I'm trying to solve a problem here, it asks to find the size of the biggest common subsquare between two matrices.
e.g.
Matrix #1
3 3
1 2 0
1 2 1
1 2 3
Matrix #2
3 3
0 1 2
1 1 2
3 1 2
Answer: 2
Biggest common subsquare is:
1 2
1 2
I know that Rabin-Karp algorithm can be extended to work on a 2D matrix, but I can't understand how exactly can we do that, I tried to understand the author's code in the editorial, but its too complicated, I also did some search for a good explanation, but I couldn't find a clear one.
Can anyone simply explain how can I use Rabin-Karp algorithm to hash a matrix, I know I will hash rows and columns, but I can't see how to mix their hashes together to come up with a hashed matrix, and how the rolling hash function will be handled in this case ?
Related
I want to reduce a two dimensional matrix to row vector.
But using reshape with large matrices is really slow. The other alternative is to use colon, but i want matrix's transpose to be colon and not the matrix itself.
e.g.
A=magic(3)
A =
8 1 6
3 5 7
4 9 2
A(:) will stack up all the columns one by one. but i am looking for something like this:
AA=A(2:3,:)';
and then reshape or colon AA instead of A.
The issue is i dont want to define additional variable like AA.
Is there anyway to reduce dimension of two dimensional matrix without reshape?
You can avoid the additional variable by linear indexing. For your example:
A([2 5 8 3 6 9])
which gives
3 5 7 4 9 2
What's happening here is that you treat A as if it was already transformed into a vector, and the elements of this one-dimensional array are accessed through indices 1 through 9. Using the colon is a special case of linear indexing, A(:) is the same as A(1 : end).
Figuring out the right linear indices can be tricky, but sub2ind can help with that.
It is possible that this slightly speeds up the code, mainly because (as #Shai wrote) you avoid writing data to an intermediate variable. I wouldn't expect too much, though.
Try looking at subsref. For your example, you could use it as follows:
subsref(A',struct('type','()','subs',{{2:3,':'}}))
Update: I misunderstood the original question; I thought the OP wanted to select rows from 2:3 from the transposed A matrix keeping the columns as is. I will keep the previous answer in case it could be useful to others.
I think he/she could use the following to slice and flatten a matrix:
subsref(A(2:3,:)', struct('type','()','subs',{{':'}}))
This would give as output:
[3 5 7 4 9 2]'
I have data that is output from a computational chemistry program (Gaussian09) which contains sets of Force Constant data. The data is arranged with indexes as the first 2-4 columns (quadratic, cubic and quartic FC's are calculated). As an example the cubic FC's look something like this, and MatLab has read them in successfully so I have the correct matrix:
cube=[
1 1 1 5 5 5
1 1 2 6 6 6
.
.
4 1 1 8 8 8
4 2 1 9 9 9
4 3 1 7 7 7 ]
I need a way to access the last 3 columns when feeding in the indices of the first 3 columns. Something along the lines of
>>index=find([cube(:,1)==4 && cube(:,2)==3 && cube(:,3)==1]);
Which would give me the row number of the data that is index [ 4 3 1 ] and allow me to read out the values [7 7 7] which I need within loops to calculate anharmonic frequencies.
Is there a way to do this without a bunch of loops?
Thanks in advance,
Ben
You have already found one way to solve this, by using & in your expression (allowing you to make non-scalar comparisons).
Another way is to use ismember:
index = find(ismember(cube(:,1:3),[4 3 1]));
Note that in many cases, you may not even need the call to find: the binary vector returned by the comparisons or ismember can directly be used to index into another array.
A matrix has 2 rows and several columns, and the first contains alternating strings of 1's and 0's. I want to use this binary as a decision to copy the information below it into one of two cell arrays. I understand that this can be done through iteration with the use of the IF conditional, or with while loops, but I'm having trouble cleaning it up.
For example, for
mat = [ 1 1 1 0 0 1 1 0 0 0 0 1 1 1 ;...
1 2 3 4 5 6 7 1 2 3 4 5 6 7 ]
I would like to output two cell arrays, one for the '1s', and one for the '0s':
1 2 3
6 7
5 6 7
and:
4 5
1 2 3 4
There are a couple of ways of doing this, I am sure. One might be using a loop; however, you can also make good use with the built-in find function. Below is a sample solution based on your example.
