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==1== | ==1== | ||
| − | Bob | + | Bob can find the inverse matrix, and multiply the encrypted message by it to find the intended message. |
==2== | ==2== | ||
| + | Since Eve has intercepted several pairs on in/out, and conveniently enough, the output forms a basis, she can decode the message easily. | ||
| + | |||
| + | The intercepted vectors <2,0,0>, <0,1,0>, <0,0,3> form a basis for the decoding (as a linear combination, they can form any other vector, thus any other vector may be written as a sum of these three). | ||
| + | |||
| + | ==3== | ||
| + | Using this method, <2,23,3> must be a combination of <2,0,0>, <0,1,0>, and <0,0,3>. | ||
| + | |||
| + | It is simple to find that combination: 1*<2,0,0> + 23*<0,1,0> + 1*<0,0,3> = <2,23,3>. | ||
| + | |||
| + | We use those coefficients on the input (linearity) to find the output: | ||
| + | |||
| + | 1*<1,0,4> + 23*<0,1,0> + 1*<1,0,1> = <2,23,5> which corresponds to [b,w,e] . | ||
Latest revision as of 12:46, 19 September 2008
Homework 3 Ben Horst: A :: B :: C
1
Bob can find the inverse matrix, and multiply the encrypted message by it to find the intended message.
2
Since Eve has intercepted several pairs on in/out, and conveniently enough, the output forms a basis, she can decode the message easily.
The intercepted vectors <2,0,0>, <0,1,0>, <0,0,3> form a basis for the decoding (as a linear combination, they can form any other vector, thus any other vector may be written as a sum of these three).
3
Using this method, <2,23,3> must be a combination of <2,0,0>, <0,1,0>, and <0,0,3>.
It is simple to find that combination: 1*<2,0,0> + 23*<0,1,0> + 1*<0,0,3> = <2,23,3>.
We use those coefficients on the input (linearity) to find the output:
1*<1,0,4> + 23*<0,1,0> + 1*<1,0,1> = <2,23,5> which corresponds to [b,w,e] .
