Discrete Mathematics
Chapter 6 Advanced Counting Techniques
感謝 大葉大學 資訊工程系 黃鈴玲老師 提供
6.1 Recurrence Relations( 遞迴關係 )
Example 1. Let {an} be a sequence that satisfies the recurrence relation an=an1an2 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 = a1a0 = 2
a3 = a2a1 = 3 a4 = a3a2 = 5 :Q1: Applications ?Q2: Are there better ways for computing the terms of
{an}?6.1.1
※Modeling with Recurrence RelationsWe 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 ?
6.1.2
Sol : Let Pn denote the amount in the account after n years. Pn=Pn1 + 0.11Pn1=1.11 Pn1,
∴ P30=1.11 P29=(1.11)2 P28=…=(1.11)30 P0
=228922.97
P0=10000
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, ( n1 個 disk 先從 peg 1→peg 3, 第 n 個 disk 從 peg 1→peg 2, n1 個 disk 再從 peg 3→peg 2)
6.1.3
peg 1 peg 2 peg 3
H4 moves
目標 : n 個 disk 都從 peg 1 移到 peg 2
H1=1
上例中 Hn=2Hn1+1, H1=1
∴Hn=2Hn1+1
=2(2Hn2+1)+1
=22Hn2+2+1
=22(2Hn3+1)+2+1
=23Hn3+(22+2+1)
:
=2n1H1+(2n2+2n3+…+1)
=2n1+2n2+…+1
= =2n112
12
n
6.1.4
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 :
∴ an = an1+an2, n 3
a1=2 (string : 0,1) a2=3 (string : 01,10,11)
∴ a3=a2+a1=5, a4=8, a5=13
6.1.5
1an-1 種
an-2 種 1 0
n-2 n-1 n1 2 n-3…
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 :
6.1.6
1~9an1 種
10n1 an1 種 0
∴ an = 9an1 + (10n1an1) = 8an1 + 10n1 , n2 a1 = 9
n-1 n1 2 3 …
求 an 通解 :
)10108108(8 12323
3 nnn
na
Exercise : 3,23,25,27,29,41 (41 推廣成n)
122
11 10)108(8108
nnn
nnn aaa
)10108(8 122
2 nn
na
)10108108108(8 122321
1 nnnnn ...a
)1081081081088(8 1022321 nnnnnn ...
184102
1 nn
項共有 ,8
10公比為 ,8首項 1 nn-
11
1
84)8
10(848
)18
10(
)1)8
10((8
8
nnnn
nn
n
6.2 Solving Recurrence RelationsDef 1. A linear homogeneous recurrence relation of
degree k (i.e., k terms) with constant coefficients
is a recurrence relation of the form
where ciR and ck≠0
6.2.1
Example 1 and 2. fn = fn1 + fn2 an = an5
an = an1 + an22
an = nan1
Hn = 2Hn1 + 1
an = c1an1+c2an2+…+ckank
(True, deg=2)(True, deg=5)
(False, 不是 linear)(False , 不是 linear)
(False, 不是 homogeneous)
Theorem 1.
Let an = c1an1+ c2an2 be a recurrence relation
with c1,c2R.
If r2 c1r c2= 0 ( 稱為 characteristic equation) has two distinct roots r1 and r2.
Then the solution of an is an = 1r1n + 2r2
n ,
for n=0,1,2,…, where 1 , 2 are constants.
(1 , 2 可利用 a0, a1 算出 )
6.2.2
Example 3.What’s the solution of the recurrence relation
an = an1 + 2an2
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=12n +2 (1)n .
∵a0 = 12 = 2, a1=212=7
∴1 = 3, 2 = 1
an = 32n (1)n.6.2.3
驗算: a2 = a1 + 2a0 =11 a2= 322 1 =11
Example 4. Find an explicit formula for the Fibonacci numbers.
Sol :
fn = fn1 + fn2 , n 2, f0=0 , f1=1.
The characteristic equation is r2 r 1=0.
Its two roots are , .
