I have written a script in Octave below based off of a post:
How to draw vectors (physical 2D/3D vectors) in MATLAB?
I have the following code in Octave and I get the error
' "vector" undefined near line 3 column 1'
Can anyone help me? I have written a few other *.m files for Octave with no problem and now I am stuck.
a = [0 0 1]; %# Vector 1
b = [0.894 0 0.447]; %# Vector 2
c = [0.276 0.851 0.447]; %# Vector 3
d = [-0.724 0.526 0.447]; %# Vector 4
e = [-0.724 -0.526 0.447]; %# Vector 5
f = [0.276 -0.851 0.447]; %# Vector 6
g = [0.724 0.526 -0.447]; %# Vector 7
h = [-0.276 0.851 -0.447]; %# Vector 8
i = [-0.894 0 -0.447]; %# Vector 9
j = [-0.276 -0.851 -0.447] %# Vector 10
k = [0.724 -0.526 -0.447]; %# Vector 11
l = [0 0 -1]; %# Vector 12
starts = zeros(3,3);
ends = [a;b;c;d;e;f;g;h;i;j;k;l];
quiver3(starts(:,1), starts(:,2), starts(:,3), starts(:,4), starts(:,5),
starts(:,6), starts(:,7), starts(:,8), starts(:,9), starts(:,10),
starts(:,11), starts(:,12), ends(:,1), ends(:,2), ends(:,3),
ends(:,4), ends(:,5), ends(:,6), ends(:,7), ends(:,8), ends(:,9),
ends(:,10), ends(:,11), ends(:,12), 0)
axis equal
This doesn't work because starts is of size 3x3 and you are indexing it up to 12 on the column index. The error message I get is:
error: A(I,J): column index out of bounds; value 4 out of bound 3
Related
7|8|9
6|5|4
1|2|3
1 -> (1,1)
2 -> (2,1)
3 -> (3,1)
4 -> (3,2)
5 -> (2,2)
6 -> (1,2)
7 -> (1,3)
8 -> (2,3)
9 -> (3,3)
In this grid, the mapping of the numbers to coordinates is shown above.
I'm struggling to come up with a formula where given the number of the grid and the number of rows and columns in the grid, it outputs the coordinates of the grid.
I tried following the logic in this question but in this question, the coordinate system starts from 0 and the rows are not alternating.
If there was no alternating and the numbers were all starting at 0 and not 1, then you could apply Euclidean division directly:
x = n % 3
y = n // 3
where // gives the quotient of the Euclidean division, and % gives the remainder of the Euclidean division.
If there was no alternating, but the numbers all start at 1 instead of 0, then you can fix the above formula by removing 1 from n to make it start at 0, then adding 1 to x and y to make them start at 1:
x = ((n - 1) % 3) + 1
y = ((n - 1) // 3) + 1
Now all we have to change to take the alternating into account is to flip the x values on the right-to-left rows.
y remains unchanged, and x remains unchanged on the left-to-right rows.
The right-to-left rows are the rows with an even y, and you can flip x symmetrically around 2 by removing it from 4:
if y % 2 == 0:
x = 4 - x
Putting it all together in a function and testing it, in python:
def coord(n):
y = ((n-1) // 3) + 1
x = ((n-1) % 3) + 1
if y % 2 == 0: # right-to-left row
x = 4 - x # flip horizontally
return (x, y)
for n in range(1, 9+1):
x, y = coord(n)
print(f'{n} -> ({x},{y})')
Output:
1 -> (1,1)
2 -> (2,1)
3 -> (3,1)
4 -> (3,2)
5 -> (2,2)
6 -> (1,2)
7 -> (1,3)
8 -> (2,3)
9 -> (3,3)
Inverse function
The inverse operation of a Euclidean division is a multiplication and an addition:
if y % 2 == 1:
n = 3 * (y-1) + x
else:
n = 3 * (y-1) + 4 - x
I'm trying to implement the following code from here but it won't work correctly.
What I want is the shortest path distances from a source to all nodes and also the predecessors. Also, I want the input of the graph to be an adjacency matrix which contains all of the edge weights.
I'm trying to make it work in just one function so I have to rewrite it. If I'm right the original code calls other functions (from graph.jl for example).
