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Programming Languages 2nd edition Tucker and Noonan

Programming Languages 2nd edition Tucker and Noonan. Chapter 14 Functional Programming It is better to have 100 functions operate one one data structure, than 10 functions on 10 data structures. A. Perlis. Contents. 14.1 Functions and the Lambda Calculus 14.2 Scheme

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Programming Languages 2nd edition Tucker and Noonan

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  1. Programming Languages2nd editionTucker and Noonan Chapter 14 Functional Programming It is better to have 100 functions operate one one data structure, than 10 functions on 10 data structures. A. Perlis

  2. Contents 14.1 Functions and the Lambda Calculus 14.2 Scheme 14.2.1 Expressions 14.2.2 Expression Evaluation 14.2.3 Lists 14.2.4 Elementary Values 14.2.5 Control Flow 14.2.6 Defining Functions 14.2.7 Let Expressions 14.2.8 Example: Semantics of Clite 14.2.9 Example: Symbolic Differentiation 14.2.10 Example: Eight Queens 14.3 Haskell

  3. 14.2.8 Example: Semantics of Clite • Program state can be modeled as a list of pairs. E.g., • ((x 1) (y 5)) • Function to retrieve the value of a variable from the state: • (define (get id state) • (if (equal? id (caar state)) (cadar state) • (get id (cdr state)) • )) • E.g., (get ‘y ‘((x 5) (y 3) (z 1))) • = (get ‘y ‘((y 3) (z 1))) • = 3

  4. State transformation • Function to store a new value for a variable in the state: • (define (onion id val state) • (if (equal? id (caar state)) • (cons (list id val) (cdr state)) • (cons (car state) (onion id val (cdr state))) • )) • E.g., • (onion ‘y 4 ‘((x 5) (y 3) (z 1))) • = (cons ‘(x 5) (onion ‘y 4 ‘((y 3) z 1))) • = (cons ‘(x 5) (cons ‘(y 4) ‘((z 1)))) • = ‘((x 5) (y 4) (z 1))

  5. Modeling Clite Abstract Syntax • Skip (skip) • Assignment (assignment target source) • Block (block s1 s2 … sn) • Loop (loop test body) • Conditional (conditional test thenbranch elsebranch) • Expression • Value (value val) • Variable (variable id) • Binary (operator term1 term2)

  6. Semantics of Statements • (define (m-statement statement state) • (case (car statement) • ((skip) (m-skip statement state)) • ((assignment) (m-assignment statement state)) • ((block) (m-block (cdr statement) state)) • ((loop) (m-loop statement state) • ((conditional) (m-conditional statement state)) • (else ()) • ))

  7. Skip, Block, and Loop • (define (m-skip statement state) state) • (define (m-block alist state) • (if (null? alist) state • (m-block (cdr alist) (m-statement (car alist) state)) • )) • (define (m-loop) statement state) • (if (m-expression (car statement) state) • (m-loop statement (m-statement (cdr statement) state)) • state • ))

  8. Expression • (define (m-expression expr state) • (case (car expr) • ((value) (cadr expr)) • ((variable) (get (cadr expr) state)) • (else (applyBinary (car expr) (cadr expr) (caddr expr) state)) • )) • (define (applyBinary) op left right state) • (let ((leftval (m-expression left state)) • ((rightval (m-expression right state))) • (case op • ((plus) (+ leftval rightval)) • … • ))

  9. To Do: • 1. Show that these definitions give 5 as the meaning of y+2 in the state ((x 5) (y 3) (z 1)). I.e., show that • (m-expression ‘(plus (variable y) (value 2)) ‘((x 5) (y 3) (z 1))) • … • = 5 • 2. Give a definition of m-assignment. • 3. What about defining m-conditional?

