How do I generate a given Matrix in one command?

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marciuc
marciuc el 23 de Feb. de 2011
Comentada: DGM el 23 de Feb. de 2023
I have to generate a matrix A = [1 1 0 0 0; 1 1 1 0 0; 0 1 1 1 0; 0 0 1 1 1; 0 0 0 1 1]
writing a single command.
Teacher told as some helpful commands would be
EYE(m,n)
ONES(m,n)
ZEROS(m,n)
RAND(m,n)
I give you a beer
  4 comentarios
Paulo Silva
Paulo Silva el 24 de Feb. de 2011
+1 vote to marciuc just because of the beer :)
Jan
Jan el 25 de Feb. de 2011
+1: If you can explain how all solutions given here work, you will pass the complete Matlab course based on just one question. Every single answer is funny, but all answers together are serious.

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Respuestas (17)

Jan
Jan el 23 de Feb. de 2011
Refering to your former post, which has been deleted now: I know also, who you are: you are marciuc.
At first I suggest this:
A = [1 1 0 0 0;1 1 1 0 0;0 1 1 1 0;0 0 1 1 1;0 0 0 1 1]
This is a single command and it is the most efficient solution: No temporary memory, no overhead for calling commnad, and easy to debug. There is no better solution.

Sean de Wolski
Sean de Wolski el 23 de Feb. de 2011
Probably the most compact:
A = toeplitz([1 1 0 0 0])
  1 comentario
Jan
Jan el 23 de Feb. de 2011
This is the most compact command, except for the ambitious RAND apporach. I vote it.

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Jan
Jan el 23 de Feb. de 2011
Give the beer to your teacher. He is obviously funny if he suggests RAND - but it really works:
A = round(rand(5))
There is at least a certain chance to get the correct answer.
  5 comentarios
Andrew Newell
Andrew Newell el 23 de Feb. de 2011
A small price to pay for elegance!
Jiro Doke
Jiro Doke el 23 de Feb. de 2011
Or if you have MATLAB R2008b or newer,
A = randi([0 1], 5)

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Jan
Jan el 23 de Feb. de 2011
A = dec2bin([24 28 14 7 3]) - '0'
  4 comentarios
David Young
David Young el 24 de Feb. de 2011
dec2bin('vzlea'-'^')-'0'
Jan
Jan el 24 de Feb. de 2011
There is the vuvuzela again. The question is obviously more diabolic than I've expected. Again you see: Subtracting zero can reveal the clandestine information.

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Matt Tearle
Matt Tearle el 23 de Feb. de 2011
A = full(gallery('tridiag',ones(1,4),ones(1,5),ones(1,4)))
But my current favorite:
A = 1-reshape(mod(floor((1:25)/3),2),5,5)

Kenneth Eaton
Kenneth Eaton el 28 de Feb. de 2011
I can't believe no one suggested dilation:
A = imdilate(eye(5),ones(2));
Or convolution:
A = sign(conv2(eye(5),ones(2),'same'));
A = sign(filter2(ones(2),eye(5)));
  1 comentario
Jan
Jan el 28 de Feb. de 2011
*You* have suggested it now. So I can't believe it now, too. +1

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Jan
Jan el 23 de Feb. de 2011
A general method to create a diagonal matrix is using DIAG (as the example in "help diag" explains):
A = diag(ones(1, 5)) + diag(ones(1,4), 1) + diag(ones(1,4), -1);
You can discuss, if this is still "a single command".
I do not drink beer. But you can ask your teacher to send me the points gained by solving this homework.
  2 comentarios
Sean de Wolski
Sean de Wolski el 23 de Feb. de 2011
Can I have your beer ?!
Matt Tearle
Matt Tearle el 23 de Feb. de 2011
You can have all the beer that marciuc's teacher sends me. How's that for a deal?

