I'm writing an optimal path planning algorithm for one of my projects, and am a bit stuck on how to generate the map. In order to test my algorithm, I want to generate random maps that the object will pass through with obstacles in the way. To generate the maps I am making a logical array with "1" for wall or obstacle and "0" when there is nothing in the way. I am having some issues placing the obstacles however.
For now, this is what I've written in MATLAB (though I think the core idea can be solved in any language):
% lenx = length in x-direction of map [m]
% leny = length in y-direction of map [m]
% numObs = number of obstacles
% obsLoc = 2D array with the central location(s) of the obstacle(s)
% [x1, y1; x2 y2; ...]
% obsSize = size of the obstacles (all same size) [m]
% assume the obstacles are square.
% res = resolution of the map [cell/m]
%
% Outputs:
% map = 2D logical array which translates into a map
%
%---------------------------------------------------------
function map = createMap(lenx, leny, numObs, obsLoc, obsSize, res)
%Making map of the given size
map = zeros(leny*res, lenx*res);
%Adding walls
map(:,1) = 1;
map(:, lenx*res) = 1;
map(1,:) = 1;
map(leny*res, :) = 1;
%Recalculating obstacle size for this grid
obsSize = obsSize - 1;
%Adding obstacles
for i = 1:1:numObs
map(obsLoc(i,1)*res-obsSize*res:obsLoc(i,1)*res+obsSize*res,...
obsLoc(i,2)*res-obsSize*res:obsLoc(i,2)*res+obsSize*res) = 1;
end
end
The obsSize parameter is representing the dimension of one of the sides of the square obstacle. Here is a sample solution I ran:
>> createMap(10,10,1,[5,5],1,1)
ans =
1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 1
1 0 0 0 1 0 0 0 0 1
1 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1
Looks good! But the issue begins for obstacle sizes and resolutions that are not 1 or 2...
>> createMap(10,10,1,[5,5],0.5,2)
ans =
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
>> createMap(10,10,1,[5,5],3,2)
ans =
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
For the first one, we would expect there to be a 1 somewhere close to the middle, but it's nowhere to be found. In the second one, we'd expect to find 6 ones in each direction (3m * 3cell/m = 6 cell), but we are getting 8 (It's also possible that this is not what my code is calculating, but this is what I'm hoping it does. Neither case is working, and I think it has to do with this line:
Recalculating obstacle size for this grid
obsSize = obsSize - 1;
I'm pretty sure I do have to recalculate the object size so it fits to the grid, but how can I do this for any grid and obstacle? Is there a way to scale that "-1" such that it acts according to the size of the grid/obstacle?
Let me know what you come up with, and if there's any other glaring issues in my code?
Thanks!
Related
i have a matrix 20 rows and 20 columns,
If the value 1 in the row 5 the column take 0
matric=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ;
0 1 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 1;
0 1 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 1 0 1;
0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;];
if (matric(5,:)==1)
matric(1:5,1:end)=0;end
I try to compare the second row and the 5 row
If we have "1" in row 2 and row 5
The row 2 take 0
if (matric(5,:)==matric(2,:)==1)
matric(2,1:end)=0;end
do you have an idea
Thank you
The desired output is:
matric=[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
1 0 1 1 0 0 1 1 1 1 0 1 0 1 1 1 1 0 1 0 ;2row will change
0 1 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 1;
0 1 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 1 0 1;
0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1; % 5 row
You can use logical indexing.
I strongly encourage you to read the following references:
Mathworks documentation page: Find Array Elements That Meet a Condition
Loren on the Art of MATLAB blog entry: Logical Indexing – Multiple
Conditions
Use the following lines of code:
Put 0 in all those columns that have a value of 1 in row 5:
matric(:, matric(5, :) == 1) = 0;
Put 0 in all those columns of row 2 that have a value of 1 both in rows 2 and 5:
matric(2, matric(2, :) == matric(5, :)) = 0;
You can use logical indexing to achieve this. Now I must say I am a bit confused by what exactly you want to achieve based on your description, but based on the your code the first statement can be done as follow:
matric(1:5,matric(5,:)==1) = 0;
and the second would look like:
matric(2,matric(5,:)==1 & matric(2,:)==1)=0;
I want to create a MxN matrix as shown below:
[1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0;
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
I have window size, let's say, 5 and it moves 3 in every row. Is it possible to create such a matrix without using for loops? Or is there any optimum way to do it?
This is a one line solution:
reshape([reshape([ones(5,6);zeros(21,6)], 1,[]), ones(1,5)],[],7).'
note:
The desired matrix can be seen as concatenation of a [6, 5+21] matrix:
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
and a [1 ,5] matrix:
1 1 1 1 1
that reshaped to a [7 , 23] matrix.