  10. 14.2.9 Example: Symbolic Differentiation • Symbolic Differentiation Rules • Fig 14.2

  11. Scheme Encoding • Uses Cambridge Prefix notation E.g., 2x + 1 is written as (+ (* 2 x) 1) • Function diff incorporates these rules. E.g., (diff ‘x ‘(+ (* 2 x) 1)) should give an answer. • However, no simplification is performed. E.g. the answer for (diff ‘x ‘(+ (* 2 x) 1)) is (+ (+ (* 2 1) (* x 0)) 0) which is equivalent to the simplified answer, 2.

  12. Scheme Program (define (diff x expr) (if (not (list? Expr)) (if (equal? x expr) 1 0) (let ((u (cadr expr)) (v (caddr expr))) (case (car expr) ((+) (list ‘+ (diff x u) (diff x v))) ((-) (list ‘- (diff x u) (diff x v))) ((*) (list ‘+ (list ‘* u (diff x v)) (list ‘* v (diff x u)))) ((/) (list ‘div (list ‘- (list ‘* v (diff x u)) (list ‘* u (diff x v))) (list ‘* u v))) ))))

  13. Trace of the Example (diff ‘x ‘(+ ‘(* 2 x) 1)) = (list ‘+ (diff ‘x ‘(*2 x)) (diff ‘x 1)) = (list ‘+ (list ‘+ (list ‘* 2 (diff ‘x ‘x)) (list ‘* x (diff ‘x 2))) (diff ‘x 1)) = (list ‘+ (list ‘+ (list ‘* 2 1) (list ‘* x (diff ‘x 2))) (diff ‘x 1)) = (list ‘+ (list ‘+ ‘(* 2 1) (list ‘* x (diff ‘x 2))) (diff ‘x 1)) = (list ‘+ (list ‘+ ‘(* 2 1) (list ‘* x (diff ‘x 2))) (diff ‘x 1)) = (list ‘+ (list ‘+ ‘(* 2 1) (list ‘* x 0)) 0) = ‘(+ (+ (* 2 1) (* x 0)) 0)

  14. 14.2.10 Example: Eight Queens • A backtracking algorithm for which • each trial move’s: • Row must not be occupied, • Row and column’s SW diagonal must not be occupied, and • Row and column’s SE diagonal must not be occupied. If a trial move fails any of these tests, the program backtracks and tries another. The process continues until each row has a queen (or until all moves have been tried).

  15. Checking for a Valid Move • (define (valid move soln) • (let ((col (length (cons move soln)))) • (and (not (member move soln)) • (not (member (seDiag move col) (selist soln))) • (not (member (swDiag move col) (swlist soln))) • ))) • Note: the and encodes the three rules listed on the previous slide.

  16. Representing the Developing Solution • Positions of the queens kept in a list soln whose nth entry gives the row position of the queen in column n, in reverse order. E.g., soln = (5 3 1) represents queens in (row, column) positions (1,1), (3,2), and (5,3); i.e., see previous slide. • End of the game occurs when soln has N (= 8) entries: • (define (done soln) (>= (length soln) N)) • Continuing the game tests hasmore and generates nextmove: (define (hasmore move) (<= move N)) • (define (nextmove move) (+ move 1)

  17. Generating Trial Moves • (define (trywh move soln) • (if (and (hasmore move) (not (car soln))) • (let ((atry (tryone move (cdr soln)))) • (if (car atry) atry (trywh (nextmove move) soln)) • ) • soln • )) • The try function sets up the first move: • (define (try soln) (trywh 1 (cons #f soln)))

  18. Trying One Move • (define (tryone move soln) • (let ((xsoln (cons move soln))) • (if (valid move soln) • (if (done xsoln) • (cons #t xsoln) • (try xsoln)) • (cons #f soln)) • )) • Note: the #t or #f reveals whether the solution is complete.

  19. SW and SE Diagonals • (define (swdiag row col) (+ row col)) • (define (sediag row col) (- row col)) • (define (swlist alist) • (if (null? Alist) ‘() • (cons (swDiag (car alist) (length alist)) • (swlist (cdr alist))))) • (define (selist alist) • (if (null? Alist) ‘() • (cons (seDiag (car alist) (length alist)) • (selist (cdr alist)))))

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