i need to get value which first 16 characters are TZxy2o2h2I2NMVR+ for which I have a formula. The formula goes like this: Base64(XOR("KonstantaZaLDAP", MD5(521009)) + XOR(521009, "KonstantaZaLDAP")) or in a word:
I have two values:
int radID = 521009
String konst = "KonstantaZaLDAP"
The first step is to apply XOR operation to konst and MD5 hash value of konst >>XOR(kost, MD5(radID))
Second, I need to apply XOR operation to radID and konst >> XOR(radID, konst).
After that i should concatenate values from first and second step >> XOR(kost, MD5(radID)) + XOR(radID, konst) and finaly Base64 encode concatenated value.
That is Base64(XOR(konst, MD5(radID)) + XOR(radID, konst)).
I have tried to achieve wanted value, and whatever I do, I get first 13 characters right, and after that it's all wrong. The value I get is TZxy2o2h2l2NMfUfpPmJNA==
Can anyone help!?
Related
I have the following Powershell variable
$var = "AB-0045"
I would like to increase the number in the string to become "AB-0046".
I can do:
$newNumber = [int]$var.Substring($var.length -4,4) + 1
Which will give me the desired number 46, but then I have to append that 46 as a string to a new string "AB-00".
Is there a better way to do that?
Now that you have the integer, you'll have to convert back to string formatted in the way you'd like and concatenate.
I'd recommend adding to "AB-" rather than "AB-00" in case your number goes over 100.
To pad leading zeros, you can use the -f operator.
e.g. "{0:d4}" -f 45
You'll still need to get the integer first (45 in the example) from your original string.
I tested with regex class Replace() method and string class Split() method with string formatter. Split() seems faster provided your string is always in the same format. The Replace() method does not care what happens before the last 4 numbers:
# Replace Method
[regex]::Replace($var,'\d{4}$',{([int]$args[0].Value+1).ToString('0000')})
# Split method
$a,[int]$b = $var.split('-'); "{0}-{1:0000}" -f $a,++$b
I have a text file with 1 million decimal digits of "e" number with 80 digits on each line excluding the first and the last line which have 76 and 4 digits and the file has 12501 lines. I want to convert it into a vector in matlab with each digit on each row. I tried num2str function, but the problem is that it gets converted like for example '7.1828e79' (13 characters). What can I do?
P.S.1: The first two lines of the text file (76 and 80 digits) are:
7182818284590452353602874713526624977572470936999595749669676277240766303535 47594571382178525166427427466391932003059921817413596629043572900334295260595630
P.S.2: I used "dlmread" and got a 12501x1 vector, with the first and second row of 7.18281828459045e+75 and 4.75945713821785e+79 and the problem is that when I use num2str for example for the first row value, I get: '7.182818284590453e+75' as a string and not the whole 76 digits. My aim was to do something like this:
e1=dlmread('e.txt');
es1=num2str(e1);
for i=1:12501
for j=1:length(es1(1,:))
a1((i-1)*length(es1(1,:))+j)=es1(i,j);
end
end
e_digits=a1.';
but I get a string like this:
a1='7.182818284590453e+754.759457138217852e+797.381323286279435e+799.244761460668082e+796.133138458300076e+791.416928368190255e+79 5...'
with 262521 characters instead of 1 million digits.
P.S.3: I think the problem might be solved if I can manipulate the text file in a way that I have one digit on each line and simply use dlmread.
Well, this is not hard, there are many ways to do it.
So first you want to load in your file as a Char Array using something simple like (you want a Char Array so that you can easily manipulate it to forget about the lines breaks) :
C = fileread('yourfile.txt'); %loads file as Char Array
D = C(~isspace(C)); %Removes SPACES which are line-breaks
Next, you want to actually append a SPACE between each char (this is because you want to use the num2str transform - and matlab needs to see the space), you can do this using a RESHAPE, a STRTRIM or simply a REGEX:
E = strtrim(regexprep(D, '.{1}', '$0 ')); %Add a SPACE after each Numeric Digit
Now you can transform it using str2num into a Vector:
str2num(E)'; %Converts the Char Array back to Vector
which will give you a single digit each row.
Using your example, I get a vector of 156 x 1, with 1 digit each row as you required.
You can get a digit per row like this
fid = fopen('e.txt','r');
c = textscan(fid,'%s');
c=cat(1,c{:});
c = cellfun(#(x) str2num(reshape(x,[],1)),c,'un',0);
c=cat(1,c{:});
And it is not the only possible way.
Could you please tell what is the final task, how do you plan using the array of e digits?
I'd like to have a function generate(n) that generates the first n lowercase characters of the alphabet appended in a string (therefore: 1<=n<=26)
For example:
generate(3) --> 'abc'
generate(5) --> 'abcde'
generate(9) --> 'abcdefghi'
I'm new to Matlab and I'd be happy if someone could show me an approach of how to write the function. For sure this will involve doing arithmetic with the ASCII-codes of the characters - but I've no idea how to do this and which types that Matlab provides to do this.