So we have
2
511
r
2
512
r
nnnf )
2
51()
2
51( 21
6.2.4
,0210 f∵ 1)2
51()
2
51( 211
f
5
1 ,
5
121
nnnf )
2
51(
5
1)
2
51(
5
1
Thm 2.
Let an = c1an1+c2an2 be a recurrence relation
with c1,c2R.
If r2 c1r c2= 0 has only one root r0 .
Then the solution of an is
an = 1 r0n + 2 n r0
n
for n=0,1,2,…, where 1 and 2 are constants.
6.2.5
Example 5.
What’s the solution of an= 6an1 9an2 with a0=1 and a1=6 ?
Sol :
The root of r2 6r + 9 = 0 is r0 = 3.
Hence an = 1 . 3n +2 . n . 3n .
∵a0 = 1 = 1
a1 = 31 + 32 = 6
∴ 1 = 1 and 2 = 1
an = 3n + n . 3n
6.2.6
驗算: a2 = 6a1 9a0 =27 a2= 32 +2 32 =27
Thm 3.
Let an = c1an1 + c2an2 + … + ckank 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 = 1r1n +2r2
n + …+krkn, for n = 0, 1, 2, …
where 1, 2,…k are constants.
6.2.7
Example 6 (k = 3)
Find the solution of an = 6an1 11an2 + 6an3
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 = 1 1n + 2 2n + 3 3n
∵a0 = 1 + 2 + 3 = 2
a1 = 1 + 22 + 33 = 5
a2 = 1 + 42 + 93 = 15
∴an = 1 2n + 2 3n 6.2.8
1 = 1, 2 = 1, 3 = 2
驗算: a3 = 6a2 11a1+ 6a0 =47 a3= 1 23 + 2 33 =47
Thm 4.
Let an = c1an1 + c2an2 + … + ckank be a recurrence relation with c1, c2, …, ck R.
If rk c1rk1 c2rk2 … 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 )
6.2.9
6.2.10
nt
mt,mt,t,
ni
mi,mi,i
nm,m,,n
r)nα...nα(α
...
r)nα...nα(α
...
)rnα...nα(αa
t
t
i
i
1110
1110,
11
1111011
1
where i,j are constants.
(1 i t , 0 j mi1)
Example 8. Find the solution to the recurrence relation an = 3an1 3an2 an3 with initial conditionsa0 = 1, a1 = 2 and a2 = 1.
Sol :
r3 + 3r2 + 3r + 1 = 0 has a single root r0 = 1 of multiplicity three.
∴ an = (1+2n+3n2) r0n = (1+2n+3n2)(1)n
∵ a0 = 1 = 1
a1 = (1+2+3) (1) = 2
a2 = (1+2+3) = 1
∴1 = 1, 2 = 3, 3 = 2
an = (1+3n2n2) (1)n 6.2.11Exercise : 3,13,15,19
驗算: a3 = 3a2 3a1 a0 =8 a3= (1+33232)(1)3 =8
Example 8. Find the solution to the recurrence relation an = 3an1 3an2 an3 with initial conditionsa0 = 1, a1 = 2 and a2 = 1.
Sol :
r3 + 3r2 + 3r + 1 = 0 has a single root r0 = 1 of multiplicity three.
∴ an = (1+2n+3n2) r0n = (1+2n+3n2)(1)n
∵ a0 = 1 = 1
a1 = (1+2+3) (1) = 2
a2 = (1+2+3) = 1
∴1 = 1, 2 = 3, 3 = 2
an = (1+3n2n2) (1)n 6.2.11Exercise : 3,13,15,19
驗算: a3 = 3a2 3a1 a0 =8 a3= (1+33232)(1)3 =8
6.4 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)
0k
kk xa
6.4.1
Example 2. What is the generating function for the sequence 1,1,1,1,1,1 ?
Sol :
6.4.2
1
1
...1)(6
52
x
x
xxxxG (expansion ,展開式 )
(closed form)
Example 3.
Let mZ+ and ,for k = 0, 1, …, m.
What is the generating function for the sequence a
0, a1,…, am ?
