Ten parity bytes recover five corrupted bytes, and that halving is the part worth carrying into a design review. RSCodec(10) has to locate the damage before it can correct it, so the tutorial's own budget line is 10 // 2 = 5 [8]. A storage tier never pays for location. A dead disk announces itself, the fragment is absent rather than wrong, and the position comes free. So RS 6-3 keeps three parity fragments and survives three simultaneous losses [12][2], where the byte-level demo keeps ten and survives five [8][8]. Size m from the code sample and you are out by a factor of two, in whichever direction is more expensive.
Two lines carry the rest of the decision. Overhead is m/k, and tolerance is m [2][3]. HDFS's default of three copies per block tolerates two losses [1][3]. RS 6-3 tolerates three and pays 50% [1][2]. RS 6-2 tolerates the same two as replication and pays 33%, a sixth of replication's storage penalty [5]. The article promises to tell you when RAID 6 fits, when HDFS with erasure coding fits, and when simple replication fits [11]. The excerpt never reaches the criteria. Overhead, tolerance and repair reads are the criteria.
The word doing the work in "any k" is any, and it comes from the matrix. Encoding multiplies the k data fragments by a Vandermonde or Cauchy generator matrix, both chosen because any subset of their rows is invertible [5]. Decoding assembles a square submatrix from whatever survived, inverts it in the finite field, and multiplies the missing fragments back [6]. Because every k-subset inverts, the code is indifferent to which fragments died, and to whether the thing that died was a disk, a node or a whole rack [3]. A mirrored RAID is not indifferent, since it protects one specific disk with one specific partner [14].
The demo library is pure Python [7]. That is the right call for making field arithmetic visible on a laptop, and I would not put a pure-Python Galois multiply anywhere near a write path. The runnable snippet also flips bits inside a message, while the objective it illustrates is losing whole fragments and rebuilding with no identical copy anywhere [16]. Those are two different failure models, and the second is the one your rack produces.
For the 50% to transfer, RS 6-3's nine fragments need nine places that fail independently. Spread them across five hosts and at least one host is holding two of the nine, so one host failure consumes two thirds of the three-loss budget [7]. The code is then protecting against fragment loss while the hardware fails in host-sized units, and no amount of finite-field algebra reconciles those two units for you.
