function \(1-F\) (i.e., the map
\(x \mapsto 1-F(x)\)) is in \(\mathbf{NP}\) as well. In fact, it is believed
that there do exist functions \(F\) satisfying \(F\in \mathbf{NP}\) but
\(1-F \not\in \mathbf{NP}\).*not* satisfiable is not
known (nor believed) to be in \(\mathbf{NP}\).*does* satisfy that if
\(F\in \mathbf{P}\) then \(1-F\) is in \(\mathbf{P}\) as well.

\(3SAT\) is in \(\mathbf{NP}\) since for every \(\ell\)-variable formula \(\varphi\), \(3SAT(\varphi)=1\) if and only if there exists a satisfying assignment \(x \in \{0,1\}^\ell\) such that \(\varphi(x)=1\), and we can check this condition in polynomial time.

The above reasoning explains why \(3SAT\) is in \(\mathbf{NP}\), but since this is our first example, we will now belabor the point and expand out in full formality what is the precise representation of the witness \(w\) and the algorithm \(V\) that demonstrate that \(3SAT\) is in \(\mathbf{NP}\).

Specifically, we can represent a 3CNF formula \(\varphi\) with \(k\) variables and \(m\) clauses as a string of length \(n=O(m\log k)\), since every one of the \(m\) clauses involves three variables and their negation, and the identity of each variable can be represented using \(\lceil \log_2 k \rceil\). We assume that every variable participates in some clause (as otherwise it can be ignored) and hence that \(m \geq k\), which in particular means that \(n\) is larger than both \(m\) and \(k\).

We can represent an assignment to the \(k\) variables using a \(k\)-length string, which, since \(n > k\), can be “padded” to a string \(w\in \{0,1\}^n\) in some standard way. (For example, if \(y\in \{0,1\}^k\) is the assignment, we can let \(w=y10^{n-k-1}\); given the string \(w\) we can “read off” \(y\), by chopping off all the zeroes at the end of \(w\) until we encounter the first \(1\), which we remove as well.)

Now checking whether a given assignment \(y\in \{0,1\}^k\) satisfies a given \(k\)-variable 3CNF \(\varphi\) can be done in polynomial time through the following algorithm \(V\):

Algorithm \(V\):

Input:

- 3CNF formula \(\varphi\) with \(k\) variables and \(m\) clauses (encoded as a string of length \(n=O(m\log k))\)
- Assignment \(y\in \{0,1\}^k\) to the variables of \(\varphi\) (encoded using padding as a string \(w \in \{0,1\}^n\))

Output:\(1\) if and only if \(y\) satisfies \(\varphi\).

Operation:

- For every clause \(C = (\ell_1 \vee \ell_2 \vee \ell_3)\) of \(\varphi\) (where \(\ell_1,\ell_2,\ell_3\) are literals), if all three literals evaluate to
falseunder the assignment \(y\) then halt and output \(0\).- Output \(1\).

Algorithm \(V\) runs in time polynomial in the length \(n\) of \(\varphi\)’s description as a string. Indeed there are \(m\) clauses, and checking the evaluation of a literal of the form \(y_i\) or \(\neg y_j\) can be done by scanning the \(k\)-length string \(y\), and hence the running time of Algorithm \(V\) is at most \(O(mk)=O(n^2)\), as both \(k\) and \(m\) are smaller than \(n\).

By its definition the algorithm outputs \(1\) if and only if the assignment \(y\) satisfies all the clauses of the 3CNF formula \(\varphi\), which means that \(3SAT(\varphi)=1\) if and only if there exists some \(w\in \{0,1\}^n\) such that \(V(\varphi w)=1\) which is precisely the condition needed to show that \(3SAT \in \mathbf{NP}\) per NP-def.

The “padding trick” we used in threesatinnpex can always be
used to expand a witness of length smaller than \(an^b\) to a witness of
exactly that length. Therefore one can think of the condition
\eqref{{NP:eq}} in NP-defas simply stipulating that the
“solution” \(w\) to the problem \(x\) is of length *at most* polynomial in
\(|x|\).

