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[Atom_Force]

Math part is:

[danteIL]

Okay, well I have decoded part of this new code, but I can't seem to figure out how to use the 6x6 folded paper -- which seems to be the decoder -- to fill in the gaps.

Basically, this is a variant on the same code that we've already seen before.

The 'math' part at the top tells you how many characters to decode using a particular key. For example:

[konamouse]

[enaxor]

I think the last part should read

[danteIL]

That works enaxor!

So even though I still haven't worked out all the code, this is what I think the message says:

[rowan]

Well, I've discovered how part of the decoder works - at least in determining which letters stay the same. Still haven't figured out how to change letters, but there's gotta be a formula.

If you look closely at the the decoder, you'll notice that there are 8 flaps that work around the outside. The number determines which flap to fold down. So, if you fold down flap 1, you leave the three right hand columns visible. These are the letters that stay the same (so in a '1' transformation, D will stay 5, E will stay R and F will stay F).

Flaps are numbered clockwise. The 3 flap will cover the top half of the grid, 5 will cover the right side starting from the bottom. For the 1+8, you fold the 1 flap and the 8 flap, leaving only 6 characters unchanged.

So that's half of the solution. Now to figure out what the rule is for determining the covered letters.

[danteIL]

[rowan]

Update on the decoder box.

I haven't been able to figure out how the letters are assigned to a particular square, but you can recreate a large portion of what the letters are on each flap.

One of the main things that I've found is that the letters stay in their 9x9 squares. So the I-S-2-7-G-P-6-C-0 grouping will always be assigned to A-B-C-G-H-I-M-N-O quadrant, just rearranged in different ways. This helps in spotting errors in the cipher text (like TUUB - technically there is no way the B could be an N because they aren't in the same quadrants. It should read TUUS.) This also lets you fill in some of the missing letters in a grouping (like we never decode a B with a 1 flap, but since all the other letters are taken, it must be '6')

One other thing you have to take into account is the order that the flaps are given. You should fold them down in the order that they are given. So for 1+7, you would fold down flap 1 and then flap 7 over it. If it were 7+1 you would fold 7 and then 1.

Based on the text we have right now, you can fill in probably about half of the flaps (the hardest to fill in is the VWX quadrant since we dont have most of those letters and very few numbers).

[danteIL]

[rowan]

It's a typo. There are actually quite a few in the last code we got.

[Sylvia]

UPDATES

[Jas0n]

you can use this as a key to using the above decoder




note that you have to make a 6x6 with the alphabet the way it is shown in the black box above.

As rowan said - when the code calls for flaps 1+7+3 then fold flap 1 down, then 7, then 3... works fabulously

edit: note that the flap with the G 6 0 (zero) as the line closest to the middle is flap 1... so basically if you take that last image and put it on the back of the first image - then turn it around so that the first image is facing you ... top left is flap 1.

[Jas0n]

[danteIL]

[enaxor]

[Update] Feb 28th - Journal Entries 28 & 29