Thanks doranchak,
I'm not sure how good the conditions are here but it shows that it's possible. For some while I've noticed that higher multiplicities can be attempted for shorter ciphers and the 6 row 408 solve is another example of that.
Sure, I'll run through the permutations. It's something that can be revisited at later times also when improvements to the solver are made.
My work
Re: My work
by doranchak » Fri Oct 16, 2015 10:59 am
Dan Olson's idea was to take lines 1 and 11, and write them next to each other to form a new line of length 34.
Then take lines 2 and 12, and write them next to each other to form a new line of length 34.
Then take lines 3 and 13, and write them next to each other to form a new line of length 34.
He thinks the message is in those new lines, rather than the rest of the cipher.
So, I wanted to look at all of the different unique ways to perform these operations. Perhaps we should combine lines 1 and 13, 2 and 12, 3 and 11. That's one of 720 possible ways to make selections of those 6 lines.
Then I thought, we should add more variety. After selecting line 1, say, let's write it down backwards before combining it with another line.
Even more variety: Instead of just writing two lines next to each other, why not interleave them? Such as this:
Before: ZODIAC CIPHER
After: ZCOIDPIHAECR
So I generated cipher texts for all possible combinations of these choices. Download here: http://zodiackillerciphers.com/dan-olso ... ons.txt.gz (3.3MB gzipped)
The downside is that there are 368,640 total variations based on the above approach.
Here is how to interpret the entries in the file:
There's an array of row choices: C = [0,1,2, 10,11,12]. These correspond to the "zero-repeat" rows (lines 0-2, 10-12, where numbering begins at row 0).
[5 true 4 true true] <== [line1, do reverse?, line2, do reverse? do interleave?]
So, 5 means 5th element of C which is 12. 4 means 4th element of C which is 11.
5 true <== means reverse line 5
4 true <== means reverse line 4
Then, the last true means to interleave the lines instead of just copying them next to each other.
Because there are so many ciphers, I'm going to check bigram statistics to see if any permutations seem more promising than others.
Then take lines 2 and 12, and write them next to each other to form a new line of length 34.
Then take lines 3 and 13, and write them next to each other to form a new line of length 34.
He thinks the message is in those new lines, rather than the rest of the cipher.
So, I wanted to look at all of the different unique ways to perform these operations. Perhaps we should combine lines 1 and 13, 2 and 12, 3 and 11. That's one of 720 possible ways to make selections of those 6 lines.
Then I thought, we should add more variety. After selecting line 1, say, let's write it down backwards before combining it with another line.
Even more variety: Instead of just writing two lines next to each other, why not interleave them? Such as this:
Before: ZODIAC CIPHER
After: ZCOIDPIHAECR
So I generated cipher texts for all possible combinations of these choices. Download here: http://zodiackillerciphers.com/dan-olso ... ons.txt.gz (3.3MB gzipped)
The downside is that there are 368,640 total variations based on the above approach.
Here is how to interpret the entries in the file:
- Code: Select all
[5 true 4 true true] [3 true 2 true true] [0 false 1 true true]
RF8B;&(+c.BMF45T|5N*+|#JyRSl9c6<z2OJKHMzT#bLp)B(YWDG|ZEUt+5M/cR:+yUBH)EfRK>*p<l.^YVWPDk%|O1#L(TBG+2pdN
There's an array of row choices: C = [0,1,2, 10,11,12]. These correspond to the "zero-repeat" rows (lines 0-2, 10-12, where numbering begins at row 0).
[5 true 4 true true] <== [line1, do reverse?, line2, do reverse? do interleave?]
So, 5 means 5th element of C which is 12. 4 means 4th element of C which is 11.
5 true <== means reverse line 5
4 true <== means reverse line 4
Then, the last true means to interleave the lines instead of just copying them next to each other.
Because there are so many ciphers, I'm going to check bigram statistics to see if any permutations seem more promising than others.
-
doranchak - Posts: 2360
- Joined: Thu Mar 28, 2013 5:26 am
Re: My work
by doranchak » Fri Oct 16, 2015 11:09 am
Here are the results sorted by bigram repeat counts:
http://zodiackillerciphers.com/dan-olso ... ted.txt.gz
The best ones have 8 bigram repeats, and end up making a quadgram repeat. One example is the sequence "FUB+", appearing twice in this cipher:
Here are the permutations that had 6 or more bigram repeats:
(For comparison, there is only 1 bigram repeat in the cipher resulting from Dan Olson's original suggestion)
A futher note: Might be worth exploring more varieties of interleaving operations. The one I used alternates between the two combined lines at every letter. But you could alternate every 2 letters, 3 letters, 4 letters, etc.
