# Exponential cipher

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• Aug 10th 2011, 05:36 AM
vidhi96
Exponential cipher
I tried to work through this ex in my book which is not typical of the ex I have done and I am confused after first part
for part a I worked out that inverse of 5 in Z36 is 29. hope this is correct
but I have no clue as to where to and how to start part b - which is attached here. Can someone please guide me in detail as I am studying this for the first type.
• Aug 10th 2011, 07:34 AM
pece
Re: Exponential cipher
Quote:

Originally Posted by vidhi96
I tried to work through this ex in my book which is not typical of the ex I have done and I am confused after first part
for part a I worked out that inverse of 5 in Z36 is 29. hope this is correct

This is correct : $\displaystyle 5\times 29 = 145 = 4\times 36 + 1$

Quote:

Originally Posted by vidhi96
but I have no clue as to where to and how to start part b - which is attached here. Can someone please guide me in detail as I am studying this for the first type.

The table gives you a correspondence $\displaystyle C : \{A,\dots,Z\} \rightarrow \{1,\dots,26\}$. So for a letter $\displaystyle \alpha$, the cipher letter is $\displaystyle C^{-1}(E(C(\alpha)))$ (I didn't check, but in order to apply $\displaystyle C^{-1}$, one must have $\displaystyle m^5 (\mathrm{mod}\, 37) \in \{1,\dots,26\},\ \forall m\in\{1,\dots,26\}$).

When you have a word, it's a sequence of letter : you encrypt every letter of this sequence and have this way a sequence of cipher letter. It's your cipher text.
• Aug 10th 2011, 08:07 AM
vidhi96
Re: Exponential cipher
i am more confused. What I figured out by now is message ext is <23 8 5 14> spelling WHEN.

so now I need to find c - which is chipher text is that correct

Quote:

Originally Posted by pece
This is correct : $\displaystyle 5\times 29 = 145 = 4\times 36 + 1$

The table gives you a correspondence $\displaystyle C : \{A,\dots,Z\} \rightarrow \{1,\dots,26\}$. So for a letter $\displaystyle \alpha$, the cipher letter is $\displaystyle C^{-1}(E(C(\alpha)))$ (I didn't check, but in order to apply $\displaystyle C^{-1}$, one must have $\displaystyle m^5 (\mathrm{mod}\, 37) \in \{1,\dots,26\},\ \forall m\in\{1,\dots,26\}$).

When you have a word, it's a sequence of letter : you encrypt every letter of this sequence and have this way a sequence of cipher letter. It's your cipher text.

• Aug 10th 2011, 11:36 AM
pece
Re: Exponential cipher
Quote:

Originally Posted by vidhi96
i am more confused. What I figured out by now is message ext is <23 8 5 14> spelling WHEN.

Now you have to apply $\displaystyle E$ to each of these numbers and then retranslate the result into letters.
• Aug 10th 2011, 11:42 AM
vidhi96
Re: Exponential cipher
Quote:

Originally Posted by pece
Now you have to apply $\displaystyle E$ to each of these numbers and then retranslate the result into letters.

Ok I'll try and type it and send it to you tomorrow