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The manuscript below corresponds to my talk presented to
the mathematics Colloquium, but with references added
later. The references are by no means complete,
particularly with regard to books that I could not
conveniently locate. I want to thank Dan Burghelea for
inviting me to give a talk on P = NP (as part of a series
of lectures on the Clay Millenium Problems at Ohio State
University) as this is not something that I would normally
do. CLAY MILLENIUM PROBLEM:
P = NP
by
Harvey M. Friedman
Mathematics Colloquium
Ohio State University
October 20, 2005
The equation P = NP concerns algorithms for deciding
membership in sets. The consensus is that P ≠ NP, although
some prominent experts guess otherwise.
P = NP is
algorithm
specified
algorithm one of many questions of the form: If there is an
for determining membership in a given set using
limited resources, is there automatically such an
using yet more limited specified resources? P = NP sits in discrete complexity theory  based on sets
of finite strings in a finite alphabet of letters. Sets of
integers, sets of rationals, and sets of finite sequences
of such, are all treated as special cases of sets of finite
strings.
Discrete complexity theory is dependent on a formal
treatment of algorithms  initiated in the 1930’s by Alan
Turing.
The official problem description [Co] contains much
detailed information that will not be mentioned here.
1.
2.
3.
4.
5. TURING MACHINES.
SOME R.E. COMPLETENESS.
TIME, SPACE, NONDETERMINISM.
SOME NP COMPLETENESS.
OPINIONS. 2
1. TURING MACHINES.
There are many sources that discuss Turing machines. E.g.,
[Tu36], [Tu37], [Ro67], [Co71], [Da73], [AHU74], [HU79],
[GJ79], [Od89], [Pa94], [Da00], [CLR01], [Si05], [Co], and
many dozens of others.
Turing machines form the most common primitive model of
computation used in theoretical studies. Micro Center does
not have Turing machines for sale.
TM’s process symbols from a finite alphabet. When given a
finite string from a finite alphabet as input, it grinds
away, perhaps halting after finitely many steps, or perhaps
going on forever.
TM’s have a two way infinite tape divided into squares
indexed by the integers, and a reading head that always
sits on exactly one square of tape.
Every TM comes with a finite set of distinct symbols
a0,...,ak,b0,...,bn, k,n ≥ 0. {a0,...,ak} is the input
alphabet. b0,...,bn are the auxiliary symbols. b0 is the
“blank”.
Every TM also comes with a finite set of distinct states
q0,...,qm, m ≥ 2. q0 is the starting state, q1 is “accept”,
and q2 is “reject”.
Every TM comes already programmed. The program consists of
a finite list of instructions that tell the reading head
what to do.
Each instruction takes the following form:
If the reading head sees symbol x and TM is in state q,
then replace x with symbol x’, put TM into state q’, and
move the reading head one square to the left (right).
Note that x = x’ and q = q’ are allowed.
TM’s are deterministic  there is exactly one such
instruction for each symbol and state other than q1,q2.
(Nondeterministic Turing machines are called NDTM’s).
For nonempty A, A* is the set of all finite strings from A
(empty string allowed). 3 A finite input string a Œ {a0,..., ak}*, is given.
The TM is starts by writing a on the tape starting with
square 1. The remaining squares have the blank, b0. The
reading head is placed on square 1. The starting state is
q0.
Instructions are successively followed. If final state is q1
then TM “accepts” the input string a. If final state is q2
then TM “reject” a. If there is no final state (q1 or q2)
then TM neither accepts nor rejects a.
DEFINITION. Let S Õ A*. S is recursively enumerable (r.e.)
if and only if there is a TM with input alphabet A, such
that S = {x Œ A*: x is accepted by the TM}.
DEFINITION. Let S Õ A*. S is recursive if and only if there
is a TM with input alphabet A, such that S = {x Œ A*: x is
accepted by the TM} and every x Œ A* is accepted or
rejected by the TM.
THEOREM 1.1. Every recursive subset of A* is r.e., but not
vice versa.
THEOREM 1.2. S Õ A* is recursive if and only if S and A*\S
are r.e.
The above two Theorems are discussed in virtually any place
that discusses Turing machines without resource bounds
(before computational complexity theory with resource
bounds).
Adaptation to Z is by identifying nonzero integers with
their base 2 expansion as a string in {0,1}, and 0 with the
empty string. Related identifications work for Q, and for
finite sequences from Z or Q.