http://zodiackillerciphers.com/dan-olso ... ted.txt.gz
The best ones have 8 bigram repeats, and end up making a quadgram repeat. One example is the sequence "FUB+", appearing twice in this cipher:
- Code: Select all
zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBHNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)FUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2O
Here are the permutations that had 6 or more bigram repeats:
- Code: Select all
8, 2, zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBHNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)FUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2O
8, 2, zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBFUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2OHNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)
8, 2, O2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUFBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jz)df2KG*T<L.1Y|WkDP%VO^#l(pB>+RpENH
8, 2, O2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUF)df2KG*T<L.1Y|WkDP%VO^#l(pB>+RpENHBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jz
8, 2, HNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBFUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2O
8, 2, HNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)FUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2OzJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRB
8, 2, FUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2OzJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBHNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)
8, 2, FUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2OHNEpR+>Bp(l#^OV%PDkW|Y1.L<T*GK2fd)zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRB
8, 2, BRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6JzO2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUF)df2KG*T<L.1Y|WkDP%VO^#l(pB>+RpENH
8, 2, BRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jz)df2KG*T<L.1Y|WkDP%VO^#l(pB>+RpENHO2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUF
8, 2, )df2KG*T<L.1Y|WkDP%VO^#l(pB>+RpENHO2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUFBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jz
8, 2, )df2KG*T<L.1Y|WkDP%VO^#l(pB>+RpENHBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6JzO2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUF
6, 2, zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBNHpE+RB>(p#lO^%VDPWkY|.1<L*TKGf2)dFUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2O
6, 2, zJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBFUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2ONHpE+RB>(p#lO^%VDPWkY|.1<L*TKGf2)d
6, 2, d2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, d2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, d2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2
6, 2, d2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FURB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ
6, 2, d2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pNUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, d2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pNRB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, d2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pNJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2
6, 2, d2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pN2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FURB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ
6, 2, d)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHNUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, d)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHNO2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUFBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jz
6, 2, d)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHNJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2
6, 2, d)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHNBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6JzO2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUF
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2d2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2d2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pNJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2d)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHNJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2Np+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REHJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2Np+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2dJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRd2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRd2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pN
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRd)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHN
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRNp+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REH
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRNp+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2d
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRHER>pl^VPk|1LTG2dNp+B(#O%DWY.<*Kf)
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRHER>pl^VPk|1LTG2d)fK*<.YWD%O#(B+pN
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR)fK*<.YWD%O#(B+pNd2GTL1|kPV^lp>REH
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR)fK*<.YWD%O#(B+pNHER>pl^VPk|1LTG2d
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2HER>pl^VPk|1LTG2dNp+B(#O%DWY.<*Kf)JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2HER>pl^VPk|1LTG2d)fK*<.YWD%O#(B+pNJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2)fK*<.YWD%O#(B+pNd2GTL1|kPV^lp>REHJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, UF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2)fK*<.YWD%O#(B+pNHER>pl^VPk|1LTG2dJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJd2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJd2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pN2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJNp+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REH2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJNp+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2d2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJNHpE+RB>(p#lO^%VDPWkY|.1<L*TKGf2)d2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJHER>pl^VPk|1LTG2dNp+B(#O%DWY.<*Kf)2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJHER>pl^VPk|1LTG2d)fK*<.YWD%O#(B+pN2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUd2GTL1|kPV^lp>REHNp+B(#O%DWY.<*Kf)
