
Known error locations help in the usual way.
If error positions are known, the
code corrects any erasure pattern that does not contain a logical operator. Any two
erasures are correctable. Three are correctable unless they form a face or a claw, which
is 512 of the 1,161,280 triples at
𝐿 = 4
, or
4.4 × 10
−4
.
What a working construction would need.
Raising the distance means changing
the chain complex so that faces stop being cycles of
𝐻
𝑍
and claws stop being cycles of
𝐻
𝑋
. By Theorem 1 and Eq. (6) the cost comes out of the same
2𝐿
3
− 1
classes. That is
the budget for any construction on this lattice.
8 Conclusion
The weight-3 logical operators of the FCC code are the
4𝐿
3
triangular faces in the
𝑋
sector and the
4𝐿
3
claws in the
𝑍
sector. The two families are duals, and each carries
2𝐿
3
−1
of the
2𝐿
3
+2
logical classes, a count that follows from the homology of the 3-torus
for every even
𝐿
. A single geometric fact drives the obstruction: every triangular face is
shared between one tetrahedron and one octahedron, so an operator supported on a face
has odd overlap with an octahedral stabilizer. The high rate of the FCC code is a count
of its weight-3 operators, and removing them returns
𝑘 = 3
.
Data Availability
The verication script reproducing every number in this paper is given in Appendix A.
It uses only the Python standard library and runs in under one second at
𝐿 = 10
. No
experimental data were used.
A Verication Code
The script builds the FCC lattice, assembles
𝐻
𝑍
and
𝐻
𝑋
, enumerates the triangular faces
and the claws, computes the cross-sector pairing rank, and checks every claim by exact
GF(2) elimination. Vectors are Python integers used as bitmasks.
NN = [(1,1,0),(1,-1,0),(-1,1,0),(-1,-1,0),
(1,0,1),(1,0,-1),(-1,0,1),(-1,0,-1),
(0,1,1),(0,1,-1),(0,-1,1),(0,-1,-1)]
OCT = [(1,0,0),(-1,0,0),(0,1,0),(0,-1,0),(0,0,1),(0,0,-1)]
def build(L):
nidx, nodes = {}, []
for x in range(L):
for y in range(L):
for z in range(L):
if (x+y+z) % 2 == 0:
10