4.1. Self-Consistent Computability
The ability of Single Symbol Time-Traveling Turing Machines (SSTTMs) to send tape symbols into the past introduces non-determinism and potential timeline bifurcations. We formalize the notion of a self-consistent computation and output, drawing inspiration from Novikov’s self-consistency principle, which states that the only possible time travel scenarios are those that do not lead to logical contradictions or paradoxes [
3].
Definition 1. Let M be an SSTTM, and let be the sequence of symbols sent to the past by M during its computation on input w. Sending each symbol results in either 0 or 1 new bifurcations in the computation history of M, denoted as:
Let be the set of all possible outputs printed on tape 2 from the r many bifurcations caused by sending symbols into the past during the computation of .
Then, the computation of is said to be self-consistent if:
, i.e., no symbols are sent to the past by , and halts normally with output .
, i.e., symbols are sent to the past, causing bifurcations, but . That is, all branches halt with the same output on tape 2. We consider this a halting computation with the common output.
If there exist l and m such that the outputs of halting branches are not equal, , then we consider this computation to be non-halting.
Intuitively, a self-consistent computation has only one possible output, despite any apparent non-determinism from time travel. This definition ensures that the computation respects Novikov’s self-consistency principle, avoiding paradoxes or contradictions that could arise from sending symbols back in time.
To illustrate these concepts, consider the following examples:
Example 1 (Consistent Computation). Let M be an SSTTM with the following behavior on input "01":
At step 3, M sends the symbol "1" back to step 1, overwriting the original "0".
M continues its computation and halts at step 5 with output "11".
This computation is self-consistent because there is only one possible output, "11", regardless of the bifurcation caused by sending the symbol back in time. The branch that starts with the original input "01" will also halt with output "11" after receiving the symbol "1" from the future.
Example 2 (Inconsistent Computation). Let M be an SSTTM with the following behavior on input "01":
At step 3, M sends the symbol "0" back to step 1, overwriting the original "0".
At step 4, M sends the symbol "1" back to step 2, overwriting the original "1".
The branch that starts with the original input "01" halts at step 5 with output "00".
The branch that receives the symbol "0" at step 1 halts at step 5 with output "01".
This computation is inconsistent because the two branches halt with different outputs, "00" and "01". The sending of symbols back in time has created a paradox, violating Novikov’s self-consistency principle.
We can now define the output of a self-consistent computation. For a given SSTTM M:
If M has a unique timeline with no time travel, the output is the tape 2 contents upon halting.
If M bifurcates timelines by time travel, the output is the common tape 2 contents across all branches when halting.
If M bifurcates timelines by time travel but some branches have different traditional outputs, we consider the self-consistent computation to be non-halting or entering an error or rejecting state.
Theorem 3. Checking that an output has been found in a self-consistent SSTTM computation within n finite steps can be finitely decided by a traditional Turing Machine within also a finite number of steps.
Proof. Let M be an SSTTM, and w be an input. We will construct a traditional Turing Machine T that can decide in finite time whether M has a consistent output on w across all bifurcations with n or fewer steps.
T will simulate the computation tree of M on w in a breadth-first manner, tracking tape contents at each timestep in each branch.
Formally, T operates as follows:
Create an initial configuration of M on input w.
Set list to track configurations.
-
While L is non-empty:
- (a)
Remove the first configuration C from L.
- (b)
Simulate one step of C to generate configurations .
- (c)
-
If C halts, compare the output tape to C’s previously saved output tape.
- (d)
Add all to the end of L.
If L empties with no rejection, accept.
Observe:
Each C depends only on its parent configuration, so T can correctly recreate C.
The branching factor of M is finite, so is at most exponential in runtime.
Therefore, as L is finite, T will eventually simulate all configurations of M on w and detect any inconsistencies or accept. As T only needs to track a finite number of configurations, this process terminates in finite time.
This shows T can decide finitely if M has a consistent output across bifurcations within finite n steps. □
This formalization of self-consistency, along with the concrete examples and the connection to Novikov’s self-consistency principle, provides a clearer understanding of what constitutes a valid output for an SSTTM computation involving time travel. The theorem and proof demonstrate that checking for self-consistency within a finite number of steps is decidable by a traditional Turing Machine, ensuring that the notion of self-consistent computability is well-defined and computable.
