Summation of Perceptron not working properly. Getting large summation - perceptron

So I have a run method which summates the weights of the edges in the artificial neural network with the threshold values of the input nodes.
Sort of like this:
Now my test perceptron should produce a summation of -3, but I am getting a value of 1176!!! What is going on here?
Here is the code that I have written for my run() method, constructor, and my main method.
Constructor:
public class Perceptron {
//We want to create a variable which will represent the number of weighted edges
//in the 2-dimensional array.
protected int num_weighted_Edges;
//Inside this class we want to create a data field which is a
//2-D array of WeightedEdges. Since the weightedEdges will be in
//double data type, we will create a double type 2-dimensional
//array.
protected WeightedEdge[][] weightedEdges;
protected int[] weights;
//We set a double field named eta equal to 0.05.
protected double eta = 0.05;
//We initialize a constructor which only takes a parameter int n.
public Perceptron(int n){
//We want to create a new graph which will have n + 1 vertices
//, where we also want vertex 0 to act like the output node
//as in a neural network.
this.num_weighted_Edges = n;
weights = new int[num_weighted_Edges];
//First we need to verify that n is a positive real number
if (num_weighted_Edges < 0){
throw new RuntimeException("You cannot have a perceptron of negative value");
}
else {
//Test code for testing if this code works.
System.out.println("A perceptron of " + num_weighted_Edges + " input nodes, and 1 output node was created");
}
//Now we create a graph object with "n" number of vertices.
weightedEdges = new WeightedEdge[num_weighted_Edges + 1][num_weighted_Edges + 1];
//Create a for loop that will iterate the weightedEdges array.
//We want to create the weighted edges from vertex 1 and not vertex 0
//since vertex 0 will be the output node, so we set i = 1.
for (int i = 1; i < weightedEdges.length; i++){
for (int j = 0; j < weightedEdges[i].length; j++){
//This will create a weighted edge in between [1][0]...[2][0]...[3][0]
//The weighted edge will have a random value between -1 and 1 assigned to it.
weightedEdges[i][0] = new WeightedEdge(i, j, 1);
}
}
}
This is my run() method:
//This method will take the input nodes, do a quick verification check on it and
//sum up the weights using the simple threshold function described in class to return
//either a 1 or -1. 1 meaning fire, and -1 not firing.
public int run(int[] weights){
//So this method will act like the summation function. It will take the int parameters
//you put into the parameter field and multiply it times the input nodes in the
//weighted edge 2 d array.
//Setup a summation counter.
int sum = 0;
if (weights.length != num_weighted_Edges){
throw new RuntimeException("Array coming in has to equal the number of input nodes");
}
else {
//We iterate the weights array and use the sum counter to sum up weights.
for (int i = 0; i < weights.length; i++){
//Create a nested for loop which will iterate over the input nodes
for ( int j = 1; j < weightedEdges.length; j++){
for (int k = 0; k < weightedEdges[j].length; k++){
//This takes the weights and multiplies it times the value in the
//input nodes. The sum should equal greater than 0 or less than 0.
sum += (int) ((weightedEdges[j][0].getWeight()) * i);
//Here the plus equals sign takes the product of (weightedEdges[j][0] * i) and
//then adds it to the previous value.
}
}
}
}
System.out.println(sum);
//If the sum is greater than 0, we fire the neuron by returning 1.
if (sum > 0){
//System.out.println(1); test code
return 1;
}
//Else we don't fire and return -1.
else {
//System.out.println(-1); test code
return -1;
}
}
This is my main method:
//Main method which will stimulate the artificial neuron (perceptron, which is the
//simplest type of neuron in an artificial network).
public static void main(String[] args){
//Create a test perceptron with a user defined set number of nodes.
Perceptron perceptron = new Perceptron(7);
//Create a weight object that creates an edge between vertices 1 and 2
//with a weight of 1.5
WeightedEdge weight = new WeightedEdge(1, 2, 1.5);
//These methods work fine.
weight.getStart();
weight.getEnd();
weight.setWeight(2.0);
//Test to see if the run class works. (Previously was giving a null pointer, but
//fixed now)
int[] test_weight_Array = {-1, -1, -1, -1, -1, 1, 1};
//Tested and works to return output of 1 or -1. Also catches exceptions.
perceptron.run(test_weight_Array);
//Testing a 2-d array to see if the train method works.
int[][] test_train_Array = {{1}, {-1}, {1}, {1}, {1}, {1}, {1}, {1}};
//Works and catches exceptions.
perceptron.train(test_train_Array);
}
}

