Playfair Cipher (aka. Playfair square) is a symmetric encryption technique and was the first literal digraph substitution cipher. It was invented by Charles Wheatstone and popularized by Lyon Playfair.

The technique encrypts pairs of letters instead of single letters, which means it is a type of digraph cipher.

The Playfair Cipher is significantly harder to break since the frequency analysis used for simple substitution ciphers does not work with it. So its no wonder, that is was used during both the first and second world war.

Lets take a closer look. First we have to choose a key, as example we will use “JULEMANDEN”.
The Playfair cipher uses a 5 by 5 table containing a key word or phrase, where the phrase normally will be the alphabet. So the key “JULEMANDEN” will become:

I	U	L	E	M
A	N	D	B	C
F	G	H	K	O
P	Q	R	S	T
V	W	X	Y	Z

How did this key be generated? Well first we took JULEMANDEN and removed duplicated letters – after this we where left with JULEMAND.

Next we replaced J with I. J & I are represented by the same letter.

Finally we split it up into a 5 x 5 square, and filled the missing field values with letters from the alphabet.

This is how it could be done in Navision:

//convert j to i
inKey := CONVERTSTR(inKey,’j',’i');

//Remove duplicates
FOR i := 1 TO STRLEN(inKey) DO BEGIN
……IF STRPOS(outKey,FORMAT(Key[i])) = 0 THEN
………..outKey += FORMAT(inKey[i]);
END;

//Put in fill chars
alphabet := ‘abcdefghiiklmnopqrstuvwxyz’;
aIndex := 1;
WHILE STRLEN(outKey) < 25 DO BEGIN
……IF STRPOS(outKey, FORMAT(alphabet[aIndex])) = 0 THEN
………..outKey += FORMAT(alphabet[aIndex]);
……aIndex += 1;
END;

Now that the key is in place, we will focus on the encryption.

Lets try to encrypt the text: “this is encrypted”. Before the text can be encrypt it has to prepared. The preparation will go through the following steps:

• Remove spaces
• Remove unsupported letters. Ex. dot, numbers and localized letters ex. æøå
• Replace j with i
• If two letters are the same, then add a X. Ex. “this show” would be “this xshow”
• If text length is odd then add a X to the end

Finally break the message into digraphs (groups of 2 letters). So before starting the encryption the prepared text would look like this:

Text:	this is encrypted
Prepared Text:	th is is en cr yp te dx

And as an Navision example:

orgText := LOWERCASE(orgText);

//Remove not supported signs – only letters are supported!
orgText := DELCHR(orgText,’=',’ .,0123456789æøå’);
orgText := CONVERTSTR(orgText,’j',’i');

//If the seond letter of a pair is the same as the first letter, then use x.
i := 1;
WHILE i <= STRLEN(orgText) DO BEGIN

……nextChar := FORMAT(orgText[i]);
……IF (i > 1) THEN
………..IF orgText[i] = orgText[i-1] THEN
…………….nextChar := ‘x’ + FORMAT(orgText[i]);

……preText += nextChar;
……i += 1;
END;

//If string length is odd then add x
IF (STRLEN(preText) MOD 2) = 1 THEN
……preText += ‘x’;

Now lets do the encryption by looking on the individual groups and their placement in the key.
To encrypt a group we use the following rules:

• Is both letters in the same Row? If yes then take the next letter in the row
• Is both letters in the same Column? If yes then take the next letter in the column
• If none off the above occur, the rule differs. To encrypt the first letter, look along its row until the column containing the second letter is reached – the letter at this intersection replaces the first letter. To encrypt the second letter, look along its row until the column containing the first letter is reached – the letter at this intersection replaces the second letter. Hence, ‘th’ becomes ‘ro’.

