El Aleph, Or a Monster Lurks in the Belly of Computer Science

Abstract

In 1962 Tibor Radó uncovered a monster hiding in the midst of innocent-looking Turing machines.

Appendices

Appendix

Technicalities Galore

We now definitively change the language to best operate the weapons we have to try to tame, or at least to control, our monster functions. And we will now dive into the obscurities of a formal language. Portions of these technical details have already been presented in Costa and Doria (2016).

We deal here some possible formalizations for $P=NP$ and $P<NP$; we have called the unusual formalizations (there are infinitely many) the “exotic formalizations.” They are intuitively equivalent, but if we formalize things there are difficulties to be dealt with when trying to establish their equivalence.

Remark 1

The reason we actually have such a plethora of (not always equivalent) definitions has to do with the fact that when we say we have a polynomial bound we are talking about some polynomial bound, and not, say, a minimal bound. That apparently minor fact is the source of many complications, as we will soon see. $\quad \square$

Let $t_m(x)$ be the primitive recursive function that gives the operation time of $\{m\}$ over an input x of length |x|.

Recall that the operation time of a Turing machine is given as follows: if $\{m \}$ stops over an input x, then:

$$t_{m}(x) = |x| + [\text {number of cycles of the machine until it stops}].$$

$t_m$ is primitive recursive and can in fact be defined out of Kleene’s T predicate.

Definition 2

(Standard formalization for $P=NP$.)

$[P=NP]\leftrightarrow _\mathrm{Def}\exists m,a\in \omega \, \forall x\in \omega \,[(t_{m}(x)\le |x|^a + a)\wedge \,R(x,m)]$. $\quad \square$

R(x, y) is a polynomial predicate; as its interpretation we can say that it formalizes a kind of “verifying machine” that checks whether or not x is satisfied by the output of $\{m\}$. (There is an equivalent formalization for $[P=NP]$ where again one uses Kleene’s T predicate to get the time measure $t_m$.)

Definition 3

$[P<NP]\leftrightarrow _\mathrm{Def}\lnot [P=NP]$. $\quad \square$

Now suppose that $\{e_\mathsf{f} \}= \mathsf{f}$ is total recursive and strictly increasing:

Remark 4

The naïve version for the exotic formalization is:

$$[P=NP]^\mathsf{f}\leftrightarrow \exists m\in \omega ,a\,\forall x\in \omega \,[(t_{m}(x)\le |x|^{\mathsf{f}(a)} + \mathsf{f}(a))\wedge \,R(x,m)].$$

However as we will soon see, there is no reason why we should ask that $\mathsf{f}$ be total; on the contrary, there will be interesting situations where such a function may be partial and yet it may provide a reasonable exotic formalization for $P<NP$ (Guillaume 2003). $\quad \square$

So, for the next definitions and results let $\mathsf{f}$ be in general a (possibly partial) recursive function which is strictly increasing over its domain, and let $e_\mathsf{f}$ be the Gödel number of an algorithm that computes $\mathsf{f}$. Let $p(\langle e_\mathsf{f},b,c\rangle , x_1,$ $x_2,\ldots , x_k)$ be an universal Diophantine polynomial with parameters $e_\mathsf{f},b,c$; that polynomial has integer roots if and only if $\{e_f\}(b) = c$. We may if needed suppose that polynomial to be ${\ge }0$. We omit the “$\in \omega$” in the quantifiers, since they all refer to natural numbers.

Definition 5

$M_\mathsf{f}(x, y)\leftrightarrow _\mathrm{Def}\exists \,x_1,\ldots , x_k\,[p(\langle e_\mathsf{f}, x, y\rangle , x_1,\ldots , x_k) = 0]$. $\quad \square$

Actually $M_\mathsf{f}(x, y)$ stands for $M_{e_\mathsf{f}}(x, y)$, or better, $M(e_\mathsf{f}, x, y)$, as dependence is on the Gödel number $e_\mathsf{f}$.

Definition 6

$\lnot Q(m, a, x)\leftrightarrow _\mathrm{Def} [(t_{m}(x)\le |x|^a + a)\rightarrow \,\lnot R(x,m)]$. $\quad \square$

Proposition 7

(standard formalization, again.)

$$[P<NP]\leftrightarrow \,\forall m, a\,\exists x\,\lnot Q(m, a, x). $$

$\quad \square$

Definition 8

$\lnot Q_\mathsf{f}(m, a, x)\leftrightarrow _\mathrm{Def}\exists a'\,[M_\mathsf{f}(a, a')\wedge \lnot Q(m, a', x)]$. $\quad \square$

Remark 9

We will sometimes write $\lnot Q (m, \mathsf{f}(a), x)$ for $\lnot Q_\mathsf{f}(m, a,$ x), whenever $\mathsf{f}$ is—safely, in some sense—total. $\quad \square$

Definition 10

(Exotic formalization.)

$$[P<NP]^\mathsf{f}\leftrightarrow _\mathrm{Def}\,\forall m, a\,\exists x\,\lnot Q_\mathsf{f}(m, a, x).$$

$\quad \square$

Notice that again this is a $\Pi _2$ arithmetic sentence:

$$\forall m,a\,\exists x,a',x_1,\ldots , x_k\,\{[p(\langle e_\mathsf{f}, a, a'\rangle ,\ldots ,x_1,\ldots ,x_k)=0]\wedge \,\lnot Q(m, a',x)\}. $$

(Recall that Q is primitive recursive.)

Definition 11

$[P=NP]^\mathsf{f}\leftrightarrow _\mathrm{Def}\lnot [P<NP]^\mathsf{f}$. $\quad \square$

The Monster Shows Glimpses of Its Face

For the definition of Sat see Machtey and Young (1979); for the BGS recursive set of poly Turing machines see Baker et al. (1975). In a nutshell, sat is the set of all Boolean expressions in conjunctive normal form (cnf) that are satisfiable, and BGS is a recursive set of poly Turing machines that contains emulations of every conceivable poly Turing machines.

The full counterexample function f is defined as follows; let $\omega$ be also a set of codes for an enumeration of the Turing machines (see on what we mean by “standard coding,” Mendelson 1997, p. 320ff). Similarly we code by an analogous standard code sat onto $\omega$:

• If $n\in \omega$ isn’t a poly machine, $f(n) = 0$.

• If $n\in \omega$ codes a poly machine:

  • $f(n) =$ first instance x of Sat so that the machine fails to output a satisfying line for x, plus 1, that is, $f(n) = x + 1$.

  • Otherwise f(n) is undefined, that is, if $P=NP$ holds for $n, f(n) =$ undefined.

As defined, f is non computable. It will also turn out to be at least as fast growing as the Busy Beaver function, since in its peaks it tops all intuitively total recursive functions.

The idea in the proof of that fact goes as follows:

• Use the s–m–n theorem to obtain Gödel numbers for an infinite family of “quasi-trivial machines”—soon to be defined. The table for those Turing machines involves very large numbers, and the goal is to get a compact code for that value in each quasi-trivial machine so that their Gödel numbers are in a sequence $\mathsf{g}(0), \mathsf{g}(1), \mathsf{g}(2), \ldots$, where $\mathsf{g}$ is primitive recursive.

• Then add the required clocks as in the BGS sequence of poly machines, and get the Gödel numbers for the pairs machine $+$ clock. We can embed the sequence we obtain into the sequence of all Turing machines.

• Notice that the subsets of poly machines we are dealing with are (intuitive) recursive subsets of the set of all Turing machines. More precisely: if we formalize everything in some theory S, then the formalized version of the sentence “the set of Gödel numbers for these quasi-trivial Turing machines is a recursive subset of the set of Gödel numbers for Turing machines” holds of the standard model for arithmetic in S, and vice versa.

  However S may not be able to prove or disprove that assertion, that is to say, it will be formally independent of S.

• We can thus define the counterexample functions over the desired set(s) of poly machines, and compare them to fast-growing total recursive functions over similar restrictions.

Recall:

Definition 12

For $f,g:\omega \rightarrow \omega$,

$$f\;\mathrm {\mathbf {dominates}}\;g\leftrightarrow _\mathrm{Def} \exists y\,\forall x\,(x > y\rightarrow f(x) \ge g(x)).$$

We write $f\succ g$ for f dominates g. $\quad \square$

Quasi-trivial Machines

Recall that the operation time of a Turing machine is given as follows: if M stops over an input x, then the operation time over x,

$$t_\mathsf{M} = |x| + \text {number}~\text {of}~\text {cycles}~\text {of}~\text {the}~\text {machine}~\text {until}~\text {it}~\text {stops}.$$

Example 13

• First trivial machine. Note it $\mathsf{O}$. $\mathsf{O}$ inputs x and stops.

  $$t_\mathsf{O} = |x| + \text {moves}~\text {to}~\text {halting}~\text {state}~+ \text {stops}.$$

  So, operation time of $\mathsf{O}$ has a linear bound.

• Second trivial machine. Call it $\mathsf{O}'$. It inputs x, always outputs 0 (zero) and stops. Again operation time of $\mathsf{O}'$ has a linear bound.