I think you should change
sum += (int) ((weightedEdges[j][0].getWeight()) * i);
to
sum += (int) ((weightedEdges[j][k].getWeight()) * i);

Related

Autodiff for Jacobian derivative with respect to individual joint angles

I am trying to compute $\partial{J}{q_i}$ in drake C++ for manipulator and as per my search, the best approach seems to be using autodiff function. I was not able to fully understand autodiff approach from the resources that I found, so I apologize if my approach is not clear enough. I have used my understanding from some already asked questions mentioned on the forum regarding auto diff as well as https://drake.mit.edu/doxygen_cxx/classdrake_1_1multibody_1_1_multibody_plant.html as reference.
As I want to calculate $\partial{J}{q_i}$, the return type will be a tensor i.e. 3 * 7 * 7(or 6 * 7 * 7 depending on the spatial jacobian). I can think of using std::vectorEigen::MatrixXd to allocate the output or alternatively just doing one $q_i$ at a time and computing the respective jacobian for the auto diff. In either case, I was struggling to pass it in the initializing the jacobian function.
I did the following to initialize autodiff
std::unique_ptr<multibody::MultibodyPlant<AutoDiffXd>> mplant_autodiff = systems::System<double>::ToAutoDiffXd(mplant);
std::unique_ptr<systems::Context<AutoDiffXd>> mContext_autodiff = mplant_autodiff->CreateDefaultContext();
mContext_autodiff->SetTimeStateAndParametersFrom(*mContext);
const multibody::Frame<AutoDiffXd>* mFrame_EE_autodiff = &mplant_autodiff->GetBodyByName(mEE_link).body_frame();
const multibody::Frame<AutoDiffXd>* mWorld_Frame_autodiff = &(mplant_autodiff->world_frame());
//Initialize the q as autodiff vector
drake::AutoDiffVecXd q_autodiff = drake::math::InitializeAutoDiff(mq_robot);
MatrixX<AutoDiffXd> mJacobian_autodiff; // Linear Jacobian matrix.
mplant_autodiff->SetPositions(context_autodiff.get(), q_autodiff);
mplant_autodiff->CalcJacobianTranslationalVelocity(*mContext_autodiff,
multibody::JacobianWrtVariable::kQDot,
*mFrame_EE_autodiff,
Eigen::Vector3d::Zero(),
*mWorld_Frame_autodiff,
*mWorld_Frame_autodiff,
&mJacobian_autodiff
);
However, as far as I understand, InitializeAutoDiff initializes to the identity matrix, whereas I want to $\partial{J}{q_i}$, so is there is a better way to do it. In addition, I get error messages when I try to call the jacobian matrix. Is there a way to address this problem both for $\partial{J}{q_i}$ for each q_i and changing q_i in a for loop or directly getting the result in a tensor. My apologies if I am doing something total tangent to the correct approach. I thank you in anticipation.
However, as far as I understand, InitializeAutoDiff initializes to the identity matrix, whereas I want to $\partial{J}{q_i}$, so is there is a better way to do it
That is correct. When you call InitializeAutoDiff and compute mJacobian_autodiff, you get a matrix of AutoDiffXd. Each AutoDiffXd has a value() function that stores the double value, and a derivatives() storing the gradient as an Eigen::VectorXd. We have
mJacobian(i, j).value() = J(i, j)
mJacobian_autodiff(i, j).derivatives()(k) = ∂J(i, j)/∂q(k)
So if you want to create a std::vecot<Eigen::MatrixXd> such that the k'th entry of this vector stores the matrix ∂J/∂q(k), then here is a code
std::vector<Eigen::MatrixXd> dJdq(q_autodiff.rows());
for (int i = 0; i < q_autodiff.rows(); ++i) {
dJdq[i].resize(mJacobian_autodiff.rows(), mJacobian_autodiff.cols());
}
for (int i = 0; i < q_autodiff.rows(); ++i) {
// dJidq stores the gradient of the ∂J.col(i)/∂q, namely dJidq(j, k) = ∂J(j, i)/∂q(k)
auto dJidq = ExtractGradient(mJacobian_autodiff.col(i));
for (int j = 0; j < static_cast<int>(dJdq.size()); ++j) {
dJdq[j].col(i) = dJidq.col(j);
}
}
Compute ∂J/∂q(i) for a single i
If you do not want to compute ∂J/∂q(i) for all i, but only for one specific i, you can change the initialization of q_autodiff from InitializeAutoDiff to this
AutoDiffVecXd q_autodiff(q.rows());
for (int k = 0; k < q_autodiff.rows(); ++k) {
q_autodiff(k).value() = q(k)
q_autodiff(k).derivatives() = Vector1d::Zero();
if (k == i) {
q_autodiff(k).derivatives()(0) = 1;
}
}
namely q_autodiff stores the gradient ∂q/∂q(i), which is 0 for all k != i and 1 when k == i. And then you can compute mJacobian_autodiff using your current code. Now mJacobian_autodiff(m, n).derivatives() store the gradient of ∂J(m, m)/∂q(i) for that specific i. You can extract this gradient as
Eigen::Matrix dJdqi(mJacobian_autodiff.rows(), mJacobian_autodiff.cols());
for (int m = 0; m < dJdqi.rows(); ++m) {
for (int n = 0; n < dJdqi.cols(); ++n) {
dJdqi(m, n) = mJacobian_autodiff(m, n).derivatives()(0);
}
}

