For loop with 2 iterators in Julia/JuMP - optimization

I need to implement the following pseudocode to JuMP/Julia:
forall{i in M, j in Ni[i]}: x[i] <= y[j];
I imagine something like:
for i in M and j in Ni[i]
#constraint(model, x[i] <= y[j])
end
How do I properly implement the 2 iterators in the for loop?

I don't know if you want one iteration with both values, or the Cartesian product of the iterators, but here are example for both:
julia> M = 1:3; N = 4:6;
julia> for (m, n) in zip(M, N) # single iterator over both M and N
#show m, n
end
(m, n) = (1, 4)
(m, n) = (2, 5)
(m, n) = (3, 6)
julia> for m in M, n in N # Cartesian product
#show m, n
end
(m, n) = (1, 4)
(m, n) = (1, 5)
(m, n) = (1, 6)
(m, n) = (2, 4)
(m, n) = (2, 5)
(m, n) = (2, 6)
(m, n) = (3, 4)
(m, n) = (3, 5)
(m, n) = (3, 6)

You want
#constraint(model, [i = M, j = Ni[i]], x[i] <= y[j])
Here is the relevant documentation: https://www.juliaopt.org/JuMP.jl/stable/constraints/#Constraint-containers-1

Related

Numpy: How To Vectorize Operations?

I have the following vectors
shape u_w: (50,)
shape Vt: (6, 50)
shape v: (50,)
and with them I perform the following calculations
w = np.tanh(u_w + Vt[0])
w_squared = w ** 2
z = np.dot(v, w)
s = sigmoid(np.dot(v, w))
J = -np.log(sigmoid(z))
dv = np.dot(sigmoid(z) - 1, w)
du_w = np.dot(s - 1, v, (1 - w_squared))
dVt = np.dot(s - 1, v, (1 - w_squared))
for vt in Vt[1:]:
t = np.tanh(u_w + vt)
svt = sigmoid(np.dot(-v, t))
J -= np.log(svt)
dv -= np.dot((svt - 1), t)
du_w -= np.dot((svt - 1), v, (1 - t**2))
dVt = np.vstack((dVt, -np.dot(svt - 1, v, (1 - t**2))))
How do I vectorize the calculations for J, dv, du_w and dVt, so that they work for a batch of S items with the following shapes?
shape(u_w) => (512, 50)
shape(Vt) => (512, 6, 50)
shape(v) => (50,)

Efficient way to calculate 3D matrix multiplication using numpy

How can I efficiently write and calculate this multiplication using numpy:
for k in range(K):
for i in range(SIZE):
for j in range(SIZE):
for i_b in range(B_SIZE):
for j_b in range(B_SIZE):
for k_b in range(k+1):
data[k, i * w + i_b, j * h + j_b] += arr1[k_b, i_b, j_b] * arr2[k_b, i, j]
For example:
SIZE, B_SIZE = 32, 8
arr1.shape -> (8, 8, 8)
arr2.shape -> (8, 32, 32)
data.shape -> (K, 256, 256)
Thank you.
You can use Numba for such kind of non-trivial case and rework the loops to use efficiently the CPU cache. Here is an example:
import numba as nb
#nb.njit
def compute(data, arr1, arr2):
for k in range(K):
for k_b in range(k+1):
for i in range(SIZE):
for j in range(SIZE):
tmp = arr2[k_b, i, j]
for i_b in range(B_SIZE):
for j_b in range(B_SIZE):
data[k, i * w + i_b, j * h + j_b] += arr1[k_b, i_b, j_b] * tmp
If you do this operation once, then you can pre-compile the Numba code by providing the types of the arrays. If K is big, then you can parallelize the code using #nb.njit(parallel=True) and use for k in nb.prange(K) rather than for k in range(K). This should be several order of magnitude fater.

