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5h
revised Show that four codewords is the maximal size for a code in V^8 = {(a1,…a8) | ai is in {0,1}} that corrects 2 errors
added 11 characters in body
5h
comment Show that four codewords is the maximal size for a code in V^8 = {(a1,…a8) | ai is in {0,1}} that corrects 2 errors
@ztforster I guess that we are speaking here of linear codes, so that the number of codewords must be of the form $2^k$.
16h
answered Show that four codewords is the maximal size for a code in V^8 = {(a1,…a8) | ai is in {0,1}} that corrects 2 errors
22h
revised A probabilty of error calculation
edited tags
1d
comment Highest pairwise Hamming distance between k bitvectors of length n
Not very clear for me. You mean you want to find the $k$ vectors so as to maximize the smallest distance?
1d
revised Highest pairwise Hamming distance between k bitvectors of length n
added 13 characters in body
1d
revised what is $e$ really? what is its meaning?
added 14 characters in body; edited title
1d
revised Different approaches to N balls and m boxes problem
added 19 characters in body
1d
answered Different approaches to N balls and m boxes problem
Oct
27
comment Counting binary words distance one from codewords
Yes. You only need to answer the last question to assert that you are not counting duplicates, and that $15\times 2048$ is the right answer.
Oct
26
answered Counting binary words distance one from codewords
Oct
26
revised How to determine whether there is a linear dependence between rows or columns in a matrix?
edited title
Oct
24
comment Find the range of function $f(x) =\cos(\sin(\ln(\frac{x^2+e}{x^2+1})))+\sin(\cos(\ln(\frac{x^2+e}{x^2+1})))$…
"maximum value of the function is when denominator term is minimum " of which function?
Oct
24
comment The definition of NMSE (normalized mean square error)
Where did you find the "strange definitions"? They both look quite nonsensical to me
Oct
23
revised How many $4$ digit numbers can be formed using $0,0,2,2,2,2,3,3$?
edited title
Oct
23
comment Combination and Permutation
So the funal solution is $C(9,3)-C(4,3)=84-4=80$
Oct
23
comment Combination and Permutation
Yes, but you should compute that as another combination C(4,3)
Oct
23
comment Combination and Permutation
No. You need to understand what $C(9,3)=84$ is: it's the total number of possible selections of 3 people (imagine them). Of those 84 selections (say, teams) the forbidden one are those who are all men (no?). Now, how many ways are of selecting 3 mens from a total of 4 men?
Oct
23
answered Combination and Permutation
Oct
23
comment Combination and Permutation
That's better, I copied that to the question body.