Other solution using repelem + bsxfun + accumarray:
r = repelem (1:7,5);
c= bsxfun(#plus, ((1:5)-3).',3*(1:7));
out = accumarray([r(:) c(:)] ,1)
Indices of rows and columns of 1 s can be generated and accumarray can be used to create the desired matrix.
My first question on stackoverflow! The title is vague, so let me elaborate: I have a NxN lower triangular logical matrix
N = 10 % for example
L = tril(true(N),-1)
L =
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0
1 1 1 1 1 1 0 0 0 0
1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 1 0 0
1 1 1 1 1 1 1 1 1 0
with all trues below the diagonal. For a m=2^p a power of 2, I want to end up with m NxN lower triangular logical matrices L_1, ..., L_m such that each column of L_i contains the i-th 1/m-th (rounded) number of the Trues in the corresponding column in L. One consequence is that \sum_i(L_i) == L again.
For example, for m = 2 I know that
L_2 = L(:,ceil((N:2*N-1)/2))
L_2 =
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0
1 1 1 1 1 0 0 0 0 0
1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 1 1 0
L_1 = L - L_2
L_1 =
0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 0 0
0 1 1 1 1 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 1 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0
will do the trick, but this trick does not generalize to higher powers of 2 for m. Any ideas how to do this reasonably fast for general N and m = 2^p?
(Context: each column of L are logical indices for a bisection type algorithm. Every next power p of m = 2^p corresponds to a deeper level of the bisection algorithm)
I am confusing about exponentiation of matrix in Galois Field 2. Assume that I have a matrix that is represented in Galois Field 2 (GF2). I want to take exponentiation of 30. That is,
A^30
In the matlab we have two way to do it.
First, We perform A power 30 and take mod of 2 (Note that A is a double matrix)
A1=mod(A^30,2)
Second way, we convert A to galois matrix and take exponentiation
A=gf(A,2)
A2=A^30
Actually, two way must same result. However, when I check A1 and A2. They have a different result. What is happen in here? Thanks
Let see my A matrix
A =
1 1 0 1 1 1 0 1 0 0 0 1 0 1 0
0 0 0 1 0 1 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 1 0 0 0 1
1 0 1 1 0 0 0 0 0 0 1 1 1 0 0
1 0 1 0 0 0 0 1 0 0 0 0 0 0 0
0 1 0 1 1 1 1 1 0 1 1 1 1 0 1
0 0 0 1 0 0 0 1 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 1 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 1 0 0 1 1
1 0 1 0 0 0 0 0 0 0 1 0 0 0 1
0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
1 0 1 1 1 1 1 1 1 0 1 1 1 0 1
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
1 0 0 1 0 1 0 0 0 1 0 0 0 1 1
Method1:
A1=
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Method 2
A2 =
1 1 0 1 1 1 0 1 0 0 0 1 0 1 0
0 0 0 1 0 1 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1 0 1 0 0 0 1
1 0 1 1 0 0 0 0 0 0 1 1 1 0 0
1 0 1 0 0 0 0 1 0 0 0 0 0 0 0
0 1 0 1 1 1 1 1 0 1 1 1 1 0 1
0 0 0 1 0 0 0 1 0 0 1 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 1 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 1 0 0 1 1
1 0 1 0 0 0 0 0 0 0 1 0 0 0 1
0 1 0 0 0 0 0 1 0 0 0 0 0 0 0
1 0 1 1 1 1 1 1 1 0 1 1 1 0 1
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
1 0 0 1 0 1 0 0 0 1 0 0 0 1 1
I want to detect digist with hopfield.for any digit let 96 cell ( 8*12).
my data are
zero=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
one=[0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 ];
two=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
three=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
four=[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ];
five=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
six=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
seven=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ];
eight=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
nine=[0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 ];
target =[zero;one;two;three;four;five;six;seven;eight;nine];
and create a hopfield with newhop
net = newhop(target);
[Y,Pf,Af] = sim(net,10,[],target);
I conclusion this from matlab website
but really I don't know How I achive this
my project is detect digits by hopfield network
appreciate any orgency help
Try:
target =[zero;one;two;three;four;five;six;seven;eight;nine]';
Then, for a test:
test={three'};
[Y,Pf,Af] = sim(net,{1 5},{},test);
Y{1}
You'll notice that Y{1}-three' is very close to zero, which means that your network does its job. Now, in order to recognize the digit you enter, you need to subtract Y{1} from the matrix of digits and check the row the elements of which are closest to zero. This will indicate you the digit you entered.