Math | The Hoa\Math library | Math library
kandi X-RAY | Math Summary
kandi X-RAY | Math Summary
This library provides tools around mathematical operations.
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- Visits an element
- Main function .
- Returns an integer from a range .
- Initializes function objects .
- Initializes the constants
- Get float value
- Initializes the parameters .
- Moves forward to the next set .
- Rewind to the array
- Show the usage .
Math Key Features
Math Examples and Code Snippets
const lcm = (...arr) => {
const gcd = (x, y) => (!y ? x : gcd(y, x % y));
const _lcm = (x, y) => (x * y) / gcd(x, y);
return [...arr].reduce((a, b) => _lcm(a, b));
};
lcm(12, 7); // 84
lcm(...[1, 3, 4, 5]); // 60
const gcd = (...arr) => {
const _gcd = (x, y) => (!y ? x : gcd(y, x % y));
return [...arr].reduce((a, b) => _gcd(a, b));
};
gcd(8, 36); // 4
gcd(...[12, 8, 32]); // 4
const isNumber = val => typeof val === 'number' && val === val;
isNumber(1); // true
isNumber('1'); // false
isNumber(NaN); // false
Community Discussions
Trending Discussions on Math
QUESTION
ANSWER
Answered 2022-Mar-21 at 14:44Does this work for you?
QUESTION
Given a list of Strings:
...ANSWER
Answered 2022-Mar-22 at 07:13This problem should be solved easily using a trie.
The trie node should basically keep a track of 2 things:
- Child nodes
- Count of prefixes ending at current node
Insert all strings in the trie, which will be done in O(string length * number of strings)
. After that, simply traversing the trie, you can hash the prefixes based on the count as per your use case. For suffixes, you can use the same approach, just start traversing the strings in reverse order.
Edit:
On second thought, trie might be the most efficient way, but a simple hashmap implementation should also work here. Here's an example to generate all prefixes with count > 1.
QUESTION
It seems that JDK 8 and JDK 13 have different floating points.
I get on JDK 8, using Math:
ANSWER
Answered 2022-Mar-20 at 18:16This seems to be caused by a JVM intrinsic function for Math.cos
, which is described in the related issue JDK-8242461. The behavior experienced there is not considered an issue:
The returned results reported in this bug are indeed adjacent floating-point values [this is the case here as well]
[...]
Therefore, while it is possible one or the other of the returned values is outside of the accuracy bounds, just have different return values for Math.cos is not in and of itself evidence of a problem.
For reproducible results, use the StrictMath.cos instead.
And indeed, disabling the intrinsics using -XX:+UnlockDiagnosticVMOptions -XX:DisableIntrinsic=_dcos
(as proposed in the linked issue), causes Math.cos
to have the same (expected) result as StrictMath.cos
.
So it appears the behavior you are seeing here is most likely compliant with the Math
documentation as well.
QUESTION
I have written a Raku script to call erf
function in C standard library:
ANSWER
Answered 2022-Feb-04 at 12:56For the C code, change %f
to %.99g
to show more digits. This reveals erf(4)
returns 0.9999999845827420852373279558378271758556365966796875.
%f
requests six digits after the decimal point. The value is rounded to fit that format. %.numberf
requests number
digits after the decimal point and always used the “fixed” format. %.numberg
requests number
significant digits and uses a a “general” format that switches to exponential notation when appropriate.
For the Raku code, if you want output of “1.0” or “1.000000”, you will need to apply some formatting request to the output. I do not practice Raku, but a brief search shows Raku has printf
-like features you can use, so requesting the %f
format with it should duplicate the C output.
QUESTION
I have an array of positive integers. For example:
...ANSWER
Answered 2022-Feb-27 at 22:44This problem has a fun O(n) solution.
If you draw a graph of cumulative sum vs index, then:
The average value in the subarray between any two indexes is the slope of the line between those points on the graph.
The first highest-average-prefix will end at the point that makes the highest angle from 0. The next highest-average-prefix must then have a smaller average, and it will end at the point that makes the highest angle from the first ending. Continuing to the end of the array, we find that...
These segments of highest average are exactly the segments in the upper convex hull of the cumulative sum graph.
Find these segments using the monotone chain algorithm. Since the points are already sorted, it takes O(n) time.
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
- Is there a short-hand for nth root of x in Python?
- 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 run sample JHM benchmark which suppose to show dead code elimination. Code is rewritten for conciseness from jhm github sample.
...ANSWER
Answered 2022-Feb-09 at 17:17Those samples depend on JDK internals.
Looks like since JDK 9 and JDK-8152907, Math.log
is no longer intrinsified into C2 intermediate representation. Instead, a direct call to a quick LIBM-backed stub is made. This is usually faster for the code that actually uses the result. Notice how measureCorrect
is faster in JDK 17 output in your case.
But for JMH samples, it limits the the compiler optimizations around the Math.log
, and dead code / folding samples do not work properly. The fix it to make samples that do not rely on JDK internals without a good reason, and instead use a custom written payload.
This is being done in JMH here:
QUESTION
Whenever I am trying to run the docker images, it is exiting in immediately.
...ANSWER
Answered 2021-Aug-22 at 15:41Since you're already using Docker
, I'd suggest using a multi-stage build. Using a standard docker image like golang
one can build an executable asset which is guaranteed to work with other docker linux images:
QUESTION
I wrote some code in https://github.com/p6steve/raku-Physics-Measure that looks for a Measure type in each maths operation and hands off the work to non-standard methods that adjust Unit and Error aspects alongside returning the new value:
...ANSWER
Answered 2021-Dec-30 at 03:53There are a few ways to approach this but what I'd probably do – and a generally useful pattern – is to use a subset to create a slightly over-inclusive multi and then redispatch the case you shouldn't have included. For the example you provided, that might look a bit like:
QUESTION
I want to generate a rank 5 100x600 matrix in numpy with all the entries sampled from np.random.uniform(0, 20), so that all the entries will be uniformly distributed between [0, 20). What will be the best way to do so in python?
I see there is an SVD-inspired way to do so here (https://math.stackexchange.com/questions/3567510/how-to-generate-a-rank-r-matrix-with-entries-uniform), but I am not sure how to code it up. I am looking for a working example of this SVD-inspired way to get uniformly distributed entries.
I have actually managed to code up a rank 5 100x100 matrix by vertically stacking five 20x100 rank 1 matrices, then shuffling the vertical indices. However, the resulting 100x100 matrix does not have uniformly distributed entries [0, 20).
Here is my code (my best attempt):
...ANSWER
Answered 2022-Jan-24 at 15:05Not a perfect solution, I must admit. But it's simple and comes pretty close.
I create 5 vectors that are gonna span the space of the matrix and create random linear combinations to fill the rest of the matrix.
My initial thought was that a trivial solution will be to copy those vectors 20 times.
To improve that, I created linear combinations of them with weights drawn from a uniform distribution, but then the distribution of the entries in the matrix becomes normal because the weighted mean basically causes the central limit theorm to take effect.
A middle point between the trivial approach and the second approach that doesn't work is to use sets of weights that favor one of the vectors over the others. And you can generate these sorts of weight vectors by passing any vector through the softmax function with an appropriately high temperature parameter.
The distribution is almost uniform, but the vectors are still very close to the base vectors. You can play with the temperature parameter to find a sweet spot that suits your purpose.
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