We Keep Coding

Square root line doesn’t work

I wrote a quick code for factorizing formulas, however the line that takes creates the square root of D doesn’t work. The line is line 10. Any help is appreciated.
using System;
public class MainClass {
//Source Numbers
public int A = 1;
public int B = 3;
public int C = 9;
//Calculation Numbers
public float Di;
public static double Sqrt(double Di); //This is the faulted line.
//Answers
public float X;
public float X1;
public float X2;
public static void Main() {
Console.Writeline("D=", Di);
//Calculation for the Square root of D
// (DSq)Math.Sqrt(Di);
Di = B^2-4*A*C;
//Calculation for the answers
if(Di>0) {
X1 = ((0-B)-DSq)/(A*2);
X2 = ((0-B)+DSq)/(A*2);
Console.Writeline("X=", X1, " or X=", X2);
}
else if(Di=0) {
X = 0-B;
Console.Writeline("X=", X);
}
else {
Console.Writeline("The formula cannot be solved.");
}
}
}

You are using a method definition with no body. In any case you dont need to invent the wheel, since Math has already a Math.Sqrt(), method. Try:
........
Di = B^2-4*A*C;
if (Di>0)
{
var sqrDi = Math.Sqrt(Di);
.....
}
...

You have several errors in your code, like the spelling of WriteLine, and comparing in if statements (use ==). This returns a list of valid solutions (X-values):
public IList<double> factorizeABC(double a, double b, double c)
{
var solutions = new List<double>();
var Di = b * b - 4 * a * c;
if (Di > 0)
{
var rtDi = Math.Sqrt(Di);
var X1 = (-b - rtDi) / (a * 2);
var X2 = (-b + rtDi) / (a * 2);
solutions.Add(X1);
solutions.Add(X2);
}
else if (Di == 0)
{
var X = -b / (a * 2);
solutions.Add(X);
}
return solutions;
}
usage:
var results = factorizeABC(1, 2, -8);
if (results.Count() == 0)
Console.WriteLine("The formula cannot be solved.");
if (results.Count() == 1)
Console.WriteLine("X=" + results[0].ToString());
if (results.Count() == 2)
Console.WriteLine("X=" + results[0].ToString() + " or X=" + results[1].ToString());

How can I generate a "random constant colour" for a given string?

how can I generate a "random constant colour" for a given string at runtime?
So a given string value will always have the same colour but different strings will have different colours.
Like how gmail assigns colours to the sender names.
Thanks
Responses to comments:
Thinking to generate the colour from a hashcode.
The colours won't be stored but generated from a hashcode.

I don't know any dedicated method for this, but here is a simple method generating Hexadecimal values with MD5 based on a given string:
using System.Security.Cryptography;
using System.Text;
static string GetColor(string raw)
{
using (MD5 md5Hash = MD5.Create())
{
byte[] data = md5Hash.ComputeHash(Encoding.UTF8.GetBytes(raw));
return BitConverter.ToString(data).Replace("-", string.Empty).Substring(0, 6);
}
}
Examples:
example#example.com
-> 23463B
info#google.com
-> 3C9015
stack#exchange.com
-> 7CA5E8
Edit:
I didn't tested it enough, so you may want to tweak it a little bit to get more different and unique values.
Edit2:
If you want transparency, check out this question/answer. By setting the Substring to Substring(0,8) , you should return a string with the alpha channel.

