5.3 Recursive definitions and structural induction

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1 /8/ Recursive defiitios ad structural iductio CSE03 Discrete Computatioal Structures Lecture 6 A recursively defied picture Recursive defiitios e sequece of powers of is give by a = for =0,,, Ca also be defied by a 0 =, ad a rule for fidig a term of te sequece from te previous oe, i.e., a + =a Ca use iductio to prove results about te sequece Structural iductio: We defie a set recursively by specifyig some iitial elemets i a basis step ad provide a rule for costructig ew elemets from tose already i te recursive step 3 Recursively defied fuctios Use two steps to defie a fuctio wit te set of o-egative itegers as its domai Basis step: specify te value for te fuctio at zero Recursive step: give a rule for fidig its value at a iteger from its values at smaller itegers Suc a defiitio is called a recursive or iductive defiitio 4

2 /8/05 Suppose f is defied recursively by f0=3 f+=f+3 Fid f, f, f3, ad f4 f=f0+3=*3+3=9 f=f+3=*9+3= f3=f+3=*+3=45 f4=f3+3=*45+3=93 Give a iductive defiitio of te factorial fuctio f=! Note tat +!=+! We ca defie f0= ad f+=+f o determie a value, e.g., f5=5!, we ca use te recursive fuctio f5=5 f4=5 4 f3=5 4 3 f=5 4 3 f =5 4 3 f0=5 4 3 =0 5 6 Recursive fuctios Recursively defied fuctios are well defied For every positive iteger, te value of te fuctio is determied i a uambiguous way Give ay positive iteger, we ca use te two parts of te defiitio to fid te value of te fuctio at tat iteger We obtai te same value o matter ow we apply two parts of te defiitio 7 Give a recursive defiitio of a, were a is a o-zero real umber ad is a o-egative iteger Note tat a + =a a ad a 0 = ese two equatios uiquely defie a for all o-egative iteger 8

3 /8/05 Fiboacci umbers Give a recursive defiitio of a k k0 e first part of te recursive defiitio 0 k 0 a k a 0 e secod part is a k k0 k0 a a k Fiboacci umbers f 0, f, f, are defied by te equatios, f 0 =0, f =, ad f =f - +f - for =, 3, 4, By defiitio f =f +f 0 =+0= f 3 =f +f =+= f 4 =f 3 +f =+=3 f 5 =f 4 +f 3 =3+=5 f 6 =f 5 +f 4 =5+3=8 9 0 Recursively defied sets ad structures Cosider te subset S of te set of itegers defied by Basis step: 3 S Recursive step: if x S ad y S, te x+y S e ew elemets formed by tis are 3+3=6, 3+6=9, 6+6=, We will sow tat S is te set of all positive multiples of 3 usig structural iductio 3 Strig e set * of strigs over te alpabet ca be defied recursively by Basis step: λ * were λ is te empty strig cotaiig o symbols Recursive step: if w * ad x te wx * e basis step defies tat te empty strig belogs to strig e recursive step states ew strigs are produced by addig a symbol from to te ed of stigs i * At eac applicatio of te recursive step, strigs cotaiig oe additioal symbol are geerated 4 3

4 /8/05 If ={0, }, te strigs foud to be i *, te set of all bit strigs, are λ, specified to be i * i te basis step 0 ad foud i te st recursive step 00, 0, 0, ad are foud i te d recursive step, ad so o Cocateatio wo strigs ca be combied via te operatio of cocateatio Let be a set of symbols ad * be te set of strigs formed from symbols i We ca defie te cocateatio for two strigs by recursive steps Basis step: if w *, te w λ=w, were λ is te empty strig Recursive step: If w *, w * ad x, te w w x=w w x Oftetimes w w is rewritte as w w e.g., w =abra, ad w =cadabra, w w =abracadabra 5 6 Legt of a strig Give a recursive defiitio of lw, te legt of a strig w e legt of a strig is defied by lλ=0 lwx=lw+ if w * ad x Well-formed formulae We ca defie te set of well-formed formulae for compoud statemet forms ivolvig, F, propositio variables ad operators from te set {,,,, } Basis step:, F, ad s, were s is a propositioal variable are well-formed formulae Recursive step: If E ad F are well-formed formulae, te E, E F, E F, E F, E F are well-formed formulae From a iitial applicatio of te recursive step, we kow tat p q, p F, F q ad q F are well-formed formulae A secod applicatio of te recursive step sows tat p q q F, q p q, ad p F are well-formed formulae 7 8 4

