When p=0;
I have to use a loop to flip the value of p between 0 and 1 and each time it has to be different in a single line of code.
Thank you

2 件のコメント

Image Analyst
Image Analyst 2013 年 2 月 15 日
What does the "When p=0;" mean? You only have to flip when p=0???
When you say "flip the value of p between 0 and 1" does that mean that p is either 0 or 1 only, or does it mean p can have a value between 0 and 1, like p=.426973 and flipping means inverting like flipped_p = 1-p?
Rooy
Rooy 2013 年 2 月 15 日
This was the question given : Write a MATLAB program that will start with the variable p=0. After the initial assignment to p, have a loop flip the value of p between 0 and 1, changing it once each loop iteration USING A SINGLE LINE OF CODE. [HINT: This is a math challenge.] Display it each iteration, using the pause() function to stop the loop temporarily each iteration
Thank you everyone for all the help

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 採用された回答

Image Analyst
Image Analyst 2013 年 2 月 15 日

0 投票

Is this what you're looking for?
p=0;
for k = 1 : 5 % Here's the iterative loop it requested
clc;
% Here's the single line
p = ~p
pause(1);
end
fprintf('Done with demo!\n');

3 件のコメント

Youssef  Khmou
Youssef Khmou 2013 年 2 月 15 日
right !
Jan
Jan 2013 年 2 月 15 日
Alternatively, if p should keep its type DOUBLE:
p = 1 - p;
Image Analyst
Image Analyst 2013 年 2 月 15 日
That also works for any value or p in between 0 and 1, like I asked about in my comment way up at the top. So perhaps this is the most general solution.

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その他の回答 (2 件)

Youssef  Khmou
Youssef Khmou 2013 年 2 月 14 日
編集済み: Youssef Khmou 2013 年 2 月 14 日

1 投票

your_vector=mod(0:1000,2); % single line code
for i=1:length(your_vector)
% YOUR CODE
fprintf('%d\n',your_vector(i))
pause(0.3)
end
Youssef  Khmou
Youssef Khmou 2013 年 2 月 14 日

0 投票

for i=1:N
if mod(i,2)==0
R(i)=1;
end
end

2 件のコメント

Rooy
Rooy 2013 年 2 月 14 日
Would you count this a one line of code ? Thank you
Youssef  Khmou
Youssef Khmou 2013 年 2 月 14 日
ok look at the 2nd answer

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