Similar to what the other answers are suggesting (hash the string in some form then use that hash to pick the color), but instead of using the hash to directly calculate the color use it as the index to an array of "Acceptable" colors.
class ColorPicker
{
public ColorPicker(int colorCount)
{
//The ".Skip(2)" makes it skip pure white and pure black.
// If you want those two, take out the +2 and the skip.
_colors = ColorGenerator.Generate(colorCount + 2).Skip(2).ToArray();
}
private readonly Color[] _colors;
public Color StringToColor(string message)
{
int someHash = CalculateHashOfStringSomehow(message);
return _colors[someHash % _colors.Length];
}
private int CalculateHashOfStringSomehow(string message)
{
//TODO: I would not use "message.GetHashCode()" as you are not
// guaranteed the same value between runs of the program.
// Make up your own algorithom or use a existing one that has a fixed
// output for a given input, like MD5.
}
}
This prevents issues like getting a white color when you plan on showing the text with a white background and other similar problems.
To populate your Color[] see this answer for the ColorGenerator class or just make your own pre-defined list of colors that look good on whatever background they will be used on.
Appendix:
In case the link goes down, here is a copy of the ColorGenerator class
public static class ColorGenerator
{
// RYB color space
private static class RYB
{
private static readonly double[] White = { 1, 1, 1 };
private static readonly double[] Red = { 1, 0, 0 };
private static readonly double[] Yellow = { 1, 1, 0 };
private static readonly double[] Blue = { 0.163, 0.373, 0.6 };
private static readonly double[] Violet = { 0.5, 0, 0.5 };
private static readonly double[] Green = { 0, 0.66, 0.2 };
private static readonly double[] Orange = { 1, 0.5, 0 };
private static readonly double[] Black = { 0.2, 0.094, 0.0 };
public static double[] ToRgb(double r, double y, double b)
{
var rgb = new double[3];
for (int i = 0; i < 3; i++)
{
rgb[i] = White[i] * (1.0 - r) * (1.0 - b) * (1.0 - y) +
Red[i] * r * (1.0 - b) * (1.0 - y) +
Blue[i] * (1.0 - r) * b * (1.0 - y) +
Violet[i] * r * b * (1.0 - y) +
Yellow[i] * (1.0 - r) * (1.0 - b) * y +
Orange[i] * r * (1.0 - b) * y +
Green[i] * (1.0 - r) * b * y +
Black[i] * r * b * y;
}
return rgb;
}
}
private class Points : IEnumerable<double[]>
{
private readonly int pointsCount;
private double[] picked;
private int pickedCount;
private readonly List<double[]> points = new List<double[]>();
public Points(int count)
{
pointsCount = count;
}
private void Generate()
{
points.Clear();
var numBase = (int)Math.Ceiling(Math.Pow(pointsCount, 1.0 / 3.0));
var ceil = (int)Math.Pow(numBase, 3.0);
for (int i = 0; i < ceil; i++)
{
points.Add(new[]
{
Math.Floor(i/(double)(numBase*numBase))/ (numBase - 1.0),
Math.Floor((i/(double)numBase) % numBase)/ (numBase - 1.0),
Math.Floor((double)(i % numBase))/ (numBase - 1.0),
});
}
}
private double Distance(double[] p1)
{
double distance = 0;
for (int i = 0; i < 3; i++)
{
distance += Math.Pow(p1[i] - picked[i], 2.0);
}
return distance;
}
private double[] Pick()
{
if (picked == null)
{
picked = points[0];
points.RemoveAt(0);
pickedCount = 1;
return picked;
}
var d1 = Distance(points[0]);
int i1 = 0, i2 = 0;
foreach (var point in points)
{
var d2 = Distance(point);
if (d1 < d2)
{
i1 = i2;
d1 = d2;
}
i2 += 1;
}
var pick = points[i1];
points.RemoveAt(i1);
for (int i = 0; i < 3; i++)
{
picked[i] = (pickedCount * picked[i] + pick[i]) / (pickedCount + 1.0);
}
pickedCount += 1;
return pick;
}
public IEnumerator<double[]> GetEnumerator()
{
Generate();
for (int i = 0; i < pointsCount; i++)
{
yield return Pick();
}
}
IEnumerator IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
public static IEnumerable<Color> Generate(int numOfColors)
{
var points = new Points(numOfColors);
foreach (var point in points)
{
var rgb = RYB.ToRgb(point[0], point[1], point[2]);
yield return Color.FromArgb(
(int)Math.Floor(255 * rgb[0]),
(int)Math.Floor(255 * rgb[1]),
(int)Math.Floor(255 * rgb[2]));
}
}
}

3 integer variables, r,g and b.
Loop through each character in the string in steps of 3 and add the character code.
r += n + 0
g += n + 1
b += n + 2
after the loop take r,g, and b modulo 255 and create a color using Color.FromARGB.
No guarantees the color will be pretty though, and some strings may happen to have colors very close to each other.

