How do I generate a given Matrix in one command?

42 Ansichten (letzte 30 Tage)
marciuc
marciuc am 23 Feb. 2011
Kommentiert: DGM am 23 Feb. 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 Kommentare
Paulo Silva
Paulo Silva am 24 Feb. 2011
+1 vote to marciuc just because of the beer :)
Jan
Jan am 25 Feb. 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.

Melden Sie sich an, um zu kommentieren.

Antworten (17)

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

Melden Sie sich an, um zu kommentieren.


Jan
Jan am 23 Feb. 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 Kommentare
Andrew Newell
Andrew Newell am 23 Feb. 2011
A small price to pay for elegance!
Jiro Doke
Jiro Doke am 23 Feb. 2011
Or if you have MATLAB R2008b or newer,
A = randi([0 1], 5)

Melden Sie sich an, um zu kommentieren.


Jan
Jan am 23 Feb. 2011
A = dec2bin([24 28 14 7 3]) - '0'
  4 Kommentare
David Young
David Young am 24 Feb. 2011
dec2bin('vzlea'-'^')-'0'
Jan
Jan am 24 Feb. 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.

Melden Sie sich an, um zu kommentieren.


Matt Tearle
Matt Tearle am 23 Feb. 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 am 28 Feb. 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 Kommentar
Jan
Jan am 28 Feb. 2011
*You* have suggested it now. So I can't believe it now, too. +1

Melden Sie sich an, um zu kommentieren.


Jan
Jan am 23 Feb. 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 Kommentare
Sean de Wolski
Sean de Wolski am 23 Feb. 2011
Can I have your beer ?!
Matt Tearle
Matt Tearle am 23 Feb. 2011
You can have all the beer that marciuc's teacher sends me. How's that for a deal?

Melden Sie sich an, um zu kommentieren.


Paulo Silva
Paulo Silva am 23 Feb. 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 Kommentar
Sean de Wolski
Sean de Wolski am 23 Feb. 2011
Yay! The infinite monkey makes its second appearance in a week.

Melden Sie sich an, um zu kommentieren.


Walter Roberson
Walter Roberson am 25 Feb. 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 Kommentar
Jan
Jan am 25 Feb. 2011
Thanks for line breaking. This solution will drive marciuc's teacher crazy. Such ugly! +1

Melden Sie sich an, um zu kommentieren.


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

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

Jan
Jan am 24 Feb. 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 am 24 Feb. 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 am 24 Feb. 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 Kommentare
Jan
Jan am 25 Feb. 2011
Brr.
Matt Fig
Matt Fig am 25 Feb. 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.

Melden Sie sich an, um zu kommentieren.


DGM
DGM am 23 Feb. 2023
Bearbeitet: DGM am 23 Feb. 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 Kommentare
Stephen23
Stephen23 am 23 Feb. 2023
+1 very nice. What magic did you use to reverse engineer that?
DGM
DGM am 23 Feb. 2023
I made use of the classic "commit to brute force and then go to bed" algorithm.

Melden Sie sich an, um zu kommentieren.


Paulo Silva
Paulo Silva am 23 Feb. 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 am 24 Feb. 2011
Thank you guys...regards from Romania
  1 Kommentar
Jan
Jan am 24 Feb. 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.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Historical Contests finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by