I want to extract linear combinations from matrices but by performing combinations in modulus.
Let us consider the calculation modulus 5, we then have the following for the addition:
+ | 0 1 2 3 4
--+-----------
0 | 0 1 2 3 4
1 | 1 2 3 4 0
2 | 2 3 4 0 1
3 | 3 4 0 1 2
4 | 4 0 1 2 3
and this table for the multiplication:
* | 0 1 2 3 4
--+-----------
0 | 0 0 0 0 0
1 | 0 1 2 3 4
2 | 0 2 4 1 3
3 | 0 3 1 4 2
4 | 0 4 3 2 1
So let us take an example:
Let us consider the following matrix:
E = 2 1 3 2 0
4 3 0 1 1
Then we can obtain the triangulation matrix by applying a LU decomposition (https://en.wikipedia.org/wiki/LU_decomposition) (or a Gaussian Elimination), which is the following:
T = 1 0 0 0 0
2 1 0 0 0
and finally, the matrix that I want to extract, which is the one storing the linear combinations:
CL = 3 2 3 0 3
0 1 1 3 4
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
So basically, the algorithm should work as follows:
Input: a matrix E with n rows and m columns, and p, a prime number.
* We perform a Gaussian elimination/LU decomposition to obtain the lower-triangulation matrix T.
But all the calculus are done modulo 'p'.
Output: T (with the same size as E, n rows m columns).
CL (with a size m rows, m columns),
which is basically the identity matrix on which we
applied all the modifications that were performed on E to obtain T.
Alright, so now we have the context, let me explain the problem.
I started to do it using the Armadillo library (http://arma.sourceforge.net/), but I did not find any solution on the library to perform the calculus on a mathematical Field p. I easily found the LU method to obtain the lower-triangle matrix, but the calculations are performed in the real.
#include <iostream>
#include <armadillo>
using namespace arma;
using namespace std;
int main(int argc,char** argv)
{
mat A = mat({{2,1,3,2,0},{4,3,0,1,1}});
mat L, U, P;
lu(L, U, P, A);
cout << L << endl;
return 0;
}
With the following, you obtain the lower-triangle matrix 'L' but in the real calculus. Thus you obtain:
T' = 1 0
1/2 1
Is there any technique to perform the computation in a modulus way?
EDIT The Armadillo library is not able to do it. I developed my own LU decomposition in modulus but there is still a bug there. I asked a new question here Linear Combination C++ in modulus, hoping to solve it.
First of all: drop the using namespaces, code can become completely unreadable if you do that.
I haven't used Armadillo yet. But I have looked at the documentation, and it seems templated to me.
Now things are getting a bit wild. The type you use, arma::mat seems to be a typedef on arma::Mat<double>.
The high-level function arma::lu isn't properly documented. It obviously does an LU-decomposition, but I don't know if the function is templated. If it is, i.e., you cannot just call it with double mats but also other types, you might have a shot using a custom type representing the field (since 5 is prime, otherwise you'd be completely lost) of calculations modulo 5. Meaning you write a class, let's call it IntMod5 and define all required operators for this class, meaning all operators that IntMod5 uses. For example, you'd need to define operator/(), e.g. by making a table of inverses of 4 of the 5 elements of the field (0 has none), i.e. 1->1, 2->3, 3->2, 4->4, and define
IntMod5 operator/(const IntMod5& o) const
{
return IntMod5((this->value*inverse(o.value))%5);
}
This is just one example, you likely need to define all arithmetic operators, binary and unary, and possibly more such as comparison (LU decomposition might use finding good pivot elements). If you're lucky and the library is written in a way that it works for any field, not just floating point, you have a chance.
Before you go through all the work, you should use a trivial class simply wrapping double and check if arma::Mat or arma::lu do any type checks blocking you out.
If either of these fails, you'll likely have to write your own LU decomposition modulo 5 or find another library that supports it.