I see some pretty good answeers but though it whould contribute with a little fun solutuion to generate colors from string, the Hash version looks like the best way to go but if this gives any one some inspiration to bould off, have at it
ConsoleKeyInfo ch = new ConsoleKeyInfo();
while (ch.KeyChar != 'e')
{
Console.WriteLine("type string to seed color");
string s = Console.ReadLine(); // gets text from input, in this case the command line
double d=0;
foreach(char cha in s.ToCharArray())
{
d=+ (int)cha; // get the value and adds it
}
d= (255/(Math.Pow(0.2,-0.002 *d))); // Generates a seed like value from i where 255 is the maximum. basicly 255/0.2^(-0.002*d)
int i = Convert.ToInt32(d); //then convets and get rid of the decimels
Color c = Color.FromArgb(i, i, i);// add a bit more calculation for varieng colers.
Console.WriteLine(c.Name);
Console.WriteLine("To Exit press e");
ch = Console.ReadKey()
}
Edit1: It definantly needs some tweeking, since the longer the string the ligther the color, but i think something can come from it with a little work :)

How to attain overloading without copying the complete method if only a minor change is needed?

Consider these methods (int used as parameter input to simplify the Example. But it can be any datatypes and must not be same)
public int Multiply(int a, int b)
{
int c=1;
return (a * b * c);
}
public int Multiply(int a, int b, int c)
{
return (a * b * c);
}
Suppose these methods were written long, Then how could I avoid copying the complete method?

Overloaded methods can call each other so you can avoid copying the complete method by extending the functionality, eg:
public int Multiply(int a, int b)
{
return Multiply(a,b,1);
}
public int Multiply(int a, int b, int c)
{
return (a * b * c);
}

In C# you can use optional arguments with default values:
public int Multiply(int a, int b, int c = 1)
{
return (a * b * c);
}
And can be used like:
int a = Multiply(2, 3); // 2 * 3 * 1 = 6
int b = Multiply(2, 3, 4); // 2 * 3 * 4 = 24

If one call cal be reduced to another more general call with appropriate parameters, then this is the preferred option. (See Inisheer and Jeremy Thompson's suggestions).
In the general case, however, when this can't be done, I suggest breaking the large operation into smaller meaningful ones. This way, one can implement different variants of the operation in subclasses, without repeating too much code.

Something like :
private static int Multiply(params int[] nums)
{
int result = 1;
foreach (int num in nums)
{
result *= num;
}
return result;
}
The above method can be written using Linq
private static int Multiply(params int[] nums)
{
return nums.Aggregate(1, (current, num) => current*num);
}
Example :
public static void Main()
{
int ans1 = Multiply(1, 2, 3, 4); // 24
int ans2 = Multiply(1, 2, 3, 4, 5); // 120
int ans3 = Multiply(1, 2, 3, 4, 5, 6); // 720
}

Well you can override it using method with varg argument.
Since already available methods cover 2 and 3 argument , any number of inputs can represented by using 1,2,3.
Heres a sample code that whipped up , which reuses the overloaded method.
Code
/**
* #author Sudhakar Duraiswamy
*
*/
public class Reuse {
public static void main(String[] args) {
Reuse r = new Reuse();
int[] rr = {1,2,3};
System.out.println(r.Multiply(rr));
int[] rr1 = {1,2,3,3};
System.out.println(r.Multiply(rr1));
int[] rr2 = {1,2,3,3,2};
System.out.println(r.Multiply(rr2));
}
public int Multiply(int... arg){
int result = 1;
if(null != arg && arg.length >0){
if(arg.length ==1){
result = arg[0];
}else if(arg.length ==2){
result = Multiply(arg[0],arg[1]);
}else if(arg.length >=3){
int remainingSize = arg.length % 3;
for(int i=0;i< arg.length-remainingSize ;i= i+3){
result = result * Multiply(arg[i], arg[i+1], arg[i+2]);
}
if(remainingSize == 2){
result =result * Multiply(arg[arg.length-2], arg[arg.length-1]);
}else if(remainingSize == 1){
result = result * arg[arg.length-1];
}
}
}
return result;
}
public int Multiply(int a, int b)
{
int c=1;
return (a * b * c);
}
public int Multiply(int a, int b, int c)
{
return (a * b * c);
}
}

