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To solve this problem, we need a way for the robot to remember which way it was turning the last time it saw the line This will normally be the direction the robot should turn if it has lost the line Each time the robot makes a normal turn (the if-decisions in Fig 77) the subroutine stores the turn direction in the variable LastTurn to remember which direction the robot was turning Later in the routine, if none of the sensors are on (indicating we probably have lost the line), the robot will be able to head back toward the line Extremely sharp turns are still a problem even for this algorithm In this algorithm, since it is acceptable to lose the line, we can speed up the robot s progress by moving it forward 2 pixels at a time.

players will produce different final answers. The private key and the public key are also related; they are partners related mathematically. If that relationship is not there if the public key used to verify is not the partner to the private key used to sign the two agents will produce different final answers. The security of DSA lies in the discrete log problem, the same problem that gives DH its security. So the size of DSA keys will be the same as that of DH keys. As always, you can find more detailed information in the RSA Labs FAQ on the accompanying CD.

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A simple sensor setup is shown in Fig. 5.46. The resistive element must be heated to become sensitive. The sensor incorporates its own heating unit, which is separately powered. The heater requires a regulated 5 V for proper operation and draws about 130 mA. The resistive element can be read like any other resistive sensor used thus far. The potential for these gas sensors is greater than what is implied in the simple schematic. The gas sensors are not precise instruments. In other words, their response varies slightly from device to device. This analog property can be used to create a more sensitive smell detector. Let s arrange eight sensors. The resistive element from each sensor is connected to an A/D convertor. A comparator circuit wouldn t do in this situation because precise and subtle variations in response are what we are looking for. To calibrate the device, a small amount of a known gas (smell) is released by the eight sensors. The response of each detector is measured by the A/D convertor and recorded by the main computer. Since the responses of the detectors will vary, an eight-number pattern is created for each smell. Pattern matching is well established in neural networks. A neural network can be built using the information gathered that can not only measure but recognize different smells.

While the robot is still over the line it will turn only 1 at a time to stay on it, but if the line is lost, the robot will take a 3 turns to help it get back on course Notice that these choices for how much to move or turn are somewhat arbitrary With a little experimentation, you can determine the optimum values for your situation This reminds us again of the advantage of using a simulation With RobotBASIC you can change the values and see how the robot responds to your changes very quickly 734 VERY SHARP TURNS The routine in Fig 76 stays on the line nicely and even handles 90 turns However, if the line turns much more than 90 , the robot can still lose the line There are many ways to solve this problem Figure 78 shows a potential solution..

This algorithm looks a lot like DSA. The signer has three inputs: the digest, k, and the private key. The output is r and s. The verifier has the digest, s, and the public key. The output is v. If v and r are the same, the signature is verified; if they re not the same, something went wrong. What went wrong Was it the wrong digest The wrong public key Was the signature mangled in transmission You probably can t know exactly what happened, but you do know that something went wrong. The math underlying ECDSA is elliptic curves, so key size is the same as with ECDH.

FollowLine: while true rForward 1 if rSense() = 3 rForward 20 //move the centre while rSense() = 0 rTurn 1 //turn back to the wend endif if rSense() = 6 rForward 20 //move the centre while rSense() = 0 rTurn -1 //turn back to the wend endif //-- reposition over the line while rSense() & 1 rTurn 1 wend while rSense() & 4 rturn -1 wend wend Return

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