CRYPTOLOGY COPMUTER SCIENCE Suppose that a particular iris scan systems generate
ID: 3810423 • Letter: C
Question
CRYPTOLOGY COPMUTER SCIENCE
Suppose that a particular iris scan systems generates 64-bit iris co instead of the standard 2048-bit iris codes mentioned in this chap During the enrollment phase, the following iris codes (in hex) are termined. During the recognition phase, the following iris codes are obtained. User the iris codes above to answer the following questions. a. Use the formula in equation (7.1) to compute the following d(Alice, Bob), d(Alice, Charlie), d(Bob, Charlie). b. Assuming that the same statistics apply to these iris codes the iris codes discussed in Section 7.4.2.3, which of the U, V, W, X, Y, is most likely Alice? Bob? Charlie? None of above?Explanation / Answer
1.
Alice Iris Code in Binary - 1011 1110 0100 0011 1001 1010 1101 0101 1001 1000 1110 1111 0101 0001 0100 0111
Bob Iris Code in Binary - 1001 1100 1000 1011 0111 1010 0001 0100 0010 0101 0011 0110 1001 0101 1000 0100
Charlie Iris Code in Binary - 1000 1000 0101 0101 0010 0010 0011 0011 0110 0110 1001 1001 1100 1100 1011 1011
d(x, y) = number of non-match bits/number of bits compared
a)
d(Alice,Bob) =
To find number of non-match bits between Alice and Bob
1011 1110 0100 0011 1001 1010 1101 0101 1001 1000 1110 1111 0101 0001 0100 0111 -> Alice
1001 1100 1000 1011 0111 1010 0001 0100 0010 0101 0011 0110 1001 0101 1000 0100 -> Bob
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0010 0010 1100 1000 1110 0000 1100 0001 1011 1101 1101 1001 1100 0100 1100 0011
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Total unmatched bits = 29
Total bits compared = 64
d(Alice,Bob) = 29/64 = 0.453125
b)
d(Alice,Charlie) =
To find number of non-match bits between Alice and Charlie
1011 1110 0100 0011 1001 1010 1101 0101 1001 1000 1110 1111 0101 0001 0100 0111 -> Alice
1000 1000 0101 0101 0010 0010 0011 0011 0110 0110 1001 1001 1100 1100 1011 1011 -> Charlie
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0011 0110 0001 0110 1011 1000 1110 0110 1111 1110 0111 0110 1001 1101 1111 1100
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Total unmatched bits = 39
Total bits compared = 64
d(Alice,Charlie) = 39/64 = 0.609375
c)
d(Bob, Charlie) =
To find number of non-match bits between Bob and Charlie
1001 1100 1000 1011 0111 1010 0001 0100 0010 0101 0011 0110 1001 0101 1000 0100 -> Bob
1000 1000 0101 0101 0010 0010 0011 0011 0110 0110 1001 1001 1100 1100 1011 1011 -> Charlie
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0001 0100 1101 1110 0101 1000 0010 0111 0100 0011 1010 1111 0101 1001 0011 1111
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Total unmatched bits = 34
Total bits compared = 64
d(Bob,Charlie) = 34/64 = 0.53125
2.
Here for each of the Alice, Bob and Charlie, we have to find
For Alice -> find d(Alice, U), d(Alice,V), d(Alice,W), d(Alice,X), d(Alice,Y), d(Alice,Z) on similar lines as above.
if the value of d(x,y) is found to be less than 0.32, then the respective is a match.
For example, if d(Alice, U) is say less than 0.32 then U is same as Alice.
Repeat the same for Bob and Charlie