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class Encode { static void Main() { char ch1 = 'H'; char ch2 = 'i'; char ch3 = '!'; int key = 88; ConsoleWriteLine("Original message: " + ch1 + ch2 + ch3); // Encode the message ch1 = (char) (ch1 ^ key); ch2 = (char) (ch2 ^ key); ch3 = (char) (ch3 ^ key); ConsoleWriteLine("Encoded message: " + ch1 + ch2 + ch3); // Decode the message ch1 = (char) (ch1 ^ key); ch2 = (char) (ch2 ^ key); ch3 = (char) (ch3 ^ key); ConsoleWriteLine("Decoded message: " + ch1 + ch2 + ch3); } }

Here is the output:

Original message: Hi! Encoded message: 1y Decoded message: Hi!

4x 2

As you can see, the result of two XORs using the same key produces the decoded message (Remember, this simple XOR cipher is not suitable for any real-world, practical use because it is inherently insecure) The unary one s complement (NOT) operator reverses the state of all the bits of the operand For example, if some integer called A has the bit pattern 1001 0110, then ~A produces a result with the bit pattern 0110 1001 The following program demonstrates the NOT operator by displaying a number and its complement in binary:

// Demonstrate the bitwise NOT using System; class NotDemo { static void Main() { sbyte b = -34; for(int t=128; t > 0; t = t/2) { if((b & t) != 0) ConsoleWrite("1 "); if((b & t) == 0) ConsoleWrite("0 "); } ConsoleWriteLine();

4:

// reverse all bits b = (sbyte) ~b;

The first integral equals ( 1/25) log |x + 2|, the second integral equals ( 1/25) log |x 3|, and the third integral equals ( 1/5) /( x 3) In summary, we have found that log |x 3| 1 log |x + 2| 1 + C dx = 25 25 5( x 3) x 3 4x 2 3x + 18 We see that our integral of the reciprocal of a cubic polynomial leads to the sum of three factors; two of these are logarithmic, but one is not

that makes the point Note that the cost of correcting an error increases geometrically, rising sharply the longer it takes to identify the problem An error caught on the computer monitor, for example, before anything s been printed, costs your company about $5 If the error is caught once the document has been printed, xing it costs your company about $50 If the document has been sent to a commercial printer and lm has been created before the error is caught, correcting it costs the company close to $500 And if the item has been printed, the error costs a minimum of $5,000 to x Proo ng as you write, and at every step of the revision process, helps you catch errors earlier rather than later

for(int t=128; t > 0; t = t/2) { if((b & t) != 0) ConsoleWrite("1 "); if((b & t) == 0) ConsoleWrite("0 "); } } }

Here is the output:

1 1 0 1 1 1 1 0 0 0 1 0 0 0 0 1

Since the denominator is a cubic polynomial, it must factor The factors of the constant term are 1 and 2 After some experimentation, we find that x = 2 is a root and in fact the polynomial factors as x 3 + 2x 2 + x + 2 = ( x + 2) ( x 2 + 1) Thus we wish to write the integrand as the sum of a factor with denominator ( x + 2) and another factor with denominator ( x 2 + 1) The correct way to do this is Bx + C x x A + 2 = = x+2 x 3 + 2x 2 + x + 2 ( x + 2) ( x 2 + 1) x +1 We put the right-hand side over a common denominator to obtain x x 3 + 2x 2 + x + 2 A( x 2 + 1) + ( B x + C ) ( x + 2) x 3 + 2x 2 + x + 2

In C# it is possible to shift the bits that comprise an integer value to the left or to the right by a specified amount C# defines the two bit-shift operators shown here: