Polynomial creativity

Intuitively, a set is creative when there is a polynomial-time algorithm that creates a counterexample for any candidate fast nondeterministic recognition algorithm for its complement.

The classes of fast nondeterministic recognition algorithms are formalized by Joseph and Young as the sets ${\displaystyle \mathrm {NP} ^{(k)}}$ of nondeterministic Turing machine programs ${\displaystyle p}$ that, for inputs ${\displaystyle x}$ that they accept, have an accepting path with a number of steps that is at most ${\displaystyle |p|(|x|^{k}+1)}$. This notation should be distinguished with that for the complexity class NP. The complexity class NP is a set of formal languages, while ${\displaystyle \mathrm {NP} ^{(k)}}$ is instead a set of programs that accept some of these languages. Every language in NP is recognized by a program in one of the sets ${\displaystyle \mathrm {NP} ^{(k)}}$, with a parameter ${\displaystyle k}$ that is (up to the factor ${\displaystyle |p|}$ in the bound on the number of steps) the exponent in the polynomial running time of the program.[2]

According to Joseph and Young's theory, a language ${\displaystyle L}$ in NP is ${\displaystyle k}$-creative if it is possible to find a witness showing that the complement of ${\displaystyle L}$ is not recognized by any program in ${\displaystyle \mathrm {NP} ^{(k)}}$. More formally, there should exist a polynomially computable function ${\displaystyle f}$ that maps programs in this class to inputs on which they fail. When given a nondeterministic program ${\displaystyle p}$ in ${\displaystyle \mathrm {NP} ^{(k)}}$, the function ${\displaystyle f}$ should produce an input string ${\displaystyle x=f(p)}$ that either belongs to ${\displaystyle L}$ and causes the program to accept ${\displaystyle x}$, or does not belong to ${\displaystyle L}$ and causes the program to reject ${\displaystyle x}$. The function ${\displaystyle f}$ is called a productive function for ${\displaystyle L}$. If this productive function exists, the given program does not produce the behavior on input ${\displaystyle x}$ that would be expected of a program for recognizing the complement of ${\displaystyle L}$.[2]

Joseph and Young define a polynomial-time function ${\displaystyle f}$ to be polynomially honest if its running time is at most a polynomial function of its output length. This disallows, for instance, functions that take polynomial time but produce outputs of less than polynomial length. As they show, every one-to-one polynomially-honest function ${\displaystyle f}$ is the productive function for a ${\displaystyle k}$-creative language ${\displaystyle K_{f}^{k}}$.[2]

Given ${\displaystyle f}$, Joseph and Young define ${\textstyle K_{f}^{k}}$ to be the set of values ${\displaystyle f(p)}$ for nondeterministic programs ${\displaystyle p}$ that have an accepting path for ${\displaystyle f(p)}$ using at most ${\textstyle |p|(|f(p)|^{k}+1)}$ steps. This number of steps (on that input) would be consistent with ${\displaystyle p}$ belonging to ${\textstyle \mathrm {NP} ^{(k)}}$. Then ${\displaystyle K_{f}^{k}}$ belongs to NP, for given an input ${\displaystyle f(p)}$ one can nondeterministically guess both ${\displaystyle p}$ and its accepting path, and then verify that the input equals ${\displaystyle f(p)}$ and that the path is valid for ${\displaystyle p}$.[2]

Language ${\textstyle K_{f}^{k}}$ is ${\displaystyle k}$-creative, with ${\displaystyle f}$ as its productive function, because every program ${\displaystyle p}$ in ${\displaystyle \mathrm {NP} ^{(k)}}$ is mapped by ${\displaystyle f}$ to a value ${\displaystyle f(p)}$ that is either accepted by ${\displaystyle p}$ (and therefore also belongs to ${\displaystyle K_{f}^{k}}$) or rejected by ${\displaystyle p}$ (and therefore also does not belong to ${\displaystyle K_{f}^{k}}$).[2]

Every ${\displaystyle k}$-creative set with a polynomially honest productive function is NP-complete. For any other language ${\displaystyle X}$ in NP, by the definition of NP, one can translate any input ${\displaystyle x}$ for ${\displaystyle X}$ into a nondeterministic program ${\displaystyle p_{x}}$ that ignores its own input and instead searches for a witness for ${\displaystyle x}$, accepting its input if it finds one and rejecting otherwise. The length of ${\displaystyle p_{x}}$ is polynomial in the size of ${\displaystyle x}$ and a padding argument can be used to make ${\displaystyle p_{x}}$ long enough (but still polynomial) for its running time to qualify for membership in ${\displaystyle \mathrm {NP} ^{(k)}}$. Let ${\displaystyle f}$ be the productive function used to define a given ${\displaystyle k}$-creative set ${\displaystyle L}$, and let ${\displaystyle g}$ be the translation from ${\displaystyle x}$ to ${\displaystyle p_{x}}$. Then the composition of ${\displaystyle g}$ with ${\displaystyle f}$ maps inputs of ${\displaystyle X}$ into counterexamples for the algorithms that test those inputs. This composition maps inputs that belong to ${\displaystyle X}$ into strings that belong to ${\displaystyle L}$, and inputs that do not belong to ${\displaystyle X}$ into strings that do not belong to ${\displaystyle L}$. Thus, it is a polynomial-time many-one reduction from ${\displaystyle X}$ to ${\displaystyle L}$. Since ${\displaystyle L}$ is (by definition) in NP, and every other language in NP has a reduction to it, it must be NP-complete.[2]

It is also possible to prove more strongly that there exists an invertible parsimonious reduction to the ${\displaystyle k}$-creative set.[2]

The Berman–Hartmanis conjecture states that there exists a polynomial-time isomorphism between any two NP-complete sets: a function that maps yes-instances of one such set one-to-one into yes-instances of the other, takes polynomial time, and whose inverse function can also be computed in polynomial time. It was formulated by Leonard C. Berman and Juris Hartmanis in 1977, based on the observation that all NP-complete sets known at that time were isomorphic. An equivalent formulation of the conjecture is that every NP-complete set is paddable. This means that there exists a polynomial-time and polynomial-time-invertible one-to-one transformation ${\displaystyle h(x,y)}$ from yes-instances ${\displaystyle x}$ to larger yes-instances that encode the "irrelevant" information ${\displaystyle y}$.[4]

However, it is unknown how to find such a padding transformation for a ${\displaystyle k}$-creative language whose productive function is not polynomial-time-invertible. Therefore, if one-way permutations exist, the ${\displaystyle k}$-creative languages having these permutations as their productive functions provide candidate counterexamples to the Berman–Hartmanis conjecture.[2]

The (unproven) Joseph–Young conjecture formalizes this reasoning. The conjecture states that there exists a one-way length-increasing function ${\displaystyle f}$ such that ${\displaystyle K_{f}^{k}}$ is not paddable.[2] Alan Selman observed that this would imply a simpler conjecture, the encrypted complete set conjecture: there exists a one-way function ${\displaystyle f}$ such that ${\displaystyle \mathrm {SAT} }$ (the set of yes-instances for the satisfiability problem) and ${\displaystyle f(\mathrm {SAT} )}$ are non-isomorphic.[5] There exists an oracle relative to which one-way functions exist, both of these conjectures are false, and the Berman–Hartmanis conjecture is true.[6]