uz.yj permutowania wpisywanego has?a:

Suppose one enters some letters of
length K using my method. The password is of the same length and contains
the same set of letters. To crack the password with a chance of 30% one
needs (1/3)*K! attempts, because my method is nothing else but a
permutation. This is what I do. I write down my password on a piece of
paper and under each letter I write a number in a random way.

Suppose my password is :
c - a - v - a - l - i - e - r - d - e - f - u - b - l - a - s
12 - 8 - 2 - 7 - 1 - 4 - 6 - 3 - 11 - 9 - 10 - 14 - 5 - 13 - 15 - 16

Numbers tell me the order of letters I type into the passphrase window.
That is simple. So what the keylogger gets in this particular example is:
lvribeaaefdcluas

Now I need to know where in the created string I need to enter next
character. So after entering each character I cross out that character.
While entering next character I count the number of crossed out letters on
the left of that character (or on the right, if it is a smaller number, so
easier to count). What I get is the current position to which I should move
the caret with a mouse. This I repeat until entering all the string. I
have made a test. It took me four minutes to type in the 18-letters
password