When I use the following line on sage
A=random_matrix(QQ,4,5,algorithm='echelonizable',rank=3,upper_bound=10); latex(A); A
I get:
\left(\begin{array}{rrrrr}
1 & 1 & 7 & -4 & 7 \\
0 & 1 & 4 & -3 & 4 \\
0 & 2 & 9 & -6 & 9 \\
0 & 0 & -4 & 0 & -4
\end{array}\right)
[ 1 1 7 -4 7]
[ 0 1 4 -3 4]
[ 0 2 9 -6 9]
[ 0 0 -4 0 -4]
How can I get the following format instead?:
1 , 1 , 7 , -4 , 7 ;
0 , 1 , 4 , -3 , 4 ;
0 , 2 , 9 , -6 , 9 ;
0 , 0 , -4 , 0 , -4;
I thought that it could be an easy way to modify latex() however it is getting too hard to do. Maybe there is a way to get the desired display format without using latex().
Best,
Chilote
I'm not sure what program you are trying to emulate here - we have initializers for matrices in most scientific software you might want to import it into (Octave/Matlab, Mathematica, Maxima, Numpy, ...). But here is a hack from one of them that should work:
sage: M = random_matrix(ZZ,3,3)
sage: M._scilab_init_()[1:-1]
'-1,1,-2;11,0,5;-1,-19,1'
You are correct that you shouldn't try to mess with the latex function; better to do a search and replace or something in one of the existing methods. Good luck.
Related
Hi I'm new to nim and dont get how evaluations of arithmetics work?
3 and 1 # Outputs 1
and
4 and 1 # Outputs 0
Which logic is going on here?
Thanks
I think, in binary: 3 = 0b'11, 1 = 0b'01, so 3 and 1 = 0b'01, while 4 = 0b'100 and 1 = 0b'001, so 4 and 1 = 0b'00.
Suppose I have the following sample data file.
0 1 2
0 3 4
0 1 9
0 9 2
0 19 0
0 6 1
0 11 0
1 3 2
1 3 4
1 1 6
1 9 2
1 15 0
1 6 6
1 11 1
2 3 2
2 4 4
2 1 6
2 9 6
2 15 0
2 6 6
2 11 1
first column gives value of time. Second gives values of x and 3rd column y. I wish to plot graphs of y as functions of x from this data file at different times,
i.e, for t=0, I shall plot using 2:3 with lines up to t=0 index. Then same thing I shall do for the variables at t=1.
At the end of the day, I want to get a gif, i.e, an animation of how the y vs x graph changes shape as time goes on. How can I do this in gnuplot?
What have you tried so far? (Check help ternary and help gif)
You need to filter your data with the ternary operator and then create the animation.
Code:
### plot filtered data and animate
reset session
$Data <<EOD
0 1 2
0 3 4
0 1 9
0 9 2
0 19 0
0 6 1
0 11 0
1 3 2
1 3 4
1 1 6
1 9 2
1 15 0
1 6 6
1 11 1
2 3 2
2 4 4
2 1 6
2 9 6
2 15 0
2 6
2 11 1
EOD
set terminal gif animate delay 50 optimize
set output "myAnimation.gif"
set xrange[0:20]
set yrange[0:10]
do for [i=0:2] {
plot $Data u 2:($1==i?$3:NaN) w lp pt 7 ti sprintf("Time: %g",i)
}
set output
### end of code
Result:
Addition:
The meaning of $1==i?$3:NaN in words:
If the value in the first column is equal to i then the result is the value in the third column else it will be NaN ("Not a Number").
It is the first time I deal with column-compress storage (CCS) format to store matrices. After googling a bit, if I am right, in a matrix having n nonzero elements the CCS is as follows:
-we define a vector A_v of dimensions n x 1 storing the n non-zero elements
of the matrix
- we define a second vector A_ir of dimensions n x 1 storing the rows of the
non-zero elements of the matrix
-we finally define a third vector A_jc whose elements are the indices of the
elements of A_v which corresponds to the beginning of new column, plus a
final value which is by convention equal t0 n+1, and identifies the end of
the matrix (pointing theoretically to a virtual extra-column).
So for instance,
if
M = [1 0 4 0 0;
0 3 5 2 0;
2 0 0 4 6;
0 0 7 0 8]
we get
A_v = [1 2 3 4 5 7 2 4 6 8];
A_ir = [1 3 2 1 2 4 2 3 3 4];
A_jc = [1 3 4 7 9 11];
my questions are
I) is what I wrote correct, or I misunderstood anything?
II) what if I want to represent a matri with some columns which are zeroes, e.g.,
M2 = [0 1 0 0 4 0 0;
0 0 3 0 5 2 0;
0 2 0 0 0 4 6;
0 0 0 0 7 0 8]
wouldn't the representation of M2 in CCS be identical to the one of M?
Thanks for the help!
I) is what I wrote correct, or I misunderstood anything?
You are perfectly correct. However, you have to take care that if you use a C or C++ library offsets and indices should start at 0. Here, I guess you read some Fortran doc for which indices are starting at 1. To be clear, here is below the C version, which simply translates the indices of your Fortran-style correct answer:
A_v = unmodified
A_ir = [0 2 1 0 1 3 1 2 2 4] (in short [1 3 2 1 2 4 2 3 3 4] - 1)
A_jc = [0 2 3 6 8 10] (in short [1 3 4 7 9 11] - 1)
II) what if I want to represent a matri with some columns which are
zeroes, e.g., M2 = [0 1 0 0 4 0 0;
0 0 3 0 5 2 0;
0 2 0 0 0 4 6;
0 0 0 0 7 0 8]
wouldn't the representation of M2 in CCS be identical to the one of M?
