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Let's say I have the following (always binary) options:
import numpy as np
a=np.array([1, 1, 0, 0, 1, 1, 1])
b=np.array([1, 1, 0, 0, 1, 0, 1])
c=np.array([1, 0, 0, 1, 0, 0, 0])
d=np.array([1, 0, 1, 1, 0, 0, 0])
And I want to find the optimal combination of the above that get's me to at least, with minimal above:
req = np.array([50,50,20,20,100,40,10])
For example:
final = X1*a + X2*b + X3*c + X4*d
Does this map to a known operational research problem? Or does it fall under mathematical programming?
Is this NP-hard, or exactly solveable in a reasonable amount of time (I've assumed it's combinatorally impossible to solve exactly)
Are there know solutions to this?
Note: The actual length of arrays are longer - think ~50, and the number of options are ~20
My current research has led me to some combination of the assignment problem and knapsack, but not too sure.
It's a covering problem, easily solvable using an integer program solver (I used OR-Tools below). If the X variables can be fractional, substitute NumVar for IntVar. If the X variables are 0--1, substitute BoolVar.
import numpy as np
a = np.array([1, 1, 0, 0, 1, 1, 1])
b = np.array([1, 1, 0, 0, 1, 0, 1])
c = np.array([1, 0, 0, 1, 0, 0, 0])
d = np.array([1, 0, 1, 1, 0, 0, 0])
opt = [a, b, c, d]
req = np.array([50, 50, 20, 20, 100, 40, 10])
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver.CreateSolver("SCIP")
x = [solver.IntVar(0, solver.infinity(), "x{}".format(i)) for i in range(len(opt))]
extra = [solver.NumVar(0, solver.infinity(), "y{}".format(j)) for j in range(len(req))]
for j, (req_j, extra_j) in enumerate(zip(req, extra)):
solver.Add(extra_j == sum(opt_i[j] * x_i for (opt_i, x_i) in zip(opt, x)) - req_j)
solver.Minimize(sum(extra))
status = solver.Solve()
if status == pywraplp.Solver.OPTIMAL:
print("Solution:")
print("Objective value =", solver.Objective().Value())
for i, x_i in enumerate(x):
print("x{} = {}".format(i, x[i].solution_value()))
else:
print("The problem does not have an optimal solution.")
Output:
Solution:
Objective value = 210.0
x0 = 40.0
x1 = 60.0
x2 = -0.0
x3 = 20.0
As an application of Eulers method, I'm trying to implement a code which would compute the recursive matrix product Yn = Yn-1 + A(Yn-1), where Y is a vector and A is a matrix such that the product is defined. This is the current code I have
def f(A, y):
return A.dot(y)
def euler(f, t0, y0, T, dt):
t = np.arange(t0, T + dt, dt)
y = [0,0,0,0]*len(t)
y[0] = y0
for i in range(1, len(t)):
y[i] = y[i - 1] + f(A, y[i - 1])*dt
return t, y
# Define problem specific values
A = np.array([[0, 0, 1, 0],
[0, 0, 0, 1],
[-2, -3, 0, 0],
[-3, -2, 0, 0]])
y1_0 = 1
y2_0 = 2
y3_0 = 0
y4_0 = 0
y0 = [y1_0, y2_0, y3_0, y4_0]
t,y = euler(f,0,y0,2,1)
print(t,y)
For example, the result for points in the range t0 = 0, T = 2 should be the vectors Y1 and Y2. Instead I have
[0 1 2] [[1, 2, 0, 0], array([ 1, 2, -8, -7]), 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
Something is wrong here. While Y1 = [1, 2, -8, -7 ] does show up, there is all of this unnecessary stuff. And Y2 is not printed at all. I suspect this is due to how I define the variable y. For every point in the range of t, I need a vector of 4 zeros - which is then filled up by the function euler, I think. How should correct this?
The computer always does what you tell it to do. In your case y is constructed by repeating 4 zeros len(t) times, giving a list of 12 zeros. The first list entry is replaced by the list y0. The second list entry is replaced by the result of the numpy operations which is a numpy.array. Then the return statement at the level of the loop instructions breaks the loop and returns the t and y arrays. y still contains 10 zeros from its construction that were not replaced.
So construct
y = np.zeros([len(t), len(y0)])
and repair the indentation level.
