[yters]

for limited runtime you can

[ukj]

So you were unable to compress the stream given the allocated time.

Perhaps you could've compressed it had you kept running for just a second longer?

[yters]

i mean run all the shorter bitstrings until time runs out or they terminate.

[ukj]

How does “time run out” in practice other than you putting an upper bound on your computation?

Obviously, the code will not halt until it compresses the stream successfully.

[yters]

enumerate all bitstrings of length 10.

run them for N steps.

if none terminate on the target bitstring of length 100, then we've eliminated the computation hypothesis for 10 bits and N runtime

if we continue this approach and eliminate all the available storage and time available, then we eliminate the computation hypothesis altogether for our scenario

[ukj]

You understand that "Number of bitstrings of length 10" is a function of the cardinality of your alphabet, right?

A binary alphabet has 2^10 such strings. A decimal alphabet has 10^10 such strings.

So before you can make the assertion you've made you first have to answer two question:

1. How many symbols does your alphabet contain? 2. How did you choose that number?

Down that path you arrive the Linear speedup theorem. https://en.wikipedia.org/wiki/Linear_speedup_theorem

[yters]

bit is base 2 as far as i know

yes, there is always a turing machine that can generate the target with any performance characteristics

so, the key is to keep the reference turing machine fixed