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Text & Language Processing
What Fraction of Possible Letter Sequences Are Words?
Example Notebook
Open in Cloud
Download Notebook
Get the lengths of words in the word list for English:
I
n
[
1
]
:
=
w
o
r
d
l
e
n
g
t
h
s
=
S
t
r
i
n
g
L
e
n
g
t
h
[
W
o
r
d
L
i
s
t
[
]
]
;
S
h
o
r
t
[
w
o
r
d
l
e
n
g
t
h
s
,
3
]
O
u
t
[
1
]
/
/
S
h
o
r
t
=
{
1
,
3
,
8
,
5
,
6
,
5
,
7
,
7
,
9
,
1
1
,
5
,
9
,
5
,
7
,
9
,
5
,
9
,
8
,
4
,
6
,
5
,
5
,
1
0
,
1
1
,
1
2
,
8
,
1
0
,
7
,
9
,
6
,
9
,
9
,
8
,
5
,
4
,
8
,
1
0
,
4
,
7
,
8
,
5
,
1
0
,
9
,
8
,
5
,
7
,
7
,
6
,
9
,
8
,
1
0
,
6
,
6
,
7
,
8
,
8
,
6
,
4
,
6
,
8
,
4
,
8
,
1
0
,
8
,
3
9
0
4
8
,
7
,
4
,
5
,
5
,
3
,
5
,
6
,
4
,
4
,
3
,
6
,
4
,
6
,
8
,
7
,
9
,
5
,
4
,
3
,
6
,
6
,
8
,
4
,
6
,
4
,
7
,
9
,
5
,
4
,
6
,
6
,
5
,
7
,
4
,
9
,
4
,
6
,
6
,
3
,
6
,
5
,
6
,
9
,
3
,
6
,
5
,
6
,
8
,
6
,
5
,
4
,
6
,
3
,
1
0
,
9
,
7
,
4
,
8
,
6
,
8
,
8
,
6
,
6
,
7
}
Make a histogram of their lengths:
I
n
[
2
]
:
=
H
i
s
t
o
g
r
a
m
[
w
o
r
d
l
e
n
g
t
h
s
]
O
u
t
[
2
]
=
There are only two words of length 1 in the word list:
I
n
[
3
]
:
=
S
e
l
e
c
t
[
W
o
r
d
L
i
s
t
[
]
,
S
t
r
i
n
g
L
e
n
g
t
h
[
#
]
1
&
]
O
u
t
[
3
]
=
{
a
,
I
}
Here are the words of length 2:
I
n
[
4
]
:
=
S
e
l
e
c
t
[
W
o
r
d
L
i
s
t
[
]
,
S
t
r
i
n
g
L
e
n
g
t
h
[
#
]
2
&
]
O
u
t
[
4
]
=
{
a
d
,
a
h
,
a
m
,
a
n
,
a
s
,
a
t
,
a
x
,
b
e
,
b
y
,
d
B
,
d
o
,
e
h
,
e
m
,
e
n
,
e
r
,
e
x
,
f
a
,
g
o
,
h
a
,
h
e
,
h
i
,
h
m
,
h
o
,
i
d
,
i
f
,
i
n
,
I
Q
,
i
s
,
i
t
,
k
c
,
k
W
,
l
a
,
l
o
,
m
a
,
m
e
,
m
i
,
m
s
,
m
u
,
m
y
,
n
o
,
n
u
,
o
f
,
o
h
,
o
n
,
o
w
,
o
x
,
p
a
,
p
H
,
p
i
,
r
e
,
s
h
,
s
o
,
t
a
,
t
i
,
t
o
,
T
V
,
u
m
,
u
p
,
u
s
,
w
e
,
x
i
,
y
a
,
y
e
}
(Some of them are slightly questionable words
…
)
Here is a count of the number of length-2 words in the word list:
I
n
[
5
]
:
=
L
e
n
g
t
h
[
S
e
l
e
c
t
[
W
o
r
d
L
i
s
t
[
]
,
S
t
r
i
n
g
L
e
n
g
t
h
[
#
]
2
&
]
]
O
u
t
[
5
]
=
6
3
Here is a list of all possible pairs of letters:
I
n
[
6
]
:
=
S
t
r
i
n
g
J
o
i
n
/
@
T
u
p
l
e
s
[
A
l
p
h
a
b
e
t
[
]
,
2
]
/
/
S
h
o
r
t
O
u
t
[
6
]
/
/
S
h
o
r
t
=
{
a
a
,
a
b
,
a
c
,
a
d
,
a
e
,
a
f
,
a
g
,
a
h
,
a
i
,
a
j
,
a
k
,
6
5
4
,
z
p
,
z
q
,
z
r
,
z
s
,
z
t
,
z
u
,
z
v
,
z
w
,
z
x
,
z
y
,
z
z
}
Find the number of possible pairs:
I
n
[
7
]
:
=
L
e
n
g
t
h
[
%
]
O
u
t
[
7
]
=
6
7
6
One can compute the number of possible pairs like this:
I
n
[
8
]
:
=
2
6
^
2
O
u
t
[
8
]
=
6
7
6
Note that the number of actual 2-letter words is much smaller than the number of combinations of 2 letters.