I am attempting to convert a sum of absolute deviations to a linear programming problem so that I can utilize CPLEX (or other solver). I am stuck on how the matrices are to be set up. The problem is as follows:
minimize abs(x1 - 5) + abs(x2 - 3)
s.t. x1 + x2 = 10
I have the following constraints set up to transform the problem into a linear form:
x1 - 5 <= t1
-(x1 - 5) <= t1 and
x2 - 3 <= t2
-(x2 - 3) <= t2
I've set up the objective function as
c = [0,0,1,1]
but I am lost on how to set up
Ax <= b
in matrix form. What I have so far is:
A = [[ 1, -1, 0, 0],
[-1, -1, 0, 0],
[ 0, 0, 1,-1],
[ 0, 0,-1,-1]]
b = [ 5, -5, 3,-3]
I have set up the other constraint in matrix for as:
B = [1, 1, 0, 0]
b2 = [10]
When I run the following:
linprog(c,A_ub=A,b_ub=b,A_eq=B,b_eq=b2,bounds=[(0,None),(0,None)])
I get the following error message back:
ValueError: Invalid input for linprog: A_eq must have exactly two dimensions, and the number of columns in A_eq must be equal to the size of c
I know there is a solution because when I use scipy.optimize.minimize it solves to [6,4]. I'm sure the issue is I am not formulating the input matrices correctly but I am not sure how to set them up so that it runs.
Edit - here is the code that does not run:
import numpy as np
from scipy.optimize import linprog, minimize
c = np.block([np.zeros(2),np.ones(2)])
print("c =>",c)
A = [[ 1, -1, 0, 0],
[-1, -1, 0, 0],
[ 0, 0, 1,-1],
[ 0, 0,-1,-1]]
b = [[ 5, -5, 3,-3]]
print(A)
print(np.multiply(A,b))
B = [ 1, 1, 0, 0]
b2 = [10]
print(np.multiply(B,b2))
linprog(c,A_ub=A,b_ub=b,A_eq=B,b_eq=b2,bounds=[(0,None),(0,None)],
options={'disp':True})
I think the message is quite good. B should be 2-dimensional matrix instead of a 1-dimensional vector. So:
B = [[1, 1, 0, 0]]
Secondly, the bounds array is too short.
Thirdly, your ordering of variables is inconsistent. The columns in A are x1,t1,x2,t2 while the columns in B (and c) seem to be x1,x2,t1,t2. They need to follow the same scheme.
I need to solve a problem in which I have spent hours, with the data from my excel sheet I have created a 6x36 '' zeros '' matrix of zeros and a 6x6 '' matrix_tran '' coordinate transformation matrix [image 1].
My problem is that I can't find a way to replace the zeros of the '' zeros '' matrix with the values that the matrix '' matrix_tran '' dictates, and whose location must be in the columns (4,5,6, 7,8,9) that are given by the connection vector (4,5,6,7,8,9) of element 15 of the Excel sheet, that is, the last row of the for loop iteration [image 2].
In summary: Below I show how it fits and how it should look [image 3 and 4 respectively].
I would very much appreciate your help, and excuse my English, but it is not my native language, a big greeting.
import pandas as pd
import numpy as np
ex = pd.ExcelFile('matrix_tr.xlsx')
hoja = ex.parse('Hoja1')
cols = 36
for n in range(0,len(hoja)):
A = hoja['ELEMENT #'][n]
B = hoja['1(i)'][n]
C = hoja['2(i)'][n]
D = hoja['3(i)'][n]
E = hoja['1(j)'][n]
F = hoja['2(j)'][n]
G = hoja['3(j)'][n]
H = hoja['X(i)'][n]
I = hoja['Y(i)'][n]
J = hoja['X(j)'][n]
K = hoja['Y(j)'][n]
L = np.sqrt((J-H)**2+(K-I)**2)
lx = (J-H)/L
ly = (K-I)/L
zeros = np.zeros((6, cols))
counters = hoja.loc[:, ["1(i)", "2(i)", "3(i)", "1(j)", "2(j)", "3(j)"]]
for _, i1, i2, i3, j1, j2, j3 in counters.itertuples():
matrix_tran = np.array([[lx, ly, 0, 0, 0, 0],
[-ly, lx, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, lx, ly, 0],
[0, 0, 0, -ly, lx, 0],
[0, 0, 0, 0, 0, 1]])
zeros[:, [i1 - 1, i2 - 1, i3 - 1, j1 - 1, j2 - 1 , j3 - 1]] = matrix_tran
Try with a transposed zeros matrix
import pandas as pd
import numpy as np
ex = pd.ExcelFile('c:/tmp/SO/matrix_tr.xlsx')
hoja = ex.parse('Hoja1')
counters = hoja.loc[:, ["1(i)", "2(i)", "3(i)", "1(j)", "2(j)", "3(j)"]]
# zeros matrix transposed
cols = 36
zeros_trans = np.zeros((cols,6))
# last row only
for n in range(14,len(hoja)):
Xi = hoja['X(i)'][n]
Yi = hoja['Y(i)'][n]
Xj = hoja['X(j)'][n]
Yj = hoja['Y(j)'][n]
X = Xj-Xi
Y = Yj-Yi
L = np.sqrt(X**2+Y**2)
lx = X/L
ly = Y/L
matrix_tran = np.array([[lx, ly, 0, 0, 0, 0],
[-ly, lx, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, lx, ly, 0],
[0, 0, 0, -ly, lx, 0],
[0, 0, 0, 0, 0, 1]])
i = 0
for r in counters.iloc[n]:
zeros_trans[r-1] = matrix_tran[i]
i += 1
print(np.transpose(zeros_trans))
I'm trying to implement STDP (Spike-Timing Dependent Plasticity) in tensorflow. It's a bit complicated. Any ideas (to get running entirely within a tensorflow graph)?
