I have 2 numeric values...
1. 50634031 (before encoding)
2. 7827165 (after encoding)
The value (1) is the initial value which is encoded with an unknown encoding method to the value (2).As I'm learning about encoding techniques regarding creating a validation mechanishm for an application,My question is that,can we find the encoding technique,so that I can encode another value say 50979163...
No. You can guess which encoding method is used. But there will be an infinite number of mathematical calculations that can transform one single number to another single number.
Related
I have an exponential value for e.g. 3.22122E+23
In Marklogic when I try- xs:decimal(3.22122E+23)
I get this error:
[1.0-ml] XDMP-CAST: (err:FORG0001) xs:decimal(xs:double("3.22122E23")) -- Invalid cast: xs:double("3.22122E23") cast as xs:decimal
A lower value for e.g. xs:decimal(3.22122E+18) gives me the correct result i.e. 3221220000000000000.
I see that this is because of decimal overflow and cannot be represented as a decimal data type but is there any way in Marklogic to handle and calculate such huge values?
Same question applies for the negative values(3.22122E-23) where I can handle and display data above 20 decimal places.
It would be helpful to clarify what kind of logic or calculations you are trying to accomplish and why exactly you need to convert the value to decimal. For example, to "display" the double value, you can use the standard format-number function without any conversion to decimal:
let $x := xs:double(3.22122E+23)
return format-number($x,"#,##0.00")
yields:
322,122,000,000,000,000,000,000.00
See https://docs.marklogic.com/fn:format-number for details regarding fn:format-number() usage.
See https://help.marklogic.com/Knowledgebase/Article/View/487/0/marklogic-server-and-the-decimal-type-implementation for details of the limitations of the xs:decimal type.
I am a beginner to cryptography and i have the following the question. I know that to extract a plaintext from a mono-alphabetic substitution cipher i can perform shifting the characters by n times until i have the plaintext, or i can also use the frequency letter analysis. However by doing this im getting only the plaintext and not they keyword thats been used.
The answer might include creating a table with the alphabet linked to the corresponding letters used in ciphertext by following the frequency letter analysis but im not sure how this works. Can someone explain to me how i will be able to get the keyword?
I am working with neo4r library in R. When i use this function
call_neo4j(con, type = "graph")
I get the error
Fehler in readBin(content, character()) : R character strings are limited to 2^31-1 bytes
Anyone have any idea about it?
As the error would suggest, you reached the limit on the size of character strings. From the documentation:
The number of bytes in a character string is limited to 2^31 - 1 ~ 2*10^9, which is also the limit on each dimension of an array.
Without more information, we can't help you solve this problem. See here to create a minimal reproducible example that could help us solve your problem.
I'm running into some problems with the R function as.character() and paste(): they do not give back what they're being fed...
as.character(1415584236544311111)
## [1] "1415584236544311040"
paste(1415584236544311111)
## [1] "1415584236544311040"
what could be the problem or a workaround to paste my number as a string?
update
I found that using the bit64 library allowed me to retain the extra digits I needed with the function as.integer64().
Remember that numbers are stored in a fixed number of bytes based upon the hardware you are running on. Can you show that your very big integer is treated properly by normal arithmetic operations? If not, you're probably trying to store a number to large to store in your R install's integer # of bytes. The number you see is just what could fit.
You could try storing the number as a double which is technically less precise but can store larger numbers in scientific notation.
EDIT
Consider the answers in long/bigint/decimal equivalent datatype in R which list solutions including arbitrary precision packages.
I have a very long expression that I need to take the derivative of:
D(expression(-4750000+(((14400*(((x/25)*1)-7.2))+(0*((x*1.05)-7.2))+(144*
((x*0.6)-7.2))*30.41667)-2500)/((0.1+1)^((1-0.5)/12))+(((13216.5802942644*
(((x/25)*1)-7.2))+(0*((x*1.05)-7.2))+(132.165802942644*
((x*0.6)-7.2))*30.41667)-2500)/((0.1+1)^((2-0.5)/12))+.........),'x')
When the expression is less than 4000 characters I get a solution. When it is more than 4000 characters, R gives me a newline and +, expecting more input. I haven't been able to find documentation on a character limit. Does anyone know why this might be? Any workarounds or alternatives to finding this derivative? The final character length will be at least 50k.
You are probably seeing an effect of input through stdin which uses the OS newline function. Try putting it in a text file and source()-ing.