Induction Proofs. ) ( for all n greater than or equal to n. is a fixed integer. A proof by Mathematical Induction contains two steps:

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1 CMPS Algorthms ad Abstract Data Types Iducto Proofs Let P ( be a propostoal fucto, e P s a fucto whose doma s (some subset of) the set of tegers ad whose codoma s the set {True, False} Iformally, ths meas P ( s a setece, statemet, or asserto whose truth or falsty depeds o the teger Mathematcal Iducto s a proof techque whch ca be used to prove statemets of the form : P( ) ( for all greater tha or equal to, P( s true ), where s a fxed teger A proof by Mathematcal Iducto cotas two steps: ( : ( P( I Base Step: Prove drectly that the proposto P ) s true IIa Iducto Step: Prove )) To do ths pck a arbtrary, ad assume for ths that P ( s true The show as a cosequece that P ( ) s true The statemet P ( s ofte called the ducto hypothess, sce t s what s assumed the ducto step Whe I ad II are complete we coclude that P ( s true for all Iducto s sometmes explaed terms of a domo aalogy Cosder a fte set of domos whch are led up ad ready to fall Each domo s labeled by a postve teger, startg wth (It s ofte the case that, whch we assume here for the sake of defteess) Let P ( be the asserto: the th domo falls Frst prove P (), e the frst domo falls, the prove : ( P( )) whch says f ay partcular domo falls, the the ext domo must also fall Whe ths s doe we may coclude : P(, all domos fall There are a umber of varatos o the ducto step The frst s just a reparametrzato of IIa IIb Iducto Step: Prove > : ( P( ) )) Let >, assume P ( ) s true, the prove P ( s true Forms IIa ad IIb are sad to be based o the frst prcple of mathematcal ducto The valdty of ths prcple s proved the appedx of ths hadout Aother mportat varato s called the secod prcple of mathematcal ducto, or strog ducto IIc Iducto Step: Prove : (( k : P( ) )) Let, assume for all k the rage k that P ( s true The prove as a cosequece that P ( ) s true I ths case the term ducto hypothess refers to the stroger assumpto: k : P( The strog ducto form s ofte reparametrzed as IIb:

2 IId Iducto Step: Prove > : (( k < : P( ) )) Let >, assume for all k the rage k <, that P ( s true, the prove as a cosequece that P ( s true I ths case the ducto hypothess s k < : P( I terms of the Domo aalogy, the strog ducto form IId says we must show: (I) the frst domo th th falls, ad (II) for ay, f all domos up to but ot cludg the domo fall, the the domo falls From (I) ad (II) we may coclude that all domos fall Strog Iducto s most ofte parameterzed as IId, ad form IIc s ucommo We preset here a umber of examples of IIa, IIb, ad IId Example Prove that for all : ( )( ) Let P ( be the boxed equato above We beg the ducto at I ( ) ( ) Base step Clearly, showg that P () s true IIa Iducto Step Let ad assume P ( s true That s, for ths partcular value of, the boxed equato holds The ( ) ( )( ) ( ) (by the ducto hypothess) ( )( ) ( ) ( ) [ ( ) ] [ ( ) ] (by some algebra) showg that P ( ) s true We coclude that P ( s true for all /// Whe wrtg a ducto proof, always state the ducto hypothess explctly Also make ote of the pot the proof where the ducto hypothess s used Example Let x R ad x Show that for all : x x x Here we wll use form IIb Aga let P ( be the boxed equato We beg the ducto at I Base step x x x x, showg that P () s true

