Abstract
We discuss how much space is sufficient to decide whether a unary given number n is a prime. We show that O(log log n) space is sufficient for a deterministic Turing machine, if it is equipped with an additional pebble movable along the input tape, and also for an alternating machine, if the space restriction applies only to its accepting computation subtrees. In other words, the language is a prime is in pebble–DSPACE(log log n) and also in accept–ASPACE(log log n). Moreover, if the given n is composite, such machines are able to find a divisor of n. Since O(log log n) space is too small to write down a divisor, which might require Ω(log n) bits, the witness divisor is indicated by the input head position at the moment when the machine halts.
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