It works like this: say I have 2 input neurons, and they connect to 3 output neurons, via this matrix: [[1.0, 1.0, 0.0], [0.0, 0.0, 1.0]] (input neuron 0 connects to output neurons 0 and 1...).
Say I have these spikes for the input neurons (2 neurons, 7 timesteps):
Input Spikes:
[[0, 0, 1, 1, 0, 1, 0],
[1, 1, 0, 0, 0, 0, 1]]
And these spikes for the output neurons (3 neurons, 7 timesteps):
Output Spikes:
[[0, 0, 0, 1, 0, 0, 1],
[1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1]]
Now, for each non-zero weight, I want to compute a dw. For instance, for input neuron 0 connecting to output neuron 0:
The time stamps of the spikes for input neuron 0 are [2, 3, 5], and the timestamps for output neuron 0 are [3, 6]. Now, I compute all the delta times:
Delta Times = [ 2-3, 2-6, 3-3, 3-6, 5-3, 5-6 ] = [ -1, -4, 0, -3, 2, -1 ]
Then, I compute some function (the actual STDP function, which isn't important for this question - some exponential thing)
dw = SUM [ F(-1), F(-4), F(0), F(-3), F(2), F(-1) ]
And that's the dw for the weight connecting input neuron 0 to output neuron 0. Repeat for all non-zero weights.
So I can do all this in numpy, but I'd like to be able to do it entirely within a single tensorflow graph. In particular, I'm stuck on computing the delta times. And how to do all this for all non-zero weights, in parallel.
This is the actual stdp function, btw (the constants can be parameters):
def stdp_f(x):
return tf.where(
x == 0, np.zeros(x.shape), tf.where(
x > 0, 1.0 * tf.exp(-1.0 * x / 10.0), -1.0 * 1.0 * tf.exp(x / 10.0)))
A note on performance: the method given by #jdehesa, below, is both correct and clever. But it also turns out to be slow. In particular, for a real neural network of 784 input neurons feeding into 400 neurons, over 500 time steps, the spike_match = step performs multiplication of (784, 1, 500, 1) and (1, 400, 1, 500) tensors.
I am not familiar with STDP, so I hope I understood correctly what you meant. I think this does what you describe:
import tensorflow as tf
def f(x):
# STDP function
return x * 1
def stdp(input_spikes, output_spikes):
input_shape = tf.shape(input_spikes)
t = input_shape[-1]
# Compute STDP function for all possible time difference values
stdp_values = f(tf.cast(tf.range(-t + 1, t), dtype=input_spikes.dtype))
# Arrange in matrix such that position [i, j] contains f(i - j)
matrix_idx = tf.expand_dims(tf.range(t - 1, 2 * t - 1), 1) + tf.range(0, -t, -1)
stdp_matrix = tf.gather(stdp_values, matrix_idx)
# Find spike matches
spike_match = (input_spikes[:, tf.newaxis, :, tf.newaxis] *
output_spikes[tf.newaxis, :, tf.newaxis, :])
# Sum values where there are spike matches
return tf.reduce_sum(spike_match * stdp_matrix, axis=(2, 3))
# Test
input_spikes = [[0, 0, 1, 1, 0, 1, 0],
[1, 1, 0, 0, 0, 0, 1]]
output_spikes = [[0, 0, 0, 1, 0, 0, 1],
[1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 1, 1]]
with tf.Graph().as_default(), tf.Session() as sess:
ins = tf.placeholder(tf.float32, [None, None])
outs = tf.placeholder(tf.float32, [None, None])
res = stdp(ins, outs)
res_val = sess.run(res, feed_dict={ins: input_spikes, outs: output_spikes})
print(res_val)
# [[ -7. 10. -15.]
# [-13. 7. -24.]]
Here I assume that f is probably expensive (and that its value is the same for every pair of neurons), so I compute it only once for every possible time delta and then redistribute the computed values in a matrix, so I can multiply at the pairs of coordinates where the input and output spikes happen.
I used the identity function for f as a placeholder, so the resulting values are actually just the sum of time differences in this case.
EDIT: Just for reference, replacing f with the STDP function you included:
def f(x):
return tf.where(x == 0,
tf.zeros_like(x),
tf.where(x > 0,
1.0 * tf.exp(-1.0 * x / 10.0),
-1.0 * 1.0 * tf.exp(x / 10.0)))
The result is:
[[-3.4020822 2.1660795 -5.694256 ]
[-2.974073 0.45364904 -3.1197631 ]]