Ternary Repetition Codes#
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# %load '../imports.py'
import numpy as np
import pandas as pd
import matplotlib as mpl
from matplotlib.patches import Rectangle
import matplotlib.pyplot as plt
plt.style.use('ggplot');
plt.rcParams.update({'text.usetex' : True});
%matplotlib inline
from collections import defaultdict
from itertools import combinations,product
import itertools
from typing import Set
from IPython.display import display, Math
from datetime import datetime as d
import locale as l
import platform as p
import sys as s
pad = 20
print(f"{'Executed'.upper():<{pad}}: {d.now()}")
print()
print(f"{'Platform' :<{pad}}: "
f"{p.mac_ver()[0]} | "
f"{p.system()} | "
f"{p.release()} | "
f"{p.machine()}")
print(f"{'' :<{pad}}: {l.getpreferredencoding()}")
print()
print(f"{'Python' :<{pad}}: {s.version}")
print(f"{'' :<{pad}}: {s.version_info}")
print(f"{'' :<{pad}}: {p.python_implementation()}")
print()
print(f"{'Matplotlib' :<{pad}}: {mpl.__version__}")
print(f"{'NumPy' :<{pad}}: {np .__version__}")
#==================================================
def rc (q : int,
n : int) -> Set[str]:
"""Repetition Code
Generate a q-ary repetition block code of length n.
"""
S=set()
for i in range(q):
S.add(str(i)*n)
return S
def fqn (q : int,
n : int,
g : int = 0) -> Set[str]:
"""Construct a linear space of dimension n over a finite field of order q.
Parameters
==========
g : If the space is very large, opt for the first g elements of a generator object.
"""
if bool(g):
f=itertools.product(range(q),repeat=n)
return set(''.join(str(i) for i in next(f)) for _ in range(g))
else:
return {''.join(str(bit) for bit in word) for word in itertools.product(range(q),repeat=n)}
def qarycode_to_nbitstring (code={'3121','2101'},k=4):
"""Convert a q-ary code """
for n in code:
print(' '.join(format(int(i),f'0{k}b') for i in n))
def hd (a : str = '1001',
b : str = '0101') -> int:
"""HAMMING DISTANCE
Parameters
==========
x : str
y : str
Return
======
int
"""
assert len(a) == len(b), 'x and y must have the same length'
return sum(x!=y for x,y in zip(a,b))
def nbfmd (c : Set[str],
pr : bool = False) -> np.float16:
"""NAIVE BRUTE FORCE MINIMUM DISTANCE d(C)
Computes the pairwise Hamming distance for all codewords and returns the minimum value.
This is a naive (i.e., non vectorized) implementation using nested for loops.
Parameters
==========
c : code
pr : Print intermediate steps.
Returns
=======
d(C)
"""
# convert a set of string vectors to a 2D NumPy array of integers
c=np.array([list(codeword) for codeword in c],dtype=np.float16)
# intialize empty hamming distance matrix
hamming = np.empty([c.shape[0]]*2,dtype=np.float16)
for i,x in enumerate(c):
for j,y in enumerate(c):
hamming[i,j]=(x!=y).sum()
# the diagonal represents the Hamming distance of a codeword with itself, which is always 0.
np.fill_diagonal(hamming,np.inf)
if pr == True:
print(hamming)
return hamming.min().astype(np.int8)
def one_error_detecting (q : int,
code : Set[str],
p : bool = False) -> bool:
"""Verify that a code is one-error detecting.
No one-bit error equals a codeword.
"""
flag=True
alphabet=set(str(i) for i in range(q))
for codeword in code:
if p:
print()
print(f"{'orig cw : ':10}{codeword}")
for i in range(len(codeword)):
a,b,c=codeword[:i],codeword[i],codeword[i+1:]
symbols=alphabet-set(codeword[i])
for symbol in symbols:
cw=codeword[:i]+symbol+codeword[i+1:] # SINGLE ERROR
if cw in code:
flag=False
if p:
print(f"{'ERROR':10}{cw}")
else:
if p:
print(f"{'':10}{cw}")
return flag
# set(''.join(l for l in i) for i in itertools.product('10',repeat=3))
# set(''.join(l for l in i) for i in itertools.combinations_with_replacement('012',r=3))
# set(''.join(l for l in i) for i in itertools.combinations('01',r=2))
EXECUTED : 2025-01-17 17:27:54.085789
Platform : 15.2 | Darwin | 24.2.0 | arm64
: UTF-8
Python : 3.11.11 | packaged by conda-forge | (main, Dec 5 2024, 14:21:42) [Clang 18.1.8 ]
: sys.version_info(major=3, minor=11, micro=11, releaselevel='final', serial=0)
: CPython
Matplotlib : 3.9.3
NumPy : 2.0.2
Ternary Repetition Code of length \(3\)#
\( C= \begin{cases} 000\\ 111\\ 222\\ \end{cases} \)
\((3,3,3)\)-code
Equivalent codes
\( \begin{aligned} C= \begin{cases} 012\\ 120\\ 201\\ \end{cases} \,\,\,\,\, \underset{p_2}{ \left( \begin{matrix} 0&1&2\\ \downarrow&\downarrow&\downarrow\\ 2&0&1\\ \end{matrix} \right) } \,\,\,\,\, C= \begin{cases} 002\\ 110\\ 221\\ \end{cases} \,\,\,\,\, \underset{p_3}{ \left( \begin{matrix} 0&1&2\\ \downarrow&\downarrow&\downarrow\\ 1&2&0\\ \end{matrix} \right) } \,\,\,\,\, C= \begin{cases} 000\\ 111\\ 222\\ \end{cases} \end{aligned} \)
q=3
n=3
print(f"{q**n}")
f=fqn(q,n)
code=rc(q,n)
code
27
{'000', '111', '222'}
f
{'000',
'001',
'002',
'010',
'011',
'012',
'020',
'021',
'022',
'100',
'101',
'102',
'110',
'111',
'112',
'120',
'121',
'122',
'200',
'201',
'202',
'210',
'211',
'212',
'220',
'221',
'222'}