pretty-formula | small Java library to parse mathematical formulas | Math library
kandi X-RAY | pretty-formula Summary
kandi X-RAY | pretty-formula Summary
A small Java library to parse mathematical formulas to LaTeX and display them as images
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Top functions reviewed by kandi - BETA
- Initialize the components
- Matches a variable
- Matches an Atom context
- Parses a term
- Evaluate a semprepred rule
- Return true if the predicate evaluates to true
- Visit all expressions
- Translates a negative number
- Visit all child elements
- Visit number
- Implements the visitor function
- Visit all children
- Gets the text of the variable
- Sets the look and feel
- Translate a negative variable
pretty-formula Key Features
pretty-formula Examples and Code Snippets
Community Discussions
Trending Discussions on Math
QUESTION
I recently came across this problem:
You are given two strings, s1 and s2, comprised entirely of lowercase letters 'a' through 'r', and need to process a series of queries. Each query provides a subset of lowercase English letters from 'a' through 'r'. For each query, determine whether s1 and s2, when restricted only to the letters in the query, are equal. s1 and s2 can contain up to 10^5 characters, and there are up to 10^5 queries.
For instance, if s1 is "aabcd" and s2 is "caabd", and you are asked to process a query with the subset "ac", then s1 becomes "aac" while s2 becomes "caa". These don't match, so the query would return false.
I was able to solve this in O(N^2) time by doing the following: For each query, I checked if s1 and s2 would be equal by iterating through both strings, one character at a time, skipping the characters that do not lie within the subset of allowed characters, and checking to see if the "allowed" characters from both s1 and s2 match. If at some point, the characters don't match, then the strings are not equal. Otherwise, the s1 and s2 are equal when restricted only to letters in the query. Each query takes O(N) time to process, and there are N queries, for a total of O(N^2) time.
However, I was told that there was a way to solve this faster in O(N). Does anyone know how this might be done?
...ANSWER
Answered 2022-Mar-28 at 11:30The first obvious speedup is to ensure your set membership test is O(1). To do that, there's a couple of options:
- Represent every letter as a single bit -- now every character is an 18-bit value with only one bit set. The set of allowed characters is now a mask with these bits ORed together and you can test membership of a character with a bitwise-AND;
- Alternatively, you can have an 18-value array and index it by character (
c - 'a'
would give a value between 0 and 17). The test for membership is then basically the cost of an array lookup (and you can save operations by not doing the subtraction -- instead just make the array larger and index directly by character.
The next potential speedup is to recognize that any character which does not appear exactly the same number of times in both strings will instantly be a failed match. You can count all character frequencies in both strings with a histogram which can be done in O(N) time. In this way, you can prune the search space if such a character were to appear in the query, and you can test for this in constant time.
Of course, that won't help for a real stress-test which will guarantee that all possible letters have a frequency matched in both strings. So, what do you do then?
Well, you extend the above premise by recognizing that for any position of character x
in string 1 and some position of that character in string 2 that would be a valid match (i.e the same number of character x
appears in both strings up to their respective positions), then the total count of any other character up to those positions must also be equal. For any character where that is not true, it cannot possibly be compatible with character x
.
Let's start by thinking about this in terms of a technique known as memoization where you can leverage precomputed or partially-computed information and get a whole lot out of it. So consider two strings like this:
QUESTION
It is a number whose gcd of (sum of quartic power of its digits, the product of its digits) is more than 1. eg. 123 is a special number because hcf of(1+16+81, 6) is more than 1.
I have to find the count of all these numbers that are below input n. eg. for n=120 their are 57 special numbers between (1 and 120)
I have done a code but its very slow can you please tell me to do it in some good and fast way. Is there is any way to do it using some maths.
...ANSWER
Answered 2022-Mar-06 at 18:14The critical observation is that the decimal representations of special numbers constitute a regular language. Below is a finite-state recognizer in Python. Essentially we track the prime factors of the product (gcd > 1 being equivalent to having a prime factor in common) and the residue of the sum of powers mod 2×3×5×7, as well as a little bit of state to handle edge cases involving zeros.
From there, we can construct an explicit automaton and then count the number of accepting strings whose value is less than n using dynamic programming.
QUESTION
I need to calculate the square root of some numbers, for example √9 = 3
and √2 = 1.4142
. How can I do it in Python?
The inputs will probably be all positive integers, and relatively small (say less than a billion), but just in case they're not, is there anything that might break?
