# What is uniformity in Boolean circuits exactly?

I have two questions on Kaveh's answer to Definition of uniform boolean circuit :

1. Kaveh mentions that the input is in unary encoding. In the definition it says the input is $$1^n$$, afaik $$1^n$$ is a sequence of 1's repeated n times, but unary encoding is a sequence of 1's and ends with 0 there is no power of 1 that gives 0... So how is $$1^n$$ a unary encoding?
2. Why do we use a unary encoding and not the binary encoding of n? what happens if we use binary instead?

Here is the binary encoding of 5: $$101$$ And here is its unary encoding: $$11111$$ Hopefully the difference is apparent.
Let's see what happens if we switch to binary encoding in the context of the linked question. Given the binary encoding of $$n$$, we need to produce a circuit, operating on $$n$$ bits, in logspace, and so in polynomial time. How much time do we have? The binary encoding of $$n$$ has length roughly $$\log n$$, so we have time $$O(\log^C n)$$, which isn't enough to produce a circuit which reads all input bits.
In contrast, if we use unary encoding, then the length of the input is $$n$$, and so we have time $$O(n^C)$$, which is more reasonable (in addition, we are only allowed to use $$O(\log n)$$ space).