Dart: How to iterate digits of integer?

How to iterate digits of integer? for example sum of digits here, it works, but is any way to right way?
int sumOfDigits(int num) {
int sum = 0;
String numtostr = num.toString();
for (var i = 0; i < numtostr.length; i++) {
sum = sum + int.parse(numtostr[i]);
}
return sum;
}
If you're looking for a shorter way to do this, you can combine split, map and reduce
int sum = num.split('').map((e) => int.parse(e)).reduce((t, e) => t + e);
You can even do this:
int sum = num.split('').map(int.parse).reduce((t, e) => t + e);
Thank you #julemand101
It's fairly inefficient to create a string, then split the string, and parse the individual digits back to integers.
How about something like:
Iterable<int> digitsOf(int number) sync* {
do {
yield = number.remainder(10);
number ~/= 10;
} while (number != 0);
}
This iterates the digits of the (non-negative) number in base 10, from least significant to most significant, without allocating any strings along the way.
If you want the digits in the reverse order, you can either create a list from the iterable above and reverse it, or use a different approach:
Iterable<int> digitsHighToLow(int number) sync* {
var base = 1;
while (base * 10 < number) {
base = base * 10;
}
do {
var digit = number ~/ base;
yield digit;
number = (number - digit * base) * 10;
} while (number != 0);
}
(again, only works on non-negative numbers, you'll have to figure out what you want for negative numbers, either throw, or try negating the number, it's the same digits after all, or something else).

