Existence of code with parameters $[22, 15, 3]_2$ I need to verify the existence of a code with parameters as in the title ($n=22,k=15,d=3,q=2$). I have seen this post but neither of the bounds works to conclude anything for this case.
I have also tried to actually construct a concrete code with these parameters, but with no success. Attempted as well some Hamming-like codes with these parameters but I can't get the distance to be 3, always get it of 2.
Could you give some clever way on how to look at this problem? Thanks in advance.
 A: Binary linear $[n,k,d=3]$ codes are easy to construct. Let $r=n-k$ be the redundancy. All you need is a parity check matrix $H$ with $r$ rows and $n$ columns observing the rules:

*

*none of the columns of $H$ are all zeros, and

*the columns of $H$ are pairwise distinct.

This works because the absence of all zeros columns means that no word of weight one passes the parity checks, and no repetitions among the columns means that no word of weight two passes them either. If your $n$ is too large, and you cannot find enough distinct column vectors, then a code with these parameters cannot exist.
So to construct your code you need to find a list $22$ distinct, non-zero binary vectors of length $22-15=7$, and build a check matrix from those. I'm sure you can manage that (you may actually end up with a $[22,16,3]$-code or even a $[22,17,3]$-code if you only draw columns from a $5$-dimensional subspace, such as set the last two bits to zero on each and every one of them).

As @kodlu pointed out, it is still easy to get a $[22,15,4]$-code. One way of achieving that is to begin like above, and select a collection of $22$ distinct, non-zero binary vectors of $r-1=6$, and build a parity check matrix with those as columns (six check equations so far). Then add a seventh parity check that is just the overall parity check (=add an extra row of all ones to your $H$). Any word passing the first six checks must have weight at least three. Any word also passing the last check has an even weight, so the minimum weight is at least four. Again, you can actually construct a $[22,16,4]$-code this way.

On the other hand, it is impossible to find a $[22,15,5]$-code. The sum
$$
\binom{22}0+\binom{22}1+\binom{22}2=254
$$
exceeds $2^r=128$, so the Hamming bound prohibits it. The Hamming bound would allow a $[22,14,5]$-code, just barely, but I would be mildly surprised if one existed. I'm sure such a case has been studied and databases, like the one linked to by kodlu, would have the answer. Such existence questions do become difficult surprisingly soon as the parameters grow.
A: Actually, according to Markus Grassl's codetables see here a code with those parameters and distance $d=4$ exists.
Click on query form under linear block codes and enter the $n,k$ you desire. The answer comes back as below:

Bounds on linear codes [22,15] over GF(2); lower bound:4; upper bound: 4

Luckily there is even a construction given. It is a series of steps:

Regarding modification of codes such as shortening, subcodes etc, see Jon Hall's Chapter 6 of coding theory notes available online here.
Note: The Plotkin sum is also referred to as $(u|u+v)$ construction in the literature.
