Question

- Determine the key space of mono-alphabetic substitution cipher for Polish text. Assume that there are 32 letters in the Polish alphabet.

Answer 1

Key Space for an encryption can be define as all possible ways to encrypt an message.

In Mono-alphabetic substitution, we substitute a single alphabet with any available alphabet in that text but "WITHOUT REPEAT".

Let's firstly solve your doubt with English alphabet so that you can understand easily and can relate with Polish alphabet too.

• As we know, there are 26 unique alphabet in English text.
• Alphabet 'A' can be substitute with any alphabet like, A, B, C, D , ......., X, Y, Z. So total key = 26.
• Now, Alphabet 'B' can again substitute with any alphabet but in "Mono-alphabetic substitution", we can't substitute with that alphabet which were previously used to substitute other alphabet. So, Total Key Space= [26- 1(which was used for A)]= 25.
• Similarly, for C, key Space will be 26- 2( used for A and B) = 24.
• similarly for Z, Key Space will be 26-25= 1.
• So, Key space for whole English text is 26*25*24*.....2*1= 26!

Now, comes to Polish Text. As you said, There are 32 alphabet in Polish text. And total key space for Mono-alphabetic substitution for any text will be equal to factorial of unique alphabet.

Hence key space of mono-alphabetic substitution cipher for Polish text is= 32! ( factorial of 32).

I hope, this solution will help you. If you still have any doubt, you can put you question in Comment Section. I would like to sort them out. THANK YOU!