Question of the Day  What three letter word completes the first word and starts the second one: DON???CAR

Question of the Day  What three letter word completes the first word and starts the second one: DON???CAR  Key  Key completes the first word (donkey) and a key starts a car

 re-cur-sion  re-cur-sion: Solving problems by breaking into identical, smaller problems and combining solutions after base case(s) reached. Recursion

 Recursive stepbase case(s)  Recursive step simplifies problem to base case(s)  Recast using slightly easier version in recursive step 4! = 4 * 3! = 4 * (3 * 2!) = 4 * (3 * (2 * 1!))  Base case(s) handle and solve obvious cases = 4 * (3 * (2 * 1))  After base case, combine solutions in recursive steps = 4 * (3 * 2) = 4 * 6 = 24 See Recursion Work

 No different than usual coding had been doing  At some point, recursive method calls itself  Parameters & locals in frame & so specific to each call  Once complete, execution returns to calling frame NOT Writing Recursive Code

Starting Recursive Solution  Identify when solution easy with little/no work  Checking for situation involves one or two operations  Simple return statement with literal or no value base case(s)  May find more than one of these base case(s) If you cannot determine, stop immediately That are many problems without recursive solution

Base Case(s) Example Base Case(s)  Index leaves array with 0 or 1 entries  If specific item searched for in array, stop at 0 entries base case  1 entry used as base case when each item will be used  At last Node or null reference in Linked List base case(s)  Just like with arrays, base case(s) are comparable  int reaches 0 if working linearly through data base case  If multiplying int each pass, base case at 1  0 or 1 characters left in a String

Recursive Step Starting Recursive Step base case(s)  Determine state preceding each of base case(s)  These cases should be almost as easy to solve base case(s)  One step away from reaching base case(s) base case  Stand-alone base case may not have prior state base case  Identify single step advancing toward base case  Coding much easier when same step used in all cases  Know when each is used if different steps needed

Consider All Possible Inputs  After defining recursive call solution will be using  Must check that it works for all possible inputs  For each input you examine, must consider  Check that recursive call moves toward base case recursive steps  No conflicting recursive steps that cycle within space

Consider All Possible Inputs  After defining recursive call solution will be using  Must check that it works for all possible inputs  For each input you examine, must consider  Check that recursive call moves toward base case recursive steps  No conflicting recursive steps that cycle within space

Next to Last Step  Examine how method combines recursive result  No result for some recursive solutions, so can skip this  Some other recursive solutions just return result  Others require even more work to be performed  When using recursion, fairly simple actions required  If many or complex actions required, then…  First ask yourself: is this solution actually recursive?  Could it be more recursive & simpler is next question

 Do not want to be locked into recursive solution  Recursion often adds parameter to the signature  Public method only has required parameters  Linked list instance or array to be processed  User is interested in testing some value  Use private method to hold the extra variable  Value or array index where method currently working  Node in linked list to be processed by method  Range or indices where method starts & stops Last Step

public static int findMin(int[] a, int j) { if (j == a.length - 1) { return a[j]; } else { int minFollowing = findMin(a, j+1); return Math.min(a[j], minFollowing); } } Check This Out

public static int findMin(int[] a, int j) { if (j == a.length - 1) { return a[j]; } else { int minFollowing = findMin(a, j+1); return Math.min(a[j], minFollowing); } } public static int findMin(int[] a) { } Check It Out

private static int findMin(int[] a, int j) { if (j == a.length - 1) { return a[j]; } else { int minFollowing = findMin(a, j+1); return Math.min(a[j], minFollowing); } } public static int findMin(int[] a) { return findMin(a, 0); } Check It Out

Your Turn  Get into your groups and complete activity

For Next Lecture  Read GT for class on Friday  Is all code really equal?  How can we figure out which is faster?  Week #7 weekly assignment available now  Angel also has programming assignment #1  Pulls everything together and shows off your stuff NO!