RECURSION Lecture 8 CS2110 Spring 2017 Four things 2 Note: Weve covered almost everything in JavaSummary.pptx! Recursion: 7.1-7.39 slide 1-7 Base case: 7.1-7.10 slide 13

How Java stack frames work 7.8-7.10 slide 28-32 Supplemental material Pinned Piazza note @96 Java concepts Study habits Wrapper classes Recursion exercises Final exam date is set: Sunday, May 21st at 2:00pm To Understand Recursion 3

4 Recursion Real Life Examples = , or , or Example: terrible horrible no-good very bad day 5 Recursion Real Life Examples = , or , or ancestor(p) = parent(p), or parent(ancestor(p)) great great great great great great great great great great great grandmother.

0! = 1 n! = n * (n-1)! 1, 1, 2, 6, 24, 120, 720, 5050, 40320, 362880, 3628800, 39916800, 47900 Sum the digits in a non-negative integer 6 /** = sum of digits in n. * Precondition: n >= 0 */ public static int sumDigs(int n) { if (n < 10) return n; sum calls itself! // { n has at least two digits } // return first digit + sum of rest return n%10 + sum(n/10); } sum(7) = 7 sum(8703) = 3 + sum(870) = 3 + 8 + sum(70) = 3 + 8 + 7 + sum(0)

Two different questions, two different answers 7 1. How does it execute? (or, why does this even work?) 2. How do we understand recursive methods? (or, how do we write/develop recursive methods?) Stacks and Queues 8 stack grows top element 2nd element ... bottom element first

second last Americans wait in a line. The Brits wait in a queue ! Stack: list with (at least) two basic ops: * Push an element onto its top * Pop (remove) top element Last-In-First-Out (LIFO) Like a stack of trays in a cafeteria Queue: list with (at least) two basic ops: * Append an element * Remove first element First-In-First-Out (FIFO) Stack Frame 9 A frame contains information about a method call: At runtime Java maintains a

a frame stack that contains frames for all method calls that are being executed but have not completed. local variables parameters return info Method call: push a frame for call on stack. Assign argument values to parameters. Execute method body. Use the frame for the call to reference local variables and parameters. End of method call: pop its frame from the stack; if it is a function leave the return value on top of stack. Memorize method call execution! 11 A frame for a call contains parameters, local variables, and other information needed to properly execute a method call. To execute a method call: 1.

push a frame for the call on the stack, 2. assign argument values to parameters, 3. execute method body, 4. pop frame for call from stack, and (for a function) push returned value on stack When executing method body look in frame for call for parameters and local variables. Frames for methods sum main method in the system 12

public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } Frame for method in the system that calls method main frame: frame: frame: n ___ return info r ___ args ___

return info ? return info Example: Sum the digits in a non-negative integer 13 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } Frame for method in the system that calls method main: main is then called

main system r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer 14 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } Method main calls sum:

main system 824 n ___ return info r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer 15 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) {

int r= sum(824); System.out.println(r); } n >= 10 sum calls sum: 82 n ___ return info 824 n ___ main system return info r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer 16

public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } n >= 10. sum calls sum: 8 n ___ return info 82 n ___ return info 824 n ___ main system

return info r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer 17 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } n8___

8 return info 82 n ___ return info 824 n ___ main n < 10 sum stops: frame is popped system and n is put on stack: return info r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer

18 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } 8 n82 ___ 10 return info main Using return value 8 stack computes

2 + 8 = 10 pops frame from stack puts return value 10 on stack 824 n ___ return info r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer 19 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824);

System.out.println(r); } 10 main Using return value 10 stack computes 4 + 10 = 14 pops frame from stack puts return value 14 on stack n824 ___ 14 info return r ___ args ___ return info ? return info Example: Sum the digits in a non-negative integer

20 public static int sum(int n) { if (n < 10) return n; return n%10 + sum(n/10); } public static void main( String[] args) { int r= sum(824); System.out.println(r); } Using return value 14 main stores 14 in r and removes 14 from stack main 14 r ___ 14 args __ return info ? return info