Algorithm to define a 2d grid

Suppose a grid is defined by a set of grid parameters: its origin (x0, y0), an angel between one side and x axis, and increments and - please see the figure below.
There are scattered points with known coordinates on the grid but they don’t exactly fall on grid intersections. Is there an algorithm to find a set of grid parameters to define the grid so that the points are best fit to grid intersections?
Suppose the known coordinates are:
(2 , 5.464), (3.732, 6.464), (5.464, 7.464)
(3 , 3.732), (4.732, 4.732), (6.464, 5.732)
(4 , 2 ), (5.732, 3 ), (7.464, 4 ).
I expect the algorithm to find the origin (4, 2), the angle 30 degree, and the increments both 2.
You can solve the problem by finding a matrix that transforms points from positions (0, 0), (0, 1), ... (2, 2) onto the given points.
Although the grid has only 5 degrees of freedom (position of the origin + angle + scale), it is easier to define the transformation using 2x3 matrix A, because the problem can be made linear in this case.
Let a point with index (x0, y0) to be transformed into point (x0', y0') on the grid, for example (0, 0) -> (2, 5.464) and let a_ij be coefficients of matrix A. Then this pair of points results in 2 equations:
a_00 * x0 + a_01 * y0 + a_02 = x0'
a_10 * x0 + a_11 * y0 + a_12 = y0'
The unknowns are a_ij, so these equations can be written in form
a_00 * x0 + a_01 * y0 + a_02 + a_10 * 0 + a_11 * 0 + a_12 * 0 = x0'
a_00 * 0 + a_01 * 0 + a_02 * 0 + a_10 * x0 + a_11 * y0 + a_12 = y0'
or in matrix form
K0 * (a_00, a_01, a_02, a_10, a_11, a_12)^T = (x0', y0')^T
where
K0 = (
x0, y0, 1, 0, 0, 0
0, 0, 0, x0, y0, 1
)
These equations for each pair of points can be combined in a single equation
K * (a_00, a_01, a_02, a_10, a_11, a_12)^T = (x0', y0', x1', y1', ..., xn', yn')^T
or K * a = b
where
K = (
x0, y0, 1, 0, 0, 0
0, 0, 0, x0, y0, 1
x1, y1, 1, 0, 0, 0
0, 0, 0, x1, y1, 1
...
xn, yn, 1, 0, 0, 0
0, 0, 0, xn, yn, 1
)
and (xi, yi), (xi', yi') are pairs of corresponding points
This can be solved as a non-homogeneous system of linear equations. In this case the solution will minimize sum of squares of distances from each point to nearest grid intersection. This transform can be also considered to maximize overall likelihood given the assumption that points are shifted from grid intersections with normally distributed noise.
a = (K^T * K)^-1 * K^T * b
This algorithm can be easily implemented if there is a linear algebra library is available. Below is an example in Python:
import numpy as np
n_points = 9
aligned_points = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
grid_points = [(2, 5.464), (3.732, 6.464), (5.464, 7.464), (3, 3.732), (4.732, 4.732), (6.464, 5.732), (4, 2), (5.732, 3), (7.464, 4)]
K = np.zeros((n_points * 2, 6))
b = np.zeros(n_points * 2)
for i in range(n_points):
K[i * 2, 0] = aligned_points[i, 0]
K[i * 2, 1] = aligned_points[i, 1]
K[i * 2, 2] = 1
K[i * 2 + 1, 3] = aligned_points[i, 0]
K[i * 2 + 1, 4] = aligned_points[i, 1]
K[i * 2 + 1, 5] = 1
b[i * 2] = grid_points[i, 0]
b[i * 2 + 1] = grid_points[i, 1]
# operator '#' is matrix multiplication
a = np.linalg.inv(np.transpose(K) # K) # np.transpose(K) # b
A = a.reshape(2, 3)
print(A)
[[ 1. 1.732 2. ]
[-1.732 1. 5.464]]
Then the parameters can be extracted from this matrix:
theta = math.degrees(math.atan2(A[1, 0], A[0, 0]))
scale_x = math.sqrt(A[1, 0] ** 2 + A[0, 0] ** 2)
scale_y = math.sqrt(A[1, 1] ** 2 + A[0, 1] ** 2)
origin_x = A[0, 2]
origin_y = A[1, 2]
theta = -59.99927221917264
scale_x = 1.99995599951599
scale_y = 1.9999559995159895
origin_x = 1.9999999999999993
origin_y = 5.464
However there remains a minor issue: matrix A corresponds to an affine transform. This means that grid axes are not guaranteed to be perpendicular. If this is a problem, then the first two columns of the matrix can be modified in a such way that the transform preserves angles.
Update: I fixed the mistakes and resolved sign ambiguities, so now this algorithm produces the expected result. However it should be tested to see if all cases are handled correctly.
Here is another attempt to solve this problem. The idea is to decompose transformation into non-uniform scaling matrix and rotation matrix A = R * S and then solve for coefficients sx, sy, r1, r2 of these matrices given restriction that r1^2 + r2^2 = 1. The minimization problem is described here: How to find a transformation (non-uniform scaling and similarity) that maps one set of points to another?
def shift_points(points):
n_points = len(points)
shift = tuple(sum(coords) / n_points for coords in zip(*points))
shifted_points = [(point[0] - shift[0], point[1] - shift[1]) for point in points]
return shifted_points, shift
n_points = 9
aligned_points = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
grid_points = [(2, 5.464), (3.732, 6.464), (5.464, 7.464), (3, 3.732), (4.732, 4.732), (6.464, 5.732), (4, 2), (5.732, 3), (7.464, 4)]
aligned_points, aligned_shift = shift_points(aligned_points)
grid_points, grid_shift = shift_points(grid_points)
c1, c2 = 0, 0
b11, b12, b21, b22 = 0, 0, 0, 0
for i in range(n_points):
c1 += aligned_points[i][0] ** 2
c2 += aligned_points[i][0] ** 2
b11 -= 2 * aligned_points[i][0] * grid_points[i][0]
b12 -= 2 * aligned_points[i][1] * grid_points[i][0]
b21 -= 2 * aligned_points[i][0] * grid_points[i][1]
b22 -= 2 * aligned_points[i][1] * grid_points[i][1]
k = (b11 ** 2 * c2 + b22 ** 2 * c1 - b21 ** 2 * c2 - b12 ** 2 * c1) / \
(b21 * b11 * c2 - b12 * b22 * c1)
# r1_sqr and r2_sqr might need to be swapped
r1_sqr = 2 / (k ** 2 + 4 + k * math.sqrt(k ** 2 + 4))
r2_sqr = 2 / (k ** 2 + 4 - k * math.sqrt(k ** 2 + 4))
for sign1, sign2 in [(1, 1), (-1, 1), (1, -1), (-1, -1)]:
r1 = sign1 * math.sqrt(r1_sqr)
r2 = sign2 * math.sqrt(r2_sqr)
scale_x = -b11 / (2 * c1) * r1 - b21 / (2 * c1) * r2
scale_y = b12 / (2 * c2) * r2 - b22 / (2 * c2) * r1
if scale_x >= 0 and scale_y >= 0:
break
theta = math.degrees(math.atan2(r2, r1))
There might be ambiguities in choosing r1_sqr and r2_sqr. Origin point can be estimated from aligned_shift and grid_shift, but I didn't implement it yet.
theta = -59.99927221917264
scale_x = 1.9999559995159895
scale_y = 1.9999559995159895