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUd2GTL1|kPV^lp>REH)fK*<.YWD%O#(B+pN
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUNp+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REH
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUNp+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2d
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUNHpE+RB>(p#lO^%VDPWkY|.1<L*TKGf2)d
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUHER>pl^VPk|1LTG2dNp+B(#O%DWY.<*Kf)
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FUHER>pl^VPk|1LTG2d)fK*<.YWD%O#(B+pN
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU)fK*<.YWD%O#(B+pNd2GTL1|kPV^lp>REH
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU)fK*<.YWD%O#(B+pNHER>pl^VPk|1LTG2d
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ)fK*<.YWD%O#(B+pNd2GTL1|kPV^lp>REH2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, RB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ)fK*<.YWD%O#(B+pNHER>pl^VPk|1LTG2d2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, O2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUFd)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHNBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jz
6, 2, O2K<McTlbRpJB|Y*D5|TE4tM5./+R&+BUFBRy8:;c(Mc+BUFZ5G|WN(+)#Ly#Sz9H6Jzd)2fGKT*L<1.|YkWPDV%^Ol#p(>BR+EpHN
6, 2, Np+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REHUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, Np+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REHRB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, Np+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REHJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2
6, 2, Np+B(#O%DWY.<*Kf)d2GTL1|kPV^lp>REH2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FURB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ
6, 2, Np+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2dUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2JzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BR
6, 2, Np+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2dRB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FU
6, 2, Np+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2dJzH6z9#SLy)#(+WNG|Z5UF+BMcc(:;y8BRUF+BR&/+5.tME4|TD5Y*B|pJbRTlMcK<O2
6, 2, Np+B(#O%DWY.<*Kf)HER>pl^VPk|1LTG2d2O<KcMlTRbJp|B*Y5DT|4EMt.5+/&RB+FURB8y;:(ccMB+FU5Z|GNW+(#)yLS#9z6HzJ
6, 2, NHpE+RB>(p#lO^%VDPWkY|.1<L*TKGf2)dzJ6H9zS#yL#)+(NW|G5ZFUB+cM(c;:8yRBFUB+&R+/.5Mt4ET|5D*Y|BJpRblTcM<K2O
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6, 0, )Rf8K;*(<c.BYFW5D|%NO+##(yBS+9p6NzHER>pl^VPk|1LTG2dOKMTbpBYD|Et5/R+UBFyB:&c+M.+MU4ZTG5W*(|)JLR#lzcH<J2
6, 0, )Rf8K;*(<c.BYFW5D|%NO+##(yBS+9p6NzBFyB:&c+M.+MU4ZTG5W*(|)JLR#lzcH<J2OKMTbpBYD|Et5/R+UHER>pl^VPk|1LTG2d
6, 0, )Rf8K;*(<c.BYFW5D|%NO+##(yBS+9p6NzBFyB:&c+M.+MU4ZTG5W*(|)JLR#lzcH<J2HER>pl^VPk|1LTG2dOKMTbpBYD|Et5/R+U
(For comparison, there is only 1 bigram repeat in the cipher resulting from Dan Olson's original suggestion)
A futher note: Might be worth exploring more varieties of interleaving operations. The one I used alternates between the two combined lines at every letter. But you could alternate every 2 letters, 3 letters, 4 letters, etc.
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doranchak - Posts: 2360
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Re: My work
by doranchak » Fri Oct 16, 2015 11:19 am
An observation based on the bigram results:
The permuted ciphers that had the maximal 8 bigram repeats all had this in common: Each pair of lines was interleaved:
[0 false 1 false true] [4 true 3 false true] [5 false 2 true true]
[0 false 1 false true] [5 false 2 true true] [4 true 3 false true]
[1 true 0 true true] [2 false 5 true true] [3 true 4 false true]
[1 true 0 true true] [3 true 4 false true] [2 false 5 true true]
[2 false 5 true true] [1 true 0 true true] [3 true 4 false true]
[2 false 5 true true] [3 true 4 false true] [1 true 0 true true]
[3 true 4 false true] [1 true 0 true true] [2 false 5 true true]
[3 true 4 false true] [2 false 5 true true] [1 true 0 true true]
[4 true 3 false true] [0 false 1 false true] [5 false 2 true true]
[4 true 3 false true] [5 false 2 true true] [0 false 1 false true]
[5 false 2 true true] [0 false 1 false true] [4 true 3 false true]
[5 false 2 true true] [4 true 3 false true] [0 false 1 false true]
(The final boolean value in each group indicates if the pair of lines was interleaved).
Perhaps this is just another view of the "even/odds" phenomenon.
The permuted ciphers that had the maximal 8 bigram repeats all had this in common: Each pair of lines was interleaved:
[0 false 1 false true] [4 true 3 false true] [5 false 2 true true]
[0 false 1 false true] [5 false 2 true true] [4 true 3 false true]
[1 true 0 true true] [2 false 5 true true] [3 true 4 false true]
[1 true 0 true true] [3 true 4 false true] [2 false 5 true true]
[2 false 5 true true] [1 true 0 true true] [3 true 4 false true]
[2 false 5 true true] [3 true 4 false true] [1 true 0 true true]
[3 true 4 false true] [1 true 0 true true] [2 false 5 true true]
[3 true 4 false true] [2 false 5 true true] [1 true 0 true true]
[4 true 3 false true] [0 false 1 false true] [5 false 2 true true]
[4 true 3 false true] [5 false 2 true true] [0 false 1 false true]
[5 false 2 true true] [0 false 1 false true] [4 true 3 false true]
[5 false 2 true true] [4 true 3 false true] [0 false 1 false true]
(The final boolean value in each group indicates if the pair of lines was interleaved).
