I need to apply "not" operator to matrix of zeros and ones in Julia.
In Matlab I would do this:
A=not(B);
In Julia I tried doing this:
A = .~ B;
and
A = .! B;
It should turn zeros to ones and ones to zeros but I get error as a result or all matrix elements are some negative numbers that I didn't enter.
Thanks in advance!
The issue with A = .!B is that logical negation, !(::Int64), isn't defined for integers. This makes sense: What should, say, !3 reasonably give?
Since you want to perform a logical operation, is there a deeper reason why you are working with integers to begin with?
You could perhaps work with a BitArray instead which is vastly more efficient and should behave like a regular Array in most operations.
You can easily convert your integer matrix to a BitArray. Afterwards, applying a logical not works as expected.
julia> A = rand(0:1, 5,5)
5×5 Array{Int64,2}:
0 0 0 1 1
0 1 0 0 1
0 1 1 1 0
1 1 0 0 0
1 1 1 0 0
julia> B = BitArray(A)
5×5 BitArray{2}:
0 0 0 1 1
0 1 0 0 1
0 1 1 1 0
1 1 0 0 0
1 1 1 0 0
julia> .!B
5×5 BitArray{2}:
1 1 1 0 0
1 0 1 1 0
1 0 0 0 1
0 0 1 1 1
0 0 0 1 1
The crucial part here is that the element type (eltype) of a BitArray is Bool, for which negation is obviously well defined. In this sense, you could also use B = Bool.(A) to convert all the elements to booleans.
For a general solution to going from A where A is a matrix of numbers to a boolean matrix with true values where there were zeros and false values elsewhere, you can do this:
julia> A = rand(0:3, 5, 5)
5×5 Array{Int64,2}:
1 0 1 0 3
2 0 1 1 0
2 1 1 3 1
1 0 3 0 3
1 3 3 1 2
julia> (!iszero).(A)
5×5 BitArray{2}:
1 0 1 0 1
1 0 1 1 0
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1
To break down what's going on here:
iszero is a predicate that tests if a scalar value is zero
!iszero is a predicate that returns if a scalar value is not zero
(!iszero).(A) broadcasts the !iszero function over the matrix A
This returns a BitArray with the desired pattern of zeros (falses) and ones (trues). Note that in an array context, false prints as 0 and true prints as 1 (they are numerically equal). You can also compare with the number 0 like this:
julia> A .!= 0
5×5 BitArray{2}:
1 0 1 0 1
1 0 1 1 0
1 1 1 1 1
1 0 1 0 1
1 1 1 1 1
You can also roll your own:
not(x) = (x |> Bool |> !) |> Float64
defines a method that will convert x to boolean, apply not, and then convert the result back to numbers. not.(A) will act element-wise on the array A. Here |> redirects the output to the next method and works with broadcasting.
While not conceptually the cleanest, A=1.-B will do what you want. The problem with ~ is that it is performing a bitwise not on integers, which produces negative numbers. Not sure what is wrong wiht ! except it maybe should be !.B
Related
I have dug around and googled but not found an example. I'm sure Julia has a powerful function (in base?) to generate random binomial (bernoulli?) "successes" with a given probability. I can't find it or figure out how to do the equivalent to in Julia:
> rbinom(20,1,0.3)
[1] 1 1 1 0 0 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0
Thx. J
You can use Distributions and the rand function for this. Any distribution can be passed to rand. To replicate what you want:
julia> using Distributions
julia> p = Binomial(1, 0.3) # first arg is number of trials, second is probability of success
Binomial{Float64}(n=1, p=0.3)
julia> rand(p, 20)
20-element Array{Int64,1}:
0
1
1
0
1
0
0
1
0
1
1
1
0
0
1
0
1
0
0
1
I am trying to write code to detect if a matrix is permutation of a Hankel matrix but I can't think of an efficient solution other than very slow brute force. Here is the spec.
Input: An n by n matrix M whose entries are 1 or 0.
Input format: Space separated rows. One row per line. For example
0 1 1 1
0 1 0 1
0 1 0 0
1 0 1 1
Output: A permutation of the rows and columns of M so that M is a Hankel matrix if that is possible. A Hankel matrix has constant skew-diagonals (positive sloping diagonals).
When I say a permutation, I mean we can apply one permutation to the order of the rows and a possibly different one to the columns.
I would be very grateful for any ideas.
Without Loss of Generality, we will assume that there are fewer 0's than 1's. We can then find the possible diagonals in a Hankel Matrix that could be 0's to give us the appropriate number of 0's in the entire matrix. And, this will give us the possible Hankel matrices. From there, you can count the number of 0's in each column, and compare it to the number of 0's in the columns of the original matrix. Once you have done this, you have a much smaller space in which to perform a brute force search: permuting on columns and rows that have the right number of 0's.
