Adminstrivia Mathematical Preliminaries Countable Sets Uncountable sets. Why Theory of Computation?

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1 CD0 FBER Forml Lnguges utomt nd Models of Computtion Leture Mälrdlen niversity 00 Content dminstrivi Mthemtil Preliminries Countle ets nountle sets Leturer & Exminer Gordn Dodig-Crnkovi Tehing ssistnts & L upervisors Mrkus Bohlin & Lrs Brue Course Home Pge Why Theory of Computtion? kurser/d0/0_0 visit home pge regulrly!. rel omputer n e modeled y mthemtil ojet: theoretil omputer.. forml lnguge is set of strings nd n represent omputtionl prolem.. forml lnguge n e desried in mny different wys tht ultimtely prove to e identil.. imultion: the reltive power of omputing models n e sed on the ese with whih one model n simulte nother.. Roustness of omputtionl model.. The Churh-Turing thesis: nything tht n e omputed n e omputed y Turing mhine.. Nondeterminism: lnguges n e desried y the existene or nonexistene of omputtionl pths. 8. nsolvility: for some omputtionl prolems there is no orresponding lgorithm tht will unerringly solve them. Prtil pplitions. Effiient ompiltion of omputer lnguges. tring serhing. Identifying the limits; Reognizing diffiult prolems. pplitions to other res: iruit verifition eonomis nd gme theory (finite utomt s strtegy models in deision-mking); theoretil iology (L-systems s models of orgnism growth) omputer grphis (L-systems) linguistis (modeling y grmmrs) History Eulid's ttempt to xiomtize geometry (rhimedes relized during his own efforts to define the re of plnr figure tht Eulid's ttempt hd filed nd tht dditionl postultes were needed. ) Leiniz's drem of symoli logi de Morgn Boole Frege Russell Whitehed: Mthemtis s rnh of symoli logi! 900 Hilerts progrm first progrmming lnguges 9 Gödels inompleteness theorem 9 Turing mshine (showed to e equivlent with reursive funtions). Commonly epted: TM s ultimte omputer 90 utomt 9 lnguge/utomt hierrhy 8 9

2 every mthemtil truth expressed in forml lnguge onsisting of fixed lphet of dmissile symols nd expliit rules of syntx for omining those symols into meningful words nd sentenes Turing used niversl Turing mhine (TM) to prove n even more powerful inompleteness theorem euse it destroyed not one ut two of Hilert's drems:. finding finite list of xioms from whih ll mthemtil truths n e dedued. olving the entsheidungsprolem ("deision prolem ) y produing "fully utomti proedure" for deiding whether given proposition (sentene) is true or flse. Mthemtil Preliminries 0 ET et Representtions ets Funtions Reltions Grphs Tehniques set is olletion of elements = {} B = { trin us iyle irplne} We write ship B C = { d e f g h i j k } C = { k } finite set = { } infinite set = { j : j > 0 nd j = k for some k>0 } = { j : j is nonnegtive nd even } = { } et Opertions Complement 0 niversl et: ll possile elements = { 0 } 9 8 = { } B = { } nion B = { } Intersetion B B = { } Differene - B = { } B - = { } -B niversl set = { } = { } = { } = 8

3 { even integers } = { odd integers } Integers DeMorgn s Lws Empty Null et: = { } odd even 0 B = B B = B = = - = = niversl et - = 9 0 uset = { } B = { } Proper uset: B B B Disjoint ets = { } B = { } B = B et Crdinlity For finite sets = { } = Powersets Crtesin Produt FNCTION powerset is set of sets = { } Powerset of = the set of ll the susets of = { } B = { } X B = { ( ) ( ) ( ) ( ) ( ) ( ) } X B = B domin f() = rnge B = { {} {} {} { } { } { } { } } Oservtion: = ( 8 = ) Generlizes to more thn two sets X B X X Z If = domin f : -> B then f is totl funtion otherwise f is prtil funtion

4 RELTION R = {(x y ) (x y ) (x y ) } x i R y i e. g. if R = > : > > > In reltions x i n e repeted 8 Equivlene Reltions Reflexive: x R x ymmetri: x R y y R x Trnsitive: x R Y nd y R z x R z Exmple R = = x = x x = y y = x x = y nd y = z x = z 9 Equivlene Clsses For equivlene reltion R equivlene lss of x = {y : x R y} Exmple: R = { ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) } Equivlene lss of = { } Equivlene lss of = { } 0 GRPH direted grph e node edge d Nodes (Verties) V = { d e } Edges E = { ( ) ( ) ( ) ( d) (d ) (e d) } Wlk e d Wlk is sequene of djent edges (e d) (d ) ( ) Pth e d Pth is wlk where no edge is repeted imple pth: no node is repeted Cyle se e d Cyle: wlk from node (se) to itself imple yle: only the se node is repeted Euler Tour 8 se e d yle tht ontins eh edge one Hmiltonin Cyle se e d simple yle tht ontins ll nodes