This also formalizes which solutions we consider valid outputs of an SSTTM computation involving time travel. Even with timeline bifurcations, a self-consistent output requires agreement between all branches. Consistency guarantees a well-defined result, avoiding paradoxes from contradictory outputs. The problem with self-consistent computations is that they do not seem computable to decide if you are on a halting state that is part of a self-consistent computation without simulating all the bifurcations, but we will see if we can prove this is computable in our model. This means that there is a Universal SSTTM that is properly utilizing the time-traveling internal state to simulate any of the possible SSTTMs.
Theorem 4. There exists a Universal SSTTM U that has a self-consistent output indicating whether a given SSTTM M has a self-consistent computation.
Proof. Let U be a Universal SSTTM, and M be an arbitrary SSTTM that U takes as input. We will show that U can generate a self-consistent output indicating whether M has a self-consistent computation on input w.
U simulates the computation tree of in a breadth-first manner, keeping track of tape contents across all branches.
Formally, the operation of U is:
Universal SSTTM U:
1. C_0 ← initial configuration of M(w)
2. L ← [C_0]
3. while L is not empty:
4. C ← L.dequeue()
5. C’ ← simulate_one_step(C)
6. if C halts:
7. o_C ← get_output_tape(C)
8. send_to_all_prior_branches(o_C)
9. for each branch point B:
10. if o_C = o_B:
11. continue
12. else:
13. halt in state q_inconsistent
14. L.enqueue_all(C’)
15. if L is empty:
16. halt in state q_consistent
We observe:
The branching factor of M is finite as it is an SSTTM. Thus, is at most exponential in the runtime, i.e., for some constant c and runtime t.
U can correctly recreate any configuration C from its parent.
As L is finite, U will eventually simulate all possible configurations of .
The number of simulation steps performed by U is bounded by the total number of configurations in the computation tree of . In the worst case, this is exponential in the runtime of M, i.e., for some constant c and runtime t. The size of each configuration is bounded by the size of M and the length of the input w, i.e., .
Therefore, as U simulates all branches and compares outputs at each branch point, it will definitively enter either if is self-consistent, or if not. The total runtime of U is bounded by , which is exponential in the runtime of M and linear in the size of M and w.
Thus, U generates a self-consistent output indicating if is self-consistent across all bifurcations. □
This demonstrates that consistency can be checked within the SSTTM model itself. The potentially unbounded branching is handled by harnessing time travel to share information across timelines. Also, the potentially unbounded branching makes the consistency of these computations, which is akin to a Halting Problem, undecidable by a traditional deterministic Turing Machine. We can simulate time bifurcations in a traditional Turing Machine by exploring them either in a depth-first or breadth-first manner and using the dovetailing technique, but we can prove that the number of bifurcations can be unbounded and undecidable in the traditional model.
Theorem 5. Given an SSTTM M, it is undecidable by a traditional Turing Machine whether simulating M results in a finite or infinite number of bifurcations.
Proof. Let us assume for contradiction that there exists a traditional Turing Machine T that can determine if the simulation of an arbitrary SSTTM M on input w results in a finite or infinite number of bifurcations. Consider the set of all possible pairs of an SSTTM M and input w. We can enumerate this set as . We will construct a diagonalizing SSTTM N that takes as input a pair and behaves as follows:
N simulates T on input to obtain T’s prediction of whether on input will result in a finite or infinite number of bifurcations.
If T predicts that on input will result in a finite number of bifurcations, N simulates on input but introduces a time travel paradox that causes an infinite number of bifurcations. N achieves this by sending a symbol back in time that contradicts the symbol that was originally present at that position, creating a scenario similar to the grandfather paradox. This contradiction leads to an infinite number of bifurcations, each corresponding to a different resolution of the paradox.
If T predicts that on input will result in an infinite number of bifurcations, N simulates on input without introducing any additional time travel. In this case, N will have the same number of bifurcations as , which is finite.