Poll time! 21 Assume my program's main method calls sumDigs(1837420) During this call, what is the maximum number of stack frames above (not including) main's stack frame? Two different questions, two different answers 22 1. How does it execute? (or, why does this even work?) Its not magic! Trace the codes execution using the method call algorithm, drawing the stack frames as you go. Use this only to gain understanding / assurance that recursion works. 2. How do we understand recursive methods? (or, how do

wedifferent write/develop recursive This requires a totally approach. methods?) Back to Real Life Examples 23 Easy to make math definition Factorial function: into a Java function! 0! = 1 public static int fact(int n) { n! = n * (n-1)! for n > 0if (n == 0) return 1; (e.g.: 4! = 4*3*2*1=24) return n * fact(n-1); }

Exponentiation: public static int exp(int b, int c) { b0 = 1 if (c == 0) return 1; bc = b * bc-1 for c > 0 return b * exp(b, c-1); } How to understand what a call does 24 Make a copy of the method spec, replacing the parameters of the method by the arguments sumDigs(654) sum of digits of n sum of digits of 654 spec says that the value of a call equals the sum of

the digits of n /** = sum of the digits of n. * Precondition: n >= 0 */ public static int sumDigs(int n) { if (n < 10) return n; // n has at least two digits return n%10 + sumDigs(n/10); } Understanding a recursive method 25 Step 1. Have a precise spec! Step 2. Check that the method works in the base case(s): That is, Cases where the parameter is small enough that the result can be computed simply and without recursive calls. If n < 10 then n consists of a single digit. Looking at the spec we see that that digit is the

required sum. /** = sum of the digits of n. * Precondition: n >= 0 */ public static int sumDigs(int n) { if (n < 10) return n; // n has at least two digits return n%10 + sumDigs(n/10); } Understanding a recursive method 26 /** = sum of the digits of n. * Precondition: n >= 0 */ Step 2. Check that the method public static int sumDigs(int n) { works in the base case(s). if (n < 10) return n; // n has at least two digits Step 3. Look at the recursive return n%10 + sumDigs(n/10);

case(s). In your mind replace } each recursive call by what it does according to the method spec and verify that the correct result is then obtained. return n%10 + sum(n/10); Step 1. Have a precise spec! return n%10 + (sum of digits of n/10); // e.g. n = 843 Understanding a recursive method 27 /** = sum of the digits of n. * Precondition: n >= 0 */ Step 2. Check that the method public static int sumDigs(int n) { works in the base case(s). if (n < 10) return n; // n has at least two digits

Step 3. Look at the recursive return n%10 + sumDigs(n/10); case(s). In your mind replace } each recursive call by what it does acc. to the spec and verify correctness. Step 1. Have a precise spec! Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the pars of the method. n/10 < n, so it will get smaller until it has one digit Understanding a recursive method 28 Step 1. Have a precise spec! Important! Cant do step 3 without precise spec. Step 2. Check that the method

works in the base case(s). Step 3. Look at the recursive case(s). In your mind replace each recursive call by what it does according to the spec and verify correctness. Once you get the hang of it this is what makes recursion easy! This way of thinking is based on math induction which we dont cover in this course. Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the parameters of the method Writing a recursive method 29 Step 1. Have a precise spec! Step 2. Write the base case(s): Cases in which no recursive calls are needed. Generally for small values of the parameters.

Step 3. Look at all other cases. See how to define these cases in terms of smaller problems of the same kind. Then implement those definitions using recursive calls for those smaller problems of the same kind. Done suitably, point 4 (about termination) is automatically satisfied. Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the parameters of the method Two different questions, two different answers 30 2. How do we understand recursive methods? (or, how do we write/develop recursive Step 1. Have a precise spec! methods?) Step 2. Check that the method works in the base case(s). Step 3. Look at the recursive case(s). In your mind replace each recursive call by what it does according to the spec and verify correctness. Step 4. (No infinite recursion) Make sure that the args of recursive calls are in some sense smaller than the parameters of the method

