Will adding space to the Vigenère block make it truly unbreakable?

I just have gotten into cryptography and learned about the Vigenère block. I think it's a very good idea because, if used correctly, it is unbreakable. But the only downsite is, the key must be the same length as the message. Well, can't you just simply make the block 27x27 big by adding the space into it. That way you could have no space at all, or you can make spaces where ever you want. Is there anything like this already or am I overlooking something here?

• The Vigenere cipher is not unbreakable. – cygnusv Jun 5 '15 at 6:17
• @cygnusv, he stated "key must be the same length as the message", assuming he meant the key to be random, it should be as unbreakable as OTP, as OTP is a vigenere cipher with smaller table. – SEJPM Jun 5 '15 at 11:25
• @SOJPM, yes but then he said "bock 27x27", so I assumed the traditional Vigenere cipher with a bigger table. – cygnusv Jun 5 '15 at 11:28
• @cygnusv, which is still unbreakable provided the key is of the same length as the message and the key is random? – SEJPM Jun 5 '15 at 12:03
• If you use a random permutation over 27 symbols (letters + space) and you use a key of the length of the ciphertext, then you get a bloated version of OTP. All it does, is increase the already huge size of the key (since the whole permutation for each symbol requires $27$ symbols to be stored, instead of just one). – tylo Jun 8 '15 at 11:03

I believe you need a few clarifications to answer this question yourself.

The first is the one time pad (OTP). This is the only truly unbreakable system if it's used correctly.

Using correctly means that for every symbol of the message there is exactly one truly random symbol in the key. Specifically, this means that there is no chosen symbol of the key that is used twice (of course a symbol can occur twice in the key; I mean that for every symbol of the message there is a new symbol chosen for the key with its own randomness). The message includes spaces, so spaces need to be encrypted as well! This is what your thought was when you said to extend the table with a space.

The Vigenère cipher has a repeating key in its original definition. If you choose the key to be exactly as long as the message (and encrypting every symbol of the message, not just the letters) and you chose this key randomly, then the Vigenère cipher becomes the One Time Pad and is thus, as you correctly said, unbreakable.

If your message has more than 27 symbols (for instance capital letters, punctuation etc.) then your key must be extended, as well.