Recurrence Relations and their Solution
1. Let an be number of subsets of n-element set. Then a1 = 2. For n > 1,
there are an equal number of subsets that contain n and that don’t. We
obtain the recurrence relation
a1 = 1, an = 2an−1 (n ≥ 2).
Solution: an = 2n .
2. Let an be the regions of the plane formed by n mutually intersecting
circles. Then a1 = 2, and for n ≥ 2, the last circle added divides 2(n − 1)
previously existing regions into 2 each, yielding an = an−1 + 2(n − 1). We
obtain the recurrence relation
a1 = 2, an = an−1 + 2(n − 1) (n ≥ 2).
Solution:
an = 2 + 2 + 4 + 6 + · · · + 2(n − 1) = 2 + (n − 1)n = n2 − n + 2.
3. Assume you have 4 types of 1-cent stamps and 5 types of 2-cent stamps.
Let an be the number of ways to paste these stamps in a row on an envelope
to create a postage of n cents. Then a1 = 4, a2 = 21, and for n ≥ 3,
an = 4an−1 + 5an−2 . Reason: There are 4 ways to start with a penny stamp,
followed by an−1 ways to add n − 1 more cents, and there are 5 ways to start
with a 2-cent stamp, followed by an−2 ways to add n − 2 more stamp. We
obtain the recurrence relation
a1 = 4, a2 = 21, an = 4an−1 + 5an−2 (n ≥ 3).
Solution: Write Dan = an−1 . Then
(1 − 4D − 5D2 )an = 0,
(1 − 5D)(1 + D)an = 0.
The most general solution to (1−rD)an = 0 is an = prn . Hence two solutions
are an = p5n and an = q(−1)n , and an = p5n + q(−1)n is another solution.
Choosing p and q so that a1 = 4 and a2 = 21 yields p = 56 and q = 16 . Hence
5
1
an = 5n + (−1)n .
6
6
1
4. Analyzing the number of comparisons an required to sort a list of 2n
numbers into ascending order: If there is only one number, do nothing. Otherwise, separate the list into two sublists of length 2n−1 , sort the first sublist
using an−1 comparisons, sort the second sublist using an−1 comparisons, then
merge the two lists together using 2n comparisons. Computer scientists refer
to this as the Merge-Sort Algorithm and our analysis yields the recurrence
relation
a0 = 0, an = 2an−1 + 2n (n ≥ 1).
This implies
(1 − 2D)an = 2n ,
(1 − 2D)2 an = 0.
Therefore a homogeneous recurrence relation satisfied by an is
an − 4an−1 + 4an−2 = 0 (n ≥ 2).
5. Lemma: Solutions to (1 − rD)k an are
an = rn , an = nrn , an = n2 rn , ..., an = nk−1 rn .
Proof: We prove this by induction on k. The base case says that rn is a
solution to (1 − rD)an . This is true.
Now assume that solutions to (1−rD)k an are an = rn , nrn , n2 rn , . . . , nk−1 an .
Then these are all solutions to (1 − rD)k+1 an = 0 because
(1 − rD)k+1 an = (1 − rD)(1 − rD)k an = (1 − rD)0 = 0.
Now consider an = nk rn . Then
(1 − rD)k+1 an = (1 − rD)k (1 − rD)an = (1 − rD)k (an − ran−1 ).
Let’s calculate an − ran−1 . We have
an − ran−1 = nk rn − r(n − 1)k rn−1 = (nk − (n − 1)k )rn .
2
By the Binomial Theorem,
k X
k j
(n − 1) =
n (−1)k−j ,
j
j=0
k
therefore
k−1 X
k j
n − (n − 1) =
n (−1)k−j+1 .
j
j=0
k
k
For simplicity, we will write
nk − (n − 1)k = c0 + c1 n + c2 n2 + · · · + ck−1 nk−1 .
Therefore
an − ran−1 = c0 rn + c1 nrn + c2 n2 rn + · · · + ck−1 nk−1 rn ,
(1 − rD)k+1 an = (1 − rD)k (c0 rn + c1 nrn + c2 n2 rn + · · · + ck−1 nk−1 rn ) = 0.
Hence the Lemma is true by the Principle of Mathematical Induction.
6. We return to Problem 4. It satisfies the recurrence relation
a0 = 0, a1 = 2, (1 − 2D)2 an = 0 (n ≥ 2).
Two solutions to (1 − 2D)2 an = 0 are an = 2n and an = n2n , and we can
combine them to form the solution an = p2n + qn2n . We need only choose p
and q so that a0 = 0 and a1 = 2. Doing this yields p = 0, q = 1, an = n2n .