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Paulo Silva
Paulo Silva el 23 de Feb. de 2011
Here's probably the most awesome way to generate the matrix :D
disp('I dare you to try the Infinite monkey matrix')
answer=input('Press y and Enter if your dare to try','s')
if (strcmp(answer,'y'))
disp('Congratulations your are not a coward')
disp('Good luck')
pause(1)
disp('Please wait or press CTRL+C to cancel')
disp('but canceling the operation makes you a coward!!')
a=[1 1 0 0 0; 1 1 1 0 0; 0 1 1 1 0; 0 0 1 1 1; 0 0 0 1 1];
w=0;b=zeros(5,5);
while ~isequal(a,b)
b=randi([0 1],5,5);
w=w+1;
end
disp('Congratulations we found the Infinite monkey matrix for you')
b
disp('after')
w
disp('attempts')
else
disp('You are a coward!!!!')
end
  1 comentario
Sean de Wolski
Sean de Wolski el 23 de Feb. de 2011
Yay! The infinite monkey makes its second appearance in a week.

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Walter Roberson
Walter Roberson el 25 de Feb. de 2011
EDIT: line-broken per request.
eval(char(mod(1.0599.^ ...
'i<o<ZC<C<d<d<d_C<C<C<d<d_d<C<C<C<d_d<d<C<C<C_d<d<d<C<Cv', ...
96)))
  1 comentario
Jan
Jan el 25 de Feb. de 2011
Thanks for line breaking. This solution will drive marciuc's teacher crazy. Such ugly! +1

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Jan
Jan el 23 de Feb. de 2011
A = abs(bsxfun(@minus, 1:5, transpose(1:5))) < 2

Jan
Jan el 23 de Feb. de 2011
A = 1 - (tril(ones(5), -2) + triu(ones(5), 2))

Jan
Jan el 24 de Feb. de 2011
I cannot resist to post some variation of the DEC2BIN theme:
A = dec2bin('8<.''#' - 32) - '0'
A = dec2bin('FJ<51' - 46) - '0'
But finally you can even omit the first subtraction, because DEC2BIN operates on CHAR vectors also, but you cannot type the non-printables directly:
q = [100 101 99 50 98 105 110 40 39 24 28 14 7 3 39 41 45 39 48 39];
clipboard('copy', char(q))
==> Ctrl-v in the command window
>> dec2bin('#####')-'0'
Here the '#' are the non-printables with the ASCII codes [24,28,14,7,3]. You can write them even in a M-file.

Paulo Silva
Paulo Silva el 24 de Feb. de 2011
b=[0 0 1 1 1
0 0 0 1 1
1 0 0 0 1
1 1 0 0 0
1 1 1 0 0]
A=~b;

Matt Fig
Matt Fig el 24 de Feb. de 2011
One line, anyway. And since the array is at least dynamically pre-allocated, the code is fast.
for ii = 5:-1:1,for jj = min(ii+1,5):-1:max(ii-1,1),A(ii,jj) = 1;end,end
  2 comentarios
Jan
Jan el 25 de Feb. de 2011
Brr.
Matt Fig
Matt Fig el 25 de Feb. de 2011
LOL Jan. I am surprised that such an answer is faster than both the BSXFUN and EYE + DIAG + DIAG solutions, even for N=1000.

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DGM
DGM el 23 de Feb. de 2023
Editada: DGM el 23 de Feb. de 2023
I can't believe that everyone has missed the obvious solution!
rng(46783490); randi([0 1],5,5)
ans = 5×5
1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1
so simple!
  2 comentarios
Stephen23
Stephen23 el 23 de Feb. de 2023
+1 very nice. What magic did you use to reverse engineer that?
DGM
DGM el 23 de Feb. de 2023
I made use of the classic "commit to brute force and then go to bed" algorithm.

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Paulo Silva
Paulo Silva el 23 de Feb. de 2011
diag(diag(eye(4,4)),1)+diag(diag(eye(4,4)),-1)+eye(5,5)
or
diag(ones(1,4),1)+diag(ones(1,4),-1)+eye(5,5)
It's similar to the Jan solution above

marciuc
marciuc el 24 de Feb. de 2011
Thank you guys...regards from Romania
  1 comentario
Jan
Jan el 24 de Feb. de 2011
Please do us the favor and accept one of the answers. Or let your teacher choose one.
Beside the fun, this thread will be really helpful, because it describes the creation of tridiagonal matrices exhaustively.
I really hope you had some fun also.

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