3 IIb Iducto Step Let > ad assume that P ( ) s true, e assume for ths partcular that: x x x The x x x x x x (by the ducto hypothess) x x (by some algebra) showg that P ( s true Steps I ad II prove that P ( holds for all /// Exercse Prove the followg formulas usg both forms II a ad II b (a) Show that for all : (b) Show that for all : 3 ( ) 4 ( )( )(3 3 ) 3 Ofte the proposto to be proved s ot a formula, but some other type of asserto, lke a equalty, as the followg example Example 3 Defe the fucto T ( for Z by the recurrece f T ( T( / ) f Prove that for all, T ( lg(, ad therefore T ( O(lg Let P ( be the boxed equalty above I Base Step The equalty T ( ) lg() reduces to smply, whch s obvously true, so P () holds IId Iducto Step (Strog Iducto Let > ad assume for all k the rage k< that P ( s true, e T ( lg( I partcular whe / T / lg / Therefore k, we have ( ) T( / ) lg / lg( / ) (sce x T ( (by the defto of T ( ) (by the ducto hypothess) x for ay x) lg( ) lg() lg(, showg that P ( s true Therefore T ( lg( for all, as clamed /// 3

4 Exercse Defe S ( for Z by the recurrece f S ( S( / ) f Prove that for all : S( lg(, ad hece S( Ω(lg There are may other varatos o the ducto techque Occasoally double ducto s called for, whch volves a modfcato of both the base ad ducto steps Base Step: Prove P ) ad P ( ) ( Iducto Step: Prove ( ) : ( P( ) P( ) ) Whe these steps are complete, we coclude : P( I terms of our domo aalogy, we prove: (I) the frst two domos fall, ad (II) f ay two cosecutve domos fall, the the very ext domo falls, ad from (I) ad (II) we deduce that all domos fall The ext example uses double ducto ad cocers the Fboacc sequece F defed by: F, F, ad F F F, e each term the sequece s the sum of the precedg two Usg ths recurrece formula, the frst few terms of the Fboacc sequece are easly computed: F, F, F, F, F 3, F, etc 3 4 Example 4 Let a, ad b Prove that for all, F [ a b ] Let P ( deote the boxed equato above I Base Step Observe that P () ad P () are true sce [ a b ] F ad [ a b ] F II Iducto Step Let ad assume that both P ( ) ad P ( ) are true, e we assume for ths that F [ a b ] ad [ a b ] The ducto hypothess yelds F [ a ( a ) b ( b ) ] F F F Oe checks that a ad b are roots of the quadratc equato x x, whece a a, ad b b Therefore F [ a a b b ] [ a b ], showg that P ( s true Together (I) ad (II) mply that F [ a b ] for all /// 4

5 Exercse 3 Let F be the Fboacc sequece ad defe a as above Show that F a for all, ad hece F Ω( a ) (Prove ths by double ducto, ot as a cosequece of the last example) A Short Itroducto to Graphs Ofte the propostoal fucto P ( s some asserto cocerg other types of mathematcal structures, such as graphs, or trees A graph G s a par of sets G ( V, E) The elemets of V are called vertces, ad the elemets of E are called edges Each edge jos two dstct vertces, called t s eds, ad o two edges have the same eds Abstractly, a edge s a uordered par of vertces, e a - elemet subset of V Two vertces that are joed by a edge are sad to be adjacet, ad a edge s sad to be cdet wth t s two ed vertces Two edges are sad to be adjacet f they are cdet wth a commo ed vertex Thus the example below: vertex s adjacet to vertex 4, vertex s cdet wth edge, ad edge 4 s adjacet to edge V{,, 3, 4,, } E{, 4, 3, 4,,, 3, 3, 4, } Let x, y V A x-y path G s a sequece of vertces startg wth x ad edg wth y, whch each cosecutve par of vertces are adjacet We requre that all vertces other tha x ad y be dstct, ad that each edge the sequece be traversed at most oce We call x the tal vertex ad y the termal vertex If x y, the the path s called a cycle The legth of a path s the umber of edges traversed by the sequece I the above example we have: A - path of legth :,, 4,, 3, A - path of legth 3:, 4,, Aother - path of legth 3:,, 3, A cycle of legth :,,, 3,, 4, A cycle of legth 3:,,, A graph s sad to be coected f t cotas a x-y path for every x, y V, otherwse t s called dscoected The example above s clearly coected, whle the followg example s dscoected V {,, 3, 4,,, 7, 8, 9} E {,,,,, 37, 38, 78, 49} A subgraph of a graph G s a graph H whch V ( H ) V ( G), ad E( H ) E( G) I the above example ({,, }, {,, }) s a coected subgraph, whle ({, 3,, 7}, {, 37}) s a dscoected

