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- Combining Higher Order Functions in C# – Mike Hadlow performs some functional programming gymnastics with C# functions which gets him under way with his Windsor CurryFacility implementation
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- Creating a Custom Silverlight Pre-Loader

Project Euler problem 9~

`1: class Program`

2: {3: static void Main(string[] args)4: {`5: var sw = new Stopwatch();`

6:`7: Console.WriteLine("There exists exactly one Pythagorean triplet for which a + b + c = 1000, Find the product abc");`

8: sw.Start();9: var product1 = BruteForce();10: sw.Stop();11: Console.WriteLine(string.Format("plain loop brute force way(tick:{0}): {1}", sw.ElapsedTicks, product1));12:13: sw.Reset();14:`15: Console.WriteLine("There exists exactly one Pythagorean triplet for which a + b + c = 1000, Find the product abc");`

16: sw.Start();17: var product2 = ArithmeticWay();18: sw.Stop();19: Console.WriteLine(string.Format("Euclid's formula way, with k(tick:{0}): {1}", sw.ElapsedTicks, product2));20:21:`22: Console.WriteLine("Press any key to exit");`

23: Console.ReadKey();24: }25:`26: /// <summary>`

`27: /// plain loop all possibilities way`

`28: /// a or b cannot exceed 500, so set that as bound`

`29: /// and since a+b+c has to be 1000, which makes c = (1000-a-b)`

`30: /// and a^2 + b^2 = c^2, so two nested loops are enough`

`31: /// </summary>`

`32: /// <returns></returns>`

33: static int BruteForce()34: {35: for (int a = 1; a < 500; a++)36: {37: for (int b = a + 1; b < 500; b++)38: {`39: if (a * a + b * b == (1000 - a - b) * (1000 - a - b))`

40: {`41: return a * b * (1000 - a - b);`

42: }43: }44: }45:`46: return 0;`

47: }48:`49: /// <summary>`

`50: /// using the Euclid's formula with k, see http://en.wikipedia.org/wiki/Pythagorean_triple`

`51: /// a = k * (m^2 - n^2)`

`52: /// b = k * (2mn)`

`53: /// c = k * (m^2 + n^2)`

`54: /// starting with k=1`

`55: /// (m^2 - n^2) + (2mn) + (m^2 + n^2) = 1000`

`56: /// (2m^2 + 2mn) = 1000`

`57: /// m^2 + mn = 500`

`58: /// since m and n has to be integer and non-zero`

`59: /// so 500 has to be divisible by k and`

`60: /// we can set the m_bound to be less or equal to Math.Sqrt(k_bound), since n cannot be zero`

`61: /// </summary>`

`62: /// <returns></returns>`

63: static int ArithmeticWay()64: {`65: int k_bound = 0;`

`66: int m_bound = 0;`

67: for (int k = 1; k <= 500; k++)68: {`69: if (500 % k == 0)`

70: {71: k_bound = 500 / k;`72: m_bound = (int)Math.Sqrt(k_bound);`

73:74: for (int m = 1; m <= m_bound; m++)75: {`76: if (k_bound % m == 0)`

77: {`78: int n = k_bound / m - m;`

`79: if (m > n)`

`80: return k * (m * m - n * n) * k * (2 * m * n) * k * (m * m + n * n);`

81: }82: }83: }84: }`85: return 0;`

86: }87: }