Is “ignore constant” in complexity theory a scam?

Question

In quantum system, the problem size grow by the exponential of 2. If the algorithm requires N qubits, the cost is O(2^N).Say you invented a new algorithm. Required qubit is now N -> (N-3) [1]. By the 2^N rule, we see that this is equivalent to 2^3 = 8 times faster. O(0.125 2^N). This is the kind of major breakthrough in the field.Often I reimplement someone else' code and got 20X~50X speed up. The speed is equivalent to lowering qubit requirement by 5.But complexity theory tell us this speed up doesn't matter since constant term is ignored.[1] Tapering-off qubit requirement by IBM lab

Bimos · Accepted Answer

In the last 25 years, algorithmic advances in integer optimization coupled with hardware improvements have resulted in an astonishing 800 billion factor speedup in mixed-integer optimization (MIO).[1]
Still, it is just a constant.
[1] https://dspace.mit.edu/handle/1721.1/110328

pestatije · Answer

for the purposes of complexity theory it isn't, however if you extrapolate the theory one-for-one to your algorithm and data then obviously its not going to work

Is “ignore constant” in complexity theory a scam?

In the last 25 years, algorithmic advances in integer optimization coupled with hardware improvements have resulted in an astonishing 800 billion factor speedup in mixed-integer optimization (MIO).[1]Still, it is just a constant.[1] https://dspace.mit.edu/handle/1721.1/110328

for the purposes of complexity theory it isn't, however if you extrapolate the theory one-for-one to your algorithm and data then obviously its not going to work

In the last 25 years, algorithmic advances in integer optimization coupled with hardware improvements have resulted in an astonishing 800 billion factor speedup in mixed-integer optimization (MIO).[1]
Still, it is just a constant.
[1] https://dspace.mit.edu/handle/1721.1/110328