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The real identity of the user is exchanged during the protected exchanges of phase 2 Cisco PEAP uses one-time password (OTP) or logon passwords for client authentication OTP databases from RSA security or other vendors can be used Microsoft PEAP uses MS-CHAP 2 for client authentication and hence is limited to Windows NT domains..

and m = (Cpt 1 1 l)(2) (Cpt 1 1 l)T + I (2) (Cpt 1 1 l)(2) (It Bpt 1 1 l)T , (2)

file:///F|/WinDDK/resources/CPPPrimer/c++primerhtm (163 / 1065) [2001-3-29 11:32:04]

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In general, we recommend the use of the standard library class abstractions in this case, the use of the bitset class over that of direct bitwise manipulation of integral data types In our opinion, however, knowledge of both representations is still necessary because of the likely need to read and perhaps modify existing code For completeness, we illustrate both approaches In the rest of this section, we look at the use of the built-in integral types as bitvectors and the use of the bitwise operators In the following section, we look at the bitset class When we are using an integral type as a bitvector, the type can be either signed or unsigned; an unsigned type is strongly recommended How the "sign bit" is handled in a number of the bitwise operations is undefined and is therefore likely to differ across implementations; programs that work under one implementation may fail under another The bitwise NOT operator (~) flips the bits of its operand Each 1 bit is set to 0; each 0 bit is set to 1 The bitwise shift operators ( , >>) shift the bits of the left operand some number of positions either left or right The operand's excess bits are discarded The left shift operator ( ) inserts 0-valued bits in from the right The right shift operator (>>) inserts 0-valued bits in from the left if the operand is unsigned If the operand is signed, it can either insert copies of the sign bit or insert 0-valued bits; again, this behavior is implementation defined The bitwise AND operator (&) takes two integral operands For each bit position, the result is 1 if both operands contain 1; otherwise, the result is 0 (This operator should not be confused with the logical AND operator (&&); unfortunately, it seems that everyone has at one time or another done just that) The bitwise XOR (exclusive or) operator (^) takes two integral operands For each bit position, the result is 1 if either but not both operands contain 1; otherwise, the result is 0 The bitwise OR (inclusive or) operator (|) takes two integral operands For each bit position, the result is 1 if either or both operands contain 1; otherwise, the result is 0 (This operator should not be confused with the logical OR operator ( )) Let's look at a simple example A teacher has 30 students in a class Each week the class is given a pass/fail quiz A bitvector will be used to track the results of each quiz (Note that bits are numbered beginning with 0 Thus, bit 1 actually represents the second bit In our example, we are wasting the first bit in order to speak of bit 1 as the first bit and so on After all, our teacher is not a student of computer science).

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(7.12)

unsigned int quiz1 = 0;

The teacher must be able to turn on and off and to test individual bits For example, student 27 has taken a makeup quiz and passed The teacher must turn bit 27 on The first step is to set the 27th bit of an integer to 1, while all the other bits remain 0 This can be done with the left shift operator and the integer constant 1:

where I is an identity matrix of dimension 6 6. This is essentially a model-based estimation approach that requires a texturemapped 3D model of the object to be tracked. This is expected because the method works only with the analytically derived model, which cannot represent variations of identity within a single class of objects. By using our GAMs, this restriction can be overcome. Moreover, we can achieve this in a computationally ef cient manner by using the inverse compositional algorithm. As mentioned earlier, our tracking method is faster than 3DMM-based approaches [18] while sacri cing little in accuracy.

1 27;

If this value is bitwise ORed with quiz1, all except the 27th bit will remain unchanged The 27th bit is turned on:

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