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| − | [[Homework 3_ECE301Fall2008mboutin]] - | + | [[Homework 3_ECE301Fall2008mboutin]] - [[HW3.A Allen Humphreys_ECE301Fall2008mboutin|'''A''']] - [[HW3.B Allen Humphreys_ECE301Fall2008mboutin|'''B''']] - [[HW3.C Allen Humphreys_ECE301Fall2008mboutin|'''C''']] |
| − | [[HW3.A Allen Humphreys_ECE301Fall2008mboutin|'''A''']] - | + | |
| − | [[HW3.B Allen Humphreys_ECE301Fall2008mboutin|'''B''']] - | + | '''1.''' Assuming the secret matrix is invertible, Bob can find the inverse of the secret matrix and apply it to the secret message 3 elements at a time just as the original matrix is applied to the original string. |
| − | [[HW3.C Allen Humphreys_ECE301Fall2008mboutin|'''C''']] | + | |
| + | '''2.''' Yes. Eve has a known input and a known output. Because the system is linear the encoded message (2,23,3) must be a linear combination of the known output (2,0,0),(0,1,0), and (0,0,3). The algebra breaks down into 3 very simple equations. | ||
| + | |||
| + | i=1 | ||
| + | :<math> a \times 2 + b\times 0 + c \times 0 = a \times 2 = 2, \therefore a = 1</math> | ||
| + | |||
| + | i=2 | ||
| + | :<math> a \times 0 + b\times 1 + c \times 0 = b \times 23 = 23, \therefore b = 23</math> | ||
| + | |||
| + | i=3 | ||
| + | :<math> a \times 0 + b\times 0 + c \times 3 = c \times 3 = 3, \therefore c = 1</math> | ||
| + | |||
| + | Applying these coefficients, which correspond to a specific encrypted message, to the encrypted message which was used to find the coefficients will yield the specific decrypted message as follows: | ||
| + | |||
| + | :<math>a \times (1,0,4) + b \times (0,1,0) + c \times (1,0,1)</math> | ||
| + | :<math>1 \times (1,0,4) + 23 \times (0,1,0) + 1 \times (1,0,1) = (2,23,5) </math> | ||
| + | |||
| + | '''3.''' The secret message is BWE. | ||
Latest revision as of 15:29, 19 September 2008
Homework 3_ECE301Fall2008mboutin - A - B - C
1. Assuming the secret matrix is invertible, Bob can find the inverse of the secret matrix and apply it to the secret message 3 elements at a time just as the original matrix is applied to the original string.
2. Yes. Eve has a known input and a known output. Because the system is linear the encoded message (2,23,3) must be a linear combination of the known output (2,0,0),(0,1,0), and (0,0,3). The algebra breaks down into 3 very simple equations.
i=1
- $ a \times 2 + b\times 0 + c \times 0 = a \times 2 = 2, \therefore a = 1 $
i=2
- $ a \times 0 + b\times 1 + c \times 0 = b \times 23 = 23, \therefore b = 23 $
i=3
- $ a \times 0 + b\times 0 + c \times 3 = c \times 3 = 3, \therefore c = 1 $
Applying these coefficients, which correspond to a specific encrypted message, to the encrypted message which was used to find the coefficients will yield the specific decrypted message as follows:
- $ a \times (1,0,4) + b \times (0,1,0) + c \times (1,0,1) $
- $ 1 \times (1,0,4) + 23 \times (0,1,0) + 1 \times (1,0,1) = (2,23,5) $
3. The secret message is BWE.