Related
This is my try to count the contiguous subsequences of an array with product mod 4 is not equal to 2:
# include <iostream>
using namespace std;
int main() {
long long int n, i, j, s, t, count = 0;
cin>>n;
long long int arr[n];
count = 0;
for(i = 0; i<n; i++) {
cin>>arr[i];
}
for(i = 0; i<n; i++) {
s = 1;
for(j = i; j<n; j++) {
s = s*arr[j];
if(s%4!=2) {
count++;
}
}
}
cout<<count;
return 0;
}
However, I want to reduce the time taken by my code to execute. I am looking for a way to do it. Any help/hint would be appreciated.
Thank you.
What does this definition of contiguous subsequences mean?
Listing all the subsequences
Suppose we have the sequence:
A B C D E F
First of all, we should recognize that there is one substring for every unique start and end point. Let's use the notation C-F to mean all items from C through F: i.e.: C D E F.
We can list all subsequences in a triangular arrangement like this:
A B C D E F
A-B B-C C-D D-E E-F
A-C B-D C-E D-F
A-D B-E C-F
A-E B-F
A-F
The first row lists all the subsequences of length 1.
The second row lists all the subsequences of length 2.
The third row lists all the subsequences of length 3. Etc.
The last row is the full sequence.
Modular arithmetic
Computing the product MOD 4 of a set of numbers
To figure out the product of a bunch of numbers MOD 4, we just need to look at each element of the set MOD 4. Intuitively, this is because when you multiply a bunch of numbers, the last digit of the result is determined entirely by the last digit of each factor. In this case "the last digit base 4" is the number mod 4.
The identity we are using is:
(A * B) MOD N == ((A MOD N) * (B MOD N)) MOD N
The table of products
Now we also have to look at the matrix of possible multiplications that might happen. It's a fairly small table and the interesting entries are given here:
2 * 2 = 4 4 MOD 4 = 0
2 * 3 = 6 6 MOD 4 = 2
3 * 3 = 9 9 MOD 4 = 1
So the results of multiplying any 2 numbers MOD 4 are given by this table:
+--------+---+---+---+---+
| Factor | 0 | 1 | 2 | 3 |
+--------+---+---+---+---+
| 0 | 0 | / | / | / |
| 1 | 0 | 1 | / | / |
| 2 | 0 | 2 | 0 | / |
| 3 | 0 | 3 | 2 | 1 |
+--------+---+---+---+---+
The /'s are omitted because of the symmetry of multiplication (A * B = B * A)
An example sequence
Now for each subsequence, let's compute the product MOD 4 of its elements.
Consider the following list of numbers
242 497 681 685 410 795
The first thing we do is take all these numbers MOD 4 and list them as the first row of our list of all subsequences triangle.
2 0 1 1 2 3
The second row is just the product of the pairs above it.
2 0 1 1 2 3
0 0 1 2 3
In general, the Nth element of each row is the product, MOD 4, of:
the number just to its left in the row above left times and
the element in the first row that is diagonally to its right
For example C = A * B
* * * * B *
* * * / *
* A / *
* C *
* *
*
Again,
A is immediately up and left of C
B is diagonally right all the way to the top row from C
Now we can complete our triangle
2 0 1 1 2 3
0 0 1 2 3
0 0 2 3
0 0 2
0 0
0
This can be computed easily in O(n^2) time.
Optimization
These optimizations do not improve the time complexity of the algorithm in its worse case, but can cause an early exit in the computation, and should therefore be included if time is to be reduced and the input is unknown.
Contageous 0's
Furthermore, as a matter of optimization, notice how contagious the 0's are. Anything times 0 is 0, so you can skip computing products of cells below a 0. In your case those sequences will not equal 2 MOD 4 once the product of one of its subsequences is determined to be equal to 0 MOD 4.
* * * 0 * * // <-- this zero infects all cells below it
* * 0 0 *
* 0 0 0
0 0 0
0 0
0
Need a 2 to make a 2.
Look back at the table of factors and products. Notice that the only way to get a product that is equal to 2 MOD 4 is to have one of the factors be equal to 2 MOD 4. What that means is that there can only be a 2 below another 2. So we are only interested in following computing entries in the table that are below a 2. Other entries in rows below can never become a 2.
You don't have to store more than the whole rows.
