1
$\begingroup$

I am unable to understand the paper Linear Systems, Sparse Solutions, and Sudoku. I have to form a $4 \times 4$ Sudoku using the algorithm in this paper. Can somebody please provide me the algorithm for $4 \times 4$ Sudoku?

$\endgroup$

closed as too broad by Marcus Müller, Peter K. Jun 20 '17 at 10:01

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I think I saw one on the Mathworks community shared files section. Personally I never cheat at Suduko. $\endgroup$ – Stanley Pawlukiewicz Jun 20 '17 at 3:38
  • 1
    $\begingroup$ Hi! Welcome to Signals.SE. This question is too broad, as we can't really repeat a whole paper in hopes you'll understand it better this time. Please be specific about what you don't understand. Take small steps at first – ask the things you don't understand while reading, then build up an understanding for the whole thing. $\endgroup$ – Marcus Müller Jun 20 '17 at 7:08
  • $\begingroup$ Hi! I'm defining a matrix Ax=B for 4*4 sudoku. But i'm unable to get the output. can you please rectify the errors. The code is as follows A=[ ]; A1=[eye(4,4) eye(4,4) eye(4,4) eye(4,4) zeros(4,48)]; A=[A;A1]; A2=[zeros(4,16) eye(4,4) eye(4,4) eye(4,4) eye(4,4) zeros(4,32)]; A=[A;A2]; A3=[zeros(4,32) eye(4,4) eye(4,4) eye(4,4) eye(4,4) zeros(4,16)]; A=[A;A3]; A4=[zeros(4,48) eye(4,4) eye(4,4) eye(4,4) eye(4,4)]; A=[A;A4]; where A1-A4 are row constraints $\endgroup$ – user29264 Jun 26 '17 at 9:21
  • $\begingroup$ A5=[eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12)]; A=[A;A5]; A6=[zeros(4,4) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) zeros(4,8)]; A=[A;A6]; A7=[zeros(4,8) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) zeros(4,4)]; A=[A;A7]; A8=[zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4) zeros(4,12) eye(4,4)]; A=[A;A8]; these are column constraints $\endgroup$ – user29264 Jun 26 '17 at 9:24
  • $\begingroup$ A9=[eye(4,4) eye(4,4) zeros(4,4) zeros(4,4) eye(4,4) eye(4,4) zeros(4,40)]; A=[A;A9]; A10=[zeros(4,4) zeros(4,4) eye(4,4) eye(4,4) zeros(4,4) zeros(4,4) eye(4,4) eye(4,4) zeros(4,32)]; A=[A;A10]; A11=[zeros(4,32) eye(4,4) eye(4,4) zeros(4,4) zeros(4,4) eye(4,4) eye(4,4) zeros(4,4) zeros(4,4)]; A=[A;A11]; A12=[zeros(4,40) eye(4,4) eye(4,4) zeros(4,4) zeros(4,4) eye(4,4) eye(4,4)]; A=[A;A12]; these are box constraints $\endgroup$ – user29264 Jun 26 '17 at 9:25