flutter how to generate random numbers without duplication

Is there any way to generate random numbers without duplication?
For instance I want to generate 50 random numbers from 1 to 100 no duplication, any way to do this or do I have to check every time incoming number is already created or not?
you can use shuffle as following code.
import 'dart:math';
var list = new List<int>.generate(10, (int index) => index); // [0, 1, 4]
list.shuffle();
print(list);
You can use Set. Each object can occur only once when using it. Just try this:
Set<int> setOfInts = Set();
while (setOfInts.length < 50) {
setOfInts.add(Random().nextInt(range) + 1);
}
You can read the documentation here: Set Doc
Here is an alternative that avoids creating an array of all the possible values, and avoids repeatedly looping until no collision occurs. It may be useful when there is a large range to select from.
import 'dart:math';
class RandomList {
static final _random = new Random();
static List<int> uniqueSample({int limit, int n}) {
final List<int> sortedResult = [];
final List<int> result = [];
for (int i = 0; i < n; i++) {
int rn = _random.nextInt(limit - i); // We select from a smaller list of available numbers each time
// Increment the number so that it picks from the remaining list of available numbers
int j = 0;
for (; j < sortedResult.length && sortedResult[j] <= rn; j++) rn++;
sortedResult.insert(j, rn);
result.add(rn);
}
return result;
}
}
I haven't tested it exhaustively but it seems to work.

Fast implementation of BWT in Lua

local function fShallowCopy(tData)
local tOutput = {}
for k,v in ipairs(tData) do
tOutput[k] = v
end
return tOutput
end
local function fLexTblSort(tA,tB) --sorter for tables
for i=1,#tA do
if tA[i]~=tB[i] then
return tA[i]<tB[i]
end
end
return false
end
function fBWT(tData)
--setup--
local iSize = #tData
local tSolution = {}
local tSolved = {}
--key table--
for n=1,iSize do
tData[iSize] = fRemove(tData,1)
tSolution[n] = fShallowCopy(tData)
end
table.sort(tSolution,fLexTblSort)
--encode output--
for i=1,iSize do
tSolved[i] = tSolution[i][iSize]
end
--finalize--
for i=1,iSize do
if fIsEqual(tSolution[i],tData) then
return i,tSolved
end
end
return false
end
Above is my current code for achieving BWT encoding in Lua. The issue is because of the size of the tables and lengths of loops it takes a long time to run. For a 1000 character input the average encoding time is about 1.15 seconds. Does anyone have suggestions for making a faster BWT encoding function?
the biggest slowdowns appear to be in fLexTblSort and fShallowCopy. I have included both above the BWT function as well.
If I see right, your algorithm has complexity O(n^2 log n), if the sort is quicksort. The comparator function fLexTblSort takes O(n) itself for each pair of values you compare.
As I checked with my implementation from few years back, I see possible space to improve. You create all the possible rotations of the tData, which takes also a lot of time. I used only single data block and I stored only starting positions of particular rotations. You also use a lot of loops which can shrink into less.
Mine implementation was in C, but the concept can be used also in Lua. The idea in some hybrid pseudocode between your Lua and C.
function fBWT(tData)
local n = #tData
local tSolution = {}
for(i = 0; i < n; i++)
tSolution[i] = i;
--table.sort(tSolution, fLexTblSort)
quicksort(tData, n, tSolution, 0, n)
for(i = 0; i < n; i++){
tSolved[i] = tData[(tSolution[i]+n-1)%n];
if( tSolution[i] == 0 )
I = i;
}
return I, tSolved
end
You will also need your own sort function, because the standard does not offer enough flexibility for this magic. Quicksort is a good idea (you might avoid some of the arguments, but I pasted just the C version I was using):
void swap(int array[], int left, int right){
int tmp = array[right];
array[right] = array[left];
array[left] = tmp;
}
void quicksort(uint8_t data[], int length, int array[], int left, int right){
if(left < right){
int boundary = left;
for(int i = left + 1; i < right; i++){
if( offset_compare(data, length, array, i, left) < 0 ){
swap(array, i, ++boundary);
}
}
swap(array, left, boundary);
quicksort(data, length, array, left, boundary);
quicksort(data, length, array, boundary + 1, right);
}
}
The last step is your own comparator function (similar to your original, but working on the rotations, again in C):
/**
* compare one string (fixed length) with different rotations.
*/
int offset_compare(uint8_t *data, int length, int *array, int first, int second){
int res;
for(int i = 0; i < length; i++){
res = data[(array[first]+i)%length] - data[(array[second]+i)%length];
if( res != 0 ){
return res;
}
}
return 0;
}
This is the basic idea I came up with few years ago and which worked for me. Let me know if there is something not clear or some mistake.