You only need O(n) storage to implement this. Working line by line, you can compute the values in a row entirely from the values in the first row and values in the row above.
Reading the answers from the table
Now you can look at the rows of the triangle list as you generate them and read off which subsequences are to be included.
Entries with a 2 are to be excluded. All others are to be included.
2 0 1 1 3 2
0 0 1 3 2
0 0 3 2
0 0 2
0 0
0
The excluded subsequences for the example (which I will list only because there are fewer of them in my example) are:
A
F
E-F
D-F
C-F
Which remember, according to our convention refer to the elements:
A
F
E F
D E F
C D E F
Which are:
242
795
410 795
685 410 795
681 685 410 795
Hopefully it's obvious how to display the "included" sequences, rather than the "excluded" sequences, as I have shown above.
Displaying all the elements makes it take much longer.
Sadly, actually displaying all of the elements of such subsequences is still an O(N^3) operation in the worst case. (Imagine a sequence of all zeros.)
Summary
For me, I feel like an average developer could take the magic bullet observation made in the diagram below and write an implementation that has optimal time complexity.
C = A * B
* * * * B *
* * * / *
* A / *
* C *
* *
*
I am using the Eigen library in C++.
I need to insert a row and column to an existing matrix at specific index.
For example, say I need to insert a 0 row and 0 column at the 2nd index...
ORIGINAL MATRIX (A)
1 2 3
1 2 3
1 2 3
NEW MATRIX (B)
1 2 0 3
1 2 0 3
0 0 0 0
1 2 0 3
Thanks for the help in advance!
The new matrix B can be constructed from the original matrix A by using the block operations .topRows() and .bottomRows():
MatrixXd B = MatrixXd::Zero(4, 3);
B.topRows(2) = A.topRows(2);
B.bottomRows(1) = A.bottomRows(1);
This will insert a row of zeros between the second and third row. Analogous operations with .rightCols() and .leftCols() can be used to insert a column of zeros.
Sorry if this is a duplicate, but I did not find any answers which match mine.
Consider that I have a vector which contains 3 values. I want to construct another vector of a specified length from this vector. For example, let's say that the length n=3 and the vector contains the following values 0 1 2. The output that I expect is as follows:
0 0 0
0 0 1
0 0 2
0 1 0
0 1 1
0 1 2
0 2 0
0 2 1
0 2 2
1 0 0
1 0 1
1 0 2
1 1 0
1 1 1
1 1 2
1 2 0
1 2 1
1 2 2
2 0 0
2 0 1
2 0 2
2 1 0
2 1 1
2 1 2
2 2 0
2 2 1
2 2 2
My current implementation simply constructs for loops based on nand generates the expected output. I want to be able to construct output vectors of different lengths and with different values in the input vector.
I have looked at possible implementations using next_permutation, but unfortunately passing a length value does not seem to work.
Are there time and complexity algorithms that one can use for this case? Again, I might have compute this for up to n=17and sizeof vector around 6.
Below is my implementation for n=3. Here, encis the vector which contains the input.
vector<vector<int> > combo_3(vector<double>enc,int bw){
vector<vector<int> > possibles;
for (unsigned int inner=0;inner<enc.size();inner++){
for (unsigned int inner1=0;inner1<enc.size();inner1++){
for (unsigned int inner2=0;inner2<enc.size();inner2++){
cout<<inner<<" "<<inner1<<" "<<inner2<<endl;
unsigned int arr[]={inner,inner1,inner2};
vector<int>current(arr,arr+sizeof(arr)/sizeof(arr[0]));
possibles.push_back(current);
current.clear();
}
}
}
return possibles;
}
What you are doing is simple counting. Think of your output vector as a list of a list of digits (a vector of a vector). Each digit may have one of m different values where m is the size of your input vector.
This is not permutation generation. Generating every permutation means generating every possible ordering of an input vector, which is not what you're looking for at all.
If you think of this as a counting problem the answer may become clearer to you. For example, how would you generate all base 10 numbers with 5 digits? In that case, your input vector has size 10, and each vector in your output list has length 5.
