Aij | A simple Java AI library for personal use | Machine Learning library
kandi X-RAY | Aij Summary
kandi X-RAY | Aij Summary
A small, simple Java AI library with no external requirements.
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Top functions reviewed by kandi - BETA
- Fit a batch of examples
- Calculate the gradient for the given node
- Finalize the network
- Make an unrolled LSTM
- Unroll the LSTM step
- Fit the network using the given learning rate
- Performs a series of predictions
- Creates a single run step
- Reverse forward forward
- Performs a forward transformation matrix
- Fit the network
- Adds a dense layer
- Restores the attributes from a string
- Apply transform to a matrix
- Reverse a matrix
- Apply a forward transformation matrix
- Performs a forward operation on the matrix
- Accumulated gradients
- Apply a forward matrix to a matrix
- Adds a deconvolution layer
- Reverse the reverse matrix
- Performs a forward projection matrix
- Makes a run only LSTM
- Performs a forward convolution
- Adds a new convolution layer
- Reverse the inverse
Aij Key Features
Aij Examples and Code Snippets
Lines of Code : 34
License : No License
CopyGraph g = new Graph();
InputNode x = new InputNode(1, 1);
VariableNode m = new VariableNode(1, 1);
VariableNode b = new VariableNode(1, 1);
InputNode y = new InputNode(1, 1); // Target.
Node out = new AddNode(b, new MultiplyNode(m, x));
Node error = Model m = new Model(1, 2);
m.addDenseLayer(10, Model.Activation.TANH);
m.addDenseLayer(1, Model.Activation.SIGMOID);
double[][] x = new double[][] {
{0, 0},
{0, 1},
{1, 0},
{1, 1}
};
double[][] y = new double[][] { {0}, {1}, {1}, {0 Community Discussions
Trending Discussions on Aij
QUESTION
May you suggest me an efficient way to implement
...ANSWER
Answered 2021-Mar-19 at 18:23The operation you want is just sum(a*b) in Fortran. If a and b are vectors then the result is the dot product. s = a(1)*b(1)+a(2)*b(2)+.. But if they are matrices then I don't know what to call it.
Here is a quick overview of linear algebra in the two environments
Operation Matlab Fortran Element Multiplicationc = a .* b
c = a * b
Matrix Product
c = a * b
c = matmul(a,b)
Element Sum
sum(sum(c))
sum(c)
Transpose
a.'
transpose(a)
Inner Product
a.'*b
dot_product(a,b)
Outer Product
a*b.'
matmul(a,transpose(b))
Maximum Value
max(max(a))
maxval(a)
Minimum Value
min(min(a))
minval(a)
By the way, most compilers will recognize the do loops and vectorize them resulting in just as fast code as the above. If fact, in my testing the do-loop method is about 3-5% faster than the compact methods. I usually prefer the compact notation and only if I need to optimize only some specific code I would write out the loops. Remember to compile with AVX2 enabled, and that the compiler unrolls the loops allowing for some more efficient memory access.
QUESTION
I want to get all the sequences of number meeting a condition in a 3D array along axis 0. I am trying to vectorize everything if possible to avoid wildly inefficient loops.
The following code works, it takes array a defined, with shape (6, 2, 3) for example. It applies a mask on it, defined by array b (same dimensions).
Then, I slice my 3D array along axis 0, so that I have 2*3 1D slices, and thus obtain 6 1D array of shape (6,). In order to do that, I use a loop, which is obviously going to be an efficiency problem with larger arrays.
I then split my array according to the mask obtained, and select (simple print for now) the series with at least 3 consecutive values meeting the conditions given by the b array.
...ANSWER
Answered 2021-Feb-27 at 06:30Instead of looping over axis (1,2), you can loop through axis 0 to iteratively check whether the mask holds for consecutive elements. Depending on what exactly you want to achieve in the end, it will take at most n iterations (where n is the longest sequence you have).
For instance, if you simply want to identify the starting elements of any valid sequence of at least 3 consecutive elements along axis=0 that satisfy mask, you would do:
QUESTION
I'm very new to Haskell and haven't fully understood how it works yet. In the following method I would like to change a certain value in a matrix or as it is realized in Haskell, a list of lists.
