Kasiski Attack

[@NOSHEDMANTIS]

Thank you for this concise and clear video - you taught me what my University Cryptography lecturer could not

[@jeffdege4786]

Kasiski was first to publish this break, but it was used long before him. Charles Babbage's notes show him using the Kasiski examination and Kerckhoffs' method as early as the 1820s.

[@dergipfelderleere5536]

This actually helped me more than the videos in my motherlanguage xD nice work!

[@Gamix10]

Same

[@unhingedvagabond]

Dude, you're a truly genius! Thank you very much :O

[@pyotrleflegin7255]

Sir George Wheatstone broke the polyalphabetic as well, a bit like Liebnitz and Newton both discovering the calculus.

[@gabrielbreton8294]

Thanks a lot ! It help a lot me to write the code the decrypt Vigenere cipher !

[@JohnSmith-eo5sp]

0:37 Vigenere Ciphers were used heavily by The South and heavily broken by The North during America's previous Civil War from 1861 to 1865

[@matscho1580]

6:10 Why 6 and not 3 since more numbers in your list are divisible by 3 then 6? I mean i see the point that a keylength of 3 would be short but still possible.

[@JeffSuzukiPolymath]

Generally speaking, you want the largest number that divides most. Unfortunately, that's not a very precise specification, which is where the "art of the Kasiski attack" comes into play. (The good news is that if the key length was 3, then using 6 would still work)

[@afaqmnsr0]

At 7:13 , what is this E? where did it come from. Was following the video nicely but lost it at this point.

[@mikeb3717]

E is the most common letter in English texts, so it's the easiest assumption. E - 11.1607%.

[@Vaasu97]

this was fantastic

[@ManishaGeorgina]

This helped so much, thank you.

[@TheAwesomeCamo]

Great video, very helpful

[@dikesh4478]

First keyword was C since E --> G But in the second case E-->W and how do you came to S?

[@iliasp4275]

Iin the first row we have the normal alphabet and in the second, one swifted so that E matches W you can see that A matches S abcdefghijklmnopqrstuvwxyz stuvwxyzabcdefghijklmnopqr

[@alejandracaceres4683]

Would u mind telling me how did he get E? where does it como from? Ik it's a shift of 2, but why?

[@pipoypipoy7796]

@@alejandracaceres4683 because in most English texts, E is the most common letter. He assumed that all the Gs in the cipher text were originally E in the plain text since G is the most common letter in the cipher text.

[@alejandracaceres4683]

@@pipoypipoy7796 thank you :)

[@mikeb3717]

E is the 5th letter, W is the 23rd, so the difference is +18. Starting from A (1), simply because it's the first letter, if we go 18 spaces forward then we are looking at S (19)

[@maissamaissa1599]

What if we have only one occurrence of the trigram, what should be the distance ???? Please help

[@JeffSuzukiPolymath]

With just one appearance of a trigram, you won't be able to use it. You need at least two.

[@davicesalu2pe2]

this is one of the best explanations I've ever seen, thank you so much

[@w.m.4077]

Very helpful, thank you

[@vishavjeetsingh7862]

Why are you saying 5 A,F,Ws What we care about is the frequency in a column, not whole text

[@JeffSuzukiPolymath]

Once we've found (or think we've found) the key length, we can separate the text into subtexts, each with the same shift. So the "5 A, F, W" is a reference to the 5 A, F, Ws in the subtext, not the whole text.

[@ahmidahmid9303]

good . thank you

[@jasinahameedpk6807]

all that numbers are divisible by 3

[@dikesh4478]

you should take the bigger number as possible.

[@Gamix10]

This vid helped me more than the German ones