The TM model is a crude specialized model that can be
augmented in many ways. More tapes, more reading heads,
etc. RAM machines are more like what they sell at Micro
Center than TM’s.
Early work was devoted to simulating more powerful models
by the TM model. The recursive and r.e. sets don’t change. 4
CHURCH’S THESIS. Any reasonable model of discrete
computation can be simulated in the TM model.
The notions of recursive and r.e. subsets of A* are
remarkably robust.
Promising efforts to “prove” Church’s Thesis exist.
The Turing machines themselves can be identified with
finite strings in a fixed finite alphabet. Thus we can
consider
1. The set of all TM’s that halt at the empty input string.
2. The set of all TM’s that don’t halt at the empty input
string.
3. The set of all TM’s that halt at every input string.
THEOREM 1.3. Set 1 is r.e. but not recursive. Set 2 is cor.e. but not r.e. Set 3 is not a Boolean combination of
r.e. sets.
Again, the above two Theorems are discussed in virtually
any place that discusses Turing machines without resource
bounds (before computational complexity theory with
resource bounds).
How “hard” can membership in an r.e. set be?
To make sense of this, we need “recursive functions”.
Any TM provides a partial function f:A* Æ A* as follows. If
TM does not halt with input x Œ A*, then f(x) is undefined.
If TM does halt with input x Œ A* then, after halting, read
the input symbols from left to right, skipping over the
auxiliary symbols.
These partial f:A* Æ A* are called the “partial recursive
functions”. These f:A* Æ A* (i.e., dom(f) = A*) are called
the “recursive functions”.
Let S,T Õ A*. We write S £r T if and only if there exists a
recursive f:A* Æ A* such that
x Œ S ´ f(x) Œ T. 5
THEOREM 1.4. There exists an r.e. T Õ A* such that for all
r.e. S Õ A*, S £r T. Set 1 above has this property. There
exists a nonrecursive r.e. T Õ A* without this property.
Yet again, the above two Theorems are discussed in
virtually any place that discusses Turing machines without
resource bounds (before computational complexity theory
with resource bounds).
We call this property r.e. completeness. It obviously
implies nonrecursiveness (undecidability).
THEOREM 1.5. Let S,T Õ A* be complete r.e. There exists a
recursive permutation f:A* Æ A* such that f[S] = T.
See [My55], [Ro67], [Od89].
2. SOME R.E. COMPLETENESS.
The most familiar sets of strings, integers, rationals,
finite sequences of integers, or rationals, are easily
recursive.
But here is a simple example of a set of integers not known
to be recursive.
S = {nm: n,m prime}.
By standard techniques, S is r.e.
Here is an example from group theory.
Let E be a finite set of group equations in distinct
letters g1,...,gn. We ask:
for all groups G and g1,...,gn Œ G
obeying these equations,
do we have g1 = ... = gn = e?
This question asks if the group presented by the equations
is trivial.
THEOREM 2.1. This triviality problem is complete r.e.
Note that the data can live naturally in {0,1,•,,e,;,=,(,)}. Here strings of 0’s and 1’s for the 6
generators, • for the multiplication,  for the inverse, e
for the identity, and ; to separate the equations.
See [No55], [Bo59], [BDL73].
Here is a closely related example from topology.
One can construct from a group presentation of G, a 2dimensional simplicial complex whose fundamental group is
G. Since a simplicial complex is simply connected if and
only if its fundamental group is trivial, we have
it is not recursive (undecidable) whether a given 2dimensional simplicial complex is simply connected.
See [Ha73]. For further results along these lines, see,
e.g., [Mar58], [BCL73], and [BHP68].
We now discuss the existence of zeros.
THEOREM 2.2. There is an algorithm for deciding whether a
given polynomial in several variables with integer
coefficients has a zero in the reals.
See [Ta51], [Re92a], [Re92b], [Re92c], [BPR03].
THEOREM 2.3. There is no algorithm for deciding whether a
given polynomial in several variables with integer
coefficients has a zero in the integers. This decision
problem is complete r.e.
The proof was completed in 1970. This is sometimes called
the MRDP theorem, as [Ma70] relied substantially on earlier
efforts of three others. See [DPR61], [Ma70], and
expositions in [Da73], [Ma93]. Also see [Da77]. Also see
the discussion in [Ma93] concerning attempts to limit the
number of variables and the degree needed for this negative
result. Our knowledge is extremely poor in this regard.
WIDE OPEN. Is there an algorithm for deciding whether a
given polynomial in several variables with integer
coefficients has a zero in the rationals?
Now let W be the least class of functions in one variable,
that can be written in terms of 1,+,,•,sin and variables. 7
THEOREM 2.4. There is no algorithm for deciding whether a
given function from W has a zero in the reals. This problem
is complete r.e.
Ranges of polynomials are particularly relevant.
THEOREM 2.5. Let A Õ N. A is r.e. iff A = P[Zk] « N for
some polynomial P of k variables with integer coefficients.
THEOREM 2.6. If P has k variables and degree £ 2, with
integer coefficients, then P[Zk] is recursive.
This follows from [Sie72] on quadratic Diophantine
equations.
Even Theorem 2.3 is wide open for 2 variable cubics.
3. TIME, SPACE, NONDETERMINISM.
There are many sources that discuss Turing machines with
resource bounds. The key phrase to search for is
“complexity theory” or “computational complexity theory”.
One can be more specific with “discrete computational
complexity theory”, as there is a continuous theory that is
a horse of a different color. For discrete complexity
theory, see [Co71], [AHU74], [HU79], [GJ79], [Pa94],
[CLR01], [Si05], [Co], and many other places.
A TM is said to run in polynomial time if and only if there
exists n ≥ 1 such that at all input strings x,
TM cannot make > (x+2)n moves
where x is the length of x.
(We make this definition in this form so that we don’t have
to change it for the nondeterministic case. In this
deterministic case, this simply means that at all x, TM
makes £ (x+2)n moves).
A TM is said to run in polynomial space if and only if
there exists n ≥ 1 such that at all input strings x,
TM cannot visit > (x+2)n squares of tape.
(We make this definition in this form so that we don’t have
to change it for the nondeterministic case. In this 8
deterministic case, this simply means that at all x, TM
visits £ (x+2)n squares).
Let S Õ A*. We say that S is decidable in polynomial time
if and only if there is a TM running in polynomial time,
with input alphabet A, where S = {x Œ A*: x is accepted by
TM}.
We say that S is decidable in polynomial space if and only
if there is a TM running in polynomial space, with input
alphabet A, such that S = {x Œ A*: x is accepted by TM}.
WIDE OPEN. If S is decidable in polynomial space then is S
decidable in polynomial time?
P consists of all decision problems solvable in polynomial
time. PSPACE consists of all decision problems solvable in
polynomial space.
P = PSPACE?
is wide open.
Recall the TM instructions
If the reading head sees symbol x and the TM is in state q,
then replace x with symbol x’, put TM into state q’, and
move the reading head one square to the left (right).
We required: exactly one such instruction for each symbol
and state, other than q1,q2.
In a NDTM, we only require that there be at least one such
instruction for each symbol and state, other than q1,q2.
A NDTM is initialized in the same way as a TM with input
string x, but it can choose which of possibly many
applicable instructions to follow at each stage.
It is an old classical result that if we use NDTM’s instead
of TM’s in our definition of recursive and r.e. sets, then
we will have the same recursive and r.e. sets.
However, once resource bounds are placed on computation,
nondeterminism versus determinism becomes a very difficult
issue that, in many contexts, remains entirely unresolved. 9
NDTM runs in polynomial time if and only if there exists n ≥
1 such that at all input strings x,
NDTM cannot make > (x+2)n moves.
NDTM runs in polynomial space if and only if there exists n
≥ 1 such that at all input strings x,
NDTM cannot visit > (x+2)n squares of tape.
NDTM accepts the input string x iff it can reach state q1
following some path of instructions.
NDTM rejects the input string x if and only if it can reach
state q2 following some path of instructions.
Let S Õ A*. We say that S is decidable in nondeterministic
polynomial time if and only if there is an NDTM running in
polynomial time with input alphabet A such that S = {x Œ
A*: x is accepted by NDTM}.
We say that S is decidable in nondeterministic polynomial
space if and only if there is an NDTM running in polynomial
space with input alphabet A such that S = {x Œ A*: x is
accepted by NDTM}.
NP consists of all decision problems solved in
nondeterministic polynomial time.
NPSPACE consists of all decision problems solved in
nondeterministic polynomial space.
THEOREM 3.1. P Õ NP Õ PSPACE = NPSPACE.
See [Sa70], [HU79], [Pa94], [Si05].
Most experts believe that the two inclusions are proper. In
particular, that P ≠ NP.
I think most experts believe that it is substantially
easier to show P ≠ PSPACE than P ≠ NP, although even this
seems so hard that almost nobody admits they are working on
it.
How hard is it to solve an NP problem? To make sense of
this, we need polynomial time functions. 10
Any TM running in polynomial time provides a function f:A*
Æ A* as follows. Let x Œ A*. Run the TM until halting. Then
read the input symbols from left to right, skipping over
the auxiliary symbols.
These are the polynomial time computable functions.
Let S,T Õ A*. We write S £p T if and only if there exists a
polynomial time f:A* Æ A* such that
x Œ S ´ f(x) Œ T.
THEOREM 3.2. There exists an NP T Õ A* such that for all NP
S Õ A*, S £p T.
We call this property NP completeness.
Useful characterization of NP:
THEOREM 3.3. Let S Õ A*. Then S Œ NP if and only if there
exists T Õ A*, T Œ P, and n ≥ 1, such that S = {x:
($y)((x,y) Œ T Ÿ y £ (x+2)n)}.
See [Co].
THEOREM 3.4. If some NP complete problem is in P, then all
NP problems are in P.
So the NP complete problems stand and fall together.
THEOREM 3.5. Every NP, or even PSPACE problem can be solved
in deterministic exponential time. I.e., O(2n^k), for some
k.
This is a straightforward counting argument, where the
exponential deterministic algorithm searches over the
relevant configurations.
4. SOME NP COMPLETENESS.
nSAT, n ≥ 1.
±p11 ⁄ ... ⁄ ±p1n
±p21 ⁄ ... ⁄ ±p2n
...
±pk1 ⁄ ... ⁄ ±pkn. 11
In the above, there are k clauses, each of length n.
Repetitions of p’s are of course expected.
Each pij is px for some bit string x.
Is there a truth assignment to the p’s so that all clauses
come out true?
THEOREM 4.1. nSAT is NP complete if n ≥ 3. 2SAT is in P.
See [Co71], [AHU74], [HU79], [GJ79], [Pa94], [CLR01],
[Si05], [Co]..
We now turn to graph theory.
G = (V,E), where V is a finite set of vertices, and E is a
set of unordered pairs from V (edges).
kcolorability: Is there a coloring of the vertices of G
with at most k colors?
THEOREM 4.2. kCOL is NP complete if k ≥ 3. 2COL is in P.
See [Ka72], [CLR01].
CLIQUE: Does G contain a clique of size k?
Here k is NOT fixed, and is part of the data. Obviously we
can assume that k is at most the number of vertices of G,
and so it is unimportant whether k is given in unary or
binary.
THEOREM 4.3. CLIQUE is NP complete. For any fixed clique
size k, it is in P.
See [Ka72], [GJ79], [CLR01].
SUBISO: Given two graphs G,H, is G isomorphic to a subgraph
of H?
THEOREM 4.4. SUBISO is NP complete.
See [Co71].
TRAVSALES. Given m cities, with positive integer distance
function between each unordered pair of cities, and 12
positive integer n. Is there a tour of the cities where the
total distance traveled is £ n?
Here n and distances are given in binary.
THEOREM 4.5. TRAVSALES is NP complete.
See [GJ79], [CLR01].
GRAPHISO. Given two graphs, are they isomorphic?
THEOREM 4.6. GRAPHISO is in NP. Not known if it is NP
complete, or if it is in P.
THEOREM 4.7. GRAPHISO for planar graphs is in P. For all k ≥
1, GRAPHISO for graphs of valence £ k is in P.
See [HW74] and [Lu82].
Now for some number theory.
THEOREM 4.8. “n is prime”, n given in binary, is in P.
See [AKS04] and [Di04].
QUADCONG. Given integers a,b,c > 0, is there an integer 0 <
x < c such that x2 ≡ a mod b?
Here a,b,c given in binary.
THEOREM 4.9. QUADCONG is NP complete. In P for c = •.
See [MA78].
QUADEQN. Given integers a,b,c > 0, are there integers x,y >
0 such that ax2 + by = c?
THEOREM 4.10. QUADEQN is NP complete.
See [MA78].
SUBSUM. Given integers a1,…, an,b > 0, is b the sum of
distinct a’s?
THEOREM 4.11. SUBSUM is NP complete.
See [Ka72], [CLR01]. 13 SUBPROD. Given integers a1,...,an,b > 0, is b the product of
distinct a’s?
THEOREM 4.12. SUBPROD is NP complete.
See [Ya78].
We now come to factoring. Here we do not have a decision
problem.
FACTORING. Given integer n > 0, find a nontrivial factor of
n if it exists.
