applied cryptography - protocols, algorithms, and source code in c

# To set up a secure call from person to person the

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Unformatted text preview: rn which way Alice polarized the photon. If he sets his detector to measure diagonal polarization and the pulse is polarized rectilinearly, he will get a random measurement. He won’t know the difference. In this example, he might get the result: /|—\/\—/—| (3) Bob tells Alice, over an insecure channel, what settings he used. (4) Alice tells Bob which settings were correct. In our example, the detector was correctly set for pulses 2, 6, 7, and 9. (5) Alice and Bob keep only those polarizations that were correctly measured. In our example, they keep: *|***\—*—* Using a prearranged code, Alice and Bob each translate those polarization measurements into bits. For example, horizontal and left-diagonal might equal one, and vertical and right-diagonal might equal zero. In our example, they both have: 0011 So, Alice and Bob have generated four bits. They can generate as many as they like using this system. On the average, Bob will guess the correct setting 50 percent of the time, so Alice has to send 2n photon pulses to generate n bits. They can use these bits as a secret key for a symmetric algorithm or they can guarantee perfect secrecy and generate enough bits for a one-time pad. The really cool thing is that Eve cannot eavesdrop. Just like Bob, she has to guess which type of polarization to measure; and like Bob, half of her guesses will be wrong. Since wrong guesses change the polarization of the photons, she can’t help introducing errors in the pulses as she eavesdrops. If she does, Alice and Bob will end up with different bit strings. So,Alice and Bob finish the protocol like this: (6) Alice and Bob compare a few bits in their strings. If there are discrepancies, they know they are being bugged. If there are none, they discard the bits they used for comparison and use the rest. Enhancements to this protocol allow Alice and Bob to use their bits even in the presence of Eve [133,134,192]. They could compare only the parity of subsets of the bits. Then, if no discrepancies are fou...
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