Example: OP's suggested a 4x4 matrix with 7 0's. We need to partition this using the set {4,3,3,2,2,1,1}. So, or partitions would be:
{4,3}
{4,2,1} (2 of these matrices)
{3,3,1}
{3,2,2}
{3,2,1,1} (2 of these matrices)
And this gives us the Hankel Matrices (excluding symmetries)
1 1 0 0 1 1 1 0 0 1 1 0 1 1 0 1
1 0 0 1 1 1 0 1 1 1 0 1 1 0 1 0
0 0 1 1 1 0 1 0 1 0 1 0 0 1 0 1
0 1 1 1 0 1 0 0 0 1 0 1 1 0 1 0
1 0 0 1 0 1 1 1 0 1 0 1
0 0 1 1 1 1 1 0 1 0 1 1
0 1 1 0 1 1 0 0 0 1 1 0
1 1 0 1 1 0 0 0 1 1 0 0
The original matrix had columns with 3, 1, 2, and 1 0's in its four columns. Comparing this to the 7 possible Hankel matrices gives us 2 possibilities
1 1 1 0 0 1 1 1
1 1 0 1 1 1 1 0
1 0 1 0 1 1 0 0
0 1 0 0 1 0 0 0
Now, there are only 4 possible permutations that could map the original matrix to each of these: we have only 1 choice based on the columns with 2 and 3 0's, but 2 choices for the columns with 1 0's, and also 2 choices for the rows with 1 0's. Checking those permutations, we see that the following Hankel matrix is a permutation of the original
0 1 1 1
1 1 1 0
1 1 0 0
1 0 0 0
The one thing which the first answer to this question got right is that permuting the rows and columns doesn't change the row sums or column sums.
Another easy observation is that in a Hankel matrix, the difference in row sum between two consecutive rows is -1, 0, or 1, and each case gives us a constraint on the rows. If the difference is 0 then the entering variable is equal to the exiting variable; otherwise we know which is 0 and which is 1.
0 1 1 1
0 1 0 1
0 1 0 0
1 0 1 1
has row sums 3, 2, 1, 3. The orders which respect the difference requirement are 1 2 3 3 and 3 3 2 1, and wlog we can discard reversals because reversing the row and column permutations just rotates the matrix by 180 degrees. Therefore we reduce to considering four permuted matrices (two possible orderings of the 3s in the row sums, and two in the column sums):
0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0
1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1
We could actually have taken the analysis further by observing that by forcing the initial rows to have sums 1 and 2 we constrain the order of the columns with sum 3, since
0 0 1 0
0 0 1 1
is not a valid initial two rows of a Hankel matrix. Whether or not this kind of reasoning is easy to implement depends on your programming paradigm.
Note that in the worst case this kind of reasoning still doesn't leave a polynomial number of cases to brute force through.
Here are some ideas.
1)
Row and Column permutations preserve the row and column sums:
1 0 1 0 - 2
0 0 0 1 - 1 row sums
1 0 0 0 - 1
1 1 1 0 - 3
| | | |
3 1 2 1
column sums
Whichever way you permute the rows, the row sums will still be {2, 1, 1, 3} in some permutation; the column sums will be unchanged. And vice versa. Hankel matrices and their permutations will always have the same set of row sums as column sums. This gives you a quick test to rule out a set of non-viable matrices.
2)
I posit that Hankel matrices can always be permuted in such a way that their row and column sums are in ascending order, and the result is still a Hankel matrix:
0 1 1 0 - 2 0 0 0 1 - 1
1 1 0 0 - 2 0 0 1 1 - 2
1 0 1 1 - 3 --> 0 1 1 0 - 2
0 0 1 0 - 1 1 1 0 1 - 3
| | | | | | | |
2 2 3 1 1 2 2 3
Therefore if a matrix can be permuted into a Hankel matrix, then it can also be permuted into a Hankel matrix of ascending row and column sum. That is, we can reduce the number of permutations needed to test by only testing permutations where the row and column sums are in ascending order.
3)
I posit further that for any Hankel matrix where two or more rows have the same sum, every permutation of columns has a matching permutation of rows that also produces a Hankel matrix. That is, if a Hankel matrix exists for one permutation of columns, then it exists for every permutation of columns - since we can simply apply that same permutation to the corresponding rows and achieve a symmetrical result.
The upshot is that we only need to test permutations of rows or columns, not rows and columns.