THEOREM 4.13. If P = NP then factoring can be done in
polynomial time.
“Factoring in polynomial time” means: there exists a
factoring function computable in polynomial time. I.e., an
f:Z Æ Z such that if n > 0 is composite then f(n) > 0 is a
nontrivial factor of n.
WIDE OPEN. Can factoring be done in polynomial time? Is
factoring NP complete (in an appropriate sense)?
A suitably practical algorithm for factoring would destroy
a number of fashionable cryptographic protocols.
See [CLR01], p. 834836.
5. OPINIONS.
William Gasarch published an opinion poll on P = NP in
[Ga02]. Gasarch reported on 100 responses.
The median guess as to when it would be solved was around
2050.
61 thought P ≠ NP.
9 thought P = NP.
4 thought it could not be resolved in ZFC.
3 stated that it could be resolved using very concrete
methods.
22 offered no opinion.
Gasarch writes in [Ga02]: 14
“Of those who answered, most thought P ≠ NP. Most of the 9
who thought that P = NP were respectable members of the
community. All of them recognized that their opinion is a
minority viewpoint. A few even said they took it just to be
contrary. However, the fact that 22 people did not venture
an opinion indicates more uncertainty on this then one
would have thought”.
Here are some quotes published by Gasarch [Ga02] from some
very well known people, plus me.
Bela Bollobas: 2020, P = NP. I think that in this respect I
am on the loony fringe of the mathematical community. I I
think (not too strongly) that P = NP and this will be
proved within twenty years. Some years ago, Charles Read
and I worked on it quite a bit, and we even had a
celebratory dinner in a good restaurant before we found an
absolutely fatal mistake. I would not be astonished if very
clever geometric and combinatorial techniques gave the
result, without discovering revolutionary new tools. A bit
like Tim Gowers’s solutions to major sixtyyear old
questions of Banach. ... Sadly, we haven’t returned to the
P vs NP question since that unfortunate experience fifteen
years ago. The danger of wasting a year for no return is
rather offputting.
John Conway: 2030, P ≠ NP. In my opinion this shouldn’t
really be a hard problem; it’s just that we came late to
this theory, and haven’t yet developed any techniques for
proving computations to be hard. Eventually, it will just
be a footnote in the books.
Yuri Gurevich: 2060, P = NP. The positive solution will not
make the standard NP problems worstcase feasible in the
practical sense of the word.
Richard Karp: P ≠ NP. My intuitive belief is that P is
unequal to NP, but the only supporting arguments I can
offer are the failure of all efforts to place specific NPcomplete problems in P by constructing polynomialtime
algorithms. I believe that the traditional proof techniques
will not suffice. Something entirely novel will be
required. My hunch is that the problem will be solved by a
young researcher who is not encumbered by too much
conventional wisdom about how to attack the problem. 15
Donald Knuth: It will be solved by either 2048 or 4096. I
am currently somewhat pessimistic. The outcome will be the
truly worst case scenario: namely that someone will prove
“P = NP because there are only finitely many obstructions
to the opposite hypothesis”; hence there will exist a
polynomial time solution to SAT but we will never know its
complexity!
Laszlo Lovasz: 2017, P ≠ NP. Probably some new math
modeling the information flow through a Boolean circuit.
With luck, something like algebraic topology or algebraic
geometry will be used.
Bob Tarjan: In my view, there is no way to even make
intelligent guesses about the answer to any of these
questions. If I had to bet now, I would bet that P is not
equal to NP. I estimate the halflife of this problem at
2550 more years, but I wouldn’t bet on it being solved
before 2100. Its solution will require unforeseen new
techniques.
Jeff Ullman: 2100, P ≠ NP. I think the problem is
comparable to some of the great problems of mathematics
that lasted hundreds of years, e.g., the 4color theorem.
Thus, I’d guess 100 years. I’d bet we don’t have the
techniques, or even names for the techniques today. Again,
that would be analogous to the situation for many of the
great open problems of mathematics 30 years after they were
posed.
Avi Wigderson: I think this project is a bit premature. I
think we know too little of what is relevant to even guess
answers to your questions... The only thing I can
definitely say, is that it is one of the most important and
interesting questions ever asked by humans, and more people
and resources should participate in filling up the holes
that would allow better guesses of answers to your
questions.
Harvey Friedman: 2050, P ≠ NP. Detailed combinatorial work
on easier problems, leading up to the full result. P =
PSPACE will be refuted first.
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