Multiplying multidimensional array in python

I have two arrays:
L, M, N = 6, 31, 500
A = np.random.random((L, M, N))
B = np.random.random((L, L))
I am trying to get an array C such that:
C = B * A
C has dimension [L, M, N]
I tried answer posted at this link but it hasn't given me the desired output.
A for loop version of above code is:
L, M, N = 6, 31, 500
A = np.random.random((L, M, N))
B = np.random.random((L, L))
z1 = []
for j in range(M):
a = np.squeeze(A[:, j, :])
z1.append(np.dot(B, a))
z2 = np.stack(z1)
I think you are looking for numpy.tensordot() where you can specify along which axes to sum:
np.tensordot(B,A,axes=(1,0))

Multiply each channel by different matrix?

Is there a way, in tensorflow, to multiply each channel by a different matrix?
Imagine you have a 2D array A of dimensions (N, D1).
You can multiply it by an array B of size (D1, D2) to get output size (N, D2).
Now imagine you have a 3D array of dimensions (N, D1, 3).
Suppose you had B1, B2, B3 all of size (D1, D2). Combining the outputs A * B1, A * B2, A * B3, you could form an array of size (N, D2, 3).
But is there a way to get an output size of (N, D2, 3) by just doing multiplication once?
I looked into transpose and matmul but it doesn't seem to work for this purpose.
Thank you!
tf.einsum() could be applied here.
To make the code below easier to understand, I renamed D1 = O and D2 = P.
import tensorflow as tf
A = tf.random_normal([N, O, 3])
B = tf.random_normal([O, P, 3]) # B = tf.stack([B1, B2, B3], axis=2)
res = tf.einsum("noi,opi->npi", A, B)
You could use tf.matmul here. Its just that you will have to transpose the dimensions.
Consider, N = 2, D1 = 4, D2 = 5. First create two matrices having shapes N x D1 x 3 and D1 x D2 x 3.
a = tf.constant(np.arange(1, 25, dtype=np.int32), shape=[2,4,3])
b = tf.constant(np.arange(1, 61, dtype=np.int32), shape=[4,5,3])
Transpose the matrices so that the first dimension is the same.
a = tf.transpose(a, (2, 0, 1)) # a.shape = (3, 2, 4)
b = tf.transpose(b, (2, 0, 1)) # b.shape = (3, 4, 5)
Perform the multiplication as usual.
r = tf.matmul(a,b) # r.shape = (3, 2, 5)
r = tf.transpose(r, (1, 2, 0)) # r.shape = (2, 5, 3)
Hope this helps.