• Quasi-trivial machines. A quasi-trivial machine $\mathsf{Q}$ operates as follows: for $x\le x_0$, $x_0$ a constant, fixed value, $\mathsf{Q} = \mathsf{R}$, $\mathsf{R}$ an arbitrary total machine. For $x > x_0$, $\mathsf{Q} = \mathsf{O}$ or $\mathsf{O'}$. This machine has also a linear bound. $\quad \square$

Remark 14

Now let $\mathsf{H}$ be any fast-growing, superexponential total machine. Also let $\mathsf{H}'$ be a total Turing machine. Form the following family $\mathsf{Q}_{\ldots }$ of quasi-trivial Turing machines with subroutines $\mathsf{H}$ and $\mathsf{H}'$:

• 1. If $x\le \mathsf{H}(n)$, $\mathsf{Q}^{\mathsf{H},\mathsf{H'},n}(x) = \mathsf{H}'(x)$;

• 2. If $x > \mathsf{H}(n)$, $\mathsf{Q}^{\mathsf{H},\mathsf{H'},n}(x) = 0$. $\quad \square$

Proposition 15

There is a family $\mathsf{R}_{\mathsf{g}(n, |\mathsf{H}|,|\mathsf{H}'|)}(x) = \mathsf{Q}^{\mathsf{H},\mathsf{H'},n}(x)$, where $\mathsf{g}$ is primitive recursive, and $|\mathsf{H}|, |\mathsf{H}'|$ denotes the Gödel number of $\mathsf{H}$ and of $\mathsf{H}'$.

Proof

By the composition theorem and the s–m–n theorem. $\quad \square$

Remark 16

Very important! We are interested in quasi-trivial machines where $\mathsf{H}' = \mathsf{T}$, the standard truth-table exponential algorithm for sat. $\quad \square$

Notice that, for the counterexample function when defined over all Turing machines (with the extra condition that the counterexample function $= 0$ if $\mathsf{M}_m$ isn’t a poly machine), we have:

Proposition 17

If $\mathsf{g}(n)$ is the Gödel number of a quasi-trivial machine as in Remark 14, then $f(\mathsf{g}(n)) = \mathsf{H}'(n) + 1$.

Proof

Use the machines in Proposition 15 and Remark 16. $\quad \square$

That Hideous Strength ...

Our goal here is to prove the following result: no total recursive function dominates f.

Remark 18

We sketch below the idea of the proof. Suppose that there is a total recursive function $\mathsf{h}(n)$ that dominates f. Get a total recursive $\mathsf{k}(n)$ that dominates $\mathsf h$ and so that the relative growth speed of $\mathsf{k}$ with respect to $\mathsf h$ is faster that any primitive recursive function.

Why do we need such a condition? We use the quasi-trivial machines to reproduce $\mathsf k$ within f, that is, we (sort of) replicate function $\langle n, \mathsf{k}(n)\rangle$ within f by a sequence of machines with Gödel numbers $N(n), n = 0, 1, 2, \ldots$ (see above Proposition 17), where N is primitive recursive, so that we have that $\mathsf k$ becomes the sequence of machines $N(n), n = 0, 1, 2, \ldots$, and we get the value of f at $\mathsf k$ as $\langle N(n), \mathsf{k}(n)+ 1\rangle$ with $f(N(n))=\mathsf{k}(n)+1$.

As N can be taken to be primitive recursive, monotonic increasing on n, it slows down the growth of $\mathsf k$ by a primitive recursive function. Given our construction—which is trivially fulfilled—we have that f still overtakes $\mathsf h$ infinitely many times, as $\mathsf k$ grows faster than $\mathsf h$, and we are done. $\quad \square$

One side comment: f is extremely complex, as it contains infinitely many copies of itself (if we suppose that f is total). Just for starters... Mr. K pointed to us the fast growing property of f but gave no proof of the fact. When we (da Costa and I) first managed to sketch a proof of f’s fast growing behavior, I was shocked, and sent an email to K, with a comment, “I would never expect that out of a pedestrian, seemingly naïve question one would have to confront such a crazy growth.”

He answered me, like Darth Vader explaining the rituals surrounding the Force to Luke Skywalker, “this kind of problem has a sacred status, and should only be approached by very few people, by the high priests of the cult.”

That is, I was a heretic on the verge of facing the stake.

Proposition 19

For no total recursive function $\mathsf{h}$ does $\mathsf{h}\succ f$.

Proof

Suppose that there is a total recursive function $\mathsf{h}$ such that $\mathsf{h}\succ f$.