Applied to the original example:
1 0 1 0 - 2 0 0 0 1 0 1 0 0 - 1 0 0 0 1
0 0 0 1 - 1 1 0 0 0 0 0 0 1 - 1 0 1 0 0
1 0 0 0 - 1 --> 1 0 1 0 --> 0 0 1 1 - 2 --> 0 0 1 1 = Hankel!
1 1 1 0 - 3 1 1 1 0 1 0 1 1 - 3 1 0 1 1
| | | |
3 1 2 1 permute rows into| ditto | try swapping
ascending order | for columns | top 2 rows
4)
I posit, finally, that every Hankel matrix where there are multiple rows and columns with the same sum can be permuted into another Hankel matrix with the property that those rows and columns are in increasing order when read as binary numbers - reading left-to-right for rows and top-to-bottom for columns. That is:
0 1 1 0 0 1 0 1 0 0 1 1
1 0 0 1 0 1 1 0 0 1 0 1 New
1 0 1 0 --> 1 0 0 1 --> 1 0 1 0 Hankel
0 1 0 1 1 0 1 0 1 1 0 0
Original rows columns
Hankel ascending ascending
If this is true (and I'm still undecided), then we only ever need to create and test one permutation of any given input matrix. That permutation puts both the rows and columns in order of ascending sum, and in the case of equal sums, orders them by their binary number interpretations. If this resultant matrix is not Hankel, then there is no permutation that will make it Hankel.
Hope that gets you on the way to an algorithm!
Addendum: Counterexamples?
Trying #orlp's example:
0 0 1 0 0 0 1 0 0 0 0 1
0 1 0 1 0 1 0 1 0 1 1 0
1 0 1 1 --> 0 1 1 1 --> 0 1 1 1
0 1 1 1 1 0 1 1 1 0 1 1
(A) (B) (C)
A: Original Hankel. Row sums are 1, 2, 3, 3; Rows 3 and 4 are not in binary order.
B: Swap rows 3 and 4. Columns 3 and 4 are not in binary order.
C: Swap columns 3 and 4. Result is Hankel and satisfies all the properties.
Trying #Degustaf's example:
1 1 0 1 0 1 0 0 0 0 1 0
1 0 1 0 1 0 0 1 0 1 0 1
0 1 0 0 --> 1 0 1 0 --> 1 0 0 1
1 0 0 1 1 1 0 1 0 1 1 1
(A) (B) (C)
A: Original Hankel matrix. Row sums are 3, 2, 1, 2.
B: Rearrange so that the row sums are 1, 2, 2, 3, and the rows of sum 2 are in ascending binary order (i.e. 1001, 1010)
C: Rearrange column sums to 1, 2, 2, 3, with the two columns of sum 2 in order (0101, 1001). Result is Hankel and satisfies all the properties. Note also that the permutation on the columns matches the permutation on the rows: the new column order from the old one is {3, 4, 2, 1}, the same operation to get from A to B.
Note: I suggest the binary order (#4) only for tiebreak situations on the row or column sum, not as a replacement for the sort in (#2).
I have this piece of code
((⍳3)∘.+(⍳2))
which generates the following matrix
2 3
3 4
4 5
I want to find the occurrence of each unique element in the result i.e occurrence of 2,3,4,5 in the result.
I tried using "∘.=" with the matrix itself and then reshaping such that elements of each sub matrix is transformed into a row
using
6 6⍴ ((⍳3)∘.+(⍳2))∘.=((⍳3)∘.+(⍳2))
which gives the following result
1 0 0 0 0 0 for 2
0 1 1 0 0 0 for 3
0 1 1 0 0 0 for 3
0 0 0 1 1 0 for 4
0 0 0 1 1 0 for 4
0 0 0 0 0 1 for 5
as you can see it still contains the sum for duplicate items, and I'm lost as of now.
Any help will be appreciated.
You should do ∘.= between the unique elements in the matrix and a flat vector of all elements, like:
m ← ((⍳3)∘.+(⍳2))
(∪,m) ∘.= ,m
1 0 0 0 0 0
0 1 1 0 0 0
0 0 0 1 1 0
0 0 0 0 0 1
Then just do +/ on it to get the frequencies of ∪,m
+/ (∪,m) ∘.= ,m
1 2 2 1
∪,m
2 3 4 5
(Tested on GNU APL.)
Dyalog APL version 14.0 has the ⌸ Key operator exactly for this, you just need to ravel your data:
{≢⍵}⌸ ,((⍳3)∘.+(⍳2))
1 2 2 1
Try it online!
You can even use the left argument of ⌸'s operand function to create a table:
{⍺,≢⍵}⌸ ,((⍳3)∘.+(⍳2))
2 1
3 2
4 2
5 1
Try it online!
