There are several ways I am attempting to break the last portion of the
Kryptos puzzle. I have written several programs which i posted some of in my Yahoo group. One of them attempts to solve the last portion using a vigenere method against a dictionary. I have also used the tools on Simon Singh's site that work very well for alot of ciphers. I am trying different variations on the last section using different alphabets and running the last section with and without the "Q", as well as backwards. Programs work great for this sort of thing, but there is a downside. Imagine running the program on a 50,000 word dictionary. you've got to sift through 50,000 results looking for anything that might be a solution. I have come up with a program for this that I will post as well. I am working on several other decryption programs, as well as a transposition program that needs a few bugs worked out. The difference between my programs and other programs I've found so far is the option of running the program on a dictionary. Most likely my intention is to program every pencil and paper method possible, and hope to find a correct solution in the process of ruling out all the wrong ones.

PLAYFAIR 

Treating I & J as one.

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

At this point, we aren't sure if the last section of Kryptos is actually
97 or 98 letters, or some other amount. So I'm just working under assumptions here.

Any uneven ciphertext blocks should be padded with X's.

Let's break it up into 2 character sections for clarity:

97 letters:

From what I understand (And Ive seen people get this backwards so I'm never quite sure, check out Phreaknic codes to see what I mean):

In the Playfair Square, if your two letters are diagonal to one another, then your plaintext is equal to the letter in the same row as your
ciphertext letter, and the column of your second ciphertext letter, and vice versa.

Example:

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

XO would equal UB

you always write down the plaintext letter of the first ciphertext letter in the pair first, not the other way around.
Don't think that because O comes before X in the square that you would write down the B first. This will only result in error.

If your two letters are in the same row, then you take the letters directly to the left of each plaintext letter. If there are none to the left then you wrap
around to the right.

Example:

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

OB would equal CA (notice the wrap around from the O to the C on the right end).

If your two letters are in the same column, then you take the letters directly above each plaintext letter. If there are none above it then you wrap
around to the end.

Example:

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

UO would equal IK

As far as I know, you either keep any duplicate letters the same or use the one directly to the right and down one. (diagonal)

http://www.trumpetpower.com/Papers/Crypto/Playfair states this:

"Next, double letters—if they occur in a pair—must be divided by an X or a Z. For example, freedom in this example becomes FREXEDOM. Because of the triple s between the first and second words, they are written as CONGRESXSZ SHALL. This rule for double letters reduces the number of visible patterns in the ciphertext"

For our examples, im just going to keep duplicates the same. Since the ciphertext doesn't contain duplicates until the 19 letter (B) ,
we would know by then whether we were on the right track or not anyway. so who cares? :)

If this were our correct Key, our entire ciphertext would reveal the following:

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

CA TK IK UB FG VI AO SI MD BB

MA EV ZL NT PL DP SS CK ZY LC

QN SS DR ZZ MY KQ RI IO MO UM

MG PG RY ZR YQ WH YT XF OU XW

KQ OH ND UI DZ OW DR BS PV

(Yeah thats it, lol)