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seen Jun 12 at 14:26

Computer engineer: a lot of C, PHP, and Python, plus a bit of lots of other stuff. Linux user. Amateur cryptonerd and general science and math geek.


Mar
1
comment Understanding group presentation as a quotient
Wow, nice answer, thanks for putting in the time! I'm going to have to put in some time studying it some more to understand it, but I think I was able to follow each step, just haven't quite pulled it all together into a conclusion yet. Also, I think there's a typo in the last paragraph: you have $fef$ listed twice in $\hat{G}$.
Feb
26
comment Show that group action is homomorphism to Symmetric group
Oh, ok. So my original formulation for the condition of homomorphism was off. I think it makes sense now, thanks!
Feb
26
comment Show that group action is homomorphism to Symmetric group
Actually, I take that back, $A_g(x)$ is not a permutation of $X$, it's an element of $X$. So how can you compose a permutation/function, $A_g$ with an element of a set, $A_f(x)$?
Feb
26
comment Show that group action is homomorphism to Symmetric group
@SpamIAm: Thanks, but I think you over estimate me. I have the axioms written out as $A(e,x) = x$ for all $x$ (where $e$ is the identity of $G$) and $A(g, A(f,x)) = A(g+f, x)$, but it's still not clear to me why this defined a homomorphism.
Feb
26
comment Show that group action is homomorphism to Symmetric group
(Thanks for updating from $h$ to $A$). So the steps to change right side, $A(g+f,x)$ into $A_g * A_f(x)$ are clear, but is that necessarily equal to $A_g(x) * A_f(x)$. I know $A_g(x)$ and $A_g$ are both permutations of X, but are they they same?
Feb
26
comment Show that group action is homomorphism to Symmetric group
@ThomasAndrews: Thanks, I've updated the question to use $A$ in place of $h$: I'm not sure why I started using $h$, must be misread something somewhere. So if $A$ is the group action, we have $A : G \times X \to X$, to it $A$ takes a two-tuple $(g,x)$ with $g \in G$ and $x \in X$.
Feb
21
comment Notation for nested sigmas (summations)
Oh, that's right, I forgot you can iterate over a set of tuples. But it doesn't really resolve the question of how to neatly define the set from which they draw, which is kind of the big problem in the originally posted example.
Feb
19
comment Enumerating cases for conditional probability
Thanks. That's nearly what I came up with as described in the question, but it is not any more clear to me why you don't need $\binom{8}{4}$ included somewhere as this is the total number of ways in which you can choose which four bits are set and which four are cleared.
Sep
11
comment Name for the sense of how many items are present
Perfect, thanks!
Sep
11
comment Statistics Question.
Welcome to Stack Exchange! This is not the place to come for answers to your homework problems, but it is certainly a place to come for help with your homework. To get help, though, you'll need to first of all, make a disclaimer that this is a homework problem, and second of all show that you've attempted to solve the problem yourself by explaining your approach and what specifically you're getting stuck on.
Jul
31
comment Non-binary Hamming codes?
Thanks for all the info, it was sufficient for me to realize that I did not have as good of an understanding of the topic as I thought. "SEC" is "single error correcting", i.e., distance-3.
Jul
31
comment Variations of the Hamming code.
I think it's worth mentioning/emphasizing that simply cutting off the first $s$ bits of the codeword only works for canonical Hamming codes (i.e., where the generator matrix begins with the identity matrix). In that case, any data word beginning with $s$ zeroes is guaranteed to produce a codeword which also begins with $s$ zeroes. But for non-canonical Hamming codes, this is not necessarily the case. Non-canonical codes can still be shortened, but it is somewhat more complicated.
Jul
31
comment Non-binary Hamming codes?
Rats, looks like I have more learning to do =). When you say it wouldn't be a Hamming code over the larger-base field, do you mean it doesn't function as SEC, or just that it is not strictly a Hamming code (for instance, because it is not a perfect code)? Because I've tested it pretty thorougly and it seems to function well as SEC. But based on your $[13, 10, 3]$ code, it looks like I'm nowhere near optimal data rate.
Jul
31
comment Non-binary Hamming codes?
Does the ternary matrix you posted really work? You've got a 3x13 check matrix, so wouldn't that give you a (13,10) code? That's way more efficient than the (7,4) code you get with 3 redundant bits in binary Hamming codes. I thought I understood it, now I'm confused.
Jul
31
comment Non-binary Hamming codes?
Fair enough in this case, but I believe you can take any parity check matrix from a binary Hamming code, and use it as the parity check for any base. In this case, it will be the generator matrix that has non-binary digits in it.
Jul
31
comment Non-binary Hamming codes?
It appears that any Hamming code parity check matrix is universal, in that it will work for regardless of whether the data is in binary or some other base. And as azimut said, all Hamming codes have distance 3. The important point for non-binary codes is that when you create the Generator matrix from the parity matrix, the negation that happens has to be in the proper base.
Jul
30
comment Non-binary Hamming codes?
EDIT: I guess more to the point, do you know how to determine the minimum codeword length required to get a functioning SECSED non-binary Hamming code?
Jul
30
comment Non-binary Hamming codes?
Any idea how to determine the error correcting capabilities of a non-binary hamming code?
Jul
18
comment Mental estimate for tangent of an angle (from $0$ to $90$ degrees)
Wow, that's a pretty remarkable error, but you're right, that would be a bit much for mental math, primarily because there are so many different values to keep track of. I wonder if it can be simplified any more, allowing for more generous error bounds.
Jul
18
comment Mental estimate for tangent of an angle (from $0$ to $90$ degrees)
Great, I've seen that paper before been at that time I was learning how to approximate exponentials so I disregarded the tangent, and subsequently couldn't remember the name of the paper. Anyway, this does look like a very good method, but it will definitely take some time to get a knack for it and memorize the "magic hash table" =)