Here are some more examples for problems in \(\mathbf{NP}\). For each one of these problems we merely sketch how the witness is represented and why it is efficiently checkable, but working out the details can be a good way to get more comfortable with NP-def:

- \(QUADEQ\) is in \(\mathbf{NP}\) since for every \(\ell\)-variable instance of quadratic equations \(E\), \(QUADEQ(E)=1\) if and only if there exists an assignment \(x\in \{0,1\}^\ell\) that satisfies \(E\). We can check the condition that \(x\) satisfies \(E\) in polynomial time by enumerating over all the equations in \(E\), and for each such equation \(e\), plug in the values of \(x\) and verify that \(e\) is satisfied.
- \(ISET\) is in \(\mathbf{NP}\) since for every graph \(G\) and integer \(k\), \(ISET(G,k)=1\) if and only if there exists a set \(S\) of \(k\) vertices that contains no pair of neighbors in \(G\). We can check the condition that \(S\) is an independent set of size \(\geq k\) in polynomial time by first checking that \(|S| \geq k\) and then enumerating over all edges \(\{u,v \}\) in \(G\), and for each such edge verify that either \(u\not\in S\) or \(v\not\in S\).
- \(LONGPATH\) is in \(\mathbf{NP}\) since for every graph \(G\) and integer \(k\), \(LONGPATH(G,k)=1\) if and only if there exists a simple path \(P\) in \(G\) that is of length at least \(k\). We can check the condition that \(P\) is a simple path of length \(k\) in polynomial time by checking that it has the form \((v_0,v_1,\ldots,v_k)\) where each \(v_i\) is a vertex in \(G\), no \(v_i\) is repeated, and for every \(i \in [k]\), the edge \(\{v_i,v_{i+1}\}\) is present in the graph.
- \(MAXCUT\) is in \(\mathbf{NP}\) since for every graph \(G\) and integer \(k\), \(MAXCUT(G,k)=1\) if and only if there exists a cut \((S,\overline{S})\) in \(G\) that cuts at least \(k\) edges. We can check that condition that \((S,\overline{S})\) is a cut of value at least \(k\) in polynomial time by checking that \(S\) is a subset of \(G\)’s vertices and enumerating over all the edges \(\{u,v\}\) of \(G\), counting those edges such that \(u\in S\) and \(v\not\in S\) or vice versa.

The definition of \(\mathbf{NP}\) is one of the most important definitions of this book, and is worth while taking the time to digest and internalize. The following solved exercises establish some basic properties of this class. As usual, I highly recommend that you try to work out the solutions yourself.

Prove that \(\mathbf{P} \subseteq \mathbf{NP}\).

Suppose that \(F \in \mathbf{P}\). Define the following function \(V\): \(V(x0^n)=1\) iff \(n=|x|\) and \(F(x)=1\). (\(V\) outputs \(0\) on all other inputs.) Since \(F\in \mathbf{P}\) we can clearly compute \(V\) in polynomial time as well.

Let \(x\in \{0,1\}^n\) be some string. If \(F(x)=1\) then \(V(x0^n)=1\). On the other hand, if \(F(x)=0\) then for every \(w\in \{0,1\}^n\), \(V(xw)=0\). Therefore, setting \(a=b=1\), we see that \(V\) satisfies \eqref{{NP:eq}}, and establishes that \(F \in \mathbf{NP}\).

People sometimes think that \(\mathbf{NP}\) stands for “non polynomial time”. As PinNP shows, this is far from the truth, and in fact every polynomial-time computable function is in \(\mathbf{NP}\) as well.

If \(F\) is in \(\mathbf{NP}\) it certainly does *not* mean that \(F\) is hard
to compute (though it does not, as far as we know, necessarily mean that
it’s easy to compute either). Rather, it means that \(F\) is *easy to
verify*, in the technical sense of NP-def.

Prove that \(\mathbf{NP} \subseteq \mathbf{EXP}\).

Suppose that \(F\in \mathbf{NP}\) and let \(V\) be the polynomial-time computable function that satisfies \eqref{{NP:eq}} and \(a,b\) the corresponding constants. Then the following is an exponential-time algorithm \(A\) to compute \(F\):

Algorithm \(A\):

Input:\(x \in \{0,1\}^*\), let \(n=|x|\)

Operation:

- For every \(w\in \{0,1\}^{an^b}\), if \(V(xw)=1\) then halt and output \(1\).
- Output \(0\).

Since \(V \in \mathbf{P}\), for every \(x\in \{0,1\}^n\), Algorithm \(A\) runs in time \(poly(n)2^{an^b}\). Moreover by \eqref{{NP:eq}}, \(A\) will output \(1\) on \(x\) if and only if \(F(x)=1\).

