import java.util.concurrent.Executors; import java.util.concurrent.ScheduledExecutorService; import java.util.concurrent.TimeUnit; public class ConcurrentStuff { public static void main(String[] args) { System.out.println("begin"); ScheduledExecutorService executor = Executors.newScheduledThreadPool(1); System.out.println("scheduling a task to run after 2 seconds"); // Schedule a task to run after a delay of 2 seconds executor.schedule(new MyTask(), 2, TimeUnit.SECONDS); System.out.println("scheduling a task to delay 3 seconds then repeat every 5 seconds"); // Schedule a task to run after a delay of 3 seconds and repeat every 5 seconds executor.scheduleAtFixedRate(new MyTask(), 3, 5, TimeUnit.SECONDS); // Wait for scheduled tasks to complete try { System.out.println("sleeping to demo scheduling at work"); Thread.sleep(15000); } catch (InterruptedException e) { e.printStackTrace(); } // Shutdown the executor executor.shutdown(); System.out.println("end"); } static class MyTask implements Runnable { @Override public void run() { System.out.println("Task executed at: " + new java.util.Date()); } } }
/*
*Comments
*/
Wednesday, September 18, 2024
Java ScheduledExecutorService
Sunday, September 15, 2024
Determinant of a matrix
This little program calculates the determinant of an matrix.. using recursion. You don't even need ad - bc for a 2x2 [[a b][c d]]
https://en.wikipedia.org/wiki/Determinant
public class CodeWar { public static int[][] buildSubmatrix(int[][] matrix, int r, int c) { int size = matrix.length - 1; int result[][] = new int[size][size]; int a = 0, b = 0; for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix.length; j++) { if (i != 0 && j != c) { result[a][b] = matrix[i][j]; b++; } } if (i != r) { a++; b = 0; } } return result; } public static int determinant(int[][] matrix) { if (matrix.length == 1) return matrix[0][0]; int sign = 1; int sum = 0; for (int i = 0; i < matrix[0].length; i++) { int n = matrix[0][i]; int submatrix[][] = buildSubmatrix(matrix, 0,i); sum += sign * n * determinant(submatrix); sign = sign * -1; } return sum; } public static void main(String arg[]) { int[][] matrix = { { -2, -1,2 }, { 2,1,4 }, { -3,3,1 } }; System.out.println(determinant(matrix)); } }
Tuesday, August 20, 2024
Java: of list and set
There are multiple ways to construct a list in java. (easier in python and other languages).
To eliminate duplicates, you can use the stream distinct, or put it in a set (HashSet or LinkedHashSet to retain order) then make it back to a list.
public static void main(String arg[]) { String a[] = new String[] { "A", "B", "C", "D", "A", "C", "E" }; List<String> list = Arrays.asList(a); // constructed one way List<String> list2 = List.of(a); // constructed another way (java 9+) List<String> list3 = List.of( "A", "B", "C", "D", "A", "C", "E" ); // yet another way // using stream distinct one way List<String> noDups = list.stream().distinct().collect(Collectors.toList()); // using stream distinct another way List<String> noDups2 = list2.stream().distinct().toList(); // put on a set and make a iist out of it Set<String> set = new LinkedHashSet<>(); set.addAll(list3); List<String> noDups3 = new ArrayList<>(set); System.out.println(noDups); System.out.println(noDups2); System.out.println(noDups3); }
Wednesday, July 24, 2024
Python linear 2x2 system
This is a fun tool for any algebra 1 student. Solve a 2x2 linear system, and graph it.
import matplotlib.pyplot as plt import numpy as np import math """ Simple little 2x2 system example: x + 2y = 1 3x + 5y = 2 This can be written as matrix form [A][x] = [B]: [1 2][x] = [1] [3 5][y] [2] """ A = np.array([[1, 2], [3, 5]]) B = np.array([1, 2]) XMIN, XMAX = -5, 5 YMIN, YMAX = -5, 5 GRIDSIZE = 1 STEPS = 101 a11 = A[0][0]; a12 = A[0][1]; a21 = A[1][0]; a22 = A[1][1]; b11 = B[0]; b12 = B[1]; def f(x): return (-a11*x+b11)/a12 def g(x): return (-a21*x+b12)/a22 x = np.linspace(XMIN,XMAX,STEPS) plt.axis([XMIN,XMAX,YMIN,YMAX]) y1v = np.zeros(len(x)) y2v = np.zeros(len(x)) for i in range(len(x)): try: y1v[i] = f(x[i]) y2v[i] = g(x[i]) except: pass try: solution = np.linalg.solve(A, B) print("Solution") print(solution) solutionpoint = "(${0:.2f}, ${1:.2f})".format(solution[0], solution[1]); plt.plot(solution[0],solution[1],'bo'); # blue circle plt.annotate( solutionpoint,(solution[0],solution[1])); except: print("No solution") eq1 = "{0}x+{1}y={2}".format(a11,a12,b11); eq2 = "{0}x+{1}y={2}".format(a21,a22,b12); plt.plot(x,y1v, color='r',label=eq1) plt.plot(x,y2v, color='g',label=eq2) # make axes plt.axhline(y=0, color='k') # 'k' means black plt.axvline(x=0, color='k') # gca is get current axes plt.gca().set_aspect('equal') plt.gca().set_xticks(np.arange(XMIN,XMAX,GRIDSIZE)) plt.gca().set_yticks(np.arange(YMIN,YMAX,GRIDSIZE)) plt.grid(True) plt.legend() plt.show()
Monday, July 22, 2024
Python linear algebra
Every algebra 1 student should learn about solving a system of 2 equations and 2 unknowns. Then they will learn about 3 by 3... then gosh n by n. There is gotta be easier way to do this than combining equations, well yes there is, enter linear algebra.
But solving a linear system of equation is still hard to do by hand, painful to punch it on your calculator too. Enter computer programming! Python can effortless do this with the Numpy library. Programming this from scratch is not a very pleasant programming task.
import numpy as np """ Simple little 2x2 system example: x + 2y = 1 3x + 5y = 2 This can be written as matrix form [A][x] = [B]: [1 2][x] = [1] [3 5][y] [2] So the solution is inv(A)B Inverse matrix is painful to do by hand, but numpy can do it easy, It even has a solve method """ A = np.array([[1, 2], [3, 5]]) B = np.array([1, 2]) Ainv = np.linalg.inv(A) solution = np.matmul(Ainv,B) print("Solution by multiply inverse: ") print(solution) solution2 = np.linalg.solve(A, B) print("Solution by linalg solve: ") print(solution2)Output:
Solution by multiply inverse matrix and right hand side: [-1. 1.] or by linalg solve: [-1. 1.]That means x=-1, y =1.
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