Science:Math Exam Resources/Courses/MATH312/December 2010/Question 06

MATH312 December 2010

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Question 06

Recall that an affine cryptosystem with encryption key $\displaystyle K_{E}=(26,a,b)$ and decryption key $\displaystyle K_{D}=(26,c,d)$ works as follows. Each letter is assigned a numerical value, in alphabetical order: $\displaystyle A\mapsto 0,B\mapsto 1,...,Z\mapsto 25$ These numbers are then viewed modulo 26: They are encrypted by the formula $\displaystyle x\to y\equiv ax+b\mod {26}$ and decrypted by the formula $\displaystyle y\to x\equiv cy+d\mod {26}$ It is known that the most frequently occurring letter in the English language is E (with numerical value 4) and the second most frequently occurring letter is T (with numerical value 19). Suppose these letters are encoded as F (numerical value 5) and G (numerical value 6), respectively. (a) Find the encryption key $\displaystyle K_{E}$ (b) Find the decryption key $\displaystyle K_{D}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
For part (a), we seek to solve the system of congruences given by ${\begin{aligned}5&\equiv 4a+b\mod {26}\\6&\equiv 19a+b\mod {26}\end{aligned}}$ and for part (b), we seek to solve ${\begin{aligned}4&\equiv 5c+d\mod {26}\\19&\equiv 6c+d\mod {26}\end{aligned}}$

Hint 2
After isolating for a variable, this problem becomes an issue of finding inverses modulo 26 which can be done using the Euclidean Algorithm.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For part (a), we seek to solve the system of congruences given by ${\begin{aligned}5&\equiv 4a+b\mod {26}\\6&\equiv 19a+b\mod {26}\end{aligned}}$ Subtracting the two equations yields ${\begin{aligned}1&\equiv 15a\mod {26}\end{aligned}}$ Thus we need to find the inverse of 15 modulo 26. To do this we use the Euclidean Algorithm. ${\begin{aligned}26&=15(1)+11\\15&=11(1)+4\\11&=4(2)+3\\4&=3(1)+1\end{aligned}}$ and back substituting gives ${\begin{aligned}1&=4-3(1)\\1&=4-(11-4(2))(1)\\1&=4(3)-11(1)\\1&=(15-11(1))(3)-11(1)\\1&=15(3)-11(4)\\1&=15(3)-(26-15(1))(4)\\1&=15(7)-26(4)\\\end{aligned}}$ and so an inverse of 15 modulo 26 is given by 7. Thus as $\displaystyle 1\equiv 15a\mod {26}$, we have that $\displaystyle 7\equiv a\mod {26}$. Plugging back into the original equation gives $\displaystyle 5\equiv 4a+b\equiv 4(7)+b\mod {26}$ and isolating for b gives the value of 3 and so the encryption key is $\displaystyle K_{E}=(26,7,3)$ For part (b), we seek to solve ${\begin{aligned}4&\equiv 5c+d\mod {26}\\19&\equiv 6c+d\mod {26}\end{aligned}}$ Subtracting these two equations gives $\displaystyle 15\equiv c\mod {26}$. Thus as $\displaystyle 4\equiv 5c+d\equiv 5(15)+d\equiv 75+d\equiv -3+d\mod {26}$ (so d is 7), we have that the decryption key is given by $\displaystyle K_{D}=(26,15,7)$.

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