...ANSWER
Answered 2021-Jan-17 at 16:13First note that there isn't really any need to make this specific to Int-valued matrices: the signature could just as well be
QUESTION
I came across this problem and could not think of a good solution. The problem is as follows
There are N CS courses and N EEE courses, both numbered 1, 2, ... , N.
For each i, j (1 ≤ i, j ≤ N), the compatibility of CS course i and EEE course j is given as an integer Aij. If Aij = 1, CS course i and EEE course j do not clash; if Aij = 0, they clash.
Ravan is trying to make N pairs, each containing a CS course and a EEE course that do not clash. Here, each CS course and each EEE course must belong to exactly one pair.
Find the number of ways in which Ravan can make N pairs, modulo 10^9 + 7.
...ANSWER
Answered 2020-Oct-25 at 20:57This is just an expansion of @amit's comment that this looks like Inclusion-Exclusion.
First we take the number of ways to make N pairs. Which is simply N!. In your example, this is 6.
QUESTION
For the following pseudocode; I think that the number of flops is 2n^3. However, I am unsure that this is correct as the for loops make me doubt it. (Note: aij and xij represent entries for matrices A and X respectively)
...ANSWER
Answered 2020-Oct-12 at 02:28This isn't really a matlab question. Here is a brute force way of working it out.
The inner equation has two flops, so the k loop has 2(n-j+1) flops.
To do the rest, it helps to know the sum of q from 1 to p is p(p+1)/2 and q^2 from 1 to p is p(p+1)(2p+1)/6.
The j loop is of 2(n-j+1) for j from 1 to i, so it has 2i(n+1)-i(i+1)=2i(n+1/2)-i^2 flops.
The overall, or i, loop is a sum of 2i(n+1/2)-i^2, giving n(n+1)(n+1/2) - n(n+1)(2n+1)/6 = n(n+1)(2n+1)/3.
You can see that this is the same as the sum from 1 to n of 2i^2.
A way to check this, e.g. in matlab, is to set some n, and put f=0 at the start and replace the inner equation with f=f+2;, then the result will be f=n(n+1)(2n+1)/3.
QUESTION
I have two lists of strings, and wish to find all pairs of strings between them that contain no common characters. e.g.
...ANSWER
Answered 2020-Sep-26 at 13:59Try this and tell me if there is any improvement:
QUESTION
I have a function where a summation is calculated such that j != i as shown below:
ANSWER
Answered 2020-Sep-03 at 06:01Here is vectorize version of your function (it returns the same output as your function minus floating point error):
QUESTION
I'm solving this optimization problem where I need to figure out how many distribution centers I need to open in order to fulfill the demand of the 12 company facilities, while minimizing the transportation costs. The transportation costs are simply the distance between the distribution centers times the cost per mile, however in this problem, the cost per mile is one dollar. I have 5 choices which are Boston, Nashua, Providence, Springfield and Worcester, these 5 are part of the 12 company facilities.
I solved the problem and got the correct answer but then I tried to add two constraints to the same code and the answer I get is incorrect. The two other constraints are that the average distance from the distribution centers (DC) to the other facilities (customer) has to be less than 60 miles; and the second constraint is that the percentage of customers within 50 miles has to be greater than 80% (0.8). I know the answer to this problem, the cost has to be $66,781 dollars, the average customer distance is 15 miles and the percentage of customers within 50 miles is 90%. The output of my code is that the cost is $66289 dollars, the average customer distance is 15.36 miles and the percentage of customers within 50 miles is 179%, which doesn't make sense.
Can you help me figure out why am I getting an odd output? Thanks in advance.
...ANSWER
Answered 2020-Jun-20 at 02:42There is a message from the solver that it terminated early at 500 iterations when you set m.solve(disp=True). It returns a feasible integer solution but it may not be the best one.
QUESTION
I want to implement a command in R which generates a nxn matrix A, such that each element aij = 2^|j-k|. I have the below code but I was wondering if there was something more elegant and precise rather than a nested loop?
...ANSWER
Answered 2020-May-20 at 13:36Assuming j are rows and k are columns:
QUESTION
I'm having trouble with implementing Conv2D backpropagation using Numpy. The shape of the input is [channels, height, width]. The shape of the filters is [n_filters, channels, height, width] This is what I've done in forward propagation:
...ANSWER
Answered 2020-May-19 at 22:11I managed to find a solution, the tensordot for calculating dA is taking too much time but at least it's working.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
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Install Aij
You can use Aij like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Aij component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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