1. Chapter 7 Advanced Counting Techniques
Discrete Mathematics
Chapter Advanced Counting Techniques

2. 7.1 Recurrence Relations(遞迴關係)
Example 1. Let {an} be a sequence that satisfies the recurrence relation an=an-1-an-2 for n=2,3,…, and suppose that a0=3,and a1=5.
Here a0=3 and a1=5 are the initial conditions.
By the recurrence relation,
a2 = a1-a0 = 2
a3 = a2-a1 = -3
a4 = a3-a2 = -5 :
Q1: Applications ?
Q2: Are there better ways for computing the terms of {an}?

3. ※Modeling with Recurrence Relations
We can use recurrence relations to model (describe) a
wide variety of problems.
Example 3. Compound Interest (複利)
Suppose that a person deposits(存款) $10000 in a saving account at a bank yielding 11% per year with interest compounded annually.
How much will be in the account after 30 years ?
Sol : Let Pn denote the amount in the account after n years.
Pn=Pn Pn-1=1.11  Pn-1,
∴ P30=1.11  P29=(1.11)2  P28=…=(1.11)30  P =
P0=10000

4. Example 5. (The Tower of Hanoi)
The rules of the puzzle allow disks to be moved one at a time from one peg to another as long as a disk is never placed on top of a smaller disk. Let Hn denote the number of moves needed to solve the Tower of Hanoi problem with n disks. Set up a recurrence relation for the sequence {Hn}.
Sol : Hn=2Hn-1+1, ( n-1個 disk 先從peg 1→peg 3, 第 n 個 disk 從 peg 1→peg 2, n-1個 disk 再從 peg 3→peg 2)
目標 : n 個disk都從 peg 1 移到 peg 2
peg 1
peg 2
peg 3
H4 moves
H1=1

5. 上例中 Hn=2Hn-1+1, H1=1
∴Hn=2Hn-1+1
=2(2Hn-2+1)+1
=22Hn-2+2+1
=22(2Hn-3+1)+2+1
=23Hn-3+(22+2+1) :
=2n-1H1+(2n-2+2n-3+…+1)
=2n-1+2n-2+…+1
= =2n-1

6. Example 6. Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutive 0s. How many such bit strings are there of length 5 ?
Sol :
n-2
n-1 n 1 2
n-3 …
∴ an = an-1+an-2, n  3
a1=2 (string : 0,1)
a2=3 (string : 01,10,11) 1
an-1 種
an-2 種 1
∴ a3=a2+a1=5, a4=8, a5=13

7. Example 7. (Codeword enumeration)
A computer system considers a string of decimal digits a valid codeword if it contains an even number of 0 digits. Let an be the number of valid n-digit codewords. Find a recurrence relation for an.
Sol :
∴ an = 9an-1 + (10n-1-an-1)
= 8an n , n2
a1 = 9
n-1 n 1 2 3 …
an-1 種
1~9
10n-1 - an-1 種

8. 求an通解 :
Exercise : 3,23,25,27,29,41
(41推廣成n)

9. 7.2 Solving Recurrence Relations
Def 1. A linear homogeneous recurrence relation of
degree k (i.e., k terms) with constant coefficients
is a recurrence relation of the form
where ciR and ck≠0
an = c1an-1+c2an-2+…+ckan-k
Example 1 and 2.
fn = fn-1 + fn-2
an = an-5
an = an-1 + an-22
an = nan-1
Hn = 2Hn-1 + 1
(True, deg=2)
(True, deg=5)
(False, 不是linear)
(False , 不是linear, not constant coeff. )
(False, 不是homogeneous)

10. Theorem 1.
Let an = c1an-1+ c2an-2 be a recurrence relation
with c1,c2R.
If r2 - c1r - c2= 0 (稱為characteristic equation) has two distinct roots r1 and r2.
Then the solution of an is an = a1r1n + a2r2n ,
for n=0,1,2,…, where a1 , a2 are constants.
(a1 , a2可利用 a0, a1算出)