Checking if removing an edge in a graph will result in the graph splitting

I have a graph structure where I am removing edges one by one until some conditions are met. My brain has totally stopped and i can't find an efficient way to detect if removing an edge will result in my graph splitting in two or more graphs.
The bruteforce solution would be to do an bfs until one can reach all the nodes from a random node, but that will take too much time with large graphs...
Any ideas?
Edit: After a bit of search it seems what I am trying to do is very similar to the fleury's algorithm, where I need to find if an edge is a "bridge" or not.
Edges that make a graph disconnected when removed are called 'bridges'. You can find them in O(|V|+|E|) with a single depth-first search over the whole graph. A related algorithm finds all 'articulation points' (nodes that, if removed, makes the graph disconnected) follows. Any edge between two articulation-points is a bridge (you can test that in a second pass over all edges).
//
// g: graph; v: current vertex id;
// r_p: parents (r/w); r_a: ascents (r/w); r_ap: art. points, bool array (r/w)
// n_v: bfs order-of-visit
//
void dfs_art_i(graph *g, int v, int *r_p, int *r_v, int *r_a, int *r_ap, int *n_v) {
int i;
r_v[v] = *n_v;
r_a[v] = *n_v;
(*n_v) ++;
// printf("entering %d (nv = %d)\n", v, *n_v);
for (i=0; i<g->vertices[v].n_edges; i++) {
int w = g->vertices[v].edges[i].target;
// printf("\t evaluating %d->%d: ", v, w);
if (r_v[w] == -1) {
// printf("...\n");
// This is the first time we find this vertex
r_p[w] = v;
dfs_art_i(g, w, r_p, r_v, r_a, r_ap, n_v);
// printf("\n\t ... back in %d->%d", v, w);
if (r_a[w] >= r_v[v]) {
// printf(" - a[%d] %d >= v[%d] %d", w, r_a[w], v, r_v[v]);
// Articulation point found
r_ap[i] = 1;
}
if (r_a[w] < r_a[v]) {
// printf(" - a[%d] %d < a[%d] %d", w, r_a[w], v, r_a[v]);
r_a[v] = r_a[w];
}
// printf("\n");
}
else {
// printf("back");
// We have already found this vertex before
if (r_v[w] < r_a[v]) {
// printf(" - updating ascent to %d", r_v[w]);
r_a[v] = r_v[w];
}
// printf("\n");
}
}
}
int dfs_art(graph *g, int root, int *r_p, int *r_v, int *r_a, int *r_ap) {
int i, n_visited = 0, n_root_children = 0;
for (i=0; i<g->n_vertices; i++) {
r_p[i] = r_v[i] = r_a[i] = -1;
r_ap[i] = 0;
}
dfs_art_i(g, root, r_p, r_v, r_a, r_ap, &n_visitados);
// the root can only be an AP if it has more than 1 child
for (i=0; i<g->n_vertices; i++) {
if (r_p[i] == root) {
n_root_children ++;
}
}
r_ap[root] = n_root_children > 1 ? 1 : 0;
return 1;
}
If you remove the link between vertices A and B, can't you just check that you can still reach A from B after the edge removal? That's a little better than getting to all nodes from a random node.
How do you choose the edges to be removed?
Can you tell more about your problem domain?
Just how large Is your graph? maybe BFS is just fine!
After you wrote that you are trying to find out whether an edge is a bridge or not, I suggest
you remove edges in decreasing order of their betweenness measure.
Essentially, betweenness is a measure of an edges (or vertices) centrality in a graph.
Edges with higher value of betweenness have greater potential of being a bridge in a graph.
Look it up on the web, the algorithm is called 'Girvan-Newman algorithm'.

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