letrec Recall: Tail Recursion for rev Contents CSC324H Principle of Programming Languages (Week 5)

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1 Contents CSC324H Principle of Progrmming Lnguges (Week 5) Yiln Gu Scheme Bsic types Bsic functions Different recursions Efficiency nd til recursion Let, let* letrec More exmples of unbounded lmbd (with prrmeter list) higher-order functions Theory lmbd clculus 1 2 letrec (letrec ((vr1 expr1)... (vrn exprn)) body ) Scope: Ech binding of vrible hs the entire letrec expression s its region, vribles re visible within expressions s well -- recursion Evlution: expr1,..., exprn re evluted in n undefined order, sved, nd then ssigned to vr1,..., vrn, with the ppernce of being evluted in prllel. Let nd let* cn't be used for binding with recursion precedure Recll: Til Recursion for rev (define (rev lst) (rev-cc lst ())) (define rev-cc (lmbd (lst cc) (if (null? lst) cc (rev-cc (cdr lst) (cons (cr lst) cc))))) 3 4

2 (define reverse Letrec Exmples: rev (lmbd (lst) (letrec ((rev-cc (lmbd (lst cc) (rev-cc lst ())))) (if (null? lst) cc (rev-cc (cdr lst) (cons (cr lst) cc)))))) (define (length lst) Letrec Exmples: length (letrec ((length-til (lmbd (lst len) (length-til lst 0))) (if (null? lst) len (length-til (cdr lst) (+ len 1)))))) 5 6 Review: Unbounded Lmbd Unbounded Lmbd with Prmeter Lists OR (define sum (lmbd rg (cond ((null? rg) (disply "expecting t lest one rgument")) ((null? (cdr rg)) (cr rg)) (else (+ (cr rg) (pply sum (cdr rg))))))) (define (sum. rg) (cond ((null? rg) (disply "expecting t lest one rgument")) ((null? (cdr rg)) (cr rg)) (else (+ (cr rg) (pply sum (cdr rg)))))) 7 8

3 More Exmples of Prmeter Lists More Exmples of Prmeter Lists (define list-rgs (lmbd rg rg)) (define sum-non1-rgs (lmbd (first. rest) (pply + rest))) > (list-rgs) () >(list-rgs ') () >(list-rgs ' 'b 'c '(d e)) ( b c (d e)) > (sum-non1-rgs 1 2 4) 6 >(sum-non1-rgs 1) 0 >(sum-non1-rgs ) error 9 10 Functionl Progrmming Higher-Order Functions One of the most importnt common spects: Functions s vlues: Arguments or return vlues of "higher-order functions". A higher-order function is function tht tkes function s prmeter, returns function or does both. An exmple in mth function composition for ny two unry functions f(x), g(x), we define higher-ordered function h(f, g, x) = f(g(x)) 11 12

4 Procedures s Input Vlues (define (ll-num? list) (or (null? list) (nd (number? (cr list)) (ll-num? (cdr list))))) (define (bs-list list) (cond ((null? list) '()) (else (cons (bs (cr list)) (bs-list (cdr list)))))) Procedures s Returned Vlues (define dd-mult (lmbd (x) (cond ((nd (number? x) (> x 0)) (lmbd (y) (+ x y))) ((nd (number? x) (< x 0)) (lmbd (y) (* x y))) (else (lmbd (x) x))))) (define (ll-num-f fun list) (if (ll-num? list) (fun list) 'error)) Define Limited mp A limited version (mymp op list) Exmple > (mymp bs '(-1 0 9)) (1 0 9) > (mymp (lmbd (x) (+ 1 x)) '(1 2 3)) (2 3 4) (define (mymp f l) (cond ((null? l) '()) (else (cons (f (cr l)) (mymp f (cdr l)))))) Mp (mp proc list1... listn) All lists must be the sme length returns: A list of pplying proc to ll i th (i=1..n) elements of ech list Exmples: (mp bs '( )) => ( ) (mp * '(1 3 5) '(6 8 3)) => ( ) (mp * '(1 3 5) '(6 8 3) '(0 1 2)) => ( ) The build-in mp is more generl nd more powerful 15 16

5 (define (tomcount s) )) (cond ((null? s) 0) > (tomcount '( b)) error Wht's Wrong Here? ((not (pir? s)) 1) (else (+ (mp tomcount s))) Why? Wht's wrong? Apply (define (tomcount s) )) Fix tomcount using evl (cond ((null? s) 0) > (tomcount '( b)) 2 ((not (pir? s)) 1) (else (evl (cons '+ (mp tomcount s)))) Evl Evl The evl function tkes quoted expression or definition nd evlutes it: > (evl '(+ 1 2)) 3 The power of evl is tht n expression cn be constructed dynmiclly: (define (evl-formul formul) (evl `(let ([x 2] [y 3]),formul))) > (evl-formul '(+ x y)) 5 > (evl-formul '(+ (* x y) y)) 9 Evl (exmples) Be creful when using evl > (evl '(+ 1 2)) 3 > (evl '(ppend () (b))) Error > (evl '(ppend '() '(b))) ( b) Too complicted, nother solution: pply 19 20

6 Apply (revised) (pply proc obj1... objm list1... listn) returns: the result of pplying proc to obj... nd the elements of list pply is useful when some or ll of the rguments to be pssed to procedure re in list, since it frees the progrmmer from explicitly destructuring the list. Exmples: (pply + '(4 5)) => 9 (pply min '( )) => 1 (define (tomcount s) )) Fix tomcount using pply (cond ((null? s) 0) > (tomcount '( b)) 2 ((not (pir? s)) 1) (else (pply + (mp tomcount s))) Fold Right (foldr op id lst1 lst2... lstn) op : n binry procedure lst : list of rguments (cn be more thn one lists if op cn be pplied to multiple rguments, ll lists re sme length) pply op right-ssocitively to elements of lst (or ll the lists), nd return result of evlution the identity element id is lwys used Tht is: (one list s n exmple) (fold-right op id '())} => id (fold-right op id '(e)) => (op e id) (fold-right op id '(e1 e2... en))}=> (op e1 (op e2 (op... (op en id)))) Other dilect: In MIT Scheme: (fold-right op id lst1... lstn) Fold Right (exmples) > (foldr cons ' '(1 2 3)) ( ) (foldr cons ' '(1 2 3)) (cons '1 (foldr cons ' '(2 3))) (cons '1 (cons '2 (foldr cons ' '(3)))) (cons '1 (cons '2 (cons '3 (foldr cons ' '())))) (cons '1 (cons '2 (cons '3 ')))... ( ) 23 24

7 Fold Right (more exmples) > (foldr list ' '(1 2 3)) (1 (2 (3 ))) > (foldr cons ' '(1 2 3) '(4 5 6)).. foldr: rity mismtch, does not ccept 2 rguments #<procedure:cons> > (foldr list ' '(1 2 3) '(4 5 (k))) (1 4 (2 5 (3 (k) ))) > (foldr list ' '()) > (foldr cons ' '()) Define Limited Fold Right (define (myfoldr op id list) (if (null? list) id (op (cr list) (myfoldr op id (cdr list))))) Fold Left (two different versions) PLT scheme version: (foldl op id lst1...lstn) op : n binry procedure lst : list of rguments (ccepts multple lists) pply op left-ssocitively to elements of lst, nd return result of evlution the identity element id is lwys used Tht is: (one list s n exmple) (foldl op id '())} => id (foldl op id '(e)) => (op e id) (foldl op id '(e1 e2... en))}=> (op en... (op e2 (op e1 id))) > (foldl list ' '(1 2 3)) (3 2 1 ) >(foldl cons ' '(1 2 3)) ( ) > (foldl list ' '(1 2 3) '(4 5 6)) (3 6 (2 5 (1 4 ))) > (foldl list ' '()) > (foldl cons ' '()) foldl (PLT version) 27 28

8 Fold Left (two different versions) MIT scheme version: (fold-left op id lst) op : n binry procedure lst : list of rguments (only one list llowed) pply op left-ssocitively to elements of lst, nd return result of evlution the identity element id is lwys used Tht is: (one list s n exmple) (fold-left op id '())} => id (fold-left op id '(e)) => (op id e) (fold-left op id '(e1 e2... en))}=> (op... (op (op id e1) e2)...en ) > (fold-left list ' '(1 2 3)) ((( 1) 2) 3) >(fold-left cons ' '(1 2 3)) (((. 1). 2). 3) > (fold-left list ' '()) > (fold-left cons ' '()) fold-left (MIT version) Define fold-left using foldl nd vise vers? How to define limited fold-left using foldl in PLT scheme, nd in MIT scheme, how to define foldl using fold-left? Bonus question for ssignment 2. Reduce-right (MIT Scheme) (reduce-right op id lst) op : n binry procedure lst : list of rguments pply op right-ssocitively to elements of lst return result of evlution:if lst is empty, return id; if lst hs one element, return tht element. Tht is: (reduce-right op id '()) => id (reduce-right op id '(e)) => e (reduce-right op id '(e1 e2... en)) => (op e1 (op e2 (op... (op e_{n-1} en)))) 31 32

9 Higher-order Procedures: reduce-right (reduce-right list '() '( )) => (1 (2 (3 4))) (reduce-right list '() '( )) (list 1 (reduce-right list '() '(2 3 4))) (list 1 (list 2 (reduce-right list '() '(3 4)))) (list 1 (list 2 (list 3 (reduce-right list '() '(4))))) (list 1 (list 2 (list 3 4))) (list 1 (list 2 '(3 4))) (list 1 '(2 (3 4))) (1 (2 (3 4))) Reduce-right (exmple) Define reduce-right Exercise: define our own reduce-right (myreducer) (define myreducer (lmbd (op id lst) (cond ((null? lst) id) ((null? (cdr lst)) (cr lst)) (else (op (cr lst) (myreducer op id (cdr lst))))))) More Prctice Suppose we wnt procedure tht will test every element of list nd return list contining only those tht pss the test We wnt to be very generl: Use ny test we might give it (define (prune test lst) (cond ((null? lst) '()) ((test (cr lst)) (cons (cr lst) (prune test (cdr lst)))) (else (prune test (cdr lst))))) Other Useful Scheme: set!, Vectors, Strings, sequencing, ssoc 35 36

10 set! Globl Assignment (Generlly EVIL) void if possible Memory loction re: Mitined fter the procedure cll is complete Are used for their vlues in this or other procedure set! is useful for counters (define cons-count 0) (define (cons-co x y) (set! cons-count (+ cons-count 1)) (cons x y)) >(cons-co ' '(b c)) ;( b c) >cons-count ;1 >(cons-co ' (cons-co 'b 'c)) ;( b. c) >cons-count ;3 Strings Sequence of chrctors Written within double quotes, e.g., hi mom Useful string predicte procedures: (string=?...) (string<?...) (string<=?...) ;etc. Cse-insensitive versions (string-ci=?...) (string-ci<?...) (string-ci<=?...) ;etc. Other procedures: (string-length <string>) (string->symbol <string>) (symbol->string <sybmol>) (string->list <string>) (list->string <list>) Scheme: vectors Another compound structure (lists, pris ) Similr to lists, it cn hold heterogeneous elements Wht s the problem with lists? Access time Vectors re ccessed in constnt time. Exmple: ]=>(mke-vector 5 0) ; cretes vector of length 5 initilized to 0 ]=> #( ) ; prefix nottion # to do the sme thing ]=>(vector-length (mke-vector 150)) ; returns 150 ]=>(vector-ref #( b c) 2 ) ; returns the third element c ]=>(let ((v (vector ' 'b 'c 'd 'e ) )) (vector-set! v 2 'x)) ; #( b x d e) ]=>(vector-fill! v y ) ; replce ech element of v with y ]=>(vector->list #( b c) ) ; return list ( b c) ]=>(list->vector '( b c) ) ; return vector #( b c) ]=> (let ((v (mke-vector 5))) (for-ech (lmbd (i) (vector-set! v i (* i i))) '( )) v) sequencing Another control structure (we hve lredy seen cond nd if) Syntx: (begin exp 0 exp 1 exp n ) expressions re evluted from left to right The vlue of the rightmost expression is the vlue of the entire expression. Other expressions re for side-effects only Exmple: ]=> (define x 3) ]=> (begin (set! x (+ x 1)) (+ x x) ) ; returns 8

11 ssoc does lookup in list ssoc function Eg: ]=> (define NAMES '((Smith Pt Q) (Jones Chris J) (Wlker Kelly T) (Thompson Shelly P))) ]=> (ssoc 'Smith NAMES) (Smith Pt Q) ]=> (ssoc 'Wlker NAMES) (Wlker Kelly T) ssoc returns the first sublist if more thn one sublist with the sme key exist.