Fixed point. The function q revisited. Function q. graph F factorial graph factorial. factorial is a fixed point of F, since
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1 CS571 Notes 21 Denotatonal Semantcs of Loops (contnued 1of 25 Generalzng the soluton Parameterze the factoral functon F = λ f.λ n. n equals zero one n tmes ( f ( n mnus one Ths means fac = + 1 F ( fac.e. fac1 = F ( fac fac2 = F ( fac1 = F ( F ( fac ( ( ( fac = F F fac, tmes 2of 25 The graph of factoral The unon of the graphs of all the fac graph( fac = graph( F ( = = ( where F = F F F, and = λ n. Left-hand sde s graph(factoral ( ( = graph F ( graph factoral = 3of 25
2 Fxed pont factoral s a fxed pont of F, snce ( ( = ( graph F factoral graph factoral So, by extensonalty F factoral = factoral ( factoral s a fxed pont of F 4of 25 The functon q revsted From q = λ n. n equals zero one q( n plus one We get the functonal Q = λ g.λ n. n equals zero one g n plus one ( Then = ( λ., so ( ( = {} ( = ( ( ( ( = [ ] = ( ( = λ. ( (, so ( ( = {[ zero, one] } 2 Q n graph Q λ. λ., so {, } 1 Q n n equals zero one n n plus one 1 graph Q zero one 2 Q Q Q n n equals zero one n plus one equals zero one graph Q 5of 25 Functon q The lmt s clear (f we omt all the bottom results graph Q ( = zero, one = ( {[ ]} Call the functon wth ths graph qlmt All other solutons have graphs whch are larger, so qlmt s a fxed pont of Q, and s the smallest such fxed pont, the least fxed pont 6of 25
3 Partal orders and contnuous functons Formalzng and generalzng the method needs the theory of partal orders contanng contnuous functons A partal order s a relaton on a doman, D, that s reflexve, antsymmetrc, and transtve a b, where a, b D less defned than 7of 25 Example of a partal order The set {a,b,c,d}, such that: d b, and b c a c b d a s not related to b,c or d 8of 25 Least upper bound and greatest lower bound E.g. Powerset of {a,b,c} Elements are: {a,b,c} {a,b},{b,c},{a,c} {a},{b},{c} LUB of {a} and {b} {a,b,c} {a,b} {a,c} {b,c} {} {a} {b} {c} LUB: ü{ { a},{ b} } = { a, b} GLB: û ab,, bc, = b {} {{ } { }} { } GLB of {a,b} and {b,c} 9of 25
4 Chans If a partal order has a subset such that all pars of ts elements are n the relaton, then the subset s a chan E.g. {a,b,c,d} g a b c d e f 1 of 25 Complete partal order If all chans n the doman have a LUB, then the doman s a CPO If there s a least defned element, t s a ponted CPO E.g. Nat Example chans n Nat LUB 72 Least defned element of 25 Monotoncty and contnuousness Monotonc functons behave smoothly, e.g. add1=λn.n+1 Contnuous functons preserve chans, especally the LUB 5 add1 6 ( ü{ } = ü{ } add1 4,5 add1(4, add1( add1 5 add1(4 add1(5 12 of 25
5 Least fxed ponts IF: D s a ponted CPO A contnuous functonal F:D D exsts THEN: The least fxed pont of F s { ( } fx F = ü F, where F = F F F F, tmes 13 of 25 Proof of least fxed pont theorem (part 1 ( fx F F( ü{ F ( } ü{ ( ( } 1 ü{ F ( 2, F ( 3, F (, } ü{ F ( 1} ü { ( } { ( } ü{ F ( } F = = F F, by contnuty of F = = = = = = fx F ( F F 1, snce F (, and, F ( 14 of 25 Proof of least fxed pont theorem (part 2 Let e D be any fxed pont of F Snce ef, F e ( ( But the last term = e, whch mples { ( } fx F = ü F e Whch means that fx F s the least fxed pont 15 of 25
6 Factoral (yet agan fac : Nat Nat fac = λ n. n equals zero one n tmes ( fac( n mnus one Functonal s F : ( Nat Nat ( Nat Nat F = λ f.λ n. n equals zero one n tmes ( f ( n mnus one Least fxed pont s fx F = ü{ fac }, where fac = ( λ n. = fac = F ( fac, of 25 Operatonal expanson Use fx F = F(fx F F(fx F = λn.n equals zero one n tmes ((fx F(n mnus one = λn.n equals zero one n tmes ((F(fx F(n mnus one Each tme we use the expanson, we unfold the recurson one more level 17 of 25 ( fx F ( three = ( F( fx F ( three = (( λ f.λ n. n equals zero one n tmes ( f ( n mnus one ( fx F ( three = ( λ n. n equals zero one n tmes (( fx F ( n mnus one ( three = three tmes (( fx F ( three mnus one = three tmes (( fx F ( two = three tmes (( F ( fx F ( two = three tmes( two tmes (( fx F ( one = three tmes ( two tmes( ( F ( fx F ( one = three tmes ( two tmes ( one tmes (( fx F ( zero = three tmes ( two tmes( one tmes( ( F ( fx F ( zero = three tmes ( two tmes ( one tmes one = sx Factoral unfolded 18 of 25
7 Loop semantcs (fnally! The soluton we have been lookng for s B C = fx( λ f.λ s. B s f ( C s s C whle do B C It s statc It s non-recursve It can be unfolded operatonally 19 of 25 Unfoldng the loop C whle B do C = B fx F ( C where F = λ f.λ s. B s f C s s F = λs. 1 F = F( F = λ s. B B s ( λs. ( C C s s = λ s. B B s s 2 1 F = F( F = λ s. B B s ( λ s. B B s s ( C C s s = λ s. B B s ( B B s1 s1 s where s1 = C C s 2 of 25 B ( B ( B ( C 1 1 ( ( s1 = C s s2 = C ( C s 3 2 F = F( F =λ s. B s λ s. B s B s s s C s s = λ s. B B s B B s B B s s s s where C, C C general case: F = λ s. B B s (λ s. B B s1 (λ s. B B s2... (λ s. B B s 1 s 1... s2 s1 where s = C C ( C C ( C C s, tmes s Unfoldng the loop 21 of 25
8 A smple loop dervaton The loop s C whle A > do B := B + 1;A := A - B The ntal store s s = A sx B zero {} Functonal s C F = λ f.λ s. B A > s B := B + 1; A := A - B s s 22 of 25 Loop dervaton (contnued ( fx F s = ( F( fx F s = B A > s ( fx F ( C B := B + 1;A := A - B s s = ( fx F ( C B := B + 1;A := A - B s = ( fx F s1, where s1 = { A fve, B one } = ( F( fx F s1 = B A > s1 ( fx F ( C B := B + 1;A := A - B s1 s1 = ( fx F ( C B := B + 1;A := A - B s1 = ( fx F s2, where s2 = { A three, B two } 23 of 25 Loop dervaton (concluded = ( F( fx F s2 = B A > s2 ( fx F ( C B := B + 1;A := A - B s2 s2 = ( fx F ( C B := B + 1;A := A - B s2 = ( fx F s3, where s3 = { A zero, B three } = ( F( fx F s3 = B A > s3 ( fx F ( C B := B + 1;A := A - B s3 s3 = s 3 24 of 25
9 The nfnte loop The semantcs of the nfnte loop agree wth the semantcs of dverge n the prevous language. e.g. C whle true do nop = fx( λ f.λ s. B true s f ( C nop s s reduces to fx( λ f.λ s. f s Snce fx F = ü{ F ( } we have F = λ s. 1 F = FF ( = ( λ f.λ sfs. ( λ s. = λ s. ( λ s. s= λ s. F = λ s. So the semantcs agree 25 of 25
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