PinNP and NPinEXP together imply that

\[\mathbf{P} \subseteq \mathbf{NP} \subseteq \mathbf{EXP}\;.\]

The time hierarchy theorem (time-hierarchy-thm) implies that
\(\mathbf{P} \subsetneq \mathbf{EXP}\) and hence at least one of the two
inclusions \(\mathbf{P} \subseteq \mathbf{NP}\) or
\(\mathbf{NP} \subseteq \mathbf{EXP}\) is *strict*. It is believed that
both of them are in fact strict inclusions. That is, it is believed that
there are functions in \(\mathbf{NP}\) that cannot be computed in
polynomial time (this is the \(\mathbf{P} \neq \mathbf{NP}\) conjecture)
and that there are functions \(F\) in \(\mathbf{EXP}\) for which we cannot
even efficiently *certify* that \(F(x)=1\) for a given input \(x\).

We have previously informally equated the notion of \(F \leq_p G\) with \(F\) being “no harder than \(G\)” and in particular have seen in reductionsandP that if \(G \in \mathbf{P}\) and \(F \leq_p G\), then \(F \in \mathbf{P}\) as well. The following exercise shows that if \(F \leq_p G\) then it is also “no harder to verify” than \(G\). That is, regardless of whether or not it is in \(\mathbf{P}\), if \(G\) has the property that solutions to it can be efficiently verified, then so does \(F\).

Let \(F,G:\{0,1\}^* \rightarrow \{0,1\}\). Show that if \(F \leq_p G\) and \(G\in \mathbf{NP}\) then \(F \in \mathbf{NP}\).

Suppose that \(G\) is in \(\mathbf{NP}\) and in particular there exists \(a,b\) and \(V \in \mathbf{P}\) such that for every \(y \in \{0,1\}^*\), \(G(y)=1 \Leftrightarrow \exists_{w\in \{0,1\}^{a|y|^b}} V(yw)=1\). Define \(V'(x,w)=1\) iff \(V(R(x)w)=1\) where \(R\) is the polynomial-time reduction demonstrating that \(F \leq_p G\). Then for every \(x\in \{0,1\}^*\),

\[F(x)=1 \Leftrightarrow G(R(x)) =1 \Leftrightarrow \exists_{w \in \{0,1\}^{a|R(x)|^b} V(R(x)w) = 1 \Leftrightarrow \exists_{w\in \{0,1\}^{a|R(x)|^b} } V'(x,w)=1 }\]

Since there are some constants \(a',b'\) such that \(|R(x)| \leq a'|x|^{b'}\) for every \(x\in \{0,1\}^*\), by simple padding we can modify \(V'\) to an algorithm that certifies that \(F \in \mathbf{NP}\).

We have seen everal example of problems for which we do not know if
their best algorithm is polynomial or exponential, but we can show that
they are in \(\mathbf{NP}\). That is, we don’t know if they are easy to
*solve*, but we do know that it is easy to *verify* a given solution.
There are many, many, *many*, more examples of interesting functions we
would like to compute that are easily shown to be in \(\mathbf{NP}\). What
is quite amazing is that if we can solve 3SAT then we can solve all of
them!

The following is one of the most fundamental theorems in Computer Science:

For every \(F\in \mathbf{NP}\), \(F \leq_p 3SAT\).

We will soon show the proof of cook-levin-thm, but note that
it immediately implies that \(QUADEQ\), \(LONGPATH\), and \(MAXCUT\) all
reduce to \(3SAT\). Combining it with the reductions we’ve seen in
reductionchap, it implies that all these problems are
*equivalent!* For example, to reduce \(QUADEQ\) to \(LONGPATH\), we can
first reduce \(QUADEQ\) to \(3SAT\) using cook-levin-thm and use
the reduction we’ve seen in longpaththm from \(3SAT\) to
\(LONGPATH\). That is, since \(QUADEQ \in \mathbf{NP}\),
cook-levin-thm implies that \(QUADEQ \leq_p 3SAT\), and
longpaththm implies that \(3SAT \leq_p LONGPATH\), which by the
transitivity of reductions (transitivitylem) means that
\(QUADEQ \leq_p LONGPATH\). Similarly, since \(LONGPATH \in \mathbf{NP}\),
we can use cook-levin-thm and quadeq-thm to show
that \(LONGPATH \leq_p 3SAT \leq_p QUADEQ\), concluding that \(LONGPATH\)
and \(QUADEQ\) are computationally equivalent.

There is of course nothing special about \(QUADEQ\) and \(LONGPATH\) here:
by combining \eqref{cook-levin-thm} with the reductions we saw, we
see that just like \(3SAT\), *every* \(F\in \mathbf{NP}\) reduces to
\(LONGPATH\), and the same is true for \(QUADEQ\) and \(MAXCUT\). All these
problems are in some sense “the hardest in \(\mathbf{NP}\)” since an
efficient algorithm for any one of them would imply an efficient
algorithm for *all* the problems in \(\mathbf{NP}\). This motivates the
following definition:

We say that \(G:\{0,1\}^* \rightarrow \{0,1\}\) is *\(\mathbf{NP}\) hard* if
for every \(F\in \mathbf{NP}\), \(F \leq_p G\).

We say that \(G:\{0,1\}^* \rightarrow \{0,1\}\) is *\(\mathbf{NP}\)
complete* if \(G\) is \(\mathbf{NP}\) hard and \(G\) is in \(\mathbf{NP}\).

The Cook-Levin Theorem (cook-levin-thm) can be rephrased as
saying that \(3SAT\) is \(\mathbf{NP}\) hard, and since it is also in
\(\mathbf{NP}\), this means that \(3SAT\) is \(\mathbf{NP}\) complete.
Together with the reductions of reductionchap,
cook-levin-thm shows that despite their superficial
differences, 3SAT, quadratic equations, longest path, independent set,
and maximum cut, are all \(\mathbf{NP}\)-complete. Many thousands of
additional problems have been shown to be \(\mathbf{NP}\)-complete,
arising from all the sciences, mathematics, economics, engineering and
many other fields.

As we’ve seen in PinNP, \(\mathbf{P} \subseteq \mathbf{NP}\).
*The* most famous conjecture in Computer Science is that this
containment is *strict*. That is, it is widely conjectured that
\(\mathbf{P} \neq \mathbf{NP}\). One way to refute the conjecture that
\(\mathbf{P} \neq \mathbf{NP}\) is to give a polynomial-time algorithm for
even a single one of the \(\mathbf{NP}\)-complete problems such as 3SAT,
Max Cut, or the thousands of others that have been studied in all fields
of human endeavors. The fact that these problems have been studied by so
many people, and yet not a single polynomial-time algorithm for any of
them has been found, supports that conjecture that indeed
\(\mathbf{P} \neq \mathbf{NP}\). In fact, for many of these problems
(including all the ones we mentioned above), we don’t even know of a
\(2^{o(n)}\)-time algorithm! However, to the frustration of computer
scientists, we have not yet been able to prove that
\(\mathbf{P}\neq\mathbf{NP}\) or even rule out the existence of an
\(O(n)\)-time algorithm for 3SAT. Resolving whether or not
\(\mathbf{P}=\mathbf{NP}\) is known as the \(\mathbf{P}\) vs \(\mathbf{NP}\)
problem. A
million-dollar prize has been
offered
for the solution of this problem, a popular
book has been written, and every
year a new paper comes out claiming a proof of \(\mathbf{P}=\mathbf{NP}\)
or \(\mathbf{P}\neq\mathbf{NP}\), only to wither under scrutiny.

One of the mysteries of computation is that people have observed a certain empirical “zero-one law” or “dichotomy” in the computational complexity of natural problems, in the sense that many natural problems are either in \(\mathbf{P}\) (often in \(TIME(O(n))\) or \(TIME(O(n^2))\)), or they are are \(\mathbf{NP}\) hard. This is related to the fact that for most natural problems, the best known algorithm is either exponential or polynomial, with not too many examples where the best running time is some strange intermediate complexity such as \(2^{2^{\sqrt{\log n}}}\). However, it is believed that there exist problems in \(\mathbf{NP}\) that are neither in \(\mathbf{P}\) nor are \(\mathbf{NP}\)-complete, and in fact a result known as “Ladner’s Theorem” shows that if \(\mathbf{P} \neq \mathbf{NP}\) then this is indeed the case (see also ladner-ex and PNPscenariosfig).

We will now prove the Cook-Levin Theorem, which is the underpinning to a great web of reductions from 3SAT to thousands of problems across great many fields. Some problems that have been shown to be \(\mathbf{NP}\)-complete include: minimum-energy protein folding, minimum surface-area foam configuration, map coloring, optimal Nash equilibrium, quantum state entanglement, minimum supersequence of a genome, minimum codeword problem, shortest vector in a lattice, minimum genus knots, positive Diophantine equations, integer programming, and many many more. The worst-case complexity of all these problems is (up to polynomial factors) equivalent to that of 3SAT, and through the Cook-Levin Theorem, to all problems in \(\mathbf{NP}\).