Notice:

• Given such a function $\mathsf{h}$, we can obtain another total recursive function $\mathsf{h}'$ which satisfies:

  • 1. $\mathsf{h}'$ is strictly increasing.

  • 2. For $n > n_0$, $\mathsf{h}'(n) > \mathsf{h}(\mathsf{g}(n))$, with $\mathsf g$ as in Proposition 17. $\quad \square$

• Given a total recursive $\mathsf{h}$, there is a total recursive $\mathsf{h}'$ that satisfies the previous conditions.

For given $\mathsf{h}$, we obtain out of that total recursive function by the usual constructions a strictly increasing total recursive $\mathsf{h}^*$. Then if, for instance, $\mathsf{F}_{\omega }$ is Ackermann’s function, $\mathsf{h}'=\mathsf{h}^*\circ \mathsf{F}_{\omega }$ will do. (The idea is that $\mathsf{F}_{\omega }$ dominates all primitive recursive functions, and therefore $\mathsf{h}^*$ composed with it dominates $\mathsf{g}(n)$.)

We have that the Gödel numbers of the quasi-trivial machines $\mathsf{Q}$ are given by $\mathsf{g}(n)$. Choose adequate quasi-trivial machines, so that $f(\mathsf{g}(n)) = \mathsf{h}'(n) + 1$, from Proposition 17. We now conclude our argument. If we make explicit the computations, for $\mathsf{g}(n)$ (as the argument holds for any strictly increasing primitive recursive $\mathsf{g}$):

$$f(\mathsf{g}(n)) = \mathsf{h}'(n) + 1 = \mathsf{h}^*(\mathsf{F}_{\omega }(n)) + 1,$$

and

$$\mathsf{h}^*(\mathsf{F}_{\omega }(n)) > \mathsf{h}^*(\mathsf{g}(n)).$$

For $N = \mathsf{g}(n)$,

$$f(N) > \mathsf{h}^*(N)\ge \mathsf{h}(N), \text{ all } \text{ N }.$$

Therefore no such $\mathsf{h}$ can dominate f. $\quad \square$

Corollary 20

No total recursive function dominates f. $\quad \square$

BGS-Like Sets

We use here the BGS (Baker et al. 1975) set of poly machines:

$$\langle \mathsf{M}_m, |x|^{a} + a\rangle ,$$

where we couple a Turing machine $\mathsf{M}_m$ to a clock regulated by the polynomial $|x|^a + a$, that is, it stops $\mathsf{M}_m$ after $|x|^a + a$ steps in the operation over x, where x is the machine’s binary input and |x| its bit-length.

A more general machine-clock couple will also be dealt with here:

$$\langle \mathsf{M}_m, |x|^{\mathsf{}(a)} + \mathsf{f }(a)\rangle \mapsto \mathsf{M}_{\mathsf{c}(m, |\mathsf{f}|,a)},$$

Its Gödel number is given by $\mathsf{c}(m,|\mathsf{f}|,a)$, with $\mathsf{c}$ primitive recursive by the s–m–n theorem.

Remark 21

Notice that we can have $\mathsf c$ such that, for parameters a, b, if $a<b$, then $\mathsf{c}(\ldots a \ldots ) < \mathsf{c}(\ldots b \ldots )$. $\quad \square$

$P<NP$ is given by a $\Pi _2$ arithmetic sentence, that is, a sentence of the form “for every x there is an y so that P(x, y),” where P(x, y) is a very simple kind of relation.⁵ Now given a theory S with enough arithmetic in it, S proves a $\Pi _2$ sentence $\xi$ if and only if the associated Skolem function $\mathsf{f}_{\xi }$ is proved to be total recursive by S. For $P<NP$, the Skolem function is what we have been calling the counterexample function.

However there are infinitely many counterexample functions we may consider, an embarras de choix, as they say in French. Why is it so? For many adequate, reasonable theories S, we can build a recursive (computable) scale of functions ⁶ $\mathsf{f}_0, \mathsf{f}_1, \ldots , \mathsf{f}_k, \ldots$ with an infinite set of S-rovably total recursive functions so that $\mathsf{f}_0$ is dominated by $\mathsf{f}_1$ which is then dominated by $\mathsf{f}_2$, ..., and so on, up to the corresponding function $\mathsf{F}_S$.

Given each function $\mathsf{f}_k$, we can form a BGS-like set $\mathrm{BGS}^k$, where clocks in the time-polynomial Turing machines are bounded by a polynomial:

$$|x|^{\mathsf{f}_{k}(n)} + \mathsf{f}_{k}(n),$$

where |x| denotes the length of the binary input x to the machine. We can then consider the recursive set:

$$\bigcup _k \mathrm{BGS}^k$$

of all such sets.