5 root Trees root Level 0 Binry Trees prent Level lef hild lef Level Height Trees hve no yles Level 8 9 PROOF TECHNIQE y onstrution y indution y ontrdition 0 Constrution We define grph to e k-regulr if every node in the grph hs degree k. Theorem. For eh even numer n > there exists -regulr grph with n nodes. y Constrution Construt grph G = (V E) with n > nodes. V= { 0 n- } E = { {i i+} for 0 i n-} {{n-0}} (*) {{i i+n/ for 0 i n/ } (**) The nodes of this grph n e written onseutively round the irle. 0 (*) edges etween djent pirs of nodes 0 (**) edges etween nodes on opposite sides n = n = END OF PROOF Indution We hve sttements P P P y Indution Indutive sis Find P P P k whih re true Exmple Theorem inry tree of height n hs t most n leves. If we know for some k tht P P P k re true for ny n k tht P P P n imply P n+ Then Every P i is true Indutive hypothesis let L(i) e the numer of leves t level i Let s ssume P P P n re true for ny n k L(0) = Indutive step L() = 8 how tht P n+ is true

6 We wnt to show: L(i) i Indution tep Indutive sis Indution tep L(0) = (the root node) Level n hypothesis: L(n) n Indutive hypothesis Let s ssume L(i) i for ll i = 0 n Level n+ n hypothesis: L(n) n n+ Indution step L(n+) * L(n) * n = n+ we need to show tht L(n + ) n+ END OF PROOF 8 Indutionsevis: Potensmängdens krdinlitet Påstående En mängd med n element hr n delmängder Kontroll Tomm mängden {} (med noll element) hr r en delmängd: {}. Mängden {} (med ett element) hr två delmängder: {} oh {} Mängden { } (med två element) hr fyr delmängder: {} {} {} oh {} Mängden { } (med tre element) hr ått delmängder: {} {} {} {} oh {} {} {} {} Påstående stämmer så här långt. Bssteg Enklste fllet är en mängd med noll element (det finns r en sådn) som hr 0 = delmängder. Induktionssteg ntg tt påståendet gäller för ll mängder med k element dvs ntg tt vrje mängd med k element hr k delmängder. Vis tt påståendet I så fll okså gäller för ll mängder med k+ element dvs vis tt vrje mängd med k+ element hr k+ delmängder. Vi etrktr en godtyklig mängd med k+ element. Delmängdern till mängden kn dels upp i två sorter: Delmängder som inte innehåller element nr k+: En sådn delmängd är en delmängd till mängden med de k först elementen oh delmängder till en mängd med k element finns det (enligt ntgndet) k styken. Delmängder som innehåller element nr k+: En sådn delmängd kn mn skp genom tt t en delmängd som inte innehåller element nr k+ oh lägg till dett element. Eftersom det finns k delmängder utn element nr p+ kn mn även skp k delmängder med dett element. Totlt hr mn k + k =. k = k+ delmängder till den etrktde mängden. END OF PROOF (Exempel från oken: Diskret mtemtik oh diskret modeller K Eriksson H. Gvel) 9 0 y Contrdition Exmple = n/m m = n We wnt to prove tht sttement P is true Theorem is not rtionl Therefore n is even n is even n = k we ssume tht P is flse then we rrive t onlusion tht ontrdits our ssumptions therefore sttement P must e true ssume y ontrdition tht it is rtionl = n/m n nd m hve no ommon ftors We will show tht this is impossile m is even m = k m = k m = p Thus m nd n hve ommon ftor Contrdition! END OF PROOF

7 Countle ets Infinite sets re either Countle or nountle Countle set There is one to one orrespondene etween elements of the set nd nturl numers We strted with the nturl numers then dd infinitely mny negtive whole numers to get the integers then dd infinitely mny rtionl frtions to get the rtionls then dded infinitely mny irrtionl frtions to get the rels. Eh infinite ddition seem to inrese rdinlity: N < Z < Q < R But is this true? NO! Exmple The set of integers is ountle Integers: 0 K Correspondene: Nturl numers: 0 K f ( n) = { n / n even;( n + ) / n odd Exmple Positive Rtionl numers: The set of rtionl numers is ountle 8 K Nive Ide Rtionl numers: Correspondene: Nturl numers: Doesn t work! we will never ount numers with nomintor : K K K Better pproh Rows: onstnt numertor (täljre) Columns: onstnt denomintor

8 We proved: the set of rtionl numers is ountle y desriing n enumertion proedure Definition Let e set of strings n enumertion proedure for is n lgorithm tht genertes ll strings of one y one Oservtion set is ountle if there is n enumertion proedure for it Exmple The set of ll finite strings is ountle { } We will desrie the enumertion proedure + Nive proedure: Produe the strings in lexiogrphi order: Doesn t work! trings strting with will never e produed Better proedure Proper Order. Produe ll strings of length. Produe ll strings of length. Produe ll strings of length. Produe ll strings of length. 8 9 Produe strings in Proper Order length length length 0 Theorem The set of ll finite strings is ountle ny finite string n e enoded with inry string of 0 s nd s Find n enumertion proedure for the set of finite strings Produe strings in Proper Order length length length tring = progrm Nturl numer 0.