To prove cook-levin-thm we need to show that \(F \leq_p 3SAT\) for every \(F\in \mathbf{NP}\). We will do so in three stages. We define two intermediate problems: \(NANDSAT\) and \(3NAND\). We will shortly show the definitions of these two problems, but cook-levin-thm will follow from combining the following three results:

- \(NANDSAT\) is \(\mathbf{NP}\) hard (nand-thm).
- \(NANDSAT \leq_p 3NAND\) (threenand-thm).
- \(3NAND \leq_p 3SAT\) (threenand-sat-thm).

By the transitivity of reductions, it will follow that for every \(F \in \mathbf{NP}\),

\[ F \leq_p NANDSAT \leq_p 3NAND \leq_p 3SAT \]

hence establishing cook-levin-thm.

We will prove these three results nand-thm, threenand-thm and threenand-sat-thm one by one, providing the requisite definitions as we go along.

We define the \(NANDSAT\) problem as follows. On input a string
\(Q\in \{0,1\}^*\), we define \(NANDSAT(Q)=1\) if and only if \(Q\) is a valid
representation of an \(n\)-input and single-output NAND program and there
exists some \(w\in \{0,1\}^n\) such that \(Q(w)=1\). While we don’t need
this to prove nand-thm, note that \(NANDSAT\) is in
\(\mathbf{NP}\) since we can verify that \(Q(w)=1\) using the
polyonmial-time algorithm for evaluating NAND programs.

\(NANDSAT\) is \(\mathbf{NP}\) hard.

To prove nand-thm we need to show that for every
\(F\in \mathbf{NP}\), \(F \leq_p NANDSAT\). The high-level idea is that by
the definition of \(\mathbf{NP}\), there is some NAND++ program \(P^*\) and
some polynomial \(T(\cdot)\) such that \(F(x)=1\) if and only if there
exists some \(w \in \{0,1\}^{a|x|^b}\) such that \(P^*(xw)\) outputs \(1\)
within \(T(|x|)\) steps. Now by “unrolling the loop” of the NAND++ program
\(P^*\) we can convert it into an \(O(T(n))\) NAND program \(Q'\) with
\(n + an^b\) inputs and a single output such that for every
\(x\in \{0,1\}^n\) and \(w\in \{0,1\}^{an^b}\), \(Q'(xw)=P^*(xw)\). on input
\(x \in \{0,1\}\) that on input \(w\) will simulate \(P^*(xw)\) for \(T(|x|)\)
steps. The next step is to *hardwire* the input \(x\) to \(Q'\) to obtain an
\(O(T(n))\) line NAND program \(Q\) with \(m=an^b\) inputs such that for every
\(w\in \{0,1\}^m\), \(Q'(w)=Q(xw)\). By construction it will be the case
that for every \(x\in \{0,1\}^n\), \(F(x)=1\) if and only if there exists
\(w\in \{0,1\}^{an^b}\) such that \(Q(w)=1\), and hence this shows that
\(F \leq_p NANDSAT\).

The proof is a little bit technical but ultimately follows quite
directly from the definition of \(\mathbf{NP}\), as well as of NAND and
NAND++ programs. If you find it confusing, try to pause here and work
out the proof yourself from these definitions, using the idea of
“unrolling the loop” of a NAND++ program. It might also be useful for
you to think how you would implement in your favorite programming
language the function `unroll`

which on input a NAND++ program \(P\) and
numbers \(T,n\) would output an \(n\)-input NAND program \(Q\) of \(O(|T|)\)
lines such that for every input \(z\in \{0,1\}^n\), if \(P\) halts on \(z\)
within at most \(T\) steps and outputs \(y\), then \(Q(z)=y\).

We now present the details. Let \(F \in \mathbf{NP}\). To prove nand-thm we need to give a polynomial-time computable function that will map every \(x^* \in \{0,1\}^*\) to a NAND program \(Q\) such that \(F(x)=NANDSAT(Q)\).

Let \(x^* \in \{0,1\}^*\) be such a string and let \(n=|x^*|\) be its length. By NP-def there exists \(V \in \mathbf{P}\) and \(a,b \in \N\) such that \(F(x^*)=1\) if and only if there exists \(w\in \{0,1\}^{an^b}\) such that \(V(x^*w)=1\).

