yesterday wasn't very bad conditioning wise, except one person tossed their cookies. also both mantids are level 3s. Today conditioning was a bit more well conditioning... 6 17-60s for me, while yesterday there was only 4 (on the last 17-60 we don't do the 60 part).
Yesterday I made a method to decrypt a message pretty well. Let's use "The cat runs" as an example
Caution: Math content in the following section! (It's not like everyone likes math... :P)
"The cat runs". now to do the first method the original message will be in bold for now. to make the second half we write the message backwards
The cat runs snur tac eht
but you don't stop here, then we merge the two starting with the first letter of the original message, then you use the first letter in the message backwards. repeat with the second letter. We keep the spaces in the original message, discarding the ones in the message backwards. now for the the combination, the original message will be in caps
The cat runs snur tac eht
TsHnEu CrAtTa RcUeNhSt now with out the caps the message appears tshneu cratta rcuenhst
That's the first step...
the second step is two parts
first we assign each letter a number
space=0 A=1 B=2 C=3 ect... up until Z=26
but we shift this sequence over... In an actual message I sent so far it was a shift by 3
so space=3 A=4 B=5 C=6....V=25 W=26 X=0 Y=1 Z=2 so it wraps around after you get to 26.
Now the message would appear
remember 3 = a space
23 22 11 17 8 24 3 6 21 4 23 23 4 3 21 6 24 8 17 11 22 23
now if you noticed without the threes it's symetrical that's because the first method made it that way...
Now for the final method... you'll need a good calculator... one like the TI-84 that you can plug in for the X's
I used an equation to make points out of these numbers, further disguising them... I use the X points as placeholders to tell you the order of the Ys, but in the first one I put 4 (in the first message I gave like this). This 4 represent what A equals.. which gives you the position of all the letters.
The Y point is what you get from the equation
the equation I used was: Y=(X^3)-(16X^2)+(40X)-14
Now for this I'll give the key
0 would become -14
1 would become 11
2 would become 10
3 would become -11
4 would become -46
5 would become -89
6 would become -134
7 would become -175
8 would become -206
9 would become -221
10 would become -214
11 would become -179
12 would become -110
13 would become -1
14 would become 154
15 would become 361
16 would become 626
17 would become 955
18 would become 1354
19 would become 1829
20 would become 2386
21 would become 3031
22 would become 3770
23 would become 4609
24 would become 5554
25 would become 6611
26 would become 7786
so the message would go from
23 22 11 17 8 24 3 6 21 4 23 23 4 3 21 6 24 8 17 11 22 23
to
(4, 4609) (5, 3770) (6, -179) (7, 955) (8, -206) (9, 5554) (10, -11) (11, -134) (12, 3031) (13, -46) (14, 4609) (15, 4609) (16, -46) (17,-11) (18, 3031) (19, -134) (20, 5554) (21, -206) (22, 955) (23, -179) (24, 3770) (25, 4609)
tada! it's been encrypted 3 ways. Though if the message is long enough it could be figured out though they'd have to still figure out the first method. I'm working on two ways, I just haven't written them down yet. keyword, yet...
now if you still don't get the shift then here...
0=x 1=y 2=z 3= a space 4=a 5=b 6=c 7=d 8=e 9=f 10=g 11=h 12=i 13=j 14=k 15=l 16=m 17=n 18=o 19=p 20=q 21=r 22=s 23=t 24=u 25=v 26=w
that when A like I said is 4...
now for an experimental method 4 Type-7 ("yet" in "I haven't written it down yet" no longer applies :D) here's what the first two points become
dkfxmgcsqlfjantr
here's the key to that...
0 1 2 3 4 5 6 7 8 9 ( ) space
g h i j k l m a b c d e f
t u v w x y z n o p q r s
how did i get that?
I specified type 7, that meant I moved the letters over by 7
the original would have looked like this
0 1 2 3 4 5 6 7 8 9 ( ) space
a b c d e f g h i j k l m
n o p q r s t u v w x y z
see how a in the type 7 key resides under the 7
the original key is type 0
now to encrypt this just repeat method 2, but no shift of A is required since we were encrypting numbers, now if it were words, it would be different! This is the 5th and last method to this process, which might I add is a bit tiresome... trust me I know... it took around 2-3 hours to create and send the first message (that was before I created the 4th and 5th method today, since that was yesterday)
so dkfxmgcsqlfjantr becomes
4 11 6 24 16 7 3 19 17 12 6 10 1 14 20 18
now you could use that formula in step 3 but it's not necessary. because if anyone tries to use probability to figure out a longer message (I know you, dad, have done this before) it won't work. since these are numbers that have been encoded, and to be safe, I made it so two letters represent one thing, to further mess up probability.
heh I went a little crazy on this... but it was worth it :D
I enjoyed my first trial with it hehe... though I didn't enjoy the part where the test subject fired back...
etiam risus magne me (that was my response to it)
still haven't figured it out *cough* hint, hint *cough*
