The Link Between Bounded Halting And Factor Finding
Defining the Bounded Halting Problem
The bounded halting problem refers to determining if a Turing machine will halt within a set number of computation steps on a given input. Formally, we define the bounded halting problem as follows: Given a Turing machine M, input w, and computation bound t(n), where t is a function of the input length n, determine if M halts on w within t(n) steps.
As an example, consider the following simple Turing machine that determines if a binary input string contains more 0s or 1s:
q0: Read input symbol If 0, transition to state q1 If 1, transition to state q2 q1: Increment counter for 0s Move head right Transition to state q0 q2: Increment counter for 1s Move head right Transition to state q0 q3: Halt
If we set a computation bound of n steps for an input of length n, this Turing machine will always halt within the allotted computation. However, for a general Turing machine, determining if it halts within a specified number of steps is a challenging problem.
The Computational Complexity of Bounded Halting
The bounded halting problem is complete for the complexity class NTIME(t(n)). This means it is among the hardest problems that can be solved by a Turing machine using O(t(n)) computation time. Some key complexity classes related to bounded halting include:
- P – Polynomial time. Problems solvable in nO(1) time.
- EXPTIME – Exponential time. Problems solvable in 2O(n) time.
- NEXPTIME – Nondeterministic exponential time. Problems with nondeterministic solutions verifiable in exponential time.
Several important problems have been shown to be complete for NEXPTIME under linear-time reductions, most notably the tiling problem. Intuitively, this means that the hardest problems in NEXPTIME are no more difficult than the bounded halting problem.
Relating Bounded Halting to Integer Factoring
Integer factoring refers to the decomposition of a composite integer into its prime factors. For example, factoring 15 = 3 x 5. The presumed difficulty of integer factoring is fundamental to modern cryptography and the security of systems like RSA.
There exists a linear-time many-one reduction from the bounded halting problem for NEXPTIME Turing machines to integer factoring. This means any instance of bounded halting can be transformed into an equivalent instance of integer factoring in linear time.
More formally: There exists a function f computable in O(n) time, such that a Turing machine M halts in time t(n) on input w if and only if f(M, w) = z for some composite integer z with factors revealing whether M halts on w in time.
Thus, integer factoring is at least as hard as the challenging problem of determining if arbitrary Turing machines halt, implying it is very difficult as well.
Implications for Computational Complexity
This reduction shows integer factoring resides in a rare complexity class between P problems and problems like circuit minimization that require exponential time. Resolving whether integer factoring is truly harder than P would resolve the P vs NP problem – a major open question in computer science.
Additionally, if integer factoring could be solved in polynomial time, NEXPTIME would collapse to just exponential time. The relationship between these classes remains a significant open problem closely tied now to the difficulty of integer factoring.
Practical Applications
The RSA cryptosystem, used pervasively in securing internet communications protocols like HTTPS, relies on the challenge of factoring large integers. An efficient factoring algorithm could break most RSA implementations. Hence, the link between factoring and bounded halting problems suggests RSA may be secure.
However, increased knowledge of the complexity landscape for factoring could motivate development of new cryptographic primitives. Classically held assumptions in cryptography are frequently evolving, so reduced reliance on specific problems like factoring is advantageous.