Related
- Integer square root in python
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- Difference between **(1/2), math.sqrt and cmath.sqrt?
- Why is math.sqrt() incorrect for large numbers?
- Python sqrt limit for very large numbers?
- Which is faster in Python: x**.5 or math.sqrt(x)?
- Why does Python give the "wrong" answer for square root? (specific to Python 2)
- calculating n-th roots using Python 3's decimal module
- How can I take the square root of -1 using python? (focused on NumPy)
- Arbitrary precision of square roots
Note: This is an attempt at a canonical question after a discussion on Meta about an existing question with the same title.
...ANSWER
Answered 2022-Feb-04 at 19:44math.sqrt()
The math
module from the standard library has a sqrt
function to calculate the square root of a number. It takes any type that can be converted to float
(which includes int
) as an argument and returns a float
.
QUESTION
I write a mathematical function to be benchmark function in my optimization algorithm.
...ANSWER
Answered 2022-Feb-12 at 13:14In the platform that produces “-4,09139395927863E+154”, the Math.Cos
routine is broken. It apparently uses a processor instruction that does not support operands outside [−2−63, +2−63].
Since I do not use C#, here is a C program that reproduces the correct behavior:
QUESTION
In particular, it must work with NaNs as std::copysign
does. Similarly, I need a constexpr std::signbit
.
ANSWER
Answered 2021-Sep-20 at 19:54If you can use std::bit_cast
, you can manipulate floating point types cast to integer types. The portability is limited to the representation of double
, but if you can assume the IEEE 754 double-precision binary floating-point format, cast to uint64_t and using sign bit should work.
QUESTION
I'm working on some heavy algorithm, and now I'm trying to make it multithreaded. It has a loop with 2 nested loops:
...ANSWER
Answered 2021-Dec-20 at 09:25A third attempt:
I've taken your code, and at last got it to run properly (in python):
QUESTION
I would like to make the following sequence in R, by using rep
or any other function.
ANSWER
Answered 2022-Jan-04 at 15:43Use sequence
.
QUESTION
I am trying to achieve a calculation involving geometric progression (split). Is there any effective/efficient way of doing it. The data set has millions of rows. I need the column "Traded_quantity"
Marker Action Traded_quantity 2019-11-05 09:25 0 0 09:35 2 BUY 3 09:45 0 0 09:55 1 BUY 4 10:05 0 0 10:15 3 BUY 56 10:24 6 BUY 8128turtle = 2 (User defined)
base_quantity = 1 (User defined)
...ANSWER
Answered 2022-Jan-22 at 10:09This should work
QUESTION
I want to create a polynomial ring which has float Coefficients like this. I can create with integers but, Floats does not work.
...ANSWER
Answered 2022-Jan-18 at 23:30While I do not have previous experience with this particular (from appearances, rather sophisticated) package Oscar.jl, parsing this error message tells me that the function you are trying to call is being given a BigFloat
as input, but simply does not have a method for that type.
At first this was a bit surprising given that there are no BigFloat
s in your input, but after a bit of investigation, it appears that the culprit is the following
QUESTION
I don't know if this is possible, but I am trying to take the image of a custom outdoor football field layout and have the players' GPS
coordinates correspond to the image x
and y
position. This way, it can be viewed via the app to show the players' current location on the field as a sort of live tracking.
I have also looked into this Convert GPS coordinates to coordinate plane. The problem is that I don't know if this would work and wanted to confirm beforehand. The image provided in the post was for indoor location, and it was from 11
years ago.
I used Location
and Google Maps
packages for flutter. The player's latitude
and longitude
correspond to the actual latitude
and longitude
that the simulator in the android studio shows when tested.
The layout in question and a close comparison to the result I am looking for.
Any help on this matter would be appreciated highly, and thanks in advance for all the help.
Edit:
After looking more at the matter I tried the answer of this post GPS Conversion - pixel coords to GPS coords, but it wasn't working as intended. I took some points on the image and the correspond coordinates, and followed the same logic that the answer used, but reversed it to give me the actual image X
, Y
positions.
The formula that was given in the post above:
...ANSWER
Answered 2022-Jan-12 at 08:20First of All, Yes you can do this with high accuracy if the GPS coordinates are accurate.
Second, the main problem is rotation if the field are straight with lat lng lines this would be easy and straightforward (no bun intended).
The easy way is to convert coordinate to rotated image similar to the real field then rotated every X,Y point to the new straight image. (see the image below)
Here is how to rotate x,y knowing the angel:
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