9 PROGRM = TRING (syntti wy) PROGRM = FNCTION Ν Ν (semnti wy) string PROGRM string nountle ets Definition set is unountle if it is not ountle nturl numer n Ν PROGRM nturl numer n Ν Theorem The set of ll infinite strings is unountle (y ontrdition) We ssume we hve n enumertion proedure for the set of infinite strings Cntor s digonl rgument Infinite string Enoding w = w = w = 0 0 Cntor s digonl rgument We n onstrut new string w tht is missing in our enumertion! Conlusion The set of ll infinite strings is unountle! 8 n infinite string n e seen s FNCTION Ν Ν (n:th output is n:th it in the string) Conlusion There re some integer funtions tht tht nnot e desried y finite strings (progrms/lgorithms). Exmple of unountle infinite sets Theorem Let e n infinite ountle set The powerset of is unountle ine is ountle we n write = { s s s K}

10 Elements of the powerset hve the form: We enode eh element of the power set with inry string of 0 s nd s { s s} { s s s9 s0} Powerset element { s } { s s} Enoding s s s s Let s ssume (for ontrdition) tht the powerset is ountle. Then: we n enumerte the elements of the powerset { s s s} Powerset element Enoding p p p Tke the powerset element whose its re the omplements in the digonl p p p 0 0 p 0 0 p 0 0 New element: 00K 8 8 (inry omplement of digonl) 8 The new element must e some of the powerset However tht s impossile: from definition of the i-th it of must e the omplement of itself Contrdition! p i p i p i 88 ine we hve ontrdition: The powerset of is unountle n pplition: Lnguges Exmple lphet : { } The set of ll finite strings: * = { } = { λ K} infinite nd ountle The powerset of ontins ll lnguges: = {{ λ}{ }{ }{ } K} L L L L L unountle infinite END OF PROOF 89 90

11 Finite strings (lgorithms): ountle Lnguges (power set of strings): unountle There re infinitely mny more lnguges thn finite strings. Conlusion There re some lnguges tht nnot e desried y finite strings (lgorithms). Krdinltl Krdinltl är mått på storleken v mängder. Krdinltlet för en ändlig mängd är helt enkelt ntlet element i mängden. Två mängder är lik mäktig om mn kn pr ihop elementen i den en mängden med elementen i den ndr på ett uttömmnde sätt dvs det finns en ijektion melln dem. Dett mäktighetstänknde kn utvidgs till oändlig mängder. Till exempel är mängden v positiv heltl oh mängden v heltl lik mäktig. Däremot kn mn inte pr ihop ll reell tl med heltlen på dett sätt. Mängden v reell tl hr större mäktighet än mängden v heltl. Mn kn inför krdinltl på ett sådnt sätt tt två mängder hr smm krdinltl om oh endst om de hr smm mäktighet. T ex klls krdinltlet som hör till de hel tlen för ℵ 0 (lef 0 lef är den först okstven i det hereisk lfetet). Dess oändlig krdinltl klls trnsfinit krdinltl Mer om oändligheter Georg Cntor utveklde i slutet v 800-tlet mtemtikens logisk grund mängdlärn. Cntor införde egreppet trnsfinit krdinltl. Den enklste "minst" oändligheten kllde hn ℵ 0. Det är den uppräkningsr oändlig mängdens (exempelvis mängden v ll heltl) krdinltlet. Krdinltlet v mängden punkter på en linje oh även punktern på ett pln oh i en kropp kllde Cntor ℵ. Fnns det större oändligheter? J! Cntor kunde vis tt ntlet funktioner på en linje vr ännu oändligre än punktern på linjen oh hn kllde den mängden ℵ. Cntor fnn tt det gik tt räkn med krdinltlen preis som med vnlig tl men räknereglern lev något enhnd.. ℵ 0 + = ℵ 0 ℵ 0 + ℵ 0 = ℵ 0 ℵ 0 ℵ 0 = ℵ 0. Men vid exponering hände det något: ℵ 0 ℵ0 (ℵ 0 upphöjt till ℵ 0 ) = ℵ. Mer generellt visde det sig tt ℵn ( upphöjt till ℵ n ) = ℵ n+ Det inner tt det fnns oändligt mång oändligheter den en mäktigre än den ndr! Men vr det verkligen säkert tt det inte fnns någon oändlighet melln den uppräkningsr oh punktern på linjen? Cntor försökte evis den så kllde kontinuumhypotesen. Cntor: two different infinities ℵ 0 nd ℵ Continuum Hypothesis: ℵ 0 < ℵ = ℵ0 e även: 9