Let \(m=an^b\). Since \(V\in \mathbf{P}\) there is some NAND++ program \(P^*\) that computes \(V\) on inputs of the form \(xw\) with \(x\in \{0,1\}^n\) and \(w\in \{0,1\}^m\) in at most \({(n+m)}^c\) time for some constant \(c\). Using our “unrolling the loop NAND++ to NAND compiler” of nand-compiler, we can obtain a NAND program \(Q'\) that has \(n+m\) inputs and at most \(O((n+m)^c)\) lines such that \(Q'(xw)= P^*(xw)\) for every \(x\in \{0,1\}^n\) and \(w \in \{0,1\}^m\).

Now we can use the following simple but useful “hardwiring” technique to obtain a program:

Given a \(T\)-line NAND program \(Q'\) of \(n+m\) inputs and \(x^* \in \{0,1\}^n\), we can obtain in polynomial a program \(Q\) with \(m\) inputs of \(T+3\) lines such that for ever \(w\in \{0,1\}^m\), \(Q(w)= Q'(x^*w)\).

To compute \(Q\), we simply do a “search and replace” for all references
in \(Q'\) to `X[`

\(i\)`]`

for \(i \in [n]\), and transform them to either the
variable `zero`

or `one`

depending on whether \(x^*_i\) is equal to \(0\) or
\(1\) respectively. By adding three lines to the beginning of \(Q'\), we can
ensure that the `zero`

and `one`

variables will have the correct value.
The only thing that then remains to do another search and replace to
transform all references to the variables `X[`

\(n\)`]`

,\(\ldots\),
`X[`

\(n+m-1\)`]`

to the variables `X[`

\(0\)`]`

, \(\ldots\), `X[`

\(m-1\)`]`

so
that the \(m\) inputs to the new program \(Q\) will correspond to last \(m\)
inputs of the original program \(Q'\). See hardwiringfig for an
implementation of this reduction in Python.

Using hardwiringlem, we obtain a program \(Q\) of \(m\) inputs such that \(Q(w)=Q'(x^*w)=P^*(x^*w)\) for every \(w\in \{0,1\}^m\). Since we know that \(F(x^*)=1\) if and only if there exists \(w\in \{0,1\}^m\) such that \(P^*(x^*w)=1\), this means that \(F(x^*)=1\) if and only if \(NANDSAT(Q)=1\), which is what we wanted to prove.

The \(3NAND\) problem is defined as follows: the input is a logical formula \(\varphi\) on a set of variables \(z_0,\ldots,z_{r-1}\) which is an AND of constraints of the form \(z_i = NAND(z_j,z_k)\). For example, the following is a \(3NAND\) formula with \(5\) variables and \(3\) constraints:

\[ \left( z_3 = NAND(z_0,z_2) \right) \wedge \left( z_1 = NAND(z_0,z_2) \right) \wedge \left( z_4 = NAND(z_3,z_1) \right) \]

The output of \(3NAND\) on input \(\varphi\) is \(1\) if and only if there is an assignment to the variables of \(\varphi\) that makes it evaluate to “true” (that is, there is some assignment \(z \in \{0,1\}^r\) satisfying all of the constraints of \(\varphi\)). As usual, we can represent \(\varphi\) as a string, and so think of \(3NAND\) as a function mapping \(\{0,1\}^*\) to \(\{0,1\}\). We now prove that \(3NAND\) is \(\mathbf{NP}\) hard:

\(NANDSAT \leq_p 3NAND\).

To prove threenand-thm we need to give a polynomial-time map
from every NAND program \(Q\) to a 3NAND formula \(\Psi\) such that there
exists \(w\) such that \(Q(w)=1\) if and only if there exists \(z\) satisfying
\(\Psi\). For every line \(i\) of \(Q\), we define a corresponding variable
\(z_i\) of \(\Psi\). If the line \(i\) has the form `foo = NAND(bar,blah)`

then we will add the clause \(z_i = NAND(z_j,z_k)\) where \(j\) and \(k\) are
the last lines in which `bar`

and `blah`

were written to. We will also
set variables corresponding to the input variables, as well as add a
clause to ensure that the final output is \(1\). The resulting reduction
can be implemented in about a dozen lines of Python, see
andsattothreenandfig.