6 subgraph A subgraph H s called a coected compoet of G f t s () coected, ad () maxmal wth respect to property (), e ay other subgraph of G that cotas H s dscoected The above example clearly has three coected compoets: ({,,, }, {,,,, }), ({3, 7, 8}, {37, 38, 78}), ad ({4, 9}, {49}) Obvously a graph s coected f ad oly f t has exactly oe coected compoet A graph G s called acyclc f t cotas o cycles A tree s a graph that s both coected ad acyclc The coected compoets of a acyclc graph are obvously trees For ths reaso a acyclc graph s sometmes called a forest The followg graph s a forest wth three coected compoets Observe that the umber of edges each tree of ths forest s oe less tha the umber of vertces Ths s true for all trees, as we ow show Example For all, f T s a tree o vertces, the T cotas edges Let P( be the boxed statemet above We beg at, ad use the strog ducto form IId I Base step If T has just oe vertex, the t ca have o edges, sce the defto of a graph, each edge must have dstct ed vertces Therefore P () holds IId Iducto Step Let > ad assume for all k the rage k <, that P ( s true, e for ay such k, all trees o k vertces cota k edges Now let T be a tree o vertces, pck ay edge e T, ad remove t The removal of e splts T to two subtrees, each havg fewer tha vertces (Ths follows from some elemetary facts about graphs whch we omt for the sake of brevty) Suppose for the sake of defteess that the two subtrees have k ad k vertces, respectvely Sce o vertces were removed, we must have k k By our ductve hypothess, these two subtrees have k ad k edges, respectvely Upo replacg the edge e, we see that the umber of edges orgally T must have bee k ) ( k ) k k, as requred ( By the secod prcple of mathematcal ducto, all trees o vertces have edges /// Iducto Fallaces The ext three examples llustrate some ptfalls to be avoded whe costructg ducto proofs The result Example A was proved correctly Example Here we gve a vald proof of the same fact that llustrates a argumet whch some authors have called the ducto trap

7 Example A For all, f T s a tree o vertces the T has edges (Ivald) Base Step: If the T has o edges, sce each edge must have dstct ed vertces Iducto Step: Let ad let T be a tree o vertces Assume that T has edges Add a ew vertex ad jo t to T wth a ew edge To be precse, the ew edge has the ew vertex at oe ed, ad the other ed ca be ay exstg vertex T The resultg graph has vertces ad edges, ad s clearly a tree sce coectedess s mataed ad o cycles were created By the prcple of mathematcal ducto, all trees o vertces have edges Frst ote that the base step s detcal to that Example, ad s correct For the ducto step, the argumet attempts to follow IIa, but fals to do so I ths example P ( s of the form A( B( where A ( s the statemet T s a tree o vertces, ad B ( s T has edges The ducto step should therefore be to prove, for all, that P ( ), e ( A ( B( ) ( A( ) B( )) To prove ths, we should assume A( B(, the assume A ( ), the show as a cosequece that B( ) s true I other words we should: Assume all trees o vertces have edges Assume T has vertces Show as a cosequece that T has edges The argumet dd ot follow ths format however Istead t does the followg Assume T has vertces Assume T has edges Costruct a ew tree from T havg vertces ad edges Therefore the argumet was ot a proof by ducto Some studets would evertheless hold that the argumet s stll vald, eve though t s ot a true ducto proof The ext example shows covcgly that t caot be vald Example B For all, f G s a coected graph o vertces, the G has edges (False!) We otce rght away that the above statemet s false, sce the graph below provdes a elemetary couter-example But cosder the followg proof lght of Example A (Ivald) Base Step: If the G has o edges, sce each edge must have dstct ed vertces Iducto Step: Let ad let G be a coected graph o vertces Assume that G has edges Add a ew vertex ad jo t to G wth a ew edge The resultg graph has vertces ad edges, ad s clearly coected By the prcple of mathematcal ducto, all coected graphs o vertces have edges 7