I would rely on ASCII codes for this. You can convert an integer to a character using char.
So for example if we want an "e", we could look up the ASCII code for "e" (101) and write:
char(101)
'e'
This also works for arrays:
char([101, 102])
'ef'
The nice thing in your case is that in ASCII, the lowercase letters are all the numbers between 97 ("a") and 122 ("z"). Thus the following code works by taking ASCII "a" (97) and creating an array of length n starting at 97. These numbers are then converted using char to strings. As an added bonus, the version below ensures that the array can only go to 122 (ASCII for "z").
function output = generate(n)
output = char(97:min(96 + n, 122));
end
Note: For the upper limit we use 96 + n because if n were 1, then we want 97:97 rather than 97:98 as the second would return "ab". This could be written as 97:(97 + n - 1) but the way I've written it, I've simply pulled the "-1" into the constant.
You could also make this a simple anonymous function.
generate = #(n)char(97:min(96 + n, 122));
generate(3)
'abc'
To write the most portable and robust code, I would probably not want those hard-coded ASCII codes, so I would use something like the following:
output = 'a':char(min('a' + n - 1, 'z'));
...or, you can just generate the entire alphabet and take the part you want:
function str = generate(n)
alphabet = 'a':'z';
str = alphabet(1:n);
end
Note that this will fail with an index out of bounds error for n > 26, so you might want to check for that.
You can use the char built-in function which converts an interger value (or array) into a character array.
EDIT
Bug fixed (ref. Suever's comment)
function [str]=generate(n)
a=97;
% str=char(a:a+n)
str=char(a:a+n-1)
Hope this helps.
Qapla'
I have a string and I need two characters to be returned.
I tried with strsplit but the delimiter must be a string and I don't have any delimiters in my string. Instead, I always want to get the second number in my string. The number is always 2 digits.
Example: 001a02.jpg I use the fileparts function to delete the extension of the image (jpg), so I get this string: 001a02
The expected return value is 02
Another example: 001A43a . Return values: 43
Another one: 002A12. Return values: 12
All the filenames are in a matrix 1002x1. Maybe I can use textscan but in the second example, it gives "43a" as a result.
(Just so this question doesn't remain unanswered, here's a possible approach: )
One way to go about this uses splitting with regular expressions (MATLAB's strsplit which you mentioned):
str = '001a02.jpg';
C = strsplit(str,'[a-zA-Z.]','DelimiterType','RegularExpression');
Results in:
C =
'001' '02' ''
In older versions of MATLAB, before strsplit was introduced, similar functionality was achieved using regexp(...,'split').
If you want to learn more about regular expressions (abbreviated as "regex" or "regexp"), there are many online resources (JGI..)
In your case, if you only need to take the 5th and 6th characters from the string you could use:
D = str(5:6);
... and if you want to convert those into numbers you could use:
E = str2double(str(5:6));
If your number is always at a certain position in the string, you can simply index this position.
In the examples you gave, the number is always the 5th and 6th characters in the string.
filename = '002A12';
num = str2num(filename(5:6));
Otherwise, if the formating is more complex, you may want to use a regular expression. There is a similar question matlab - extracting numbers from (odd) string. Modifying the code found there you can do the following
all_num = regexp(filename, '\d+', 'match'); %Find all numbers in the filename
num = str2num(all_num{2}) %Convert second number from str
i donot know why there is error in this coding:
hex_str1 = '5'
bin_str1 = dec2bin(hex2dec(hex_str1))
hex_str2 = '4'
bin_str2 = dec2bin(hex2dec(hex_str2))
c=xor(bin_str1,bin_str2)
the value of c is not correct when i transform the hex to binary by using the xor function.but when i used the array the value of c is correct.the coding is
e=[1 1 1 0];
f=[1 0 1 0];
g=xor(e,f)
what are the mistake in my first coding to xor of hec to binary value??anyone can help me find the solution...
Your mistake is applying xor on two strings instead of actual numerical arrays.
For the xor command, logical "0"s are represented by actual zero elements. Any non-zero elements are interpreted as logical "1"s.
When you apply xor on two strings, the numerical value of each character (element) is its ASCII value. From xor's point of view, the zeroes in your string are not really zeroes, but simply non-zero values (being equal to the ASCII value of the character '0'), which are interpreted as logical "1"s. The bottom line is that in your example you're xor-ing 111b and 111b, and so the result is 0.
The solution is to convert your strings to logical arrays:
num1 = (bin_str1 == '1');
num2 = (bin_str2 == '1');
c = xor(num1, num2);
To convert the result back into a string (of a binary number), use this:
bin_str3 = sprintf('%d', c);
... and to a hexadecimal string, add this:
hex_str3 = dec2hex(bin2dec(bin_str3));
it is really helpful, and give me the correct conversion while forming HMAC value in matlab...
but in matlab you can not convert string of length more than 52 character using bin2dec() function and similarly hex2dec() can not take hexadecimal character string more than 13 length.