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The lemma is:

Let $M$ be an LBA with $q$ states and $g$ symbols in the tape alphabet. There are exactly $qng^n$ distinct configurations of $M$ for a tape of length $n$.

I want know why LBA has $qng^n$ configurations.

Following is the proof:

A configuration of $M$ is like a snapshot in the middle of its computation. A configuration consists of the state of the control, position of the head, and contents of the tape. Here, $M$ has $q$ states. The length of its tape is $n$, so the head can be in one of $n$ positions, and $g^n$ possible strings of tape symbols appear on the tape. The product of these three quantities is the total number of different configurations of $M$ with a tape of length $n$.

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  • $\begingroup$ The proof you posted is OK, what is the question then? $\endgroup$ – diplodoc Aug 25 at 10:05
  • $\begingroup$ I can understand why there are g^n possible strings of tape. I cannot figure out the reasoning behind it $\endgroup$ – sachal Aug 25 at 10:08
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If you have $g$ symbols (including a blank) and a tape of size $n$ then there are $g^n$ words of length $n$. This is really basic combinatorics: The reasoning is that you have $g$ options for the first symbol, $g$ options for the second symbol, i.e. $g^2$ options for the first two symbols. Then again $g$ options for the third symbol, giving you $g^3$ options for the first three symbols and so on. If you are not convinced, try it with for example a $3$-symbol alphabet.

Since your tape has length $n$ there are $n$ possible head positions, so you multiply the number of strings on the tape with $n$.

Finally, there are $q$ states, so for every possible string on tape and every possible head position, we can be in either of the $q$ states, making $qng^n$ configurations of the Turing machine.

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