I don't quite understand how to rewrite the for loop which calls the adj() function.
Also, I'm not sure if the input is correct in the way the code is for now.
function dijkstra(graph, source)
node_size = size(graph, 1)
dist = ones(Float64, node_size) * Inf
dist[source] = 0.0
Q = Set{Int64}() # visited nodes
T = Set{Int64}(1:node_size) # unvisited nodes
pred = ones(Int64, node_size) * -1
while condition(T)
# node selection
untraversed_nodes = [(d, k) for (k, d) in enumerate(dist) if k in T]
if minimum(untraversed_nodes)[1] == Inf
break # Break if remaining nodes are disconnected
end
node_ind = untraversed_nodes[argmin(untraversed_nodes)][2]
push!(Q, node_ind)
delete!(T, node_ind)
# distance update
curr_node = graph.nodes[node_ind]
for (neigh, edge) in adj(graph, curr_node)
t_ind = neigh.index
weight = edge.cost
if dist[t_ind] > dist[node_ind] + weight
dist[t_ind] = dist[node_ind] + weight
pred[t_ind] = node_ind
end
end
end
return dist, pred
end
So if I'm trying it with the following matrix
A = [0 2 1 4 5 1; 1 0 4 2 3 4; 2 1 0 1 2 4; 3 5 2 0 3 3; 2 4 3 4 0 1; 3 4 7 3 1 0]
and source 2 i would like to get the distances in a vector dist and the predeccessors in anothe vectore pred.
Right now I'm getting
ERROR: type Array has no field nodes
Stacktrace: [1] getproperty(::Any, ::Symbol) at .\sysimg.jl:18
I guess I have to rewrite it a bit more.
I m thankful for any help.
Assuming that graph[i,j] is a length of path from i to j (your graph is directed looking at your data), and it is a Matrix with non-negative entries, where 0 indicates no edge from i to j, a minimal rewrite of your code should be something like:
function dijkstra(graph, source)
#assert size(graph, 1) == size(graph, 2)
node_size = size(graph, 1)
dist = fill(Inf, node_size)
dist[source] = 0.0
T = Set{Int}(1:node_size) # unvisited nodes
pred = fill(-1, node_size)
while !isempty(T)
min_val, min_idx = minimum((dist[v], v) for v in T)
if isinf(min_val)
break # Break if remaining nodes are disconnected
end
delete!(T, min_idx)
# distance update
for nei in 1:node_size
if graph[min_idx, nei] > 0 && nei in T
possible_dist = dist[min_idx] + graph[min_idx, nei]
if possible_dist < dist[nei]
dist[nei] = possible_dist
pred[nei] = min_idx
end
end
end
end
return dist, pred
end
(I have not tested it extensively, so please report if you find any bugs)
I have a m x n tensor (Tensor 1) and another k x 2 tensor (Tensor 2) and I wish to extract all the values of Tensor 1 using indices based on Tensor 2. For example;
Tensor1
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
[torch.DoubleTensor of size 4x5]
Tensor2
2 1
3 5
1 1
4 3
[torch.DoubleTensor of size 4x2]
And the function would yield;
6
15
1
18
The first solution that comes into mind is to simply loop through indexes and pick the correspoding values:
function get_elems_simple(tensor, indices)
local res = torch.Tensor(indices:size(1)):typeAs(tensor)
local i = 0
res:apply(
function ()
i = i + 1
return tensor[indices[i]:clone():storage()]
end)
return res
end
Here tensor[indices[i]:clone():storage()] is just a generic way to pick an element from a multi-dimensional tensor. In k-dimensional case this is exactly analogous to tensor[{indices[i][1], ... , indices[i][k]}].
