Introduction to Discrete Mathematics

145 Pages·2011·0.994 MB·English

by  Dragan Masulovic

Dragan Mašulovic´
Introduction to
Discrete Mathematics
Linz,October2010
Chapter 1
Words and Sets
Thischapterconfrontsuswiththemostbasicabstractstructures:
words(orstrings),whichrepresentthesimplestordered structures,and
•
sets(orcollections),whichrepresentthesimplestunordered structures.
•
Asweshallsee,apermutationisnothingbutawordoveranappropriatelychosen
alphabet, while a combination is just a subset of a finite set. It is natural to ask
whyshouldoneinventsocomplicatednamesforsuchsimpleobjects. Theanswer
is simple. In the dark past of Discrete Mathematics the terminology used to be as
obscureastheagesthatgavebirthtoit. Sincetheintroductionofthenamessuchas
permutationandcombinationmathematicshasgonealongwayandbroughtmany
simplifications,bothinterminologyandunderstandingofthephenomena.
Throughoutthecourseweshallusethefollowingnotation
N= 1,2,3,... forthesetofpositiveintegers,
{ }
N = 0,1,2,3,... forthesetofnonnegativeintegers,and
0
{ }
N¥ = 0,1,2,3,... ¥ .
0
{ }∪{ }
ThesetN¥ isausualextensionofN withthegreatestelement¥ : x+¥ =¥ +x=
0 0
x ¥ = ¥ x = ¥ for all x N¥ , and x < ¥ for all x N . Also, we define the
· · ∈ 0 ∈ 0
factorialofanintegern N asusual:
0
∈
0!=1
n!=1 2 ... n, forn>1.
· · ·
1
2 CHAPTER1. WORDSANDSETS
1.1 Words
Analphabetisanyfinitenonemptyset. ElementsofanalphabetAwillbereferred
toasletters,andawordinAisastringofsymbolsfromA. Moreprecisely,aword
oflengthkoveranalphabetAisanytuplefromAk. Wefollowasimpleconvention
toomitcommasandparentheseswhenwritingwords.
Example1.1 HerearesomewordsoveranalphabetA= a,b,n : banana,abba,
{ }
aa,orsimplyn. Thefirstofthewordshassixletters,thencomesafour-letterword,
atwo-letterwordandfinalyawordwithonlyoneletter.
We also alow words with no letters. On any alphabet there is precisely one
such word called the empty word and denoted by e . It is a word with length 0. It
is important to note that words we deal with in this course are formal words, that
is,stringsofsymbolstowhichnomeaningisattached. So,fromthispointofview
nbbaaaisjustasgoodawordasbanana. Weshallleavethemeaningofwordsto
otherbranchesofscienceandtreatwordsjustasplainandsimplestringsofletters.
Let w be a word over an alphabet A. The length of w will be denoted by w .
| |
Foralettera A,by w wedenotethenumberofoccurencesofainw.
a
∈ | |
Example1.2 Let A = a,b,c,n and let w = banana be a word over A. Then
{ }
w =6, w =3, w =1, w =0and w =2.
a b c n
| | | | | | | | | |
Thereisnotmuchstructuraltheorybehindsuchsimpleobjectsaswords. The
mostexcitingthingwecandoatthemomentistotrytocountthem.
Problem1.3 LetA = a ,a ,...,a beanalphabetwithn > 1 lettersandlet
1 2 n
{ }
k N bearbitrary.
0
∈
(a)HowmanywordswithklettersoverAarethere?
(b)HowmanywordswithklettersoverAhavethepropertythatalltheletters
inthewordaredistinct?
(c)HowmanywordsoverAhavethepropertythateveryletterfromAappears
preciselyonceintheword?
Solution. (a)ThesetofallwordsoflengthkoverAisjustAk. Therefore,thereare
precisely Ak = A ... A =nk such words. This is an instance of an important
| | | |· ·| |
k
combinatorialprinciple:
| {z }
TheProductPrinciple:IfA ,...,A arefinitesets,then
1 n
A ... A = A ... A .
1 n 1 n
| × × | | |· ·| |
1.1. WORDS 3
There is a less formal, but more useful way to see this. A word with k letters
lookslikethis:
1st 2nd 3rd kth
...
There are n candidates for the first position, n candidates for the second position,
...,ncandidatesforthekthposition:
1st 2nd 3rd kth
...
n n n n
· · · ·
Alltogether,therearen n ... n=nk possibilites.
· · ·
k
(b)Letusagaintaketheinformalpointofview. Firstly,therearencandidates
| {z }
for the first position, but only n 1 candidates for the second position, since the
−
letter used on the first position is not allowed to appear on the second position.
