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MarcoP

I have run two similar clustering algorithms (PAM Partitioning Around Medoids and EM Expectation Maximization) using 24 dimensions vectors: for each character, I considered the percentage of words in which that character appears immediately before or after each of the 12 most commons characters.
The 12 characters used are:
a C d E e k l o r S t y
a ch d ee e k l o r sh t y
where E stands for EVA:ee ee
(the other upper characters are as in the previous post).

I have rearranged the resulting clusters so that they can be compared. The percentage number is meant to provide a “weight” for the relevance of the character: it's the percentage of unique words that contain that character.

PAM EM

# Cluster 0 # Cluster 6
l l 30.90%
r r 21.46%

# Cluster 1 # Cluster 2
E E 17.35%
e e 34.70%
h # Cluster 8
h   1.95%

# Cluster 8 # Cluster 1
f f 3.83%
p p 9.60%
# Cluster 3 t 19.42%
k
t

# Cluster 7 # Cluster 4
C C 36.58%
S F 0.73%
# Cluster 5 k 27.74%
F K 3.31%
K P 1.47%
P S 15.79%
T T 2.85%
s s 13.17%

# Cluster 6 # Cluster 5
a c 1.31%
c o 64.04%
o # Cluster 0
y a 36.77%
y 44.97%

# Cluster 9 # Cluster 9
M M 7.66%
N N 1.92%
i i 8.44%
m m 3.56%
n n 1.00%

# Cluster 2 # Cluster 7
q q 12.00%

# Cluster 4 # Cluster 3
d d 35.20%

In summary:

EVA:l and r (l r) are assigned to a same cluster by both algorithms. I found this result interesting, since in other cases the characters that are grouped together are graphically similar (while l and r aren't)
E and e are clustered together: it seems that e behaves similarly when it is repeated and when it isn't; EM clusters h h together with E and e, but this character is basically irrelevant, since most occurrences make part of "bench" sequences;

the simple “benchless” gallows cluster together, with some ambiguity about “k”, which the EM algorithm classifies together with the “benched” gallows.
all “benches” sequences are grouped together in EM, with the addition of EVA: s and possibly k (see above); the results of PAM are similar, but C ch and S sh are assigned to a separate cluster;
EVA: a c o y (a c o y) are grouped together. EM identifies two subgroups: a y and o c; the number of occurrences of c is very limited, since it is usually part of one of the sequences I represented with uppercase characters (ch and the "benched" gallows)
a group of characters that usually appear at the end of words are grouped together: M N i m n. While M and N are made up of sequences of i and n, EVA:m m is graphically a distinct character.
q and d are different from each other and from each other character.

For comparison, here are the results for King James Bible. The 12 most common characters are:
a c d e h i l n o r s t

PAM EM
# Cluster 3 # Cluster 0
b b 11.87%
c c 21.31%
f f 12.04%
j j 1.01%
l p 17.41%
m # Cluster 3
p l 28.48%
w m 14.45%
# Cluster 4 r 46.20%
v v 9.06%
w 10.85%

# Cluster 1 # Cluster 4
d d 30.87%
q q 0.92%
r s 47.14%
s t 45.29%
t

# Cluster 0 # Cluster 5
a a 40.90%
e e 70.88%
i i 41.46%
o o 34.70%
u u 19.74%

# Cluster 9 # Cluster 6
g g 18.67%
# Cluster 6 # Cluster 7
n n 40.04%
# Cluster 7 # Cluster 8
h h 25.44%
# Cluster 8 # Cluster 9
k k 6.48%
# Cluster 5 # Cluster 1
y y 9.01%
# Cluster 2 # Cluster 2
x x 1.35%

Notes:

two groups with different composition (made up two distinct clusters for each algorithm) contain most consonants: b c f j l m p v w. The position of 'r' is unclear, since PAM assigns it to the next group
a third consonant group is made up of d, q, s, t and possibly r (see above)
the vowels a e i o u are recognized as distinct by both algorithms
a number of characters are clustered as singletons with just one element: g n h k y x

King James Bible results are rather opaque to me. It seems significant that the vowels are group together, but I have no idea of what the other letters have in common.

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