The encrypted text will therefor become:

Text:	this is encrypted
Encrypted:	ro ep ep ub dt vs sm hl

This is how it could be done in Navision:

i := 1;
FOR i := 1 TO STRLEN(preText) DO BEGIN

……Index1 := STRPOS(KeyIndex,FORMAT(preText[i]));
……row1 := ROUND(Index1 / 5,1,’>');
……col1 := Index1 MOD 5;
……IF col1 = 0 THEN col1 := 5;

……Index2 := STRPOS(KeyIndex,FORMAT(preText[i+1]));
……row2 := ROUND(Index2 / 5,1,’>');
……col2 := Index2 MOD 5;
……IF col2 = 0 THEN col2 := 5;

……i += 1;

……//The letters are in the same Row
……IF row1 = row2 THEN BEGIN
………..colpos := NextPos(col1);
………..Index1 := ((row1 – 1) * 5) + colpos;

………..colpos := NextPos(col2);
………..Index2 := ((row2 – 1) * 5) + colpos;

………..crypText += (FORMAT(KeyIndex[Index1]) + FORMAT(KeyIndex[Index2]));

……//The Letters are in the same Column
……END ELSE IF col1 = col2 THEN BEGIN
………..rowpos := NextPos(row1);
………..Index1 := ((rowpos – 1) * 5) + col1;

………..rowpos := NextPos(row2);
………..Index2 := ((rowpos – 1) * 5) + col2;

………..crypText += (FORMAT(KeyIndex[Index1]) + FORMAT(KeyIndex[Index2]));

……//Letters are in a square
……END ELSE BEGIN
………..rowpos := NextPos(row1 – 1);
………..Index1 := ((rowpos – 1) * 5) + col2;

………..rowpos := NextPos(row2 – 1);
………..Index2 := ((rowpos – 1) * 5) + col1;

………..crypText += (FORMAT(KeyIndex[Index1]) + FORMAT(KeyIndex[Index2]));
……END;

END;

Where NextPos is calculated in this way:

NewPos := (Pos + 1) MOD 5;
IF NewPos = 0 THEN
……NewPos := 5;

To decrypt – just do the inverse

Now you are able to encrypt with Playfair in Navision.

Caesars Cipher is a very simple form of Cryptation. Its method is to replace each plaintext letter with one a fixed number of places down the alphabet. The fixed number is the Key.

Ex. with a key 3 – the letter A will become D. B will become E etc.

So the word SECRET will be come VHFUHW.

This form for cryptology can be represented by the following formular:

c = p + k mod n

where c = cipher, p = plaintext, k = key, n = number of letters in the alphabet

Letters (p) will be represented as numbers, where a = 0, b = 1 etc.

If the cipher is greater than the number of letters in the alphabet, we have to start from the begin of the alphabet – this is solved by mod.

Ex. the letter K & letter Z:

K is the 10′te letter in the alphabet. We use the key 3 and there is 26 letters in the used alphabet.

……c = p + k mod n = 10 + 3 mod 26 = 13, which gives us the letter N.

Z is the 26′te letter in the alphabet.

……c = 26 + 3 mod 26 = 29 mod 26 = 3, which gives us the letter C.

Now that the theory is on place – we can look on some functions for Navision.

To Encrypt you can use this function:

Key := 3;
NoLetters := 26;

IF TextToCrypt = ” THEN
……EXIT;

// Remove spaces
ClearText := UPPERCASE(DELCHR(TextToCrypt,’=',’ ‘));

// Encrypt letter by letter
FOR i := 1 TO STRLEN(ClearText) DO BEGIN

……// Char to Int – where A = 0
……ClearChar := ClearText[i] – 65;

……//Encrypt char – +65 to get a letter
……CryptChar := ((ClearChar + Key) MOD NoLetters) + 65;

……// Add char to crypt text
……CryptText += FORMAT(CryptChar);
END;

TextToCrypt := UPPERCASE(TextToCrypt);

Ok now you can encrypt – but what about decryption. Decryption is actually the same as encryption – all you have to do, is to use a different key

So which key to use? The key to encrypt was 3 and the number of letters was 26. To decrypt you have to deduct 3 letters. This gives us the decrypt key equals 23 (number of letters – key = 23).

Decrypt example:

Key := 3;
NoLetters := 26;
DecryptKey := NoLetters – Key;

IF CryptText = ” THEN
……EXIT;

//DeCrypt letter by letter
FOR i := 1 TO STRLEN(CryptText) DO BEGIN

……//Char to Int – where A = 0
……CryptChar := CryptText[i] – 65;

……//Encrypt char – +65 to get a letter
……ClearChar := ((CryptChar + DecryptKey) MOD NoLetters) + 65;

……// Add char to crypt text
……DeCryptText += FORMAT(ClearChar);
END;

If you compare the two functions, you will see that the encryption and decryptions functions are pretty much the same – they only use different keys.

Now you can encrypt and decrypt with Caesars Cipher in Navision