To prove threenand-thm we need to give a reduction from
\(NANDSAT\) to \(3NAND\). Let \(Q\) be a NAND program with \(n\) inputs, one
output, and \(m\) lines. We can assume without loss of generality that \(Q\)
contains the variables `one`

and `zero`

as usual.

We map \(Q\) to a \(3NAND\) formula \(\Psi\) as follows:

- \(\Psi\) has \(m+n\) variables \(z_0,\ldots,z_{m+n-1}\).
- The first \(n\) variables \(z_0,\ldots,z_{n-1}\) will corresponds to the inputs of \(Q\). The next \(m\) variables \(z_n,\ldots,z_{n+m-1}\) will correspond to the \(m\) lines of \(Q\).
- For every \(\ell\in \{n,n+1,\ldots,n+m \}\), if the \(\ell-n\)-th line
of the program \(Q\) is
`foo = NAND(bar,blah)`

then we add to \(\Psi\) the constraint \(z_\ell = NAND(z_j,z_k)\) where \(j-n\) and \(k-n\) correspond to the last lines in which the variables`bar`

and`blah`

(respectively) were written to. If one or both of`bar`

and`blah`

was not written to before then we use \(z_{\ell_0}\) instead of the corresponding value \(z_j\) or \(z_k\) in the constraint, where \(\ell_0-n\) is the line in which`zero`

is assigned a value. If one or both of`bar`

and`blah`

is an input variable`X[i]`

then we we use \(z_i\) in the constraint. - Let \(\ell^*\) be the last line in which the output
`y_0`

is assigned a value. Then we add the constraint \(z_{\ell^*} = NAND(z_{\ell_0},z_{\ell_0})\) where \(\ell_0-n\) is as above the last line in which`zero`

is assigned a value. Note that this is effectively the constraint \(z_{\ell^*}=NAND(0,0)=1\).

To complete the proof we need to show that there exists \(w\in \{0,1\}^n\) s.t. \(Q(w)=1\) if and only if there exists \(z\in \{0,1\}^{n+m}\) that satisfies all constraints in \(\Psi\). We now show both sides of this equivalence.

**Part I: Completeness.** Suppose that there is \(w\in \{0,1\}^n\) s.t.
\(Q(w)=1\). Let \(z\in \{0,1\}^{n+m}\) be defined as follows: for
\(i\in [n]\), \(z_i=w_i\) and for \(i\in \{n,n+1,\ldots,n+m\}\) \(z_i\) equals
the value that is assigned in the \((i-n)\)-th line of \(Q\) when executed
on \(w\). Then by construction \(z\) satisfies all of the constraints of
\(\Psi\) (including the constraint that \(z_{\ell^*}=NAND(0,0)=1\) since
\(Q(w)=1\).)

**Part II: Soundness.** Suppose that there exists \(z\in \{0,1\}^{n+m}\)
satisfying \(\Psi\). Soundness will follow by showing that
\(Q(z_0,\ldots,z_{n-1})=1\) (and hence in particular there exists
\(w\in \{0,1\}^n\), namely \(w=z_0\cdots z_{n-1}\), such that \(Q(w)=1\)). To
do this we will prove the following claim \((*)\): for every
\(\ell \in [m]\), \(z_{\ell+n}\) equals the value assigned in the \(\ell\)-th
step of the execution of the program \(Q\) on \(z_0,\ldots,z_{n-1}\). Note
that because \(z\) satisfies the constraints of \(\Psi\), \((*)\) is
sufficient to prove the soundness condition since these constraints
imply that the last value assigned to the variable `y_0`

in the
execution of \(Q\) on \(z_0\cdots w_{n-1}\) is equal to \(1\). To prove \((*)\)
suppose, towards a contradiction, that it is false, and let \(\ell\) be
the smallest number such that \(z_{\ell+n}\) is *not* equal to the value
assigned in the \(\ell\)-th step of the exeuction of \(Q\) on
\(z_0,\ldots,z_{n-1}\). But since \(z\) satisfies the constraints of \(\Psi\),
we get that \(z_{\ell+n}=NAND(z_i,z_j)\) where (by the assumption above
that \(\ell\) is *smallest* with this property) these values *do*
correspond to the values last assigned to the variables on the righthand
side of the assignment operator in the \(\ell\)-th line of the program.
But this means that the value assigned in the \(\ell\)-th step is indeed
simply the NAND of \(z_i\) and \(z_j\), contradicting our assumption on the
choice of \(\ell\).