Then, there are n 2 candidates for the third position since the two letters used
−
on the first two positions are a no-no, and so on. Finally, there will be n (k 1)
− −
candidatesforthelastposition:
1st 2nd 3rd kth
...
n n 1 n 2 n (k 1)
− − − −
n!
andputtingitalltogetherwegetn (n 1) ... (n k+1)= possibilites.
· − · · − (n k)!
Ofcourse,thisreasoningisvalidaslongask6n. Ifk>nnosu−chwordexists.
(c)IfeveryletterfromAisrequiredtoappearpreciselyonceintheword,then
the length of the word is n and all the letters have to be distinct. This is a special
caseof(b)wherek=nandtherearen!suchwords. (cid:3)
Words where letters are not allowed to repeat are called permutations of sets.
Words whereletters can appearmorethan onceconstitute anotherkindofpermu-
tations — permutations of multisets — and we shall consider them in a separate
section.
Definition1.4 A permutation of a set A is a word over A where every letter from
thealphabetappearspreciselyonceintheword. Ak-permutationofasetA,where
k6 A ,isawordoverAoflengthkwhereeachletterfromthealphabetisallowed
| |
toappearatmostonce(andtherefore,allthelettersinthewordaredistinct).
4 CHAPTER1. WORDSANDSETS
Weshallnowapplycountingtechniquesdiscussedabovetodeterminethenum-
berofallthesubsetsofafiniteset. ForasetAletP(A)denotethepower-setofA,
thatis,thesetofallthesubsetsofA:
P(A)= X :X A .
{ ⊆ }
Let A =nandA= a ,...,a . TheneverysubsetBofAcanberepresentedby
1 n
| | { }
astringc (B)of0’sand1’sasfollows:
0, a / B,
c (B)= p ...p , where p = i∈
1 n i
(1, ai B.
∈
Thewordc (B)iscalledthecharacteristicvectorofB. Wordsoverthetwo-element
alphabet 0,1 willbeparticularlyusefulinthesequel. So,weshallrefertothem
{ }
as01-words.
Example1.5 Let A = a,b,c,d,e,f and B = b,d,e . Then c (B) = 010110
{ } { }
sincea / B,b B,c / Betc. Clearly, c (∅)=000000andc (A)=111111:
∈ ∈ ∈
a b c d e f
∅ 0 0 0 0 0 0
B 0 1 0 1 1 0
A 1 1 1 1 1 1
Theorem1.6 LetAbeafinitesetwithnelements.Then P(A) =2n.
| |
Proof. The mapping c :P(A) 0,1 n that takes a subset of A onto its char-
→{ }
acteristic vector is a bijection, so P(A) and 0,1 n have the same number of
| | |{ } |
elements. We shall use this obvious but important fact on many occasions in the
course:
TheBijectionPrinciple: Wheneverthereisabijectionbetweentwosets,
theyhavethesamenumberofelements.
Therefore, P(A) equals the number of all words over 0,1 whose length
| | { }
isn,so P(A) =2n. (cid:3)
| |
ThefollowingprincipleisasortofanegationoftheBijectionPrinciple:
The Pigeon-Hole Principle: SupposethatA andB arenonemptyfinite
setssuchthat A > B.ThenthereisnoinjectivemappingfromAintoB.
| | | |
The name comes from a simple observation concerning pigeons and holes: if
n+1pigeonshideinnholes,thenthereisatleastoneholewithatleasttwopigeons
init.
1.2. SETS 5
Theorem1.7(Erdo˝s,Szekeres1935) Everysequenceofk=mn+1distinctreal
numbershasanincreasingsubsequenceoflengthm+1 oradecreasingsubse-
quenceoflengthn+1.
Proof. Let a , ..., a be a sequence of k =mn+1 distinct real numbers and as-
1 k
sumethatithasneitheranincreasingsubsequenceoflengthm+1noradecreasing
subsequence of length n+1. Then every increasing subsequence of a , ..., a
1 k
is of length 6m and, similarly, every decreasing subsequence of a , ..., a is of
1 k
length6n.
For each i let l+ denote the length of the longest increasing subsequence of
i
a , ..., a that starts with a and let l denote the length of the longest decreas-
1 k i i−
ing subsequence of a , ..., a that starts with a. This establishes a mapping
1 k i
f : 1,...,k 1,...,m 1,...,n : i (l+,l ). Let us show that f is in-
{ }→ { }×{ } 7→ i i−
jective. Take any pair of indices i= j. Then a =a . If a <a then l+ >l+ so
6 i 6 j i j i j
f(i)=(l+,l )=(l+,l )= f(j). Similarly, if a >a then l >l and we again
i i− 6 j −j i j i− −j
conclude f(i)= f(j). This shows that i= j implies f(i)= f(j) and thus f is an
6 6 6
injectivemapfromak-elementsetintoandmn-elementset. Butk>mnandhence
bythePigeon-HolePrinciplenosuchinjectivemapcanexist. Contradiction. (cid:3)
1.2 Sets
Forhistoricalreasons,ak-elementsubsetofann-elementsetiscalledak-combination
ofaset. Thenumberofk-combinationsofann-elementsetisdenotedby
n
[read: “nchoosek”].