Use result of entity's method for filtering with lambda expression or linq query

I would like to filter my entites depending on result of a function working with their properties.
ie. I got entity like this:
public class Lorem
{
public int A {get;set;}
public int B {get;set;}
public int C {get;set;}
public double DoMath(int externalValue)
{
// real implementation is longer and more complex
if(A==B) {
// some calculations
return 0.2;
}
if(B==C) {
// some calculations
return 0.9;
}
else return 0;
}
}
Now I am querying entities and I would like to get only those with DoMath > 0.
// simplified scenario for example purpose
int someValue = 95;
// working with Entity Framework 4.1 Code First
var filtered = db.Lorems.Where(x=>x.DoMath(someValue) > 0);
I am getting this error:
LINQ to Entities does not recognize the method -----, and this method cannot be translated into a store expression.
Is it possible to make it work like this?
My skills in customizing lambda expressions and working with delegates is quite poor, so I would like to get some help.
Edit: Here is that 'DoMath' function real code (without comments, because they aren't in english):
public double VypocitejPrislusnost(int value)
{
if (A == B)
{
if (value <= C)
{
return 1;
}
if (value > D)
{
return 0;
}
else
{
double b = D - C;
double tangens = Math.Tan(1 / b);
double a = tangens * (D - value);
return Math.Round(a, 2);
}
}
if (C == D)
{
if (value >= B)
{
return 1;
}
if (value <= A)
{
return 0;
}
else
{
double b = B - A;
double tangens = Math.Tan(1 / b);
double a = tangens * (value - A);
return Math.Round(a, 2);
}
}
else
{
if (value >= B && value <= C)
{
return 1;
}
if (value <= A || value >= D)
{
return 0;
}
if (value > A && value < B)
{
double b = B - A;
double tangens = Math.Tan(1 / b);
double a = tangens * (value - A);
return Math.Round(a, 2);
}
if (value > C && value < D)
{
double b = D - C;
double tangens = Math.Tan(1 / b);
double a = tangens * (D - value);
return Math.Round(a, 2);
}
else
{
return 0;
}
}
}
It basically calculates y coordinate in triangle in different scenarios. I am using this to count how much the given value fits into a fuzzy set.

Entity Framework's Where is expecting an Expression Tree, which will get converted to a t-sql statement. Unfortunately there is no translation from your DoMath method to t-sql, so you'd have to pull the results down into memory, and then call Where as you have it. The reason is that once your results are in memory, LINQ methods work on standard delegates, not expression trees, so there are no restrictions on what can be put in there
To do this, just tack an AsEnumerable() in front of the Where—of course that will pull down your whole table into memory, so do this only if it's reasonably small
var filtered = db.Lorems.AsEnumerable().Where(x => x.DoMath(someValue) > 0);
Of course if you can identify some basic circumstances under which your DoMath will not be greater than 0, you can filter those results out up front, using an expression tree. This will trim down the results coming from the database. You can then filter the rest in memory. I have no idea what your DoMath method does (you imply it's more complicated than what's listed in your question), but hypothetically something like this should work:
var filtered = db.Lorems
.Where(x => x.A < 10 && x.B != x.C)
.AsEnumerable()
.Where(x => x.DoMath(someValue) > 0);

How do I calculate a trendline for a graph?

Google is not being my friend - it's been a long time since my stats class in college...I need to calculate the start and end points for a trendline on a graph - is there an easy way to do this? (working in C# but whatever language works for you)

Thanks to all for your help - I was off this issue for a couple of days and just came back to it - was able to cobble this together - not the most elegant code, but it works for my purposes - thought I'd share if anyone else encounters this issue:
public class Statistics
{
public Trendline CalculateLinearRegression(int[] values)
{
var yAxisValues = new List<int>();
var xAxisValues = new List<int>();
for (int i = 0; i < values.Length; i++)
{
yAxisValues.Add(values[i]);
xAxisValues.Add(i + 1);
}
return new Trendline(yAxisValues, xAxisValues);
}
}
public class Trendline
{
private readonly IList<int> xAxisValues;
private readonly IList<int> yAxisValues;
private int count;
private int xAxisValuesSum;
private int xxSum;
private int xySum;
private int yAxisValuesSum;
public Trendline(IList<int> yAxisValues, IList<int> xAxisValues)
{
this.yAxisValues = yAxisValues;
this.xAxisValues = xAxisValues;
this.Initialize();
}
public int Slope { get; private set; }
public int Intercept { get; private set; }
public int Start { get; private set; }
public int End { get; private set; }
private void Initialize()
{
this.count = this.yAxisValues.Count;
this.yAxisValuesSum = this.yAxisValues.Sum();
this.xAxisValuesSum = this.xAxisValues.Sum();
this.xxSum = 0;
this.xySum = 0;
for (int i = 0; i < this.count; i++)
{
this.xySum += (this.xAxisValues[i]*this.yAxisValues[i]);
this.xxSum += (this.xAxisValues[i]*this.xAxisValues[i]);
}
this.Slope = this.CalculateSlope();
this.Intercept = this.CalculateIntercept();
this.Start = this.CalculateStart();
this.End = this.CalculateEnd();
}
private int CalculateSlope()
{
try
{
return ((this.count*this.xySum) - (this.xAxisValuesSum*this.yAxisValuesSum))/((this.count*this.xxSum) - (this.xAxisValuesSum*this.xAxisValuesSum));
}
catch (DivideByZeroException)
{
return 0;
}
}
private int CalculateIntercept()
{
return (this.yAxisValuesSum - (this.Slope*this.xAxisValuesSum))/this.count;
}
private int CalculateStart()
{
return (this.Slope*this.xAxisValues.First()) + this.Intercept;
}
private int CalculateEnd()
{
return (this.Slope*this.xAxisValues.Last()) + this.Intercept;
}
}

