## Size limits for the input and the output

If the input is exponentially smaller than the output, the problem is not in NP. If the input is power-of-two exponentially larger than the output, the problem is in P.

In the first case, the output is exponentially longer than the input and therefore takes exponentially longer to print out, even on a non-deterministic machine. Even if you could print it out instantly, you need to check the output with a deterministic verifier that needs to run in polynomial time to the original input, and reading the exponentially larger output into the verifier would take exponential time to the original input. So exponentially longer outputs are not in NP.

In the second case, if you have a polynomial time deterministic verification algorithm, you can create a solver by adding an output space enumerator in front of it. If the output is exponentially smaller than the input, hey, you've got a polynomial time solver. If |output| = log2(|input|), enumerating the outputs takes 2^{|output|} = 2^{log2(|input|)} = |input| steps. From NP, you have the verifier that runs in p(|input|) so a solver that enumerates the output space and tries the verifier on each possible solution would run in 2^{|output|} * p(|input|), which becomes |input| * p(|input|) if the output is exponentially smaller than the input like above. (So any problems in NP where the output is a single bit are also in P.)

Due to these size limits, for P ?= NP the size of the output must be polynomial(-ish??) to the size of the input.

## Size limits for the solver

How about the size of the deterministic solver relative to the input. If the deterministic solver machine is by necessity exponentially larger than the input (by necessity meaning that for any given part of the machine there exists an input that makes the machine access that part), there also exists an input that causes the machine to move the tape a distance exponential to the size of the input, which takes exponential time relative to the input. So the deterministic solver needs to have a size at most polynomial to the size of the input.

For the lower bound, consider the case where the machine extracts zero information from the input. This machine just enumerates over the output space, checking each answer using the verifier. The runtime of the machine is 2^{|output|} * p(|input|), the code is fixed in size, and it uses |output| extra tape as memory. At the other end is the answer table machine: it's an array of precomputed solutions where the key is the input. The size of the answer table machine is |output| * 2^{|input|} (and due to the above tape movement limitations, will take exponential time to run relative to some input.)

Knowing nothing about the input is out, and knowing everything about the input is out. Every bit in the program encodes information about the input-output-relation, allowing the solution search to proceed faster than brute-force enumeration when the program encounters an input that matches the encoded information. The program can also learn from the input, but it can't learn an exponential amount of information from the input (because that'd take exponential time.) Wild guess: If the amount of information used by the program is exponentially less than the amount of information in the input-output-relation, the program would end up using brute-force search over parts of the output that are polynomial in size to the input, which would take exponential time to the input. If true, there would be no sub-polynomial-space programs for P ?= NP that run in polynomial time.

Given the existence of the deterministic polynomial time verifier, we know that you can extract information from the input-output-pair in polynomial time, even though the amount of information extracted is very small. How small? Well, at least you get |output| / 2^{|output|} bits of info. After trying every 2^{|output|} outputs, you would have |output| bits and a correct output. If there are several correct outputs, you'd gain numberOfOutputs * |output| / 2^{|output|} bits. If the numberOfOutputs grows exponentially with the output size, this would give you a polynomial time solver. But, if this were the case, someone would've noticed it. Another wild guess: the amount of information we get out of the verifier shrinks exponentially in relation to the output size (and therefore the input size, which has a polynomial size relation to the output).

The number of input enumerations is 2^{|input|}, ditto for the output 2^{p(|input|)}. We have a deterministic polynomial time verifier that can extract solution information from the input, but the amount of information extracted looks to be a logarithm of the input size. We also have a hypothetical solver program that encodes a polynomially growing amount of information about the input-output-relation, allowing it to avoid brute-force search whenever it encounters an input that matches the encoded information.

If the number of patterns that can be encoded in the input grows exponentially with the input length, then as the problem size grows towards infinity the question becomes: can you discern between an infinite number of input patterns with a zero-length solver program and a verifier that can extract zero bits of information from the input? For P to be NP, the number of patterns in the input needs to grow as a polynomial to the length of the input.

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