k
(cid:18) (cid:19)
Thepronounciationcomesfromthefactthatthisisthenumberofwaystochoose
kobjectsfromapoolofnidenticalobjects. Ifwelet
P (A)= B P(A): B =k
k
{ ∈ | | }
bethesetofallk-subsetsofAandif A =n,then,clearly,
| |
n
= P (A) .
k
k | |
(cid:18) (cid:19)
n n! n
Theorem1.8 Letn,k>0.Ifn>kthen = .Otherwise, =0.
k k!(n k)! k
(cid:18) (cid:19) − (cid:18) (cid:19)
6 CHAPTER1. WORDSANDSETS
Proof. Let n>k and let A= 1,2,...,n . Although sets seem to be simpler than
{ }
wordsduetothelackofstructure,orderedstructures(wordsinthiscase)arealways
easiertocount. LetW (A)bethesetofallk-permutationsofAandlet f :W (A)
k k
→
P (A)bethemappingdefinedby
k
f(a a ...a )= a ,a ,...,a .
1 2 k 1 2 k
{ }
Since f mapsk!differentwordsfromW (A)ontothesameelementofP (A),e.g.
k k
abcd bacd cabd dabc
abdc badc cadb dacb

acbd bcad cbad dbac
 a,b,c,d ,
acdb bcda cbda dbca 7→{ }

adbc bdac cdab dcab 
adcb bdca cdba dcba





weeasilyconcludethat 
1
P (A) = W (A) .
k k
| | k!·| |
We know that the number of k-permutations of an n-element set is n! , so we
(n k)!
finalyobtainthat −
n 1 n!
= P (A) = .
k
k | | k!·(n k)!
(cid:18) (cid:19) −
n
On the other hand, if k>n then trivially =0 since an n-element set cannot
k
(cid:18) (cid:19)
haveasubsetwithmorethannelements. (cid:3)
Problem1.9 Howmany01-wordsoflengthm+narethereiftheyarerequiredto
havepreciselymzerosandpreciselynones?
Solution. ConsiderasetA= a ,a ,...,a withm+nelements. Theneach01-
1 2 m+n
{ }
wordoflengthm+nwithmzerosandnonescorrespondstoann-elementsubset
of A. Therefore, the number of such 01-words equals the number of n-element
subsetsofA,whichis
m+n
.
n
(cid:18) (cid:19)
Hereistheotherwaytoseethis. Considerastringofm+nemptyboxeswhichare
tobefilledbymzerosandnones:
1st 2nd 3rd (m+n)th
...
1.2. SETS 7
m+n
We can choose m boxes in which to write zeros in ways. Then the re-
m
(cid:18) (cid:19)
mainingnboxeshavetobefilledbyones. (cid:3)
n n
Theorem1.10 (a) = foralln>k>0;
k n k
(cid:18) (cid:19) (cid:18) − (cid:19)
n n 1 n 1
(b) = − + − foralln>k>1(Pascal’sidentity).
k k 1 k
(cid:18) (cid:19) (cid:18) − (cid:19) (cid:18) (cid:19)
Proof. (a)Thisfollowsbyaneasycalculation:
n n! n! n
= = = .
n k (n k)!(n (n k))! (n k)!k! k
(cid:18) − (cid:19) − − − − (cid:18) (cid:19)
Suchproofsareusuallycalledalgebraicproofs.
Most combinatorial identities can be proven in another way: we find an ap-
propriatecollectionofobjectsandthencounttheelementsofthecollectionintwo
different ways. The resulting expressions have to be equal because the collection
is the same. Such proofs are usually called combinatorial proofs. The principle
behindtheisapproachiscalledDoubleCounting:
Double Counting: Ifthesamesetiscountedintwodifferentways,the
answersarethesame.
Let us provide a combinatorial proof of the same identity. Consider 01-words
n
of length n with precisely k zeros. There are ways to choose k places out
k
(cid:18) (cid:19)
of n in which to write zeros, so the number of the words under consideration is
n
. On the other hand, we can first choose n k places in which to write ones
k −
(cid:18) (cid:19)
n n
in ways, so the number of the words under consideration is .
n k n k
(cid:18) − (cid:19) (cid:18) − (cid:19)
n n
Therefore, = .
k n k
(cid:18) (cid:19) (cid:18) − (cid:19)
(b)ThealgebraicproofofthePascal’sidentityiseasy:
n 1 n 1 (n 1)! (n 1)!
− + − = − + −
k 1 k (k 1)!(n k)! k!(n k 1)!