OK, here's my best pseudo math:
The equation for your line is:
Y = a + bX
Where:
b = (sum(x*y) - sum(x)sum(y)/n) / (sum(x^2) - sum(x)^2/n)
a = sum(y)/n - b(sum(x)/n)
Where sum(xy) is the sum of all x*y etc. Not particularly clear I concede, but it's the best I can do without a sigma symbol :)
... and now with added Sigma
b = (Σ(xy) - (ΣxΣy)/n) / (Σ(x^2) - (Σx)^2/n)
a = (Σy)/n - b((Σx)/n)
Where Σ(xy) is the sum of all x*y etc. and n is the number of points

Given that the trendline is straight, find the slope by choosing any two points and calculating:
(A) slope = (y1-y2)/(x1-x2)
Then you need to find the offset for the line. The line is specified by the equation:
(B) y = offset + slope*x
So you need to solve for offset. Pick any point on the line, and solve for offset:
(C) offset = y - (slope*x)
Now you can plug slope and offset into the line equation (B) and have the equation that defines your line. If your line has noise you'll have to decide on an averaging algorithm, or use curve fitting of some sort.
If your line isn't straight then you'll need to look into Curve fitting, or Least Squares Fitting - non trivial, but do-able. You'll see the various types of curve fitting at the bottom of the least squares fitting webpage (exponential, polynomial, etc) if you know what kind of fit you'd like.
Also, if this is a one-off, use Excel.

Here is a very quick (and semi-dirty) implementation of Bedwyr Humphreys's answer. The interface should be compatible with #matt's answer as well, but uses decimal instead of int and uses more IEnumerable concepts to hopefully make it easier to use and read.
Slope is b, Intercept is a
public class Trendline
{
public Trendline(IList<decimal> yAxisValues, IList<decimal> xAxisValues)
: this(yAxisValues.Select((t, i) => new Tuple<decimal, decimal>(xAxisValues[i], t)))
{ }
public Trendline(IEnumerable<Tuple<Decimal, Decimal>> data)
{
var cachedData = data.ToList();
var n = cachedData.Count;
var sumX = cachedData.Sum(x => x.Item1);
var sumX2 = cachedData.Sum(x => x.Item1 * x.Item1);
var sumY = cachedData.Sum(x => x.Item2);
var sumXY = cachedData.Sum(x => x.Item1 * x.Item2);
//b = (sum(x*y) - sum(x)sum(y)/n)
// / (sum(x^2) - sum(x)^2/n)
Slope = (sumXY - ((sumX * sumY) / n))
/ (sumX2 - (sumX * sumX / n));
//a = sum(y)/n - b(sum(x)/n)
Intercept = (sumY / n) - (Slope * (sumX / n));
Start = GetYValue(cachedData.Min(a => a.Item1));
End = GetYValue(cachedData.Max(a => a.Item1));
}
public decimal Slope { get; private set; }
public decimal Intercept { get; private set; }
public decimal Start { get; private set; }
public decimal End { get; private set; }
public decimal GetYValue(decimal xValue)
{
return Intercept + Slope * xValue;
}
}

Regarding a previous answer
if (B) y = offset + slope*x
then (C) offset = y/(slope*x) is wrong
(C) should be:
offset = y-(slope*x)
See:
http://zedgraph.org/wiki/index.php?title=Trend

If you have access to Excel, look in the "Statistical Functions" section of the Function Reference within Help. For straight-line best-fit, you need SLOPE and INTERCEPT and the equations are right there.
Oh, hang on, they're also defined online here: http://office.microsoft.com/en-us/excel/HP052092641033.aspx for SLOPE, and there's a link to INTERCEPT. OF course, that assumes MS don't move the page, in which case try Googling for something like "SLOPE INTERCEPT EQUATION Excel site:microsoft.com" - the link given turned out third just now.

