Recurrence Relations and their Solution 1. Let an be ... - MavDISK

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
Solution: an = 2 n .
a1 = 1, an = 2an−1 (n ≥ 2).
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
Solution:
a1 = 2, an = an−1 + 2(n − 1) (n ≥ 2).
an = 2 + 2 + 4 + 6 + · · · + 2(n − 1) = 2 + (n − 1)n = n 2 − 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 − 5D 2 )an = 0,
(1 − 5D)(1 + D)an = 0.
The most general solution to (1−rD)an = 0 is an = pr n . Hence two solutions
are an = p5 n and an = q(−1) n , and an = p5 n + q(−1) n is another solution.
Choosing p and q so that a1 = 4 and a2 = 21 yields p = 5
6
an = 5
6 5n + 1
6 (−1)n .
1
and q = 1
6
. Hence

4. Analyzing the number of comparisons an required to sort a list of 2 n
numbers into ascending order: If there is only one number, do nothing. Otherwise,
separate the list into two sublists of length 2 n−1 , sort the first sublist
using an−1 comparisons, sort the second sublist using an−1 comparisons, then
merge the two lists together using 2 n comparisons. Computer scientists refer
to this as the Merge-Sort Algorithm and our analysis yields the recurrence
relation
a0 = 0, an = 2an−1 + 2 n (n ≥ 1).
This implies
(1 − 2D)an = 2 n ,
(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 = r n , an = nr n , an = n 2 r n , ..., an = n k−1 r n .
Proof: We prove this by induction on k. The base case says that r n is a
solution to (1 − rD)an. This is true.
Now assume that solutions to (1−rD) k an are an = r n , nr n , n 2 r n , . . . , n k−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 = n k r n . 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 = n k r n − r(n − 1) k r n−1 = (n k − (n − 1) k )r n .
2

By the Binomial Theorem,
therefore
For simplicity, we will write
Therefore
(n − 1) k =
k
j=0
j=0
k
n
j
j (−1) k−j ,
n k − (n − 1) k k−1
k
= n
j
j (−1) k−j+1 .
n k − (n − 1) k = c0 + c1n + c2n 2 + · · · + ck−1n k−1 .
an − ran−1 = c0r n + c1nr n + c2n 2 r n + · · · + ck−1n k−1 r n ,
(1 − rD) k+1 an = (1 − rD) k (c0r n + c1nr n + c2n 2 r n + · · · + ck−1n k−1 r n ) = 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 = 2 n and an = n2 n , and we can
combine them to form the solution an = p2 n + qn2 n . We need only choose p
and q so that a0 = 0 and a1 = 2. Doing this yields p = 0, q = 1, an = n2 n .
7. The number of comparisons required to sort a list of n numbers using
Merge-Sort is ≤ 2n log 2(2n). Reason: Let k be an integer such that
2 k−1 ≤ n < 2 k .
We will pad our list to length 2 k by adding 2 k −n zeros to the end of it. We can
sort this list using k2 k comparisons. Given that 2 k ≤ 2n and k < log 2(2n),
we have k2 k ≤ (log 2(2n))(2n) = 2n log 2(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 (A ′ 2, A ′ 3), then place
3

A1 in the correct position relative to (A ′ 2, A ′ 3). In general, to sort the list
(A1, A2, . . . , An), set A1 aside, sort the list (A2, . . . , An) into the sorted list
(A ′ 2, . . . , A ′ n), then place A1 in the correct position relative to A ′ 2, . . . , A ′ n.
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) = 1
2n2 − 1
2n. 9. Consider the recurrence relation
a0 = 5, a1 = 6, an = an−1 + 6an−2 + 2 n + 5 · 3 n + n (n ≥ 2).
We reorganize this to
an − an−1 − 6an−2 = 2 n + 5 · 3 n + n1 n ,
(1 + 2D)(1 − 3D)an = 2 n + 5 · 3 n + n1 n .
By the Lemma in Item 5 above, (1 − 2D)2 n = 0, (1 − 3D)(5 · 3 n ) = 0, and
(1 − D) 2 (n1 n ) = 0. Applying the recurrence relation above by
we obtain
(1 − 2D)(1 − 3D)(1 − D) 2
(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
2 n , (−2) n , 3 n , n3 n , 1 n , and n1 n . A solution involving 6 constants is
an = c12 n + c2(−2) n + c33 n + c4n3 n + c51 n + c6n1 n .
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, 176
45
4
, 49
20
, 3, −13,
−1
36 6 ).

Therefore
an = −2 n + 176
45 (−2)n + 49
20 3n + 3n3 n − 13
36 1n − 1
6 n1n .
10. A formula for an = 1 2 + 3 2 + · · · + (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 = 4n 2 − 4n + 1.
Applying (1 − D) 3 to both sides yields
(1 − D) 4 an = 0.
Solutions to this equation include 1, n, n 2 , n 3 . Setting
and using the 4 initial conditions
we obtain
This yields
an = c0 + c1n + c2n 2 + c3n 3
(a1, a2, a3, a4) = (1, 10, 35, 84)
(c0, c1, c2, c3) = (0, − 1 4
, 0,
3 3 ).
1 2 + 3 2 + · · · + (2n − 1) 2 = 4n3 − n
3
= n(2n − 1)(2n + 1)
.
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 − D 2 )an = 0.
5

Factoring 1 − D − D2 into (1 − rD)(1 − sD) yields r + s = −1 and rs = −1.
. Setting
Solving the system of equations yields r = −1−√ 5
2
an = p
and solving for p and q yields
an =
−1 − √ n 5
+ q
2
and s = −1+√ 5
2
−1 + √ 5
2
n
−1 − √ n+1
5 −1 +
+
2
√ n+1 5
.
2
6