# Recursion & Iteration

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1 Recursion & Iteration York University Department of Computer Science and Engineering 1

2 Overview Recursion Examples Iteration Examples Iteration vs. Recursion Example [ref.: Chap 5,6 Wilensky] 2

3 Recursion A natural programming style in LISP A function is recursive if it calls itself Boundary condition: stopping recursion Recursive condition: must be a smaller problem to converge 3

4 Example factorial (defun factorial (n) (cond ((zerop n) 1) (t (* n (factorial (1 n)))) )) The above code works if n is a positive integer. Other numbers (not positive integers) will not reach the boundary condition and a stack overflow error will be encountered. A better implementation: (defun factorial (n) (cond ((not (and (integerp n) (>= n 0))) nil) ((zerop n) 1) (t (* n (factorial (1 n)))) )) 4

5 Example length of list Using cond: (defun llist (lst) (cond ((null lst) 0) (t (1+ (llist (cdr lst)))) )) Using if: (defun llist (lst) (if (null lst) 0 (1+ (llist (cdr lst))) )) This one will return 0 if an empty list or an atom (note nil is an atom) (defun llist (lst) (if (atom lst) 0 (1+ (llist (cdr lst))) )) 5

6 Example member of list Test if an element is a member of a list (defun lmember (e lst) (cond ((null lst) nil) ((equal e (car lst)) ) (t (lmember e (cdr lst))) )) Another way of writing above is: (defun lmember (e lst) (and lst (or The second argument of or will be evaluated, only if the first is evaluated to false (equal e (car lst)) (lmember e (cdr lst)) ))) The second argument of and will be evaluated, only if the first is evaluated to true (i.e. non nil) 6

7 Example member of list Test if an element is a member of a list, return the portion of list from the point of first match (defun lmember (e lst) (cond ((null lst) nil) ((equal e (car lst)) lst) (t (lmember e (cdr lst))) )) Exercise: Write a function that looks for members inside nested lists. 7

8 Example substitution in nested lists Function lsubst(in out lst) substitutes every occurrence of out with in in lst, which can be a list or an atom. e.g. (lsubst a x (b (x x) x)) will evaluate to (B (A A) A) (defun lsubst (in out lst) (cond ((equal out lst) in) ; if lst is out, return in ((atom lst) lst) ; otherwise if atom, no change ; otherwise two recursions cons ed (t (cons (lsubst in out (car lst)) (lsubst in out (cdr lst)) )) )) Exercise: Change the function definition to only substitute if lst is a list (no change if it is an atom) 8

9 Iteration Iteration: A loop, to be executed repeatedly Boundary condition (or terminating condition) A return value upon termination Index variables, their initial value, and the modification rule upon each iteration Unlike recursion, we need special functions, such as do to implement iteration 9

10 Iteration Do General form (do ((var1 val1 rep1) (var2 val2 rep2)...) exit clause form1 form2...) In which exit clause can be nil or in the form of (test test form1 test form2...) 1. Assign all vari with corresponding (evaluated) vali in parallel. 2. Examine exit clause. If nil, return nil as value of do (and stop). Otherwise, if test evaluates to true, evaluate test formi in order. Return the value of the last form as the value of do (and stop). 3. If test evaluates to false, evaluate formi in order. 4. Assign all vari with corresponding (evaluated) repi in parallel. 5. Go to step 2. 10

11 Example Find length of list For example : > (dolength (x y z)) 3 (defun dolength(lst) (do ( (tlst lst (cdr tlst)) (sum 0 (1+ sum))) ( (atom tlst) sum ) )) Two index variables: tlst and sum They are just like formal parameters: local to do. Terminating condition: when the list is an atom (including nil) Value of sum is returned upon termination. 11

12 Examples Use do to return a list which is the same as a list lst1 without the first n elements assuming its length is greater than n (do ((x n (1 x)) (lst2 lst1 (cdr lst2))) ( (zerop x) lst2)) Use do to return a list of numbers from 1 to n (do ( (m n (1 m)) (lst nil (cons m lst))) ( (zerop m) lst)) 12

13 Do vs. Do* do: evaluates all vali first and assigns index variables in parallel > (setq n 3) 3 > (do ( (m n (1 m)) (x nil (cons m x))) ( (zerop m) x)) (1 2 3) do*: Evaluation of vali and assignment to vari are done in sequential order > (do* ( (m n (1 m)) (x nil (cons m x))) ( (zerop m) x)) (0 1 2) 13

14 Iteration other functions dolist: iterates over elements of a list (dolist (var list val return val) form1 form2...) In each iteration, var is assigned with a value from list of values list val, Loops over formi, until all done. Then return val is returned. dotimes: iteration over integer values up to a limit (dotimes (var stop val return val) form1 form2...) Initializes var to 0, In each iteration formi are evaluated var is increased by 1, until it reaches stop val, at which point return val is returned. 14

15 Example Searching for a certain element in a given list (setq mylist ( )) (setq srch 2) (dolist (i mylist nil) (cond ((equal i srch) (return t)))) What will be returned in above case? Answer. T will be returned, since 2 exists in mylist. Note: dolist goes through elements of the list without the need for us to explicitly use car and cdr 15

16 Example Delete the first n items from a list (setq mylist ( )) (setq n 2) (dotimes (i n mylist) (setq mylist (cdr mylist))) What will be returned in above case? Answer. (3 4 5) will be returned. 16

17 Iteration vs. Recursion Example: Reversing a list Using iteration: (defun do rev(lst) (do ((x lst(cdrx)) (result nil (cons (car x) result))) (null x) result))) Using recursion: Cannot add to the end of a list We therefore use an extra variable (accumulator) More overhead due to recursive calls (defun rev2 (lst acc) (cond ((null lst) acc) ( t (rev2 (cdr lst) (cons (car lst) acc))) )) (defun reverse(lst) (rev2 lst nil)) 17

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