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I'm trying to prove that $TIME(n)\neq L$ by padding technique, yet it has a trouble:
Assume that $TIME(n)= L$, let $A\in TIME(n^2)\backslash TIME(n)$ so $A\notin L$. Let $A_{pad}=\{x01^{|x|^2-|x|-1}|x\in A\}$, it is easily to see that $A_{pad}\in TIME(n)$. Thus, we could decide $x\in A$ by padding it to $x01^{|x|^2-|x|-1}\in A_{pad}$ then deciding $A_{pad}$ in $O(\log n)$ space as assuming. When padding, however, we have used $O(n^2)$ space, so the algorithm consumes more than $O(\log n)$ space, which is not a contradiction as desired.

How could this problem be solved? Thanks in advanced!

minh quý lê
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The idea is to show $A$ log-space reduces to $A_{pad}\in L$ and thus $A\in L$ which is a contradiction. Log-space reduction works in a bit of an idiosyncratic way. You are allowed for the output to be larger-than-logarithmic in size, provided it is write-only.

So, given the input $x,$ you copy $x$ to the output, followed by a 0 (uses no space), then you compute and write down $|x|^2-|x|-1$ (uses $O(\log(|x|))$ space), then, write out $|x|^2-|x|-1$ ones to the output, counting it out by decrementing the number you wrote down.

spaceisdarkgreen
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