I converted Matt's code to Java so I could use it in Android with the MPAndroidChart library. Also used double values instead of integer values:
ArrayList<Entry> yValues2 = new ArrayList<>();
ArrayList<Double > xAxisValues = new ArrayList<Double>();
ArrayList<Double> yAxisValues = new ArrayList<Double>();
for (int i = 0; i < readings.size(); i++)
{
r = readings.get(i);
yAxisValues.add(r.value);
xAxisValues.add((double)i + 1);
}
TrendLine tl = new TrendLine(yAxisValues, xAxisValues);
//Create the y values for the trend line
double currY = tl.Start;
for (int i = 0; i < readings.size(); ++ i) {
yValues2.add(new Entry(i, (float) currY));
currY = currY + tl.Slope;
}
...
public class TrendLine
{
private ArrayList<Double> xAxisValues = new ArrayList<Double>();
private ArrayList<Double> yAxisValues = new ArrayList<Double>();
private int count;
private double xAxisValuesSum;
private double xxSum;
private double xySum;
private double yAxisValuesSum;
public TrendLine(ArrayList<Double> yAxisValues, ArrayList<Double> xAxisValues)
{
this.yAxisValues = yAxisValues;
this.xAxisValues = xAxisValues;
this.Initialize();
}
public double Slope;
public double Intercept;
public double Start;
public double End;
private double getArraySum(ArrayList<Double> arr) {
double sum = 0;
for (int i = 0; i < arr.size(); ++i) {
sum = sum + arr.get(i);
}
return sum;
}
private void Initialize()
{
this.count = this.yAxisValues.size();
this.yAxisValuesSum = getArraySum(this.yAxisValues);
this.xAxisValuesSum = getArraySum(this.xAxisValues);
this.xxSum = 0;
this.xySum = 0;
for (int i = 0; i < this.count; i++)
{
this.xySum += (this.xAxisValues.get(i)*this.yAxisValues.get(i));
this.xxSum += (this.xAxisValues.get(i)*this.xAxisValues.get(i));
}
this.Slope = this.CalculateSlope();
this.Intercept = this.CalculateIntercept();
this.Start = this.CalculateStart();
this.End = this.CalculateEnd();
}
private double CalculateSlope()
{
try
{
return ((this.count*this.xySum) - (this.xAxisValuesSum*this.yAxisValuesSum))/((this.count*this.xxSum) - (this.xAxisValuesSum*this.xAxisValuesSum));
}
catch (Exception e)
{
return 0;
}
}
private double CalculateIntercept()
{
return (this.yAxisValuesSum - (this.Slope*this.xAxisValuesSum))/this.count;
}
private double CalculateStart()
{
return (this.Slope*this.xAxisValues.get(0)) + this.Intercept;
}
private double CalculateEnd()
{
return (this.Slope*this.xAxisValues.get(this.xAxisValues.size()-1)) + this.Intercept;
}
}

This is the way i calculated the slope:
Source: http://classroom.synonym.com/calculate-trendline-2709.html
class Program
{
public double CalculateTrendlineSlope(List<Point> graph)
{
int n = graph.Count;
double a = 0;
double b = 0;
double bx = 0;
double by = 0;
double c = 0;
double d = 0;
double slope = 0;
foreach (Point point in graph)
{
a += point.x * point.y;
bx = point.x;
by = point.y;
c += Math.Pow(point.x, 2);
d += point.x;
}
a *= n;
b = bx * by;
c *= n;
d = Math.Pow(d, 2);
slope = (a - b) / (c - d);
return slope;
}
}
class Point
{
public double x;
public double y;
}

