Final answer:
The problem of deciding if a Turing Machine accepts an input using at most |w|3 space is decidable by constructing a specially designed Space-bounded TM that simulates the original TM and halts if it attempts to use more space than allowed.
Step-by-step explanation:
To determine whether the problem of deciding if a Turing Machine (TM) M accepts an input w using at most |w|3 space is decidable, we must consider a specific kind of Turing machine called a Space-bounded Turing machine. A Space-bounded Turing machine is a TM that has a restriction on the amount of tape it can use during its computation.
For a given input w, we can construct another TM, let's call it M', that simulates M on input w. However, M' is designed such that it will halt and reject the input if it ever tries to use more than |w|3 space on its tape. If M' is able to accept w without exceeding the allotted space, it will do so; otherwise, M' will halt and reject. In both scenarios, M' will halt, meaning the problem is decidable.
The reasoning behind this conclusion is that since we have a bound on the space, the number of possible configurations that M' can enter is finite (there are a finite number of states in the machine, a finite number of positions up to |w|3 on the tape, and a finite number of symbols that can be written on the tape). With a finite number of configurations, M' cannot enter an infinite loop since that would require repeating a configuration (which would imply that M' has used more space or it is in an infinite loop, either of which would cause it to halt).
Therefore, by constructing M', we have a Turing machine that will always halt on input w and will only accept w if M can do so within |w|3 space, proving that the problem is decidable.