Examples of writing recursive functions 31 For the rest of the class we demo writing recursive functions using the approach outlined below. The java file we develop will be placed on the course webpage some time after the lecture. Step 1. Have a precise spec! Step 2. Write the base case(s). Step 3. Look at all other cases. See how to define these cases in terms of smaller problems of the same kind. Then implement those definitions using recursive calls for those smaller problems of the same kind. Step 4. Make sure recursive calls are smaller (no infinite recursion). Check palindrome-hood 32 A String palindrome is a String that reads the same backward and forward:

isPal(racecar) true isPal(pumpkin) false A String with at least two characters is a palindrome if (0) its first and last characters are equal and (1) chars between first & last form a palindrome: have to be the same e.g. AMANAPLANACANALPANAMA have to be a palindrome A recursive definition! 33 A man a plan a caret a ban a myriad a sum a lac a liar a hoop a pint a catalpa a gas an oil a bird a yell a vat a caw a pax a wag a tax a nay a ram a cap a yam a gay a tsar a wall a car a luger a ward a bin a woman a vassal a wolf a tuna a nit a pall a fret a watt a bay a daub a tan a cab a datum a gall a hat a fag a zap a say a jaw a lay a wet a gallop a tug a trot a trap a tram a torr a caper a top a tonk a toll a ball a fair a sax a

minim a tenor a bass a passer a capital a rut an amen a ted a cabal a tang a sun an ass a maw a sag a jam a dam a sub a salt an axon a sail an ad a wadi a radian a room a rood a rip a tad a pariah a revel a reel a reed a pool a plug a pin a peek a parabola a dog a pat a cud a nu a fan a pal a rum a nod an eta a lag an eel a batik a mug a mot a nap a maxim a mood a leek a grub a gob a gel a drab a citadel a total a cedar a tap a gag a rat a manor a bar a gal a cola a pap a yaw a tab a raj a gab a nag a pagan a bag a jar a bat a way a papa a local a gar a baron a mat a rag a gap a tar a decal a tot a led a tic a bard a leg a bog a burg a keel a doom a mix a map an atom a gum a kit a baleen a gala a ten a don a mural a pan a faun a ducat a pagoda a lob a rap a keep a nip a gulp a loop a deer a leer a lever a hair a pad a tapir a door a moor an aid a raid a wad an alias an ox an atlas a bus a madam a jag a saw a mass an anus a gnat a lab a cadet an em a natural a tip a caress a pass a baronet a minimax a sari a fall a ballot a knot a pot a rep a carrot a mart a part a tort a gut a poll a gateway a law a jay a sap a zag a fat a hall a gamut a dab a can a tabu a day a batt a waterfall a patina a nut a flow a lass a van a mow a nib a draw a regular a call a war a stay a gam a yap a cam a ray an ax a tag a wax a paw a cat a valley a drib a lion a saga a plat a catnip a pooh a rail a calamus a dairyman a bater a canal Panama Example: Is a string a palindrome? 34 /** = "s is a palindrome" */

public static boolean isPal(String s) { if (s.length() <= 1) Substring from return true; s[1] to s[n-1] // { s has at least 2 chars } int n= s.length()-1; return s.charAt(0) == s.charAt(n) && isPal(s.substring(1,n)); } The Fibonacci Function 35 Mathematical definition: fib(0) = 0 two base cases! fib(1) = 1 fib(n) = fib(n - 1) + fib(n - 2) n 2 Fibonacci sequence: 0 1 1 2 3 5 8 Fibonacci (Leonardo 13 /**= fibonacci(n). Pre: n >= 0 */

Pisano) 1170-1240? static int fib(int n) { if (n <= 1) return n; Statue in Pisa Italy // { 1 < n } Giovanni Paganucci return fib(n-1) + fib(n-2); 1863 } Example: Count the es in a string 36 /** = number of times c occurs in s */ public static int countEm(char c, String s) { if (s.length() == 0) return 0; substring s[1..] // { s has at least 1 character } if (s.charAt(0) != c) return countEm(c, s.substring(1)); i.e. s[1]

s(s.length()-1) // { first character of s is c} return 1 + countEm (c, s.substring(1)); } countEm(e, it is easy to see that this has many es) = 4 countEm(e, Mississippi) = 0