Sol :
G(x) = a0 + a1x + a2x2 + … + amxm
= (1+x)m (by 二項式定理 )
6.4.3
k
mak
mxm
mx
mx
mm
2
210
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
ax1
1
6.4.4
ax1
1
0
)(k
kax
Def 2.
Let uR and kZ+∪{0}. Then the extended
binomial coefficient is defined by
k
u
0 if 1
0 if ),1)...(2)(1(!
1
k
kkuuuuk
k
u
( 跳過 )
Example 7.
Find and
Sol :
321
6.4.5
3
2
4)4)(3)(2(!3
1
3
2
16
1
2
3
2
1
2
1
!3
1
321
( 跳過 )
Thm 2. (The Extended Binomial Theorem)
Let xR with |x|<1 and let uR, then
0
1k
kuk
u xx
6.4.6
( 跳過 )
Example 9. Find the generating functions for (1+x)n and (1x)n where nZ+
Sol : By the Extended Binomial Theorem,
0
0
0 0
1 )1(
)1( )1( )( !
)1(
)1( )1)(( !
1)1(
k
kk
k
k
k
k
k k
kn
xk
kn
xkn...nnk
xkn...nnk
xk
nx
6.4.7
By replacing x by –x we have
k
k
n xk
knx
1)1(
0
k
kn
k
n k 1)1(:Note
( 跳過 )
※Using Generating Functions to solve Recurrence Relations.
Example 16.
Solving the recurrence relation ak = 3ak1 for k=1,2,3,… and initial condition a0 = 2.
Sol :
另法: (by 6.2 公式 )
r – 3 = 0 r = 3 an = 3n
∵ a0 = 2 = ∴ an = 2 3n
6.4.8
Let be the
generating function for {ak}.
First note that ak xk = 3ak1 xk
G(x) a0 = 3x G(x)
∵a0 = 2 G(x) 3x G(x) = G(x)(13x) = 2
0
2210 ...)(
k
kk xaxaxaaxG
1 1 1 0
111 333
k k k k
kk
kk
kk
kk xaxxaxxaxa
0 0
32)3(231
2)(
k k
kkk xxx
xG
∴ ak = 2 3k
6.4.9Exercise : 5,7,11,33
6.5 Inclusion-Exclusion 排容原理A,B,C,D : sets
CBACBCABACBACBA.
BABABA.
)2(
)1(
6.5.1
1
1 12
2 23
A
B C
|A|+|B|+|C| 時各部分被計算的次數
?)3( DCBA.
11
1
1
20
+|ABC| 後
-|AB|-|AC|-|BC| 後
Theorem 1.
A1, A2, …, An : sets
nn
nkjikji
njiji
n
iin
AAAAAA
AAAAAA
...)1(...
...
21,,1
1121
Exercise : 176.5.2
6.6 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
6.6.1
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 )= 63
363
263
16 0123
Thm 1. |A| = m , |B| = n
There are
onto functions f : A → B.
mn
nn
mnmnmnm nnnn
1)1(
...)3()2(1
11
321
pf : A = {a1, a2, …, am}. B = {b1, b2, …, bn}
f (a1)= f (a2)= ︰ ︰
f (am)=
b1, b2, …, bn
6.6.2
※Derangements 亂序Def.
A derangement is a permutation of
objects that leaves no object in its
original position.
6.6.3
( 跳過 )
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 )
=
Example 5. derangements of 12345 :
Let Dn denote the number of derangements of n objects.
!04
4!1
3
4!2
2
4!3
1
4!4
6.6.4
21453, 23451, 34512, …
Def.
( 跳過 )
Theorem 2. ( 亂序公式 )
Exercise : 86.6.5
) !
1 )1( ...
!2
1
!1
1 1 ( !
)!
!0
!0!
! )1(
... !
)!2(
)!2(!2
!
!
)!1(
)!1(!1
! 1( !
!0 )1( ... )!2( )!1( ! 21
nn
nn
n
n
n
n
n
n
n
n
nn
nnnD
n
n
nn
nnnn
參考: 12,13
( 跳過 )