POEMS AND RHYME SCHEMES


ABAB

I've never seen a purple cow
I hope I never see one
But this I'll tell you anyhow
I'd rather see than be one.

CBCB

There was a young lady from Bright
Whose speed was much faster than light.
   She set out one day
   In a relative way
And returned on the previous night.

AABBA

Roses are red
Violets are blue
Sugar is
And so are you.

ABCB

AA

Fleas

Adam
Had 'em

Here Lies my wife
Here let her lie
Now she's at rest
And so am I.

ABCB


There once was a fellow from Drew
Whose limericks stopped at line two.


There once was a man from Verdun.


Sonnets:

ABBAABBACDCDCD

ABABCDCDEFEFGG


One-line poem:  A  :  1 way to do it

Two-line poem:  AA, AB : 2 ways to do it

Three-line poem:  ABC ABB AAB AAA ABA

Four-line poem:  AAAA AAAB AABA AABB AABC ABAA ABAB ABAC ABBA ABBB ABBC ABCA ABCB ABCC ABCD


Is there way to compute the number of ways to rhyme an n-line poem IF we already know the number of ways to rhyme a poem with fewer than n lines?

Let f(l,r) be the number of ways to have r DISTINCT rhymes in a poem with l rhymes.

ABCB  has three distinct rhymes.

f(4,3) = 6

Is there an easy formula for f(l,r)?

Suppose we already know f(l-1,r) that's how many ways there are to make an l-1-line poem with r rhymes.  Well, if add an additional line, and preserve the notion that there are r distinct rhymes, the additional line must match one of the r rhymes already in use.

So this gives r x f(l-1,r) ways to do it ... so far.

Let's say we already know f(l-1,r-1).  Now if I add a new line to the bottom, to have r distinct rhymes, that new line CAN'T possibly rhyme with any others.  So, that's another f(l-1,r-1) patterns.

f(1,r) = r*f(l-1,r) + f(l-1,r-1)

BASE CASES:  f(anything,1) = 1  f(l,r), where r > l is zero.

How many ways are there to rhyme a four-line poem?

	1	2	3	4    r
1	1	0	0	0			1

2	1	1	0	0			2

3	1	3	1	0			5

4	1	7	6	1			15

5	1	15	25	10	1		52


f(2,2)  2*f(1,2) + f(1,1)
f (3,2) =


ASIDE:

g(l,r) = g(l-1,r) + g(l-1,r-1)

1

1	1

1	2	1

1	3	3	1

1	4	6	4	1