This method works fine if you don't have to extract lots of values (the bottleneck is :apply method which is not able to use many optimization techniques and SIMD instructions because the function it executes is a black box). The job can be done way more efficiently: the method :index does exactly what you need... with a one-dimensional tensor. Multi-dimensional target/index tensors need to be flattened:
function flatten_indices(sp_indices, shape)
sp_indices = sp_indices - 1
local n_elem, n_dim = sp_indices:size(1), sp_indices:size(2)
local flat_ind = torch.LongTensor(n_elem):fill(1)
local mult = 1
for d = n_dim, 1, -1 do
flat_ind:add(sp_indices[{{}, d}] * mult)
mult = mult * shape[d]
end
return flat_ind
end
function get_elems_efficient(tensor, sp_indices)
local flat_indices = flatten_indices(sp_indices, tensor:size())
local flat_tensor = tensor:view(-1)
return flat_tensor:index(1, flat_indices)
end
The difference is drastic:
n = 500000
k = 100
a = torch.rand(n, k)
ind = torch.LongTensor(n, 2)
ind[{{}, 1}]:random(1, n)
ind[{{}, 2}]:random(1, k)
elems1 = get_elems_simple(a, ind) # 4.53 sec
elems2 = get_elems_efficient(a, ind) # 0.05 sec
print(torch.all(elems1:eq(elems2))) # true
I am trying to plot a 3D graph between 2 scalars and one matrix for each of its entries. On compiling it is giving me "Submatrix incorrectly defined" error on line 11. The code:
i_max= 3;
u = zeros(4,5);
a1 = 1;
a2 = 1;
a3 = 1;
b1 = 1;
hx = linspace(1D-6,1D6,13);
ht = linspace(1D-6,1D6,13);
for i = 1:i_max
for j = 2:4
u(i+1,j)=u(i,j)+(ht*(a1*u(i,j))+b1+(((a2*u(i,j+1))-(2*a2*u(i,j))+(a2*u(i,j-1)))*(hx^-2))+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx^-1)));
plot(ht,hx,u(i+1,j));
end
end
Full error message:
-->exec('C:\Users\deba123\Documents\assignments and lecture notes\Seventh Semester\UGP\Scilab\Simulation1_Plot.sce', -1)
+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx^-1)))
!--error 15
Submatrix incorrectly defined.
at line 11 of exec file called by :
emester\UGP\Scilab\Simulation1_Plot.sce', -1
Please help.
For a 3-dimensional figure, you need 2 argument vectors and a matrix for the function values. So I expanded u to a tensor.
At every operation in your code, I added the current dimension of the term. Now, a transparent handling of you calculation is given. For plotting you have to use the plot3d (single values) or surf (surface) command.
In a 3-dim plot, you want two map 2 vectors (hx,ht) with dim n and m to an scalar z. Therefore you reach a (nxm)-matrix with your results. Is this, what you want to do? Currently, you have 13 values for each u(i,j,:) - entry, but you want (13x13) for every figure. Maybe the eval3d-function can help you.
i_max= 3;
u = zeros(4,5,13);
a1 = 1;
a2 = 1;
a3 = 1;
b1 = 1;
hx = linspace(1D-6,1D6,13); // 1 x 13
ht = linspace(1D-6,1D6,13); // 1 x 13
for i = 1:i_max
for j = 2:4
u(i+1,j,:)= u(i,j)...
+ ht*(a1*u(i,j))*b1... // 1 x 13
+(((a2*u(i,j+1)) -(2*a2*u(i,j)) +(a2*u(i,j-1)))*(hx.^-2))... // 1 x 13
+(((a3*u(i,j+1))-(a3*u(i,j-1)))*(0.5*hx.^-1)) ... // 1 x 13
+ hx*ones(13,1)*ht; // added to get non-zero values
z = squeeze( u(i+1,j, : ))'; // 1x13
// for a 3d-plot: (1x13, 1x13, 13x13)
figure()
plot3d(ht,hx, z'* z ,'*' ); //
end
end
What is the Equation that on increasing the integer x returns an alternate of 0 and 1
example
x = 22
result 1
x = 23
result 0
x = 24
result 1
Based on the example data, it would be modulo 2. Assuming x is an int (and C/C++/C#):
(x + 1) % 2;
In C or C++ this would be
int y = (x+1)%2;
mathematically,
y = (x+1) modulo 2
It's called modulo. You can use mod by 2 after adding one in the value.
x = 22
result = (x+1) modulo 2
In programming languages, it's often called %:
x = 22
result = (x+1) % 2 //<< result 1
x = 23
result = (x+1) % 2 //<< result 0
and so on..