(New page: ==Part 1== Bob can decrypt the message by using the inverse of the secret matrix. Multiplying the encrypted vector three at a time by the inverse matrix will yield the original vector. ==...) |
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Eve can decrypt the message without finding the secret matrix because the system in linear. The three vectors of the encrypted vector form a basis for all vectors because they are linearly independent. | Eve can decrypt the message without finding the secret matrix because the system in linear. The three vectors of the encrypted vector form a basis for all vectors because they are linearly independent. | ||
In other words, | In other words, | ||
| − | [encrypted vector]<math>=a[2 0 0] + b[0 1 0] + c[0 0 3]</math> where a,b,c are real numbers. | + | [encrypted vector]<math>=a[2,0,0] + b[0,1,0] + c[0,0,3]</math> where a,b,c are real numbers. |
==Part 3== | ==Part 3== | ||
Applying the method from part 2, we know that: | Applying the method from part 2, we know that: | ||
| − | <math>[2 23 3]=a[2 0 0] + b[0 1 0] + c[0 0 3]</math> | + | <math>[2,23,3]=a[2,0,0] + b[0,1,0] + c[0,0,3]</math> |
| − | so | + | so |
| + | |||
<math>2=2a</math>, <math>23=1b</math>, and <math>3=3c</math>. | <math>2=2a</math>, <math>23=1b</math>, and <math>3=3c</math>. | ||
therefore, a=1, b=23, and c=1. | therefore, a=1, b=23, and c=1. | ||
| − | Now apply the same coefficients to the known message that yielded the encrypted basis, [1 0 4 0 1 0 1 0 1]. | + | Now apply the same coefficients to the known message that yielded the encrypted basis, [1,0,4,0,1,0,1,0,1]. |
| + | |||
| + | <math>1[1,0,4] + 23[0,1,0] + 1[1,0,1]=[1+1,23,4+1]=[2,23,5]</math> | ||
| − | + | [2,23,5] corresponds to BWE. | |
| − | + | The other way of doing this is to set up equations to find the 3 x 3 matrix. however, There is a greater chance of making a mistake. | |
Latest revision as of 17:52, 18 September 2008
Part 1
Bob can decrypt the message by using the inverse of the secret matrix. Multiplying the encrypted vector three at a time by the inverse matrix will yield the original vector.
Part 2
Eve can decrypt the message without finding the secret matrix because the system in linear. The three vectors of the encrypted vector form a basis for all vectors because they are linearly independent. In other words, [encrypted vector]$ =a[2,0,0] + b[0,1,0] + c[0,0,3] $ where a,b,c are real numbers.
Part 3
Applying the method from part 2, we know that: $ [2,23,3]=a[2,0,0] + b[0,1,0] + c[0,0,3] $ so
$ 2=2a $, $ 23=1b $, and $ 3=3c $. therefore, a=1, b=23, and c=1.
Now apply the same coefficients to the known message that yielded the encrypted basis, [1,0,4,0,1,0,1,0,1].
$ 1[1,0,4] + 23[0,1,0] + 1[1,0,1]=[1+1,23,4+1]=[2,23,5] $
[2,23,5] corresponds to BWE.
The other way of doing this is to set up equations to find the 3 x 3 matrix. however, There is a greater chance of making a mistake.