Here's what I ended up using.
public class DataPoint<T1,T2>
{
public DataPoint(T1 x, T2 y)
{
X = x;
Y = y;
}
[JsonProperty("x")]
public T1 X { get; }
[JsonProperty("y")]
public T2 Y { get; }
}
public class Trendline
{
public Trendline(IEnumerable<DataPoint<long, decimal>> dataPoints)
{
int count = 0;
long sumX = 0;
long sumX2 = 0;
decimal sumY = 0;
decimal sumXY = 0;
foreach (var dataPoint in dataPoints)
{
count++;
sumX += dataPoint.X;
sumX2 += dataPoint.X * dataPoint.X;
sumY += dataPoint.Y;
sumXY += dataPoint.X * dataPoint.Y;
}
Slope = (sumXY - ((sumX * sumY) / count)) / (sumX2 - ((sumX * sumX) / count));
Intercept = (sumY / count) - (Slope * (sumX / count));
}
public decimal Slope { get; private set; }
public decimal Intercept { get; private set; }
public decimal Start { get; private set; }
public decimal End { get; private set; }
public decimal GetYValue(decimal xValue)
{
return Slope * xValue + Intercept;
}
}
My data set is using a Unix timestamp for the x-axis and a decimal for the y. Change those datatypes to fit your need. I do all the sum calculations in one iteration for the best possible performance.

Thank You so much for the solution, I was scratching my head.
Here's how I applied the solution in Excel.
I successfully used the two functions given by MUHD in Excel:
a = (sum(x*y) - sum(x)sum(y)/n) / (sum(x^2) - sum(x)^2/n)
b = sum(y)/n - b(sum(x)/n)
(careful my a and b are the b and a in MUHD's solution).
- Made 4 columns, for example:
NB: my values y values are in B3:B17, so I have n=15;
my x values are 1,2,3,4...15.
1. Column B: Known x's
2. Column C: Known y's
3. Column D: The computed trend line
4. Column E: B values * C values (E3=B3*C3, E4=B4*C4, ..., E17=B17*C17)
5. Column F: x squared values
I then sum the columns B,C and E, the sums go in line 18 for me, so I have B18 as sum of Xs, C18 as sum of Ys, E18 as sum of X*Y, and F18 as sum of squares.
To compute a, enter the followin formula in any cell (F35 for me):
F35=(E18-(B18*C18)/15)/(F18-(B18*B18)/15)
To compute b (in F36 for me):
F36=C18/15-F35*(B18/15)
Column D values, computing the trend line according to the y = ax + b:
D3=$F$35*B3+$F$36, D4=$F$35*B4+$F$36 and so on (until D17 for me).
Select the column datas (C2:D17) to make the graph.
HTH.

If anyone needs the JS code for calculating the trendline of many points on a graph, here's what worked for us in the end:
/**#typedef {{
* x: Number;
* y:Number;
* }} Point
* #param {Point[]} data
* #returns {Function} */
function _getTrendlineEq(data) {
const xySum = data.reduce((acc, item) => {
const xy = item.x * item.y
acc += xy
return acc
}, 0)
const xSum = data.reduce((acc, item) => {
acc += item.x
return acc
}, 0)
const ySum = data.reduce((acc, item) => {
acc += item.y
return acc
}, 0)
const aTop = (data.length * xySum) - (xSum * ySum)
const xSquaredSum = data.reduce((acc, item) => {
const xSquared = item.x * item.x
acc += xSquared
return acc
}, 0)
const aBottom = (data.length * xSquaredSum) - (xSum * xSum)
const a = aTop / aBottom
const bTop = ySum - (a * xSum)
const b = bTop / data.length
return function trendline(x) {
return a * x + b
}
}
It takes an array of (x,y) points and returns the function of a y given a certain x
Have fun :)

Here's a working example in golang. I searched around and found this page and converted this over to what I needed. Hope someone else can find it useful.
// https://classroom.synonym.com/calculate-trendline-2709.html
package main
import (
"fmt"
"math"
)
func main() {
graph := [][]float64{
{1, 3},
{2, 5},
{3, 6.5},
}
n := len(graph)
// get the slope
var a float64
var b float64
var bx float64
var by float64
var c float64
var d float64
var slope float64
for _, point := range graph {
a += point[0] * point[1]
bx += point[0]
by += point[1]
c += math.Pow(point[0], 2)
d += point[0]
}
a *= float64(n) // 97.5
b = bx * by // 87
c *= float64(n) // 42
d = math.Pow(d, 2) // 36
slope = (a - b) / (c - d) // 1.75
// calculating the y-intercept (b) of the Trendline
var e float64
var f float64
e = by // 14.5
f = slope * bx // 10.5
intercept := (e - f) / float64(n) // (14.5 - 10.5) / 3 = 1.3
// output
fmt.Println(slope)
fmt.Println(intercept)
}