Lecture Overview. Knowledge-based systems in Bioinformatics, 1MB602. Procedural abstraction. The sum procedure. Integration as a procedure

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1 Lecture Overview Knowledge-bsed systems in Bioinformtics, MB6 Scheme lecture Procedurl bstrction Higher order procedures Procedures s rguments Procedures s returned vlues Locl vribles Dt bstrction Compound dt Principles of dt bstrction Procedurl bstrction nd dt bstrction Procedurl bstrction = ( * )/ = ( * * )/6 (define (sum-integers b) (+ (sum-integers (+ ) b)))) (define (sum-squres b) (+ (squre ) (sum-squres (+ ) b)))) k = k = Generlized: (define (sum term next b) (+ (term ) (sum term (next ) next b)))) k k The sum procedure (define (sum term next b) (+ (term ) (sum term (next ) next b)))) Wht is the type of this procedure? (number number, number, number number, number) number term proc next proc b sum procedure Higher order procedures A higher order procedure: tkes procedure s n rgument or returns one s vlue (define (sum-integers b) (sum (lmbd (x) x) (lmbd (x) (+ x )) b)) (define (sum-squres b) (sum squre (lmbd (x) (+ x )) b)) The sum procedure is higher order procedure Integrtion s procedure Integrtion under curve is given roughly by dx(f() + f( + dx) + f( + dx) ++ f(b)) dx f (define (integrl f b) (* (sum f (lmbd (x) (+ x dx)) b) dx)) (define dx.e-3) (define (tn ) (integrl (lmbd (x) (/ (+ (squre x)))) )) b

2 Procedurl bstrction Process of procedurl bstrction Define forml prmeters, cpture process in body of procedure Give procedure nme Hide implementtion detils from user, who just invokes nmes to pply procedures (bstrction brrier) Blck box bstrction Finding fixed points of functions Problem: find x tht stisfies f(x) = x Strtegy for finding fixed points: Given guess x, let new guess be f(x ) Keep computing f for lst guess, until close enough Exmple: find x tht stisfies cos(x) = x. Input: type procedure Detils of contrct for converting input to output Output: type Approx Try Finding fixed points of functions cont. Given guess x, let new guess be f(x) f(x), f(f(x)), f(f(f(x))), f(f(f(f(x)))), Keep computing f for lst guess, until close enough (define (fixed-point f guess) (define (close? u v) (< (bs (- u v)).)) (define (try g) (if (close? (f g) g) (f g) (try (f g)))) (try guess)) Using fixed points Computing the squre root of x require finding y such tht y = x, or y = x/y This is equivlent to looking for fix point of the function f(y) = x/y (define (sqrt x) (fixed-point (lmbd (y) (/ x y)) )) Using fixed points cont. Using let to crete locl vribles (sqrt ) (fixed-point (lmbd (y) (/ y)) ) (try ) (try ) (try ) (sqrt ) oscilltes between nd So dmp out the oscilltion (define (sqrt x) (fixed-point (dmp (lmbd (y) (/ x y))) )) y = x/y (y+y)/ = (x/y+y)/ y = (x/y+y)/ (define (verge x y) (/ (+ x y) )) (define (dmp f) (lmbd (x) (verge x (f x)))) Suppose we wish to compute the function: f(x,y) = x(+xy) +y(-y) + (+xy)(-y) which we lso could express s: = + xy b = y f(x,y) = x +yb+ b In Scheme: (define (f-help b) (+ (* x (squre )) (* b))) (f-help (+ (* x y)) (- y)))

3 Using let to crete locl vribles cont. (define (f-help b) (+ (* x (squre )) (* b))) (f-help (+ (* x y)) (- y))) ((lmbd ( b) (+ (* x (squre )) (* b))) (let (( (+ (* x y))) (b (- y))) (+ (* x (squre )) (* b)))) (+ (* x y)) (- y))) Syntx of let expressions Generl form (let ((<vr > <exp >) (<vr > <exp >)... (<vr n > <exp n >)) <body>) vr i cnnot depend on vr j If this is desired, use let*, e.g. (let* (( 3) (b (* 5 ))) (* b)) Compound dt Need wys of gluing dt elements together into unit tht cn be treted s simple dt element Need wys of retrieving dt elements Need contrct between the glue nd the unglue Idelly wnt the result of this gluing to hve the property of closure: the result obtined by creting compound dt structure cn itself be treted s primitive object nd thus be input to the cretion of nother compound object Pirs (cons-cells) (cons <x-exp> <y-exp>) <Pir> Where <x-exp> evlutes to vlue <x-vl>, nd <y-exp> evlutes to vlue <y-vl> (cr <Pir>) <x-vl> Returns the cr-prt of the pir <P> (cdr <Pir>) <y-vl> Returns the cdr-prt of the pir <P> Compound dt cont. Pir bstrction Tret PAIR s single unit: Cn pss pir s rgument Cn return pir s vlue (define (mke-point x y) (cons x y)) (define (point-x point) (cr point)) (define (point-y point) (cdr point)) (define (mke-seg pt pt) (cons pt pt)) (define (strt-point seg) (cr seg)) (, 3) point Constructor (cons <x> <y>) <Pir> Accessors (cr <Pir>) <x> (cdr <Pir>) <y> Predicte (pir? <z>) #t if <z> evlutes to pir, else #f 3

