# 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