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Homework 3 Ben Horst:  [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]]  ::  [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]]  ::  [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]]
 
Homework 3 Ben Horst:  [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]]  ::  [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]]  ::  [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]]
 
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==1==
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Bob can find the inverse matrix, and multiply the encrypted message by it to find the intended message.
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==2==
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Since Eve has intercepted several pairs on in/out, and conveniently enough, the output forms a basis, she can decode the message easily.
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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).
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==3==
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Using this method, <2,23,3> must be a combination of <2,0,0>, <0,1,0>, and <0,0,3>.
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It is simple to find that combination: 1*<2,0,0> + 23*<0,1,0> + 1*<0,0,3> = <2,23,3>.
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We use those coefficients on the input (linearity) to find the output:
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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 11:46, 19 September 2008

<< Back to Homework 3

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] .

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