For '1s',
Here, we would like to get the indices with '1' from the first row.
on_array= mat(2, find(mat(1,:)));
or as suggested by #H.Muster
on_array= mat(2, mat(1,:)==1);
For '0s',
Here, we would like to get the indices with '0' from the first row.
off_array = mat(2, find(mat(1,:)==0));
or as suggested by #H.Muster
off_array= mat(2, mat(1,:)==0);
For the output format(the one that you want), i am sure you know what to do. Good luck.
I'm learning Scala while going through the Coursera course Functional Programming Principles in Scala.
The first exercise says:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
The numbers at the edge of the triangle are all 1, and each number
inside the triangle is the sum of the two numbers above it. Write a
function that computes the elements of Pascal’s triangle by means of a
recursive process.
Do this exercise by implementing the pascal function in Main.scala,
which takes a column c and a row r, counting from 0 and returns the
number at that spot in the triangle. For example, pascal(0,2)=1,
pascal(1,2)=2 and pascal(1,3)=3.
At the start, I understand, as he refers to the 'numbers' we are all familiar with, but then he goes on to use the term "elements." What does he mean by this? What does he want me to compute?
I assumed that he got bored with the word "number" and thought, after defining the names of the numbers in the triangle as 'numbers' he just wanted to use something new, thus "element," but no matter how I count I cannot get the references to work.
I cannot even really understand the term 'column' seeing as the numbers are not vertically above each other.
Can you please explain how he gets pascal(1,3) == 3?
You're thinking about columns a bit wrong. By "xth column," he means the "xth entry in a given row.
So, if you are looking at the function pascal(c,r), you would want to figure out what the cth number is in the rth row.
So, for example:
pascal(1,2) corresponds to the second entry in the 3rd row
1
1 1
1 *2* 1
pascal(1,3) wants you to look at the second entry in the 4th row.
1
1 1
1 2 1
1 *3* 3 1
Just count from the left. (0,2) is the leftmost number in the row
1 2 1
so (1,3) would be the second number in
1 3 3 1
You can simply make the triangle "rectangle", and everything will become apparent:
cols-> 0 1 2 3 4
row-0 1
row-1 1 1
row-2 1 2 1
row-3 1 3 3 1
row-4 1 4 6 4 1
And you were right in that the triangle's "elements" are made of numbers, though there's a subtle difference, but insubstantial in this case.
P.S. I would personally advice to prefer the course forum for such questions:
It will avoid controversial issues on the honor code.
Your course fellows will have a quicker understanding of the problem at hand
They will have access to material which is not available to those not undertaking the course
It will help to build up a sense of membership amongst the course students, and give you all a chance to create new, possibly fruitful, relashionships
What you're asking is against the Coursera Honor Code: https://www.coursera.org/maestro/auth/normal/tos.php#honorcode
http://www.aiqus.com/questions/41299/coursera-cheating-scala-course
I loved solving this exercise.
My thought process was the following:
Understanding that the problem is a literal description of the binomial coefficient. https://en.wikipedia.org/wiki/Binomial_coefficient
Understanding that the ask is a literal plug into the fomula (!row) / ((!col) * !((row - c))) and the formula is right there in the wiki page
Now the only thing that is missing now is implementing a tail recursive function of factorial
Bonus. if you use the extension method as such
extension (int: Int) {
def ! = factorialTailRec(int)
}
// you get to write
(r.!) / ((c.!) * ((r - c).!))
You get to write almost the identical mathematical formula. And at that moment I realised the similarities between doing maths and programming. And I cried a little with the beauty of it.
I think, the question might have already been asked before. But I could not find proper answer in this forum.
Acutally, I have 2 vectors( of unequal length). I need to compare the 2 vectors. I can do it using a for loop. But it is taking a very long time.
Any obvious method which I may be missising ?
here is a small code snippet:
a=[ 1 2 3 4 5 6 7 8 1 2 3 4];
b=[ 2 3 4];
How can we compare a and b. Basically I need the index in vector a when comparison returns true.
Thanks
You can use strfind() for this (it works with doubles):
idx = strfind(a, b);
idx will contain the indices of all matches.