Can you confirm that you classified both fir and firc correctly? I expect you did but not asking caused problems last time around.
Let’s see what else happens:
clove (should be no, because love is) cater (should be no, because ater is) acir (should be yes, because air is) wac (should be yes, because wa is) Quikly (should be yes, because Quickly is) jakdaws (should be no, because jackdaws is)
cc ccc cccc ccccc
PS: A strikethrough function which works better and is easier than adding U+033c to every character would be nice. My Python function to do it for me is very fragile, too.
def crossout(s):
l = []
inCode = False
inTag = False
for c in s:
l.append(c)
if inTag:
if c == "]":
inCode = not inCode
inTag = False
else:
if c == "[":
inTag = True
if inTag or inCode:
pass
else:
l.append("\u0336")
return "".join(l)
(empty string)
0
0(a^10)
0(a^14)
0(a^16)
0(a^20)
0(a^8)
00
000
00000 (0^5)
000000 (0^6)
00000000 (0^8)
000000000 (0^9)
0123456789
0aa (0(a^2))
0aaaa (0(a^4))
1
1o
1r
2
3
4
5
525600525600
5ire
6
6h
6ire
7
8
9
99
a
A^10
a^12
a^16
a^18
A^6
AA
aa
AAAA
aaaa
aaaaaa (a^6)
aaaaaaaaaa (a^10)
aaar
ab123456789
abcdefghijklmnopqrstuvwxyz
acir
ar
ater
b
b(a^11)
b(a^15)
b(a^17)
b(a^5)
b(a^9)
ba (b(a^1))
baaa (b(a^3))
bigbig
bigbigbigbig
Boxing
c
ca
caaa (ca^3)
ca^5
ca^9
cc
cccc
d
e
earthstone
er
f
f1re
f7r7
f8re
fe
fi
fier
fir1
fir4
fir5
fir6
fir8
fira
firc
fird
FiRe
fire
firewhisky
firf
firg
Five
fr
fr7
freshwater
g
h
hotair
i
ii
iter
j
j00ckdas
j0ckdws
jaackdaaas
jackdaws
Jump
k
l
love
lovelovelove
lovelovelovelove
m
my
n
o
of
p
q
quartz
Quikly
R
r
ri
rr
s
sphinx
t
the^02
the^04
the^06
the^10
the^12
the^14
the^24
tr
u
v
w(a^07)
w(a^09)
w(a^13)
w(a^15)
w(a^19)
wa (w(a^1))
waaa (w(a^3))
waer
wate
waterzebra
watr
wr
ws
wter
ww
WW
ww
WwWw
x
xx
xy
y
z
The red text was my thoughts before I noticed the thing I asked about in green. There is obviously something incorrect about them but they could still be useful somehow.
Okay, my next thought is that c and possibly the have larger values than we’ve seen and the aⁿ pattern repeats or almost repeats. Let’s get a bit more of the pattern; this should help figure out what it is whether or not my hypothesis is correct.
We already have a^0 to a^20 and a^24
[spoiler-box=Known and suspected lowercase-a^n values] a^0 ⟼ N a^1 ⟼ N a^2 ⟼ N a^3 ⟼ Y a^4 ⟼ N a^5 ⟼ Y a^6 ⟼ N a^7 ⟼ Y a^8 ⟼ Y a^9 ⟼ Y a^10 ⟼ N a^11 ⟼ Y a^12 ⟼ N a^13 ⟼ Y a^14 ⟼ Y a^15 ⟼ Y a^16 ⟼ N a^17 ⟼ Y a^18 ⟼ N a^19 ⟼ Y a^20 ⟼ Y a^21 ⟼ Y? a^22 ⟼ N? a^23 ⟼ Y? a^24 ⟼ Y
[/spoiler-box]
a^21 (suspect yes, from evidence in the hint) a^22 (suspect no, from evidence in the hint) a^23 (suspect yes, from evidence in the hint)
a^25: yes
a^26: yes
a^27: yes
a^28: no
a^29: yes
a^30: no
a^31: yes
I found another error: for some reason, the wa^n series seems to have gotten muddled up in the koan log. It should be fixed now. Sorry for any confusion.
(empty string)
0
0(a^10)
0(a^14)
0(a^16)
0(a^20)
0(a^8)
00
000
00000 (0^5)
000000 (0^6)
00000000 (0^8)
000000000 (0^9)
0123456789
0aa (0(a^2))
0aaaa (0(a^4))
1
1o
1r
2
3
4
5
525600525600
5ire
6
6h
6ire
7
8
9
99
a
AA
aa
AAAA
aaaa
A^6
aaaaaa (a^6)
A^10
aaaaaaaaaa (a^10)
a^12
a^16
a^18
a^22
a^28
a^30
aaar
ab123456789
abcdefghijklmnopqrstuvwxyz
acir
ar
ater
b
b(a^11)
b(a^15)
b(a^17)
b(a^5)
b(a^9)
ba (b(a^1))
baaa (b(a^3))
bigbig
bigbigbigbig
Boxing
c
ca
caaa (ca^3)
ca^5
ca^9
cc
cccc
d
e
earthstone
er
f
f1re
f7r7
f8re
fe
fi
fier
fir1
fir4
fir5
fir6
fir8
fira
firc
fird
FiRe
fire
firewhisky
firf
firg
Five
fr
fr7
freshwater
g
h
hotair
i
ii
iter
j
j00ckdas
j0ckdws
jaackdaaas
jackdaws
Jump
k
l
love
lovelovelove
lovelovelovelove
m
my
n
o
of
p
q
quartz
Quikly
R
r
ri
rr
s
sphinx
t
the^02
the^04
the^06
the^10
the^12
the^14
the^24
tr
u
v
w(a^01) (wa)
w(a^03) (waaa)
w(a^07)
w(a^09)
w(a^13)
w(a^15)
w(a^19)
waer
wate
waterzebra
watr
wr
ws
wter
ww
WW
WwWw
x
xx
xy
y
z
After the initial three nos all the runs off yesses so far are separated by single nos and have odd length. In other words, all the nos except for a^1 are at even points.
I strongly suspect the yesses are at numbers one less than composites.
I'm looking at it but I don't seem to be particularly good at this sort of thing, not particularly interested in taking part, so I haven't been doing so. (Except when people ask for a spare 'z' or something.)
I think each string is converted to a number by adding up the values of each character, and that number is accepted if it's one less than a composite. I have no clue how the characters are converted to numbers, except that a seems to have value 1 and others seem to have small positive integer value. I can try writing a program to figure some out or at least checking the suspected values with known koans and non-koans.
In the meantime, the letters in the are most important for those checks probably, because we looked at a lot of powers of the already.