5 /8/05 Rooted trees e set of rooted trees, were a rooted tree cosists of a set of vertices cotaiig a distiguised vertex called te root, ad edges coectig tese vertices, ca be defied recursively by Basis step: a sigle vertex r is a rooted tree Recursive step: suppose tat,,, are disjoit rooted trees wit roots r, r,, r, respectively. e te grap formed by startig wit a root r, wic is ot i ay of te rooted trees,,,, ad addig a edge from r to eac of te vertices r, r,, r, is also a rooted tree Rooted trees 9 0 Biary trees Exteded biary trees At eac vertex, tere are at most two braces oe left subtree ad oe rigt subtree Exteded biary trees: te left subtree or te rigt subtree ca be empty Full biary trees: must ave left ad rigt subtrees e set of exteded biary trees ca be defied by Basis step: te empty set is a exteded biary tree Recursive step: If ad are disjoit exteded biary trees, tere is a exteded biary tree, deoted by, cosistig of a root r togeter wit edges coectig te root to eac of te roots of te left subtree ad rigt subtree, we tese trees are o-empty 5

6 /8/05 Exteded biary trees Full biary trees e set of full biary trees ca be defied recursively Basis step: ere is a full biary tree cosistig oly of a sigle vertex r Recursive step: If ad are disjoit full biary trees, tere is a full biary tree, deoted by, cosistig of a root r togeter wit edges coectig te root to eac of te roots of te left subtree ad rigt subtree 3 4 Full biary tree 5 6 6

7 /8/05 Structural iductio Sow tat te set S defied by 3 S ad if x S ad y S, te x+y S, is te set of multiples of 3 Let A be te set of all positive itegers divisible by 3 o prove A=S, we must sow tat A S, ad S A o sow A S, we must sow tat every positive iteger divisible by 3 is i S Use matematical iductio to prove it 7 8 Structural iductio Let p be te statemet tat 3 belogs to S Basis step: it olds as te first part of recursive defiitio of S, 3 =3 S Iductive step: assume tat pk is true, i.e., 3k is i S. As 3k S ad 3 S, it follows from te d part of te recursive defiitio of S tat 3k+3=3k+ S. So pk+ is true Structural iductio o sow tat S A, we use recursive defiitio of S e basis step of te defiitio specifies tat 3 is i S As 3=3, all elemets specified to be i S i tis step are divisible by 3, ad tere i A o fiis te proof, we eed to sow tat all itegers i S geerated usig te d part of te recursive defiitio are i A is cosists of sowig tat x+y is i A weever x ad y are elemets of S also assumed to be i A If x ad y are bot i A, it follows tat 3 x, 3 y, ad tus 3 x+y, tereby completig te proof

8 /8/05 rees ad structural iductio Heigt of biary tree o prove properties of trees wit structural iductio Basis step: sow tat te result is true for te tree cosistig of a sigle vertex Recursive step: sow tat if te result is true for te trees ad, te it is true for, cosistig of a root r, wic as as its left subtree ad as its rigt subtree We defie te eigt of a full biary tree recursively Basis step: te eigt of te full biary tree cosistig of oly a root r is =0 Recursive step: If ad are full biary trees, te te full biary tree = as eigt =+max, 3 3 Number of vertices i a biary tree If we let deote te umber of vertices i a full biary tree, we observe tat satisfies te followig recursive formula: Basis step: te umber of vertices of te full biary tree cosistig of oly a root r is = Recursive step: If ad are full biary trees, te te umber of vertices of te full biary tree = is =+ + eorem If is a full biary tree, te + - Use structural iductio to prove tis Basis step: for te full biary tree cosistig of just te root r te result is true as = ad =0, so = 0+ -= Iductive step: For te iductive ypotesis we assume tat, were ad are full biary trees

9 /8/05 9 eorem By te recursive formulae for ad, we ave =+ + ad =+max, us, is completes te iductive step 35, max, max