7. The number of comparisons required to sort a list of n numbers using
Merge-Sort is ≤ 2n log2 (2n). Reason: Let k be an integer such that
2k−1 ≤ n < 2k .
We will pad our list to length 2k by adding 2k −n zeros to the end of it. We can
sort this list using k2k comparisons. Given that 2k ≤ 2n and k < log2 (2n),
we have k2k ≤ (log2 (2n))(2n) = 2n log2 (2n).
8. Another sorting algorithm is Bubble Sort. To sort a 1-element list (A1 ),
do nothing. To sort a 2-element list (A1 , A2 ), set A1 aside, sort the list (A2 ),
then place A1 in the correct position relative to A1 . To sort a 3-element
list (A1 , A2 , A3 ), set A1 aside, sort the list (A2 , A3 ) into (A02 , A03 ), then place
3
A1 in the correct position relative to (A02 , A03 ). In general, to sort the list
(A1 , A2 , . . . , An ), set A1 aside, sort the list (A2 , . . . , An ) into the sorted list
(A02 , . . . , A0n ), then place A1 in the correct position relative to A02 , . . . , A0n .
The algorithm for placing A1 in the correct position is to slide it past all
numbers that are smaller than it (the bubbling operation). This requires at
most n−1 comparisons. Hence if an is the worst-case number of comparisons
required to sort an n-element list, then a1 = 0 and an = an−1 + (n − 1). The
solution to this recurrence relation is an = 0 + 1 + · · · + (n − 1) = 12 n2 − 12 n.
9. Consider the recurrence relation
a0 = 5, a1 = 6, an = an−1 + 6an−2 + 2n + 5 · 3n + n (n ≥ 2).
We reorganize this to
an − an−1 − 6an−2 = 2n + 5 · 3n + n1n ,
(1 + 2D)(1 − 3D)an = 2n + 5 · 3n + n1n .
By the Lemma in Item 5 above, (1 − 2D)2n = 0, (1 − 3D)(5 · 3n ) = 0, and
(1 − D)2 (n1n ) = 0. Applying the recurrence relation above by
(1 − 2D)(1 − 3D)(1 − D)2
we obtain
(1 − 2D)(1 + 2D)(1 − 3D)2 (1 − D)2 an = 0 (n ≥ 6).
To solve this recurrence relation we need need to provide 6 initial conditions,
a0 through a5 . These can be produced using the original recurrence relation.
By the Lemma in Item 5 again, solutions to the recurrence relation include
2n , (−2)n , 3n , n3n , 1n , and n1n . A solution involving 6 constants is
an = c1 2n + c2 (−2)n + c3 3n + c4 n3n + c5 1n + c6 n1n .
Using the initial conditions
(a0 , a1 , a2 , a3 , a4 , a5 ) = (5, 6, 87, 269, 1216, 4082)
we obtain a system of 6 equations in 6 unknowns whose solution is
(c1 , c2 , c3 , c4 ,5 , c6 ) = (−1,
4
13 1
176 49
, , 3, − , − ).
45 20
36 6
Therefore
an = −2n +
13
176
49
1
(−2)n + 3n + 3n3n − 1n − n1n .
45
20
36
6
10. A formula for an = 12 + 32 + · · · + (2n − 1)2 : This satisfies the recurrence
relation
an = 1, an = an−1 + (2n − 1)2 (n ≥ 2).
We can reorganize this to
an = 1, (1 − D)an = 4n2 − 4n + 1.
Applying (1 − D)3 to both sides yields
(1 − D)4 an = 0.
Solutions to this equation include 1, n, n2 , n3 . Setting
an = c0 + c1 n + c2 n2 + c3 n3
and using the 4 initial conditions
(a1 , a2 , a3 , a4 ) = (1, 10, 35, 84)
we obtain
4
1
(c0 , c1 , c2 , c3 ) = (0, − , 0, ).
3
3
This yields
12 + 32 + · · · + (2n − 1)2 =
4n3 − n
n(2n − 1)(2n + 1)
=
.
3
3
11. A formula for the Fibonacci numbers 1, 2, 3, 5, 8, ... : The recurrence
relation is
an = an−1 + an−2 (n ≥ 2),
(1 − D − D2 )an = 0.
5
Factoring 1 − D − D2 into (1 − rD)(1 − sD) yields
r + s = −1 √and rs = −1.
√
−1− 5
Solving the system of equations yields r = 2 and s = −1+2 5 . Setting
√ !n
√ !n
−1 − 5
−1 + 5
an = p
+q
2
2
and solving for p and q yields
an =
−1 −
2
√ !n+1
5
6
+
−1 +
2
√ !n+1
5
.