n be an even integer.
I be an input of length
Positions start at 0: the first bit is bit 0. The second bit is bit 1. etc.
The decision problem is, "Is bit n/2 equal to 1"?
The input is on a tape.
We have a single pointer to the input. We may move the pointer left or right. We cannot overwrite the input tape.
We have a single working-memory tape to compute with. We have a single pointer to it. We may write a 0 or a 1. We may move the pointer left or right. We may not move to the left of, or overwrite the start symbol of the working-memory tape. The working-memory tape is initialized to either the start symbol or immediately to the right.
Space-complexity is the position of the largest cell we write to. This way we "reuse" space.
One way to decide the problem is with a counter of length
O(log(n)): Initialize to 0's, then increment until the counter equals
n/2, then check the bit.
Can we do better? Or is
O(log(n)) the best space-complexity we can do?
 If space complexity was just the total in-use space at any given point in time, we could "move around" the space we're using, using a larger total range of space, but only a smaller portion at any point in time. Defining space as the max position keeps us from "moving around" the space we're using.