Using ≤ and ≥ in csv file will be shown as question marks in html. I have to use <= and >= instead. Is there a better workaround for this?
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I was trying to solve this integral x/(x-6)dx and I used substitution. u = x-6 and x = u+6.
In the end, I ended up with the answer x+6ln|x-6|-6+C, however, the answer is x+6ln|x-6|+C without the -6. Can someone help me understand why this is the case?
MY SOLUTION
You shouldn’t be asking this here, but I’ll answer you anyway. A constant term “eats” any other constants. Because think about it. +C just denotes “plus any constant” and -6 + C means “Any constant - 6” which is still…”any constant”. So the +C effectively “eats” anything that is added or subtracted to it
I am a complete noob in Julia and its syntax. I am trying to follow this article on semi-definite-programming on Julia.
I would apprecieate if someone can help me figure out what the for loop in In[4] actually does:
for i in 1:m
A[:, (i-1)*n+1:i*n] .= random_mat_create(n)
b[i] = tr(A[:, (i-1)*n+1:i*n]*X_test)
end
To my understanding it should create a vector of matrices A (m of those) as well as an m-dimensional vector b. I am totally confused though on the indexing of A and the indexing of b.
I would like an explanation of the :, (i-1)*n+1:i*n part of this code. The reason I ask here is because I also dont know what to Google or what to search for in Julia documentation.
(i-1)*n+1:i*n creates a range from (i-1)*n + 1 to i*n. For example, if i=2 and n=10, this range becomes 11:20, so A[:, (i-1)*n+1:i*n] will grab all the rows of A (that's what : does), and columns 11-20.
There are two operations there that are not clear to you:
: operator. Consider a Matrix a = zeros(3,3). You could use array slicing operator (similarly to numpy or Matlab) to select the entire second columns as: a[1:end,2]. However, when selecting everything from start to the end those two values can be omitted and hence you can write a[:,2] (this always looked the easiest way for me to remember that)
. (dot) operator. Julia is very careful about what gets vectorized and what not. In numpy or R, vectorizing operations happens kind of always automatically. In Julia you have the control - but with the control comes the responsibility. Hence trying to assign values to the second column by writing a[:, 2] = 5.0 will throw an error because there is vector on the right and a scalar on the left. If you want to vectorize you need to tell that to Julia. Hence the dot operator .= means "perform element-wise assignment". Note that any Julia function or operator, even your own functions can be decorated by such dot .. Since this is a very important language feature have a look at https://docs.julialang.org/en/v1/manual/arrays/#Broadcasting
I have a matrix A which has a size of 54x100. For some specific condition I perform an operation on each row of A. I need to save the output of this for loop. I've tried the following but it did not work.
S=zeros(54,100);
for i=1:54;
Ri=A(i,:);
answer=mean(reshape(Ri,5,20),1);
S(i)=answer;
end
Firstly, judging by your question I'd recommend some basic Matlab tutorials like this or just detailed documentation like this.
To actually help you with your issue though; you can do this:
%% Make up A (since I don't know what it actually is)
n = 54; m = 100;
A = randn(n,m); % N x m matrix of random numbers
%% Loop over each row of A
S = cell(n,1);
for j = 1:n;
Rj = A(j,:); % j'th row
answer = mean(reshape(Rj,5,20),1); % some operation
S{j} = answer; % store the answer in cell S
end
The problem was that your answer was not a single number (1x1 matrix) but a vector and so you got a dimension mismatch error. Above I'm putting the answers into a cell object of size n. The result of your operation on j'th row can then be retrieved by calling S{j}.
Also:
Do not using i as an iterator since it also represents the imaginary unit.
Do not hard-code values but reference the existing ones. For example here I referenced n in the for-loop declaration as opposed to just writing for j = 1:54 because otherwise, if I got struck by a fancy to use my code for a 53x100 array it would not work anymore.
When you post your code I reccomend adding a minimal working example - a pece of code which people can just copy and paste into their Matlab (or whatever interpreter of whatever language) and run to reproduce your problem. Here you have not included anything which tells the code what A is, for example.
This is quite a good read in general and should help you in the future
(To begin with, I am a beginner in Ruby.)
I want to rearrange variables in a linear equation (there may be 2 or more variables).
I have an equation
a + 2*b - 1 = 0
I would like Ruby to give me
a = 1 - 2*b
Or alternatively
b = (1-a)/2
Is there a way to do this in Ruby ? (It is possible in Matlab, which, in my case, seems overkill...)
Thanks in advance
Try the symbolic gem.
It should provide you with the tools you need.
I receive encoded PDF files regularly. The encoding works like this:
the PDFs can be displayed correctly in Acrobat Reader
select all and copy the test via Acrobat Reader
and paste in a text editor
will show that the content are encoded
so, examples are:
13579 -> 3579;
hello -> jgnnq
it's basically an offset (maybe swap) of ASCII characters.
The question is how can I find the offset automatically when I have access to only a few samples. I cannot be sure whether the encoding offset is changed. All I know is some text will usually (if not always) show up, e.g. "Name:", "Summary:", "Total:", inside the PDF.
Thank you!
edit: thanks for the feedback. I'd try to break the question into smaller questions:
Part 1: How to detect identical part(s) inside string?
You need to brute-force it.
If those patterns are simple like +2 character code like in your examples (which is +2 char codes)
h i j
e f g
l m n
l m n
o p q
1 2 3
3 4 5
5 6 7
7 8 9
9 : ;
You could easily implement like this to check against knowns words
>>> text='jgnnq'
>>> knowns=['hello', '13579']
>>>
>>> for i in range(-5,+5): #check -5 to +5 char code range
... rot=''.join(chr(ord(j)+i) for j in text)
... for x in knowns:
... if x in rot:
... print rot
...
hello
Is the PDF going to contain symbolic (like math or proofs) or natural language text (English, French, etc)?
If the latter, you can use a frequency chart for letters (digraphs, trigraphs and a small dictionary of words if you want to go the distance). I think there are probably a few of these online. Here's a start. And more specifically letter frequencies.
Then, if you're sure it's a Caesar shift, you can grab the first 1000 characters or so and shift them forward by increasing amounts up to (I would guess) 127 or so. Take the resulting texts and calculate how close the frequencies match the average ones you found above. Here is information on that.
The linked letter frequencies page on Wikipedia shows only letters, so you may want to exclude them in your calculation, or better find a chart with them in it. You may also want to transform the entire resulting text into lowercase or uppercase (your preference) to treat letters the same regardless of case.
Edit - saw comment about character swapping
In this case, it's a substitution cipher, which can still be broken automatically, though this time you will probably want to have a digraph chart handy to do extra analysis. This is useful because there will quite possibly be a substitution that is "closer" to average language in terms of letter analysis than the correct one, but comparing digraph frequencies will let you rule it out.
Also, I suggested shifting the characters, then seeing how close the frequencies matched the average language frequencies. You can actually just calculate the frequencies in your ciphertext first, then try to line them up with the good values. I'm not sure which is better.
Hmmm, thats a tough one.
The only thing I can suggest is using a dictionary (along with some substitution cipher algorithms) may help in decoding some of the text.
But I cannot see a solution that will decode everything for you with the scenario you describe.
Why don't you paste some sample input and we can have ago at decoding it.
It's only possible then you have a lot of examples (examples count stops then: possible to get all the combinations or just an linear values dependency or idea of the scenario).
also this question : How would I reverse engineer a cryptographic algorithm? have some advices.
Do the encoded files open correctly in PDF readers other than Acrobat Reader? If so, you could just use a PDF library (e.g. PDF Clown) and use it to programmatically extract the text you need.