I you have an empty column, simply add a new entry in the offset table A_jc. As this column contains no element this new entry value is simply the value of the previous entry. For instance for M2 (with index starting at 0) you have:
A_v = unmodified
A_ir = unmodified
A_jc = [0 0 2 3 6 8 10] (to be compared to [0 2 3 6 8 10])
Hence the two representations are differents.
If you just start learning about sparse matrices there is an excelllent free book here: http://www-users.cs.umn.edu/~saad/IterMethBook_2ndEd.pdf
I am interested in how can I add rows and columns of zeros in a matrix so that it looks like this:
1 0 2 0 3
1 2 3 0 0 0 0 0
2 3 4 => 2 0 3 0 4
5 4 3 0 0 0 0 0
5 0 4 0 3
Actually I am interested in how can I do this efficiently, because walking the matrix and adding zeros takes a lot of time if you work with a big matrix.
Update:
Thank you very much.
Now I'm trying to replace the zeroes with the sum of their neighbors:
1 0 2 0 3 1 3 2 5 3
1 2 3 0 0 0 0 0 3 8 5 12... and so on
2 3 4 => 2 0 3 0 4 =>
5 4 3 0 0 0 0 0
5 0 4 0 3
as you can see i'm considering all the 8 neighbors of an element, but again using for and walking the matrix slows me down quite a bit, is there a faster way ?
Let your little matrix be called m1. Then:
m2 = zeros(5)
m2(1:2:end,1:2:end) = m1(:,:)
Obviously this is hard-wired to your example, I'll leave it to you to generalise.
Here are two ways to do part 2 of the question. The first does the shifts explicitly, and the second uses conv2. The second way should be faster.
M=[1 2 3; 2 3 4 ; 5 4 3];
% this matrix (M expanded) has zeros inserted, but also an extra row and column of zeros
Mex = kron(M,[1 0 ; 0 0 ]);
% The sum matrix is built from shifts of the original matrix
Msum = Mex + circshift(Mex,[1 0]) + ...
circshift(Mex,[-1 0]) +...
circshift(Mex,[0 -1]) + ...
circshift(Mex,[0 1]) + ...
circshift(Mex,[1 1]) + ...
circshift(Mex,[-1 1]) + ...
circshift(Mex,[1 -1]) + ...
circshift(Mex,[-1 -1]);
% trim the extra line
Msum = Msum(1:end-1,1:end-1)
% another version, a bit more fancy:
MexTrimmed = Mex(1:end-1,1:end-1);
MsumV2 = conv2(MexTrimmed,ones(3),'same')
Output:
Msum =
1 3 2 5 3
3 8 5 12 7
2 5 3 7 4
7 14 7 14 7
5 9 4 7 3
MsumV2 =
1 3 2 5 3
3 8 5 12 7
2 5 3 7 4
7 14 7 14 7
5 9 4 7 3
I'm creating a word search and am trying to calculate quality of the generated puzzles by verifying the word set is "distributed evenly" throughout the grid. For example placing each word consecutively, filling them up row-wise is not particularly interesting because there will be clusters and the user will quickly notice a pattern.
How can I measure how 'evenly distributed' the words are?
What I'd like to do is write a program that takes in a word search as input and output a score that evaluates the 'quality' of the puzzle. I'm wondering if anyone has seen a similar problem and could refer me to some resources. Perhaps there is some concept in statistics that might help? Thanks.
The basic problem is distribution of lines in a square or rectangle. You can eighter do this geometrically or using integer arrays. I will try the integer arrays here.
Let M be a matrix of your puzzle,
A B C D
E F G H
I J K L
M N O P
Let the word "EFGH" be an existent word, as well as "CGKO". Then, create a matrix which will contain the count of membership in eighter words in each cell:
0 0 1 0
1 1 2 1
0 0 1 0
0 0 1 0
Apply a rule: the current cell value is equal to the sum of all neighbours (4-way) and multiply with the cell's original value, if the original value is 2 or higher.
0 0 1 0 1 2 2 2
1 1 2 1 -\ 1 3 8 2
0 0 1 0 -/ 1 2 3 2
0 0 1 0 0 1 1 1
And sum up all values in rows and columns the matrix:
1 2 2 2 = 7
1 3 8 2 = 14
1 2 3 2 = 8
0 1 1 1 = 3
| | | |
3 7 | 6
14
Then calculate the avarage of both result sets:
(7 + 14 + 8 + 3) / 4 = 32 / 4 = 8
(3 + 7 + 14 + 6) / 4 = 30 / 4 = 7.5
And calculate the avarage difference to the avarage of each result set:
3 <-> 7.5 = 4.5 7 <-> 8 = 1
7 <-> 7.5 = 0.5 14 <-> 8 = 6
14 <-> 7.5 = 6.5 8 <-> 8 = 0
6 <-> 7.5 = 1.5 3 <-> 8 = 5
___avg ___avg
3.25 3
And multiply them together:
3 * 3.25 = 9.75
Which you treat as a distributionscore. You might need to tweak it a little bit to make it work better, but this should calculate distributionscores quite nicely.
Here is an example of a bad distribution:
1 0 0 0 1 1 0 0 2
1 0 0 0 -\ 2 1 0 0 -\ 3 -\ C avg 2.5 -\ C avg-2-avg 0.5
1 0 0 0 -/ 2 1 0 0 -/ 3 -/ R avg 2.5 -/ R avg-2-avg 2.5
1 0 0 0 1 1 0 0 2 _____*
6 4 0 0 1.25 < score
Edit: calc. errors fixed.