To conclude the proof of cook-levin-thm, we need to show threenand-sat-thm and show that \(3NAND \leq_p 3SAT\):

\(3NAND \leq_p 3SAT\).

To prove threenand-sat-thm we need to map a 3NAND formula \(\varphi\) into a 3SAT formula \(\psi\) such that \(\varphi\) is satisfiable if and only if \(\psi\) is. The idea is that we can transform every NAND constraint of the form \(a=NAND(b,c)\) into the AND of ORs involving the variables \(a,b,c\) and their negations, where each of the ORs contains at most three terms. The construction is fairly straightforward, and the details are given below.

It is a good exercise for you to try to find a 3CNF formula \(\xi\) on three variables \(a,b,c\) such that \(\xi(a,b,c)\) is true if and only if \(a = NAND(b,c)\). Once you do so, try to see why this implies a reduction from \(3NAND\) to \(3SAT\), and hence completes the proof of threenand-sat-thm

The constraint \[ z_i = NAND(z_j,z_k) \label{eq:NANDconstraint} \] is satisfied if \(z_i=1\) whenever \((z_j,z_k) \neq (1,1)\). By going through all cases, we can verify that \eqref{eq:NANDconstraint} is equivalent to the constraint

\[ (\overline{z_i} \vee \overline{z_j} \vee\overline{z_k} ) \wedge (z_i \vee z_j ) \wedge (z_i \vee z_k) \;\;. \label{eq:CNFNAND} \]

Indeed if \(z_j=z_k=1\) then the first constraint of eq:CNFNAND
is only true if \(z_i=0\). On the other hand, if either of \(z_j\) or \(z_k\)
equals \(0\) then unless \(z_i=1\) either the second or third constraints
will fail. This means that, given any 3NAND formula \(\varphi\) over \(n\)
variables \(z_0,\ldots,z_{n-1}\), we can obtain a 3SAT formula \(\psi\) over
the same variables by replacing every \(3NAND\) constraint of \(\varphi\)
with three \(3OR\) constraints as in eq:CNFNAND.

We have shown that for every function \(F\) in \(\mathbf{NP}\), \(F \leq_p NANDSAT \leq_p 3NAND \leq_p 3SAT\), and so \(3SAT\) is \(\mathbf{NP}\)-hard. Since in reductionchap we saw that \(3SAT \leq_p QUADEQ\), \(3SAT \leq_p ISET\), \(3SAT \leq_p MAXCUT\) and \(3SAT \leq_p LONGPATH\), all these problems are \(\mathbf{NP}\)-hard as well. Finally, since all the aforementioned problems are in \(\mathbf{NP}\), they are all in fact \(\mathbf{NP}\)-complete and have equivalent complexity. There are thousands of other natural problems that are \(\mathbf{NP}\)-complete as well. Finding a polynomial-time algorithm for any one of them will imply a polynomial-time algorithm for all of them.

- Many of the problems for which we don’t know polynomial-time
algorithms are \(\mathbf{NP}\)-complete, which means that finding a
polynomial-time algorithm for one of them would imply a
polynomial-time algorithm for
*all*of them. - It is conjectured that \(\mathbf{NP}\neq \mathbf{P}\) which means that
we believe that polynomial-time algorithms for these problems are
not merely
*unknown*but are*nonexistent*. - While an \(\mathbf{NP}\)-hardness result means for example that a
full-fledged “textbook” solution to a problem such as MAX-CUT that
is as clean and general as the algorithm for MIN-CUT probably does
not exist, it does not mean that we need to give up whenever we see
a MAX-CUT instance. Later in this course we will discuss several
strategies to deal with \(\mathbf{NP}\)-hardness, including
*average-case complexity*and*approximation algorithms*.

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 that if there is no \(n^{O(\log^2 n)}\) time algorithm for \(3SAT\)
then there is some \(F\in \mathbf{NP}\) such that \(F \not\in \mathbf{P}\)
and \(F\) is not \(\mathbf{NP}\) complete.**Hint:** Use the function \(F\) that on input a formula \(\varphi\)
and a string of the form \(1^t\), outputs \(1\) if and only if \(\varphi\)
is satisfiable and \(t=|\varphi|^{\log|\varphi|}\).

Eugene Lawler’s quote on the “mystical power of twoness” was taken from the wonderful book “The Nature of Computation” by Moore and Mertens. See also this memorial essay on Lawler by Lenstra.

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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