jcchurch

function yes = symmetric(g)
    % Determins if an adjacency matrix
    % is symmetric or not.
    yes = 1;
    [height width] = size(g);
    if height ~= width
        yes = 0;
    end

    for i = 1:height

jcchurch

import java.util.*;

public class Tribute {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int datasets = scan.nextInt();
        scan.nextLine();

        for (int i = 0; i < datasets; i++) {
	    String original = scan.nextLine();

jcchurch

function [m b] = linear_regression(X, Y)
    n = length(Y);
    EX = sum(X);
    EY = sum(Y);
    EX2 = sum(X.*X);
    EXY = sum(X.*Y);
    m = (n*EXY - EX*EY) / (n*EX2 - EX*EX);
    b = (EY - m*EX) / n;
end

jcchurch

# This is not the actual Birthday Paradox Problem.
# This is something similar suggested by Taylor Smith.
#
# How many people do you need in a room to have a 50%
# chance of having every possible birthday represented
# within the room?
#
# After typing all of this out, I realized that it's
# the same as the Coupon Collector's Problem.

jcchurch

import java.util.Scanner;

// This is another problem submission from a programming practice.
// The problem description can be found here:
// http://livearchive.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=374&page=show_problem&problem=2697

// Solved this one in approximately 1 hour.

public class Honeycomb {
    public static void main(String[] args) {

jcchurch

#!/usr/bin/env python

message = "010100000110100101100111001000000111100101101111011101010010011101110010011001010010000001100001001000000110010001101001011000110110101100101110"

bytes = []
i = 0
while i < len(message):
    bytes.append( message[i:i+8] )
    i += 8

jcchurch

function distance = distancePointToLineSegment(p, a, b)
    % p is a point
    % a is the start of a line segment
    % b is the end of a line segment

    pa_distance = sum((a-p).^2);
    pb_distance = sum((b-p).^2);

    unitab = (a-b) / norm(a-b);

jcchurch

function yes = thereIsACycle(G)
    % G is an n-by-n association matrix.
    % An edge is defined by anything in that
    % matrix that is not 0 and not positive infinity.
    % Elements along the diagonal are ignored.
    % Returns 1 if there is a cycle. 0 otherwise.
    
    yes = 0;
    n = size(G, 1);
    seen = zeros(1, n);

jcchurch

function [CENTERS, RADII] = tricenters(TRI, X, Y)
    % This function takes the same parameters as the function 'trimesh' does, since they require the same inputs.
    % INPUTS:
    %    TRI: A set of triangle indices. (N-by-3, where N is the number of triangles.)
    %    X: The X coordinate of the triangle indices.
    %    Y: The Y coordinate of the triangle indices.
    %
    % OUTPUTS:
    %    CENTERS: A 2D matrix of N-by-2, where N is the number of triangles in TRI.
    %             Each entry is a X, Y coordinate of a circumcenter.

jcchurch

function drawCircles(center,radius,NOP,style)
%------------------------------------------------------------------------------
% CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%