I'm trying to do the following in CUSP:
A=[
1,1,0,0;
2,2,2,0;
0,3,3,3;
0,0,4,4];
B=[1,1,1,1]';
disp(mldivide(A,B));
which is
X=[0.9167,0.0833,-0.5000,0.7500]
On the other hand I get a strange answer from CUSP
#include <cusp/dia_matrix.h>
#include <cusp/krylov/cg.h>
#include <cusp/print.h>
int main()
{
cusp::dia_matrix<int,float,cusp::host_memory> A(4,4,10,3);
A.diagonal_offsets[0] = -1;
A.diagonal_offsets[1] = 0;
A.diagonal_offsets[2] = 1;
for (int i = 0;i <3;i++)
{
for (int q = 0 ;q < A.num_cols;q++)
{
A.values(q,i)=q+1;
}
}
//copy
cusp::dia_matrix<int,float,cusp::device_memory> AA = A;
cusp::array1d<float,cusp::device_memory> BB(A.num_rows,1);
cusp::array1d<float,cusp::device_memory> XX(A.num_rows,0);
cusp::print(AA);
cusp::print(XX);
cusp::print(BB);
cusp::krylov::cg(AA,XX,BB);\
cusp::print(XX);
return 0;
}
The result looks like
sparse matrix <4, 4> with 10 entries
0 0 1
0 1 1
1 0 2
1 1 2
1 2 2
2 1 3
2 2 3
2 3 3
3 2 4
3 3 4
array1d <4>
0
0
0
0
array1d <4>
1
1
1
1
array1d <4>
-39.9938
-53.436
87.9025
-30.1429
The last one doesn't look quite right. Anybody know what I'm doing wrong? Am I using the code wrong or are we supposed to have a really good guessed solution + use a preconditioner?
The conjugate gradient method is only valid for use in symmetric positive definite matrices. Your matrix isn't symmetric. That is why it isn't (and cannot) producing a valid solution. Either use an appropriate, well conditioned SPD matrix, or use a different numerical method.
Using eigen2, and given a matrix A
a_0_0, a_0_1, a_0_2, ...
a_1_0, a_1_0, a_1_2, ...
...
and a matrix B:
b_0_0, b_0_1, b_0_2, ...
b_1_0, b_1_1, b_1_2, ...
...
and where A and B have the same dimensions, I would like to interleave the rows, producing:
a_0_0, a_0_1, a_0_2, ...
b_0_0, b_0_1, b_0_2, ...
a_1_0, a_1_0, a_1_2, ...
b_1_0, b_1_1, b_1_2, ...
...
Obviously I can write a function that will construct an output matrix of the proper dimensions, then loop over each of the input matrices and assign elements to the result. I'd rather not re-invent the wheel though, so if eigen2 already has a mechanism to express this sort of matrix surgery elegantly I'd much prefer to use it.
I did look through the eigen2 docs and nothing jumped out at me as obviously correct. The closest thing I found was MatrixBase::select, but that does 'element from a or element from b', where what I want is 'element from a then element from b in the next row'.
Efficiency is not of paramount concern since I don't need to do this in the fast path, only at initialization.
I apologize for the formatting if there is a better way to represent matrices.
Multiply each R x C matrix by a 2R x R matrix consisting of zeroes and ones on the appropriate diagonal, then add.
Matrix 1
1 0 0 0 ...
0 0 0 0 ...
0 1 0 0 ...
0 0 0 0 ...
Matrix 2
0 0 0 0 ...
1 0 0 0 ...
0 0 0 0 ...
0 1 0 0 ...
Not sure if this is specific to Eigen3, but you can interleave rows using the Map and Stride objects.
MatrixXi C(A.rows()+B.rows(),A.cols());
Map<MatrixXi,0,Stride<Dynamic,2> >(C.data(),A.rows(),A.cols(),Stride<Dynamic,2>(2*A.rows(),2)) = A;
Map<MatrixXi,0,Stride<Dynamic,2> >(C.data()+1,B.rows(),B.cols(),Stride<Dynamic,2>(2*B.rows(),2)) = B;
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