c# qr code generator Inside Microsoft SQL Server 2008: T-SQL Querying in Visual C#

Generation QR Code 2d barcode in Visual C# Inside Microsoft SQL Server 2008: T-SQL Querying

Inside Microsoft SQL Server 2008: T-SQL Querying
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Of course, this is just an observation of a pattern based on the cases that were tested. To ensure that the pattern holds for all cases, you need a mathematical proof. You can nd one at http://en.wikipedia.org/wiki/Josephus_problem. The following T-SQL statement calculates and returns p for a given @n:
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DECLARE @n AS INT = 41; SELECT 2 * (@n - POWER(2, CAST(LOG(@n)/LOG(2) AS INT))) + 1 AS p;
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Puzzle 23: Shipping Algebra
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Here s the algebra I used in my solution to the problem: Let s = current age of ship, b = current age of boiler, and y = years passed since the age of the ship was equal to the current age of the boiler. You can translate the statements in the puzzle to the following three equations: 1. s + b = 42 2. s = 2 (b y) 3. s y = b From equations 2 and 3 you get the following equation: s = 2 (b s + b) This gives us equation 4: 4. 3 s = 4 b From equations 1 and 4 you get the following equation: 3 s = 4 (42 s) When you solve the equation for s, you get 24. And now that the age of the ship is known, you can solve equation 1 for b: b = 42 24 = 18 The solution is that the ship s current age is 24 and the boiler s current age is 18.
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Puzzle 24: Equilateral Triangles Puzzle
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You can solve this puzzle in many ways. I provided this puzzle not because it is tough but rather the contrary it is pretty simple. However, some of the solutions are simply beautiful. I ll rst provide an ordinary solution and then a more creative one. To explain the rst solution, examine the drawing in Figure A-4.
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Appendix A E
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Logic Puzzles
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h2 a
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S A F C G H
FIGURE A-4 Solution 1 to the equilateral triangles puzzle
The segment h1 has the same length as the altitude of the triangle ABC, and the segment h2 has the same length as the altitude of the triangle CEF. G is the point where h1 intersects CA, and H is the point where h2 intersects the same line. It is fairly easy to prove that H is the same point as A but not really necessary for our purposes. The triangles GBC and HEC are similar because they have two corresponding angles that are equal (both have a right angle and share another angle). |CE| is twice |CB|; therefore |HE| (which is |h2|) is twice |GB| (which is |h1|). The area of a triangle is bh (half base times altitude). Because the bases FC and CH of the triangles CEF and ABC have equal lengths but |h2| is twice |h1|, the area of CEF is twice the area of ABC. In other words, the area of CEF (as well as DEB and DAF, which are congruent to CEF) is 2S. Therefore, the area of the triangle DEF is 3 2S + S = 7S. The second solution is more creative. Examine the drawing in Figure A-5. You draw the lines EG and GF parallel to CF and EC, respectively, to form the parallelogram CEGF. Next, draw the lines BG, BH, and CE. We know that |FC| = |CB| = |BE| = |EG| = |GH| = |HF| = |HB| = |a|. Triangles ABC, CHF, and BGH are congruent because corresponding sides and the angle between them are equal. This means that |HC| = |BG| = |a|. This means that the four triangles BEG, BGH, CBH, and CHF enclosed by the parallelogram and ABC are congruent; therefore, the area of the parallelogram is 4S. The triangle CEF has exactly half the area of the parallelogram; therefore, the triangle s area is 2S. Therefore, the area of the triangle DEF is 3 2S + S = 7S.
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