Perhaps this is just another view of the "even/odds" phenomenon.
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doranchak - Posts: 2360
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Re: My work
by Jarlve » Fri Oct 16, 2015 11:27 am
Thanks for sharing,
It took about 10 minutes on 6 cores to get the 6 row 408 solve. I'll need to estabilish how many iterations and restarts the solver needs to get a sufficient recovery rate (at least 90%+) for the 6 row 408. And then I'll know how many of these I could process (very likely not 300k). Needless to say, very early work on Dan Olson's hypothesis.
Nice find, that bigram list should be very doable.
It took about 10 minutes on 6 cores to get the 6 row 408 solve. I'll need to estabilish how many iterations and restarts the solver needs to get a sufficient recovery rate (at least 90%+) for the 6 row 408. And then I'll know how many of these I could process (very likely not 300k). Needless to say, very early work on Dan Olson's hypothesis.
Perhaps this is just another view of the "even/odds" phenomenon.
Nice find, that bigram list should be very doable.
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Jarlve - Posts: 2544
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Re: My work
by Jarlve » Fri Oct 16, 2015 11:34 am
That quadgram repeat is also very interesting. Very nice finds! 
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Jarlve - Posts: 2544
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Re: My work
by daikon » Fri Oct 16, 2015 12:23 pm
glurk wrote:Just a brief comment, but I think if you are using a GB+ sized corpus, a few "hahaha" and etc. probably won't matter that much.
Must be my solver then. Before I filtered out all "hahaha" and "zzzzz" from my 7-grams, it was sometimes getting stuck on nonsensical solutions like "hahahahahaha..." all the way through the cipher. I figured it would be *very* unlikely someone would waste time encrypting "hahaha" in their cipher too.
glurk wrote:I personally got much better results in ZKD, though, when I log-weighted the scores, very simply, like:
VALUE=(int)10*log((double)VALUE);
You kind of have to, so that you are working with probabilities. I've actually converted my solver to work directly with probabilities, so that the closer the combined score to 0, the more likely it is the correct solution. Ideally (although it is only possible to achieve theoretically), the probability should be 1, and log(1)=0. It's a trivial change in the code, you just need to flip all comparisons for the score/fitness, since we are trying to minimize it now. But I think it helps the Simulated Annealing algorithm, as you are using exp(old_fitness-new_fitness) to decide whether to accept the new change. With the higher score/fitness = better solution, the closer you get to the correct solve, the bigger the difference between old_fitness and new_fitness tends to be (since the values themselves get bigger), so you have bigger "jumps" every time you try a new change. Working with probabilities, lower fitness = better solution, so the closer you get to the correct solve, the tighter the difference between old_fitness and new_fitness becomes, so you have "smoother" transitions.
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daikon - Posts: 179
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Re: My work
by daikon » Fri Oct 16, 2015 12:32 pm
Jarlve wrote:Though there appear to be not many numerals, and ciphers that contain them have problems.
Yeah, I noticed the same thing. I'm actually considering parsing the corpus and "spelling out" all numeral into words. So every 42 and 812 will be replaced with "forty two" and "eight hundred twelve", etc.. Same with all fractions: 3/16th will be replaced with three sixteenth, etc.. Currency amounts: $3.40 = three dollars forty cents. I think there must be something that already exists that does these conversions (like a perl module?) since it is necessary for screen readers, etc..
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daikon - Posts: 179
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Re: My work
by Jarlve » Fri Oct 16, 2015 1:14 pm
daikon wrote:Yeah, I noticed the same thing. I'm actually considering parsing the corpus and "spelling out" all numeral into words.
I've experimented with that and it does improve things allot for such ciphers. I considered doing it with the Reddit corpus but thought it would be hard to distinguish between sentences that have numerical information that goes with the message or not. It may also be hard to get it just perfect because there a number of formats: #1, 7th, 2015, 16/10/15, 3s, 1.000.000, 100,8, etc. Anyway you seem to be on it and I'm sure you'll do fine.
But as glurk stated, it probably doesn't have to be perfect. But since I possibly have additional plans with the sentences I wanted to get them as good as I could get them. So I decided to discard any sentence that had numerical information "0123456789".
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Jarlve - Posts: 2544
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Re: My work
by doranchak » Fri Oct 16, 2015 1:16 pm
Jarlve wrote:It took about 10 minutes on 6 cores to get the 6 row 408 solve.
368,640 ciphers, 10 minutes each. Works out to about 7 years of processing time. Yikes.
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