4 Pir bstrction cont. Note tht there is contrct between the constructor nd the ccessors (cr (cons <> <b> )) <> (cdr (cons <> <b> )) <b> Note how pirs hve the property of closure we cn use the result of pir s n element of new pir: (cons (cons ) 3) An exmple: rtionl numbers Wht re the elements of rtionl numbers n/d? numertor: n denomintor: d A rtionl number is rtio n/d /b + c/d = (d + bc)/bd /b * c/d = (c)/(bd) Rtionl number bstrction. Constructor (mke-rt <n> <d>). Accessors (numer <r>) (denom <r>) 3. Contrct (numer (mke-rt <n> <d>)) <n> (denom (mke-rt <n> <d>)) <d> 4. Lyered Opertions (print-rt <r>) (+rt x y) (*rt x y) 5. Abstrction Brrier Sy nothing bout implementtion! Rtionl number bstrction cont.. Constructor. Accessors 3. Contrct 4. Lyered Opertions 5. Abstrction Brrier Elements of dt bstrction 6. Concrete Representtion & Implementtion (cn lternte!) (define (mke-rt n d) (cons n d)) (define (numer r) (cr r)) (define (denom r) (cdr r)) Lyered rtionl numbers opertion (define (+rt x y) (mke-rt (+ (* (numer x) (denom y)) (* (numer y) (denom x))) (* (denom x) (denom y)))) (define (*rt x y) (mke-rt (* (numer x) (numer y)) (* (denom x) (denom y)))) (define (print-rt x) (newline) (disply (numer x)) (disply / ) (disply (denom x))) Testing our procedures (define one-hlf (mke-rt )) (define three-fourths (mke-rt 3 4)) (define new (+rt one-hlf three-fourths)) (print-rt new) /8 Oops should be 5/4 not /8! Rtionlize implementtion Strtegy : remove common fctors when ccessing numer nd denom Strtegy : remove common fctors when creting rtionl number 4

5 Implementtion strtegy Implementtion strtegy (define (gcd b) (if (= b ) (gcd b (reminder b)))) (define (numer r) (let ((g (gcd (cr r) (cdr r)))) (/ (cr r) g))) (define (denom r) (let ((g (gcd (cr r) (cdr r)))) (/ (cdr r) g))) (define (mke-rt n d) (cons n d)) (define (gcd b) (if (= b ) (gcd b (reminder b)))) (define (numer r) (cr r)) (define (denom r) (cdr r)) (define (mke-rt n d) (let ((g (gcd n d))) (cons (/ n g) (/ d g)))) Dt bstrction cont. Wht is pir? If we glue two objects together using cons we cn retrieve the objects using cr nd cdr We don t know the implementtion only tht Scheme supplies procedures (cons, cr, cdr) for operting on pirs We could implement cons, cr, nd cdr without using ny dt structures but only using procedures Requirement: contrct between constructor nd ccessor (cr (cons <> <b> )) <> (cdr (cons <> <b> )) <b> Dt bstrction cont. (define (cons x y) (define (disptch m) (cond ((= m ) x) ((= m ) y) (else (error Wrong rg for disp )))) disptch) (define (cr z) (z )) (define (cdr z) (z )) (cons x y) returns procedure (higher order procedure) This style of progrmming is often clled messge pssing. References H. Abelson, G.J. Sussmn, Structure nd Interprettion of Computer Progrms nd ed, The MIT Press, Cmbridge, Msschusetts,, Chp:.3,., pp: 56-76, Spring : Lecture Notes, lecture 4, 5, 6, 5

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