# Does Kasisky test for Vigenère cyphers assumption always hold?

In the notes from which I'm studying I read about Kasisky test:

"Key observation: Two identical segments of $l ≤ L$ plaintext letters, will be encrypted to the same $l ≤ L$ ciphertext letters."

Where $L$ is the length of the keyword in the sense that

$$k=(k_0, k_1, . . . , k_{L−1}) \ with \ k_j \in (0, . . . , 25)$$

Is the cypher keyword with, obviously, the number $0, ... 25$ representing each one a different letter of the english alphabet ($A=0, B=1,...,Z=25$)

But if I think about an example like this one

$ptx:$ "Our is the fury" $\Rightarrow$ "Ouristhefury"

$key:$ "$BCDEF$"

Which denoting the $ptx$ as

$$ptx = p_0p_1... , p_i,...$$

applies the encryption:

$$c_i = (p_i + k_{i \pmod L}) \pmod {26}$$

Represented with a table:

$$\begin{array}{|c|c|c|} \hline \text{ptx} & \text{O} & \text{u} & \text{r} & \text{i} & \text{s} & \text{t} & \text{h} & \text{e} & \text{f} & \text{u} & \text{r} & \text{y} \\ \hline \text{key} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{B} & \text{C} & \text{D} & \text{E} & \text{F} & \text{B} & \text{C} \\ \hline \text{ctx} & \text{p} & \text{w} & \text{u} & \text{m} & \text{x} & \text{u} & \text{j} & \text{h} & \text{j} & \text{z} & \text{s} & \text{a} \\ \hline \end{array}$$

Following the definition above the two segments "ur" should be encrypted to two equal letters in the cypertext, but instead they are encypted respectively to:

• "wu", first occurrence
• "zs", second occurrence

I think that they're encrypted into the same two letters of the $ctx$ only when the two couples of $ptx$ letters are located in positions where the same shift occurs (same letter of the keyword).

So my question is: is there some imprecision in the statement from my notes or I'm interpreting something wrong?

[abcdea]bcdeabcdeabcde[abcdea]bcdeabc

Note that the key here is $5$ letters long and the distance between corresponding letters in the two segments is $20 = (5*4)$ so the corresponding ciphertext segments will also be the same. It is also worth observing that there is no need for $l \le L$ as seen in the above example.