The list of possible 3-letter combinations:
I
n
[
9
]
:
=
S
t
r
i
n
g
J
o
i
n
/
@
T
u
p
l
e
s
[
A
l
p
h
a
b
e
t
[
]
,
3
]
O
u
t
[
9
]
=
a
a
a
,
a
a
b
,
a
a
c
,
a
a
d
,
a
a
e
,
a
a
f
,
a
a
g
,
a
a
h
,
a
a
i
,
a
a
j
,
a
a
k
,
a
a
l
,
a
a
m
,
a
a
n
,
a
a
o
,
a
a
p
,
a
a
q
,
a
a
r
,
a
a
s
,
a
a
t
,
a
a
u
,
a
a
v
,
a
a
w
,
a
a
x
,
a
a
y
,
a
a
z
,
a
b
a
,
a
b
b
,
a
b
c
,
a
b
d
,
a
b
e
,
a
b
f
,
a
b
g
,
a
b
h
,
a
b
i
,
a
b
j
,
a
b
k
,
a
b
l
,
a
b
m
,
a
b
n
,
a
b
o
,
a
b
p
,
a
b
q
,
a
b
r
,
a
b
s
,
a
b
t
,
a
b
u
,
a
b
v
,
a
b
w
,
a
b
x
,
a
b
y
,
a
b
z
,
a
c
a
,
a
c
b
,
a
c
c
,
a
c
d
,
a
c
e
,
a
c
f
,
a
c
g
,
a
c
h
,
⋯
1
7
4
5
6
⋯
,
z
x
s
,
z
x
t
,
z
x
u
,
z
x
v
,
z
x
w
,
z
x
x
,
z
x
y
,
z
x
z
,
z
y
a
,
z
y
b
,
z
y
c
,
z
y
d
,
z
y
e
,
z
y
f
,
z
y
g
,
z
y
h
,
z
y
i
,
z
y
j
,
z
y
k
,
z
y
l
,
z
y
m
,
z
y
n
,
z
y
o
,
z
y
p
,
z
y
q
,
z
y
r
,
z
y
s
,
z
y
t
,
z
y
u
,
z
y
v
,
z
y
w
,
z
y
x
,
z
y
y
,
z
y
z
,
z
z
a
,
z
z
b
,
z
z
c
,
z
z
d
,
z
z
e
,
z
z
f
,
z
z
g
,
z
z
h
,
z
z
i
,
z
z
j
,
z
z
k
,
z
z
l
,
z
z
m
,
z
z
n
,
z
z
o
,
z
z
p
,
z
z
q
,
z
z
r
,
z
z
s
,
z
z
t
,
z
z
u
,
z
z
v
,
z
z
w
,
z
z
x
,
z
z
y
,
z
z
z
F
u
l
l
e
x
p
r
e
s
s
i
o
n
n
o
t
a
v
a
i
l
a
b
l
e
(
o
r
i
g
i
n
a
l
m
e
m
o
r
y
s
i
z
e
:
0
.
7
M
B
)
A count of them:
I
n
[
1
0
]
:
=
2
6
^
3
O
u
t
[
1
0
]
=
1
7
5
7
6
Here is the number of actual 3-letter words in the word list:
I
n
[
1
1
]
:
=
L
e
n
g
t
h
[
S
e
l
e
c
t
[
W
o
r
d
L
i
s
t
[
]
,
S
t
r
i
n
g
L
e
n
g
t
h
[
#
]
3
&
]
]
O
u
t
[
1
1
]
=
5
9
6
The count of possible 4-letter sequences:
I
n
[
1
2
]
:
=
2
6
^
4
O
u
t
[
1
2
]
=
4
5
6
9
7
6
The actual number of 4-letter words:
I
n
[
1
3
]
:
=
L
e
n
g
t
h
[
S
e
l
e
c
t
[
W
o
r
d
L
i
s
t
[
]
,
S
t
r
i
n
g
L
e
n
g
t
h
[
#
]
4
&
]
]
O
u
t
[
1
3
]
=
1
9
1
4
See Also
RandomString
Related Symbols
WordList
Publisher Information
Contributed by:
Stephen Wolfram