I'm trying to find an elegant algorithm for creating an N x N matrix of 1's and 0's, under the restrictions:
each row and each column must sum to Q (to be picked freely)
the diagonal must be 0's
the matrix must be symmetrical.
It is not strictly necessary for the matrix to be random (both random and non-random solutions are interesting, however), so for Q even, simply making each row a circular shift of the vector
[0 1 1 0 ... 0 0 0 ... 0 1 1] (for Q=4)
is a valid solution.
However, how to do this for Q odd? Or how to do it for Q even, but in a random fashion?
For those curious, I'm trying to test some phenomena on abstract networks.
I apologize if this has already been answered before, but none of the questions I could find had the symmetric restriction, which seems to make it much more complicated. I don't have a proof that such a matrix always exists, but I do assume so.
The object that you're trying to construct is known more canonically as an undirected d-regular graph (where d = Q). By the handshaking theorem, N and Q cannot both be odd. If Q is even, then connect vertex v to v + k modulo N for k in {-Q/2, -Q/2 + 1, ..., -1, 1, ..., Q/2 - 1, Q/2}. If Q is odd, then N is even. Construct a (Q - 1)-regular graph as before and then add connections from v to v + N/2 modulo N.
If you want randomness, there's a Markov chain whose limiting distribution is uniform on d-regular graphs. You start with any d-regular graph. Repeatedly pick vertices v, w, x, y at random. Whenever the induced subgraph looks like
v----w
x----y ,
flip it to
v w
| |
x y .
You can perhaps always follow your circular shift algorithm, when possible.
The only condition you need to follow while using the circular shift algorithm is to maintain the symmetric nature in the first row.
i.e. keeping Q 1's in the first row so that Q[0,1] to Q[0,N-1] {Assuming 0 indexed rows and cols, Q[0,0] is 0.} is symmetric, a simple example being 110010011.
Hence, N = 10, Q = 5, you can get many possible arrangements such as:
0 1 0 0 1 1 1 0 0 1
1 0 1 0 0 1 1 1 0 0
0 1 0 1 0 0 1 1 1 0
0 0 1 0 1 0 0 1 1 1
1 0 0 1 0 1 0 0 1 1
1 1 0 0 1 0 1 0 0 1
1 1 1 0 0 1 0 1 0 0
0 1 1 1 0 0 1 0 1 0
0 0 1 1 1 0 0 1 0 1
1 0 0 1 1 1 0 0 1 0
or
0 1 1 0 0 1 0 0 1 1
1 0 1 1 0 0 1 0 0 1
1 1 0 1 1 0 0 1 0 0
0 1 1 0 1 1 0 0 1 0
0 0 1 1 0 1 1 0 0 1
1 0 0 1 1 0 1 1 0 0
0 1 0 0 1 1 0 1 1 0
0 0 1 0 0 1 1 0 1 1
1 0 0 1 0 0 1 1 0 1
1 1 0 0 1 0 0 1 1 0
But as you can see for odd N(that means even N-1) and odd Q there can't be any such symmetric distribution.. Hope it helped.
Given an m*n binary matrix A, m*p binary matrix B, where n > m what is an efficient algorithm to compute X such that AX=B?
For example:
A =
1 1 0 0 1 1 0 1 0 0
1 1 0 0 1 0 1 0 0 1
0 1 1 0 1 0 1 0 1 0
1 1 1 1 1 0 0 1 1 0
0 1 1 0 1 0 1 1 1 0
B =
0 1 0 1 1 0 1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 0 0 1 0 1 0 0
0 1 1 0 0 0 1 1 0 0 1 1 0 0
0 0 1 1 1 1 0 0 0 1 1 0 0 0
1 0 0 1 0 0 1 0 1 0 0 1 1 0
Note, when I say binary matrix I mean matrix defined over the field Z_2, that is, where all arithmetic is mod 2.
If it is of any interest, this is a problem I am facing in generating suitable matrices for a random error correction code.
You can do it with row reduction: Place B to the right of A, and then swap rows (in the whole thing) to get a 1 in row 0, col 0; then xor that row to any other row that has a '1' in column 0, so you have only a single 1 in column 0. Then move to the next column; if [1,1] is zero then swap row 1 with a later row that has a 1 there, then xor rows to make it the only 1 in the column. Assuming 'A' is a square matrix and a solution exists, then you eventually have converted A to unity, and B is replaced with the solution to Ax=B.
If n > m, you have a system with more unknowns than equations, so you can solve for some of the unknowns, and set the others to zero. During the row reduction, if there are no values in a column which have a '1' to use (below the rows already reduced) you can skip that column and make the corresponding unknown zero (you can do this at most n-m times).