2024-02-22--perfect.ipynb

Kernel: Python 3 (ipykernel)

def delta(n,q,m):
    return sum([ binomial(n,j) * (q-1)**j for j in range(m+1) ])

Let $C \subset A^n$ be a code of minimal distance $d$ and set $t = (d-1)/2$; so that $C$ "corrects up to $t$ errors".

$C$ is perfect if $|C| \cdot \delta(t) = q^n.$

For this to hold, of course, we must have $\delta(t) \mid q^n$.

Let's see some possibilities:

def test(n,q,t):
    return Mod(q^n,delta(n,q,t)) == 0

q = 2
t = 2

list(filter(lambda n: test(n,q,t),range(1,200)))

[1, 2, 5, 90]

In fact, there is no perfect binary code of length 90 with $t = 2$.

q = 2
t = 3
list(filter(lambda n: test(n,q,t), range(1,200)))

[1, 2, 3, 7, 23]

q=3
t=2
list(filter(lambda n: test(n,q,t),range(1,200)))

[1, 2, 11]