11. What’s the solution of the recurrence relation an = an-1 + 2an-2
Example 3.
What’s the solution of the recurrence relation
an = an-1 + 2an-2
with a0=2 and a1=7 ?
Sol :
The characteristic equation is r2 – r - 2=0.
Its two roots are r1= 2 and r2 = -1.
Hence an=a12n +a2 (-1)n .
∵a0 = a1+a2 = 2, a1=2a1-a2=7
∴a1 = 3, a2 = -1
 an = 32n - (-1)n.
驗算：a2 = a1 + 2a0 = a2= 3 =11

12. Example 4. Find an explicit formula for the Fibonacci numbers. Sol :
fn = fn-1 + fn-2 , n  2, f0=0 , f1=1.
The characteristic equation is r2 - r - 1=0.
Its two roots are ,
So we have

13. Thm 2.
Let an = c1an-1+c2an-2 be a recurrence relation
with c1,c2R.
If r2 - c1r - c2= 0 has only one root r0 .
Then the solution of an is
an = a1  r0n + a2  n  r0n
for n=0,1,2,…, where a1 and a2 are constants.

14. What’s the solution of an= 6an-1 - 9an-2 with a0=1 and a1=6 ? Sol :
Example 5.
What’s the solution of an= 6an-1 - 9an-2 with a0=1 and a1=6 ?
Sol :
The root of r2 - 6r + 9 = 0 is r0 = 3.
Hence an = a1．3n +a2．n．3n .
∵a0 = a1 = 1
a1 = 3a1 + 3a2 = 6
∴ a1 = 1 and a2 = 1
 an = 3n + n．3n
驗算：a2 = 6a1 - 9a0 = a2=  32 =27

15. Thm 3.
Let an = c1an-1 + c2an-2 + … + ckan-k be a recurrence relation with c1, c2, …, ck  R.
If rk - c1rk-1 - c2rk-2 -…- ck = 0 has k distinct roots r1, r2,…, rk.
Then the solution of an is
an = a1r1n +a2r2n + …+akrkn, for n = 0, 1, 2, …
where a1, a2,…ak are constants.

16. Find the solution of an = 6an-1 - 11an-2 + 6an-3
Example 6 (k = 3)
Find the solution of an = 6an an-2 + 6an-3
with initial conditions a0=2, a1=5 and a2=15 .
Sol :
The roots of r3 - 6r2 + 11r – 6 = 0 are
r1 = 1, r2 = 2, and r3 = 3
∴an = a1  1n + a2 2n + a3 3n
∵a0 = a1 + a2 + a3 = 2
a1 = a1 + 2a2 + 3a3 = 5
a2 = a1 + 4a2 + 9a3 = 15
∴an = 1 - 2n + 2  3n
a1 = 1,
a2 = -1,
a3 = 2
驗算：a3 = 6a2 - 11a1+ 6a0 = a3=  33 =47

17. Thm 4.
Let an = c1an-1 + c2an-2 + … + ckan-k be a recurrence relation with c1, c2, …, ck R.
If rk - c1rk-1 - c2rk-2 - … - ck = 0 has t distinct roots r1, r2, …, rt with multiplicities m1, m2, …, mt respectively, where mi  1,i, and m1+ m2 +…+ mt = k,
then (接下一頁)

18. where ai,j are constants.
(1  i  t , 0  j  mi-1)

19. 驗算：a3 = - 3a2 - 3a1- a0 =8 a3= (1+33-232)(-1)3 =8
Example 8. Find the solution to the recurrence relation an = -3an-1 - 3an-2 - an-3 with initial conditions a0 = 1, a1 = -2 and a2 = -1.
Sol :
r3 + 3r2 + 3r + 1 = 0 has a single root r0 = -1 of multiplicity three.
∴ an = (a1+a2n+a3n2) r0n = (a1+a2n+a3n2)(-1)n
∵ a0 = a1 = 1
a1 = (a1+a2+a3)  (-1) = -2
a2 = (a1+a2+a3) = -1
∴a1 = 1, a2 = 3, a3 = -2
 an = (1+3n-2n2)  (-1)n
驗算：a3 = - 3a2 - 3a1- a0 = a3= (1+33-232)(-1)3 =8
Exercise : 3,13,15,19