Each $\mathrm{BGS}^k$ contains representatives of all poly machines (time polynomial Turing machines). Now, what happens if:

• There is a function $\mathsf{g}$ which is total provably recursive in S and which dominates all segments $\mathsf{f}_k$ of counterexample functions over each $\mathrm{BGS}^k$.

• There is no such an $\mathsf{g}$, but there are functions $\mathsf{g}_k$ which dominate each particular $\mathsf{f}_k$, while the sequence $\mathsf{g}_0, \mathsf{g}_1, \ldots$ is unbounded in S, that is, grows as the sequence $\mathsf{F}_0, \mathsf{F}_1, \ldots$ in S?

We will take a look into these queries.

Exotic BGS$^\mathsf{F}$ Machines

Now let $\mathsf{F}$ be a fast growing, intuitively total, algorithmic function. We consider exotic BGS$^\mathsf{F}$ machines, that is, poly machines coded by the pairs $\langle m, a\rangle$, which code Turing machines $\mathsf{M}_m$ with bounds $|x|^{\mathsf{F}(a)} + \mathsf{F}(a)$. Since the bounding clock is also a Turing machine, now coupled to $\mathsf{M}_m$, there is a primitive recursive map $\mathsf{c}$ so that:

$$\langle \mathsf{M}_m, |x|^{\mathsf{F}(a)} + \mathsf{F}(a)\rangle \mapsto \mathsf{M}_{\mathsf{c}(m, |\mathsf{F}|, a)},$$

where $\mathsf{M}_{\mathsf{c}(m, |\mathsf{F}|, a)}$ is a poly machine within the sequence of all Turing machines. We similarly obtain a $\mathsf{g}$ as above, and follows:

Proposition 22

Given the counterexample function $\mathsf{f}_k$ defined over the $\mathrm{BGS}^{k}$-machines, for no ZFC-provable total recursive $\mathsf{h}$ does $\mathsf{h}\succ \mathsf{f}_k$.

Proof

As in Proposition 19; use Gödel number coding primitive recursive function $\mathsf{c}$ to give the Gödel numbers of the quasi-trivial machines we use in the proof. $\quad \square$

Remark 23

Notice that we have a perfectly reasonable formalization for our big question:

$$[P<NP]^k\,\leftrightarrow \,[P<NP]^{\mathsf{f}^k}.$$

Also, $S\vdash \,[P<NP]^k\,\leftrightarrow \,[\mathsf{f}_c^k \text { is total}]$. So our analysis will give estimates for the growth rate of each counterexample function $\mathsf{f}_c^k$.$\quad \square$

Remark 24

The previous statements have interesting consequences, which we will briefly pursue below. For the proof of the proposition choose a BGS$^k$ so that $\mathsf{f}_k$ dominates all strictly increasing fast growing provably total recursive functions that eventually appear in the proof. $\quad \square$

We can state, for total $\mathsf{f}_{c}^{k}$:

Proposition 25

For each j there is a k, $k>j+1$, so that S proves the sentence “$\mathsf{f}_k$ doesn’t dominate the BGS$^k$ counterexample function $\mathsf{f}_{c}^{k}$.” $\quad \square$

However we cannot conclude that “for all j, we have that...” since that would imply that S proves “for all j, $\mathsf{f}_j$ is total” as a scholium, which cannot be done (as that is equivalent to “$\mathsf{F}_S$ is total,” which again cannot be proved in S).

What can be concluded: let $S'$ be the theory S + $\mathsf{F}_S$ is total. Then:

Proposition 26

If S is consistent and if $\mathsf{f}_{c}^{k}$ is total in a model with standard arithmetic for each k, then $S'$ proves: there is no proof of the totality of $\mathsf{f}_{c}^{k}$, any k, in S.

Proof

See below the discussion. $\quad \square$

Remark 27

Notice that:

• $S'\vdash \, \forall k\,([P<NP]^k\,\leftrightarrow \,[\mathsf{f}_c^k \text { is total}])$, while S cannot prove it.

• $S'\vdash \, \forall k\, ([P<NP]^k\leftrightarrow \,[P<NP])$ while again S cannot prove it.

• $S'$ is S + [S is $\Sigma _1 -$ sound]. $\quad \square$

Remark 28

That means that we can conclude:

$S'$ proves that, for every k, S cannot prove $[P<NP]^k$.

Now : does the $[P<NP]^k$ adequately translate our main question? $\quad \square$

Remark 29

Notice that theory S + $\mathsf{F}_S$ is total is S + S is $\Sigma _1$-sound. This will have further consequences. $\quad \square$