(cid:18) − (cid:19) (cid:18) (cid:19) − − − −
(n 1)! 1 1
= − +
(k 1)!(n k 1)! n k k
− − − (cid:18) − (cid:19)
(n 1)! n n
= − =
(k 1)!(n k 1)!·k(n k) k
− − − − (cid:18) (cid:19)
8 CHAPTER1. WORDSANDSETS
Thecombinatorialproofusesanotherimportantcombinatorialprinciple:
TheSumPrinciple:IfA ,...,A aremutuallydisjointfinitesets,then
1 n
A ... A = A +...+ A .
1 n 1 n
| ∪ ∪ | | | | |
Let S = 1,2,...,n be an n-element set. Clearly, the number of k-element
{ }
n
subsetsis . Ontheotherhand,allk-elementsubsetsofSsplitintotwoclasses:
k
(cid:18) (cid:19)
those that contain 1, and those that do not. The number of k-element subsets of S
n 1
thatcontain1is − sincewehavetochoosek 1elementsfroman(n 1)-
k 1 − −
(cid:18) − (cid:19)
elementsetS = 2,...,n . Thenumberofk-elementsubsetsofSthatdonotcon-
′
{ }
n
tain1 is sincenow we haveto choose all k elements fromS. Therefore,
′
k 1
(cid:18) − (cid:19)
n n 1 n 1
bytheSumPrinciple, = − + − . (cid:3)
k k 1 k
(cid:18) (cid:19) (cid:18) − (cid:19) (cid:18) (cid:19)
Due to the following important result the numbers nk are often referred to as
binomialcoefficients:
Theorem1.11(Newton’sBinomialFormula) Foralln N wehave
0
∈
(a+b)n= (cid:229) n n an kbk.
−
k
k=0(cid:18) (cid:19)
Proof. Theproofproceedsbyinductiononn. Thefirstfewcasesaretrivial:
0
(a+b)0=1=
0
(cid:18) (cid:19)
1 1
(a+b)1=a+b= a+ b
0 1
(cid:18) (cid:19) (cid:18) (cid:19)
2 2 2
(a+b)2=a2+2ab+b2= a2+ ab+ b2
0 1 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
Assumethattheclaimistruefornandletuscompute(a+b)n+1. Bytheinduction
hypothesis:
(a+b)n+1=(a+b) (a+b)n=(a+b) (cid:229) n n an kbk.
−
· · k
k=0(cid:18) (cid:19)
Afterdistributingthesumandmultiplyingweobtain:
(a+b)n+1= (cid:229) n n an k+1bk+(cid:229) n n an kbk+1.
− −
k k
k=0(cid:18) (cid:19) k=0(cid:18) (cid:19)
1.2. SETS 9
Next, we take out the first summand in the first sum and the last summand in the
secondsumtoobtain:
(a+b)n+1=an+1+(cid:229) n n an k+1bk+n(cid:229)−1 n an kbk+1+bn+1
− −
k k
k=1(cid:18) (cid:19) k=0(cid:18) (cid:19)
andreindexthesecondsum,whichisastandardtrick:
(a+b)n+1=an+1+(cid:229) n n an k+1bk+ (cid:229) n n an m+1bm+bn+1.
− −
k m 1
k=1(cid:18) (cid:19) m=1(cid:18) − (cid:19)
Puttingthetwosumstogetherweobtain:
(a+b)n+1=an+1+(cid:229) n n + n an k+1bk+bn+1.
−
k k 1
k=1(cid:18)(cid:18) (cid:19) (cid:18) − (cid:19)(cid:19)
Finally,weapplythePascal’sidentityandwrapitup:
(a+b)n+1=an+1+(cid:229) n n+1 an k+1bk+bn+1=n(cid:229)+1 n+1 an k+1bk.
− −
k k
k=1(cid:18) (cid:19) k=0(cid:18) (cid:19)
ThecombinatorialproofoftheNewton’sBinomialFormulaisbasedonasim-
pleobservation. Clearly,
(a+b)n=(a+b) (a+b) ... (a+b)
· · ·
ntimes
so if one multiplies out and write|s down the s{uzmmands as w}ords of length n (that
is, without the usuall abbreviations such as a a a=a3), one obtains all possible
· ·
wordsodlengthninlettersaandb. Forexample,
(a+b)4=aaaa+aaab+aaba+aabb+abaa+abab+abba+abbb
+baaa+baab+baba+babb+bbaa+bbab+bbba+bbbb.
n
There are words that abbreviate to an kbk since this is the number of ways
−
k
(cid:18) (cid:19)
n
wecanchoosekplacesforb(Problem1.9). Therefore,an kbk appears times
−
k
(cid:18) (cid:19)
inthesum,whence(a+b)n= (cid:229) n n an kbk. (cid:3)
−
k
k=0(cid:18) (cid:19)