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結果:
Over the weekend I came across a pi approximation using durations of years and weeks (image below, Wolfram, eq. 89), accurate to 6 digits using the average Gregorian year (365.2425 days).
Here it is in MATLAB. I divided by 1 week at the end rather than multiplying by its reciprocal because you can’t divide a numeric by a duration in MATLAB (1/week).
weeks = @(n)n*days(7);
piApprox = ((years(13)-weeks(6))/years(13) + weeks(3)) / weeks(1)
% piApprox = 3.141593493469302
Here’s a breakdown
- The first argument becomes 12.885 yrs / 13 yrs or 0.99115
- Add three weeks: 0.99115 + 3 weeks = 21.991 days
- The reduced fraction becomes 21.991 days / 7 days
Now it looks a lot closer to the more familiar approximation for pi 22/7 but with greater precision!
I'm in a community conference in Boston today and see what snacks we get! The organizer said it's a coincidence, but it's definitly a good idea to have them in our MathWorks community meetings.
(Sorry - it should be 2023b by now.)
spy
I'm curious how the community uses the hold command when creating charts and graphics in MATLAB. In short, hold on sets up the axes to add new objects to the axes while hold off sets up the axes to reset when new objects are added.
When you use hold on do you always follow up with hold off? What's your reasoning on this decision?
Can't wait to discuss this here! I'd love to hear from newbies and experts alike!
The way we've solved ODEs in MATLAB has been relatively unchanged at the user-level for decades. Indeed, I consider ode45 to be as iconic as backslash! There have been a few new solvers in recent years -- ode78 and ode89 for example -- and various things have gotten much faster but if you learned how to solve ODEs in MATLAB in 1997 then your knowledge is still applicable today.
In R2023b, there's a completely new framework for solving ODEs and I love it! You might argue that I'm contractually obliged to love it since I'm a MathWorker but I can assure you this is the real thing!
I wrote it up in a tutorial style on The MATLAB Blog https://blogs.mathworks.com/matlab/2023/10/03/the-new-solution-framework-for-ordinary-differential-equations-odes-in-matlab-r2023b/
The new interface makes a lot of things a much easier to do. Its also setting us up for a future where we'll be able to do some very cool algorithmic stuff behind the scenes.
Let me know what you think of the new functionality and what you think MathWorks should be doing next in the area of ODEs.
To solve the puzzle, first unscramble each of the words on the left. Then rearrange the letters in the yellow shaded boxes to complete the sentence on the right.
If you enjoyed this puzzle let me know with a like or in the comments below and I'll post more of them. Please don't post your answer, or any hints, and spoil it for those who come across this puzzle after you!! If you want to check your answer, you can messge me your guess through the link on my profile card (click on my name, Rena Berman, above and then on the envelope icon in the top right corner of the profile card that appears).
Thats the task:
Given a square cell array:
x = {'01', '56'; '234', '789'};
return a single character array:
y = '0123456789'
I wrote a code that passes Test 1 and 2 and one that passes Test 3 but I'm searching a condition so that the code for Test 3 runs when the cell array only contains letters and the one for Test 1 and 2 in every other case. Can somebody help me?
This is my code:
y = []
[a,b]=size(x)
%%TEST 3
delimiter=zeros(1,a)
delimiter(end)=1
delimiter=repmat(delimiter,1,b)
delimiter(end)=''
delimiter=string(delimiter)
y=[]
for i=1:a*b
y = string([y x(i)])
end
y=join(y,delimiter)
y=erase(y,'0')
y=regexprep(y,'1',' ')
%%TEST 1+2
for i=1:a*b
y = string([y x(i)])
y=join(y)
end
y=erase(y,' ' )
That's the question: Given four different positive numbers, a, b, c and d, provided in increasing order: a < b < c < d, find if any three of them comprise sides of a right-angled triangle. Return true if they do, otherwise return false .
I wrote this code but it doesn't pass test 7. I don't really understand why it isn't working. Can somebody help me?
function flag = isTherePythagoreanTriple(a, b, c, d)
a2=a^2
b2=b^2
c2=c^2
d2=d^2
format shortG
if a2+b2==c2
flag=true
else if a2+b2==d2
flag=true
else if a2+c2==d2
flag=true
else if c2+b2==d2
flag=true
else flag=false
end
end
end
end
end
That's the question:
The file cars.mat contains a table named cars with variables Model, MPG, Horsepower, Weight, and Acceleration for several classic cars.
Load the MAT-file. Given an integer N, calculate the output variable mpg.
Output mpg should contain the MPG of the top N lightest cars (by Weight) in a column vector.
I wrote this code and the resulting column vector has the right values but it doesn't pass the tests. What's wrong?
function mpg = sort_cars(N)
load cars.mat
sorted=sortrows(cars,4)
mpg = sorted(1:N,2)
end
I recently have found that I am no longer able to give my difficulty rating for questions on Cody after sucessfully completing a question. This is obviously not a big deal, I was just wondering if this was an issue on my end or if there was some change that I was not aware of.
The option to rate does not pop up after solving a problem, and the rating in general does not even show up anymore when answering questions (though it is visible from problem groups).
The MATLAB Answers community is an invaluable resource for all MATLAB users, providing selfless assistance and support. However, with the emergence of AI-based chatbots, like chatGPT, there may be concerns about the future relevance and utility of the MATLAB Answer community. What are your thoughts?
This is the 6th installment of the wish-list and bug report thread.
This topic is the follow on to the first Wish-list for MATLAB Answer sections and second MATLAB Answers Wish-list #2 (and bug reports). The third started out as New design of the forum - grey on white and the fourth and fifth also grew so large they are slow to load and navigate.
Same idea as the previous ones: one wish (or bug report) per answer, so that people can vote their wishes.
What should you post where?
Next Gen threads (#1): features that would break compatibility with previous versions, but would be nice to have
@anyone posting a new thread when the last one gets too large (about 50 answers seems a reasonable limit per thread), please update this list in all last threads. (if you don't have editing privileges, just post a comment asking someone to do the edit)
When solving problems over on Cody, I can almost always view all solutions to a problem after submitting a correct solution of my own. Very rarely, however, this is not the case, and I instead get the following message:
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
You may solve another problem from Community group to unlock all the solutions to this problem.
If this happens, then again, I can almost always rectify this by submitting a (correct) solution to a different problem (I take it that the Community group is the implicit group of all problems on Cody --- is it?). But sometimes that, too, fails.
So my question is, why? What are the criteria that determine when all solutions are, in fact, unlocked?
Simple question: I noticed there's a Modeling & Simulation Challenge Master badge over on Cody, but I can't find the corresponding group. So: where is it? Does it still exist at all?
Error: The server timed out while running and assessing your solution in MATLAB CODY. How do I resolve this? My code is correct. I have run it on PC. But, when i submit in CODY the server throws an error.
Just in case, I have my license of MATLAB. I just have this question and I didn't find any information. I wouldn't like to create another account, for this reason I prefer to ask here.
I have submitted a problem in cody some days back. Now it is not showing in my profile. Initially it was accepted and some people submitted the solutions also, however It has been removed after that, are there some guidelines which I am not following?
This is not a question, but a point of discussion for the entire community. I am aware that every 1/2 months this theme comes out, but until this is not fixed it is totally necessary that this comes, indeed, out. And I said "fix" because Mathworks has to understand that a dark theme is not only a visual/aesthetic matter, it is a substantial part of the game. Most of the OS, GUIs, programs are actually in dark mode, and a vast majority of the users makes indeed use of a global dark mode. How much one does like it is personal, but the benefits to power savings and eye health is instead a fact. Mathworks being ignoring this for years is nothing but ridiculous. Of course it is not an easy task, but every minute of committment for it is worthy. And nope, Schemer is not helpful because it does not provide a real fix to this question.
I feel free to suggest something similar to the Spyder's dark theme, which came out like 2 years ago if I remember correctly.
Of course, my point is not being disrespectful (I am instead very respectful to the huge efforts of Mathworks for making this wonderful program run). But, form a user's point of view, the fact that not a single word has so far come out from Mathworks about a dark theme (meaning that for sure we will not see it in a timing of months) requires us to put a strong pressure on this.
Mathworks, please: it's time for a dark theme.
Hi everyone
I am a new of this community and I very interested in this forum and Matlab.I am trying to submit a soultion but as tiltle my code has a built-in function so the test systerm dont reconisie it.It run completely ok on my computer.
This is problem
This is my solution
function [boOut] = BoIfPointInPoly(PolyMatrix,p_test)
%Summary of this function goes here
%{
if we draw a line from test point to a central point of a side of The polygon
then we extend that line to the furthest point of the polygon ensure that
line go through all side of Polygon in 1 direction.I call that line is line_test
Next find number of intersert of line test and all sides w polyxpoly
function
num interset point is odd mean p_test inside
num interset point is even mean p_test outside
this solution go from the concept that if a line go in from a side it has go out
from other side.So if it go in but not go out that mean it start from
inside.
%}
% Detailed explanation goes here
%line from p test throuh central of a side to furthest point of polygon
%find vector
V = ((PolyMatrix(1,:) + PolyMatrix(2,:)) /2) - p_test ;
%draw that vector to furtest point
pend = p_test + V * max(PolyMatrix(:));
%with multi of V and biggest element I assume that line will go all out the
%polygon which ensure out logic will right
line_test = [p_test ; pend];
disp('Our line test\n');
disp(line_test);
%find interst point
p_inter = polyxpoly(PolyMatrix(:,1),PolyMatrix(:,2),line_test(:,1),line_test(:,2));
%find number of interset (row)
[numIntere,trash] = size(p_inter);
disp('Number of interest point:');
disp(numIntere);
%determine in or out
if (rem(numIntere,2) == 0)
boOut = 0;
else
boOut = 1;
end
end
Can anyone has solution.
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