The construction of N is analogous to the classic diagonalization technique used in proofs of undecidability, such as the proof of the undecidability of the Halting Problem. In that proof, a diagonalizing Turing Machine is constructed that takes as input the description of a Turing Machine and its input, and behaves differently than the input machine on that input. This leads to a contradiction when the diagonalizing machine is given its own description as input.
Similarly, our diagonalizing SSTTM N takes as input a pair and behaves differently than on input with respect to the number of bifurcations, based on the prediction of the assumed deciding machine T. This leads to a contradiction when N is given its own description and input as a pair .
If T predicts that N on input will result in a finite number of bifurcations, then N will introduce a time travel paradox causing an infinite number of bifurcations. Conversely, if T predicts that N on input will result in an infinite number of bifurcations, then N will not introduce any additional time travel, resulting in a finite number of bifurcations. In either case, N behaves differently than predicted by T, contradicting the assumption that T can correctly decide whether an SSTTM on a given input will result in a finite or infinite number of bifurcations.
Therefore, by contradiction, there cannot exist a traditional Turing Machine T that can determine if the simulation of an arbitrary SSTTM M on input w will result in a finite or infinite number of bifurcations. The problem of predicting whether an SSTTM will have a finite or infinite number of bifurcations is undecidable. □
The core issue is that simulating all possible branches of an SSTTM to count bifurcations is infeasible on a traditional Turing Machine. By harnessing time travel, SSTTMs can share information across branches, but this capability cannot be reproduced without it.
4.2. Complexity Constraints
Theorem 6. Let M be a traditional Turing Machine running in time on input x. Construct an SSTTM which simulates M on x, sends x paired with the output back in time, then halts also with the correct output. In the worst-case complexity, must still run in time to be self-consistent.
Proof. Let M be a traditional Turing Machine with time complexity on inputs of length n. We construct an SSTTM that operates as follows on input x of length n:
simulates for steps, recording the output as o.
At the end of the simulation, sends the pair back in time to the initial configuration.
then halts with output o.
This creates two branches in the timeline of :
The original branch where is simulated for the full steps before halts.
The new branch starting with the output pair sent back in time.
To ensure a self-consistent output o, the computation in branch must fully complete. This occurs with probability p, where p is the probability that starts in the original timeline.
The key observations are:
The worst-case runtime of branch is to simulate .
The runtime of branch is constant as it starts with the output.
To be self-consistent, the output o must be produced in full in branch .
Therefore, the worst-case expected runtime of across branches is . The potential speedup of branch does not improve this worst-case bound due to the need for self-consistency with branch .
Now, let’s consider the average-case runtime of
under some assumptions about the branch probability distribution. Suppose that the probability of starting in branch
is
p and the probability of starting in branch
is
. The average-case runtime of
can be expressed as:
If p is a constant (independent of n), then the average-case runtime of is , which is asymptotically the same as the worst-case runtime. This suggests that, under a uniform branch probability distribution, the average-case runtime of is still constrained by the need for self-consistency with branch .
However, if
p is allowed to depend on
n, then the average-case runtime of
could potentially be improved. For example, if
, then the average-case runtime of
would be:
In this case, the average-case runtime of would be constant, representing a significant speedup over the worst-case runtime. However, it is important to note that this speedup comes at the cost of a reduced probability of self-consistency, as the probability of starting in branch decreases with n. □
These observations suggest that the complexity constraints on SSTTMs can be sensitive to assumptions about the branch probability distribution. While the worst-case runtime is always constrained by the need for self-consistency, the average-case runtime may be improved under certain probability distributions that favor faster branches. However, this improvement in average-case runtime comes at the cost of a reduced probability of self-consistency.
Further research could explore the trade-offs between average-case runtime and the probability of self-consistency under different branch probability distributions. This could lead to a more nuanced understanding of the complexity constraints on SSTTMs and the potential for leveraging time travel to achieve speedups in specific scenarios.
In summary, because only branches that complete the original computation can produce self-consistent output, probabilistic speedups in worst-case complexity cannot be obtained through time travel in this model.