Why this happens… With plot(x(i),y) you are plotting 100 single points (one in each iteration) and they are not shown by default. Therefore the plot looks empty. Solution 1: Vectorized calculation and direct plot I assume you meant to draw a continuous line. In that case no for-loop is needed because you can calculate and … Read more
Why do you want to avoid Metal? Enabling its support is as easy as wrapping your LineGraphShapes into Group and modifying it with a drawingGroup(). Try it out: … Group { let gradient = LinearGradient( gradient: Gradient(colors: [Color.red, Color.blue]), startPoint: .leading, endPoint: .trailing ) LineGraphShape(points: points[selectedPointType], closePath: true) .fill(gradient) LineGraphShape(points: points[selectedPointType], closePath: false) .stroke( gradient, … Read more
You can use this simple script to do that. But you must specify the names of the output nodes. import tensorflow as tf meta_path=”model.ckpt-22480.meta” # Your .meta file output_node_names = [‘output:0’] # Output nodes with tf.Session() as sess: # Restore the graph saver = tf.train.import_meta_graph(meta_path) # Load weights saver.restore(sess,tf.train.latest_checkpoint(‘path/of/your/.meta/file’)) # Freeze the graph frozen_graph_def = … Read more
Both are valid DFS algorithms. A DFS does not specify which node you see first. It is not important because the order between edges is not defined [remember: edges are a set usually]. The difference is due to the way you handle each node’s children. In the iterative approach: You first insert all the elements … Read more
The easiest way I’ve found to create postscripts is the following, using the setEPS() command: setEPS() postscript(“whatever.eps”) plot(rnorm(100), main=”Hey Some Data”) dev.off()
You might like to look at Lissajous curves; an example of a = 5, b = 4 (5:4) is shown below. Addendum: Once you see how to plot points in xy coordinates, then you should look at converting between polar and Cartesian coordinates. public class LissajousPanel extends JPanel { private static final int SIZE = … Read more
As a caveat, remember that there can be exponentially many shortest paths between two nodes in a graph. Any algorithm for this will potentially take exponential time. That said, there are a few relatively straightforward algorithms that can find all the paths. Here’s two. BFS + Reverse DFS When running a breadth-first search over a … Read more
The correct method to use, in order to be able to draw another chart on the same canvas, is .destroy(). You must call it on the previously created chart object. You may also use the same variable for both charts. var grapharea = document.getElementById(“barChart”).getContext(“2d”); var myChart = new Chart(grapharea, { type: ‘bar’, data: barData, options: … Read more
Even though this is a somewhat old question, I thought I’d give a practical answer for anyone stumbling across this. Let’s say you get your input data for your connections as a list of tuples like so: [(‘A’, ‘B’), (‘B’, ‘C’), (‘B’, ‘D’), (‘C’, ‘D’), (‘E’, ‘F’), (‘F’, ‘C’)] The data structure I’ve found to … Read more
It depends on the problem. Adjacency Matrix Uses O(n^2) memory It is fast to lookup and check for presence or absence of a specific edge between any two nodes O(1) It is slow to iterate over all edges It is slow to add/delete a node; a complex operation O(n^2) It is fast to add a … Read more