20. 7.4 Generating Functions.
Def 1. The generating function for the sequence {an} is the infinite power series.
G(x) = a0 + a1x +… + anxn +… =
(若 {an} 是finite，可視為是infinite，但後面的term 都等於0)

21. Example 2. What is the generating function for the sequence 1,1,1,1,1,1 ?
Sol :
(expansion，展開式)
(closed form)

22. Example 3.
Let mZ+ and ,for k = 0, 1, …, m.
What is the generating function for the sequence a0, a1,…, am ?
Sol :
G(x) = a0 + a1x + a2x2 + … + amxm
= (1+x)m (by 二項式定理)

23. Example 5. The function f (x) = is the generating
function of the sequence 1, a, a2, …,
since = 1 + ax + a2x2 + …=
when |ax| < 1 for a≠0

24. Let uR and kZ+∪{0}. Then the extended
Def 2.
Let uR and kZ+∪{0}. Then the extended
binomial coefficient is defined by
(The Binomial Theorem, 二項式定理)
Let x,y be variables, and let n be a positive integer,
Then

25. Example 7.
Find and
Sol :

26. Thm 2. (The Extended Binomial Theorem)
Let xR with |x|<1 and let uR, then

27. Find the generating functions for (1+x)-n and (1-x)-n where nZ+
Example 9.
Find the generating functions for (1+x)-n and (1-x)-n where nZ+
Sol : By the Extended Binomial Theorem,
Def 2.
By replacing x by –x we have

28. ※Using Generating Functions to solve Recurrence Relations.
Example 16.
Solving the recurrence relation ak = 3ak-1 for k=1,2,3,… and initial condition a0 = 2.
Sol :
另法： (by 7.2公式 )
r – 3 = 0  r = 3  an = a  3n
∵ a0 = 2 = a
∴ an = 2  3n

29. generating function for {ak}. First note that ak xk = 3ak-1 xk 
Let be the
generating function for {ak}.
First note that ak xk = 3ak-1 xk 
G(x) - a0 = 3x  G(x)
∵a0 = 2  G(x) - 3x  G(x) = G(x)(1-3x) = 2
∴ ak = 2  3k
Exercise : 5,7,11,33

30. 7.5 Inclusion-Exclusion 排容原理
A,B,C,D : sets A B C
|A|+|B|+|C| 時
各部分被計算的次數 1 2 1 2 1 1 2 3
-|AB|-|AC|-|BC|後 1 2 1 1
+|ABC|後

31. Theorem 1.
A1, A2, …, An : sets
Exercise : 17

32. 7.6 Applications of Inclusion and Exclusion
Example 2. How many onto functions are there form set A={1, 2, 3, 4, 5, 6} to set B={a, b, c} ?
Sol : f : A → B
f (1)= {a, b, c}
f (2)=
︰ ︰
f (6)=
不同的填法造出不同的函數
如何使a,b,c都出現 ?
＃ of onto functions
= (所有函數個數) - (a,b,c中有一個沒被對應) (a,b,c中二個沒被對應) - (a,b,c都沒被對應) =

33. Thm 1. |A| = m , |B| = n There are onto functions f : A → B. pf :
A = {a1, a2, …, am}. B = {b1, b2, …, bn}
f (a1)=
f (a2)=
︰ ︰
f (am)=
b1, b2, …, bn

34. ※Derangements 亂序
Def.
A derangement is a permutation of
objects that leaves no object in its
original position.

35. Example 5. derangements of 12345 :
21453, 23451, 34512, …
Def.
Let Dn denote the number of derangements of n objects.
D4 = (所有4個元素的permutation數)
- (4個元素有一個在原位置的permutation數)
+ (4元素中有二個在原位置的個數)
- (4個元素中有三個在原位置的個數)
+ (4元素都在原位置的個數) =

36. Theorem 2. (亂序公式)
Exercise : 8
參考：12,13