8 Observe that Example B follows the format of Example A exactly Thus f A s vald, so must B be vald But the asserto proved B s false! Therefore B caot be a vald argumet, ad so ether s A Example C All horses are of the same color (Ivald) We prove that for all : f S s a set of horses, the all horses S have the same color The result follows o lettg S be the set of all horses Let P ( be the boxed statemet, ad proceed by ducto o Base Step: Let Obvously f S s a set cosstg of just oe horse, the all horses S must have the same color Thus P () s true Iducto Step: Let > ad assume that ay set of horses, all horses are of the same color Let S be a set of S h h, h, K h The the sets, 3, horses, say { } ad S { h, h3, K, h } S { h } S { h, h3, K, h } S { h} each cota exactly horses, ad so by the ducto hypothess all horses S are of oe color, ad lkewse for S Observe that h S S 3 ad that h 3 ca have oly oe color Therefore the color of the horses S s detcal to that of the horses S (Note > 3, so there s fact a thrd horse, ad he ca have oly oe color) Sce S S S t follows that all horses S are of the same color Thus P ( ) s true, showg that P ( ) for all > The result ow follows by ducto Obvously the proposto beg proved s false, so there s somethg wrog wth the proof, but what? The base step s certaly correct, ad the ducto step, as stated, s also correct The problem s that the ducto step was ot quatfed properly We should have proved : P( ) Istead we proved (correctly) that > : P( ) Ideed t s true that P( ) 3), P( 3) 4), ad P( 4) ), etc, but we ever proved (ad t s false that) P( ) ) I terms of the domo aalogy, t s as f the frst domo falls; ad f ay domo dexed or above were to fall, the the ext domo would fall; but the frst domo s ot suffcet to topple the secod domo, ad hece o domo other tha the frst actually falls Justfcato of the Iducto Prcples Here we prove the valdty of the frst ad secod prcples of mathematcal ducto Both proofs are based o the well orderg property of the postve tegers Z, whch says: Ay o-empty set of postve tegers cotas a least elemet We assume ths property wthout proof Theorem (weak ducto form IIb) For ay propostoal fucto P ( defed o the postve tegers, the followg setece s true: [ P( ) ( > : P( ) )] : P( 8

9 Assume that P () ad > : P( ) are both true Let S { Z P( s false } It s suffcet to show that S, sce the P ( s true for all Assume, to get a cotradcto, that S The, by the well orderg property of Z, S cotas a least elemet, call t m Sce P () s true, we have S Therefore m >, ad m s a postve teger Sce m s the smallest elemet S, we must have m S, whece P ( m ) s true We have assumed for all > that P( ) s true I partcular for m, we have P( m ) m) Sce both P ( m ) ad P( m ) m) are true, we must coclude that P (m) s also true Thus m S, cotradctg the very defto of m as the smallest elemet S Thus our assumpto was false, ad hece S as requred /// Theorem (strog ducto form IId) For ay propostoal fucto P ( defed o the postve tegers, the followg setece s true: [ P( ) ( > : ( k< : P( ) )] : P( Assume P () ad > : ( k< : P( ) are true, ad aga let S { Z P( s false } As before we show S, hece P ( s true for all Assume that S By the well orderg property, S cotas a least elemet m Sce P () s true, we have S Therefore m >, ad m Sce m s the smallest elemet S, we have for ay k the rage k m that k S, whece P ( s true I other words, k< m : P( s true Now we have also assumed for all >, that ( k< : P( ) s true I partcular, whe m, we have ( k< m : P( ) m) Sce both k< m : P( ad ( k< m : P( ) m) are true, we coclude P (m) s also true Thus m S, aga cotradctg the defto of m as the smallest elemet S Our assumpto was therefore false, ad hece S as requred /// Although we proved both theorems depedetly, t s possble to show that each mples the other, e theorems ad are logcally equvalet (exercse) I fact both theorems are equvalet to the well orderg property of the postve tegers (exercse) The terms strog ad weak ducto are therefore some sese msomers, sce ether theorem s really stroger tha the other The term strog ducto refers stead to the stroger assumpto beg made the ducto step: k< : P( as opposed to P ( ) 9