In our university, we were taught that since all 26 letters cant be placed together in a $5*5$ square matrix we must put $i/j$ together like this represented in the below diagram. Here our key is $monarchy$, now what would have happened if the key contained $j$ in itself? say if the key would have been $jargon$ then how do we accommodate the one extra letter which should have been left?

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    $\begingroup$ Are questions like this considered on-topic? $\endgroup$ Commented Oct 22, 2019 at 19:57

1 Answer 1


When using the convention that i and j are in the same square then they are for all intents and purposes regarded as the exact same letter, both in the key and in the plaintext. In the ciphertext, whenever it is required to write 'i/j' you can choose which one to write, while introducing no ambiguity for the decoder (in order to make the ciphertext 'look more random' it is probably best to flip a coin).

In particular, the Playfair tableau with key 'jargon' is the same as the tableau with key 'iargon': both have 'i/j' in the upper left-hand corner. HOWEVER, if you use an online solver tool it may be (the first site I checked had this implem bug) that j is not parsed, and the tableau is built as if the key were 'argon'; this is a (common?) implem bug though.

Note also that there other conventions than placing i and j in the same square.


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