# What is the runtime overhead and how many more states are needed for simulating a TM whose head may remain stationary by a standard TM?

The standard Turing Machine (TM) can only move left and right. When I simulate a TM whose head may remain stationary and I execute the stationary move, then I need to make two steps with the standard TM: go left and then right. So does this mean I need twice the amount of steps and I need twice the amount of states?

The sketch is as follows: replace every transition of the original TM that keeps the head still and changes the state to $$q$$ with two transitions: first move the head right and change the state to a new intermediate state $$q'$$, then whenever the state is $$q'$$ move the head left (without modifying the current tape symbol) and go to state $$q$$.
You can actually construct an equivalent TM that uses at most one more state and requires at most the same number of steps. The sketch is as follows: consider a transition of the original TM that keeps the head still. This transition writes $$\alpha$$ to the current tape cell and changes the state to $$q$$. Then you can find the next transition that will be executed by the machine, i.e., the transition performed when the current tape symbol is $$\alpha$$ and the current state is $$q$$. Suppose that the latter transition writes $$\beta$$ and moves to $$q'$$. If the transition also moves the head, then you can "shortcut" the first transition by directly writing $$\beta$$ and moving to $$q'$$.
If the second transition does not move the head, then the process can be repeated. Eventually (there are at most $$|\Gamma| \cdot |Q|$$ distinct transitions that can be applied, where $$\Gamma$$ is the tape alphabet) you either reach a transition that moves the head/halts the machine, or you discover that the machine must loop.