Padding argument

Example

The proof that P = NP implies EXP = NEXP uses "padding".

\mathrm{EXP} \subseteq \mathrm{NEXP} by definition, so it suffices to show \mathrm{NEXP} \subseteq \mathrm{EXP}.

Let L be a language in NEXP. Since L is in NEXP, there is a non-deterministic Turing machine M that decides L in time 2^{n^c} for some constant c. Let

: L'={x1^{2^{|x|^c}} \mid x \in L},

where '1' is a symbol not occurring in L. First we show that L' is in NP, then we will use the deterministic polynomial time machine given by P = NP to show that L is in EXP.

L' can be decided in non-deterministic polynomial time as follows. Given input x', verify that it has the form x' = x1^{2^{|x|^c}} and reject if it does not. If it has the correct form, simulate M(x). The simulation takes non-deterministic 2^{|x|^c} time, which is polynomial in the size of the input, x'. So, L' is in NP. By the assumption P = NP, there is also a deterministic machine DM that decides L' in polynomial time. We can then decide L in deterministic exponential time as follows. Given input x, simulate DM(x1^{2^{|x|^c}}). This takes only exponential time in the size of the input, x.

The 1^d is called the "padding" of the language L. This type of argument is also sometimes used for space complexity classes, alternating classes, and bounded alternating classes.

References

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