Binary Trees 2

6_6

Binary Trees 1

6_1

There are four leaves, 3 nodes have two children, and it would take 2 iterations to find out that 12 is not in it, i believe

6_2

Yes it worked.

6_3

I don't know why it took me so long to figure out where to put the code

public void add(Object newValue)

{

    TreeNode newNode = new TreeNode(newValue,null,null);

    if(root==null)//bst is empty, make root reference node referenced by newNode

    {

        root=newNode;

    }//if

    else//search for place to insert node referenced by newNode

    {

        TreeNode T=root;//T is a traversing TreeNode reference (pointer)

        TreeNode follow=null;//follows one node behind T

        while(T!=null)

        {

            follow=T;

            if (((Comparable)newNode.getValue()).compareTo(T.getValue())==0){return;}

            if (((Comparable)newNode.getValue()).compareTo(T.getValue())>0)

            {//the value in node referenced by newNode >= 

              //value in node referenced by T (go right)

                T=T.getRight();

            }//if

            else

            {//go left

                T=T.getLeft();

            }//else

        }//while

        //follow will now reference (point to) the node where insertion to be made 

        if (((Comparable)newNode.getValue()).compareTo(follow.getValue())>=0)

        {//the value in node referenced by newNode >=

          //value in node referenced by follow (link right)

            follow.setRight(newNode);

        }//if

        else

        {//link left

            follow.setLeft(newNode);

        }//else

    }//else

}//add

}//class BST

6_4

I am not sure why

6_5

public void find(Object target){

if(root==null){return;}

TreeNode T=root;//T is a traversing TreeNode reference (pointer)

    TreeNode follow=null;//follows one node behind T

    while(T!=null); {

    follow = T;

    if (((Comparable)newNode.getValue()).compareTo(T.getValue())==0) {return T;}

    if (((Comparable)newNode.getValue()).compareTo(T.getValue())>0) {T=T.getRight();}

    else {T=T.getLeft();}

    }

return null;

}

Long Overdue Stacks

4_1 done
4_2 done
4_3 done
4_4 yes, resolved
4_5 wasn't that hard, done

The recursive sum method helped visualize things, but a little redundant to me.
To create infinite recursion with a positive number, I would have to delete the base case or alter it.

4_6
This took a while because of my syntax of ArrayStackList, but i had the logic down pretty quickly

public class StringReverseDemo extends ArrayListStack

{

    //uses stacks (class ArrayStack) to reverse word. 

    public static String getReverse(String word)

    {

    ArrayListStack stack1=new ArrayListStack();

    String result = "";

    for (int i=0;i<word.length();i++){

    stack1.push(word.substring(i, i+1));

    }

    while (!stack1.isEmpty()){

    result+=stack1.pop();

    }

    return result;

    }//getReverse

    //uses recursion to reverse word

    public static String getReverseRecursive(String word)

    {

    if (word.length()==1){return word;}

    else

    return word.substring(word.length()-1,word.length())+getReverseRecursive(word.substring(0,word.length()-1));

    }//getReverseRecursive

}//class String

4_7

I believe recursion simplifies code dramatically, at least for me, as long as you understand it. The iterative approach is easier to think about but is more prone to errors.

4_8

 public static boolean isPalindrome(String word){

    return word==getReverseRecursive(word);

    }

4_9

Ha!... I'm not sure.

TOWERS OF HANOI

I am really really frustrated because it won't let me run the TowersOfHanoi application. I can still do the other implementation and alterations but I can't check that it completley works.....

4_10

            2n^(n-1)

4_11

I didn't really understand this so I got help from classmates.

public void play()

{

    setUpPegs();

    ArrayStack blue=new ArrayStack(maxPegs);

    ArrayStack green=new ArrayStack(maxPegs);

    ArrayStack yellow=new ArrayStack(maxPegs);

    int difficulty; 

    SimpleInput inputDifficulty = new SimpleInput();

    difficulty=inputDifficulty.getInt("How many disks would you like?");

    for (int i=0;i<difficulty;i++)

    {

        disk temp=new disk(20,100-(i*10),0+(i*5),180-(i*20),"red",true);

        blue.push(temp);

        temp.makeVisible();

    }//for

    reDraw(blue,green,yellow);

    while(!(blue.isEmpty() && green.isEmpty()))

    {

        char from, to;

        SimpleInput moveCharacter = new SimpleInput();

        from = moveCharacter.getChar("from: b=blue, g=green, y=yellow");

        to = moveCharacter.getChar("to: b=blue, g=green, y=yellow");

        disk carry=null;

        

        boolean valid = false;

        while (!valid)

        {

        switch (from) 

        {

            case 'b': carry=(disk)blue.pop(); valid = true; break;

            case 'g': carry=(disk)green.pop(); valid = true; break;

            case 'y': carry=(disk)yellow.pop(); valid = true; break;

            default : System.out.println("error, try again");

        }//switch

        }

        boolean nextValid = false;

        while (!nextValid)

        {

        switch (to)

        {

        case 'b': if (blue.isEmpty() || carry.getWidth()<((disk)blue.peekTop()).getWidth())

          {

          carry.changePosition(carry.getXPosition()%100,180-(stackHeight(blue)*20));

          blue.push(carry); nextValid = true; break;

          }

          else

          {

        System.out.println("Cannot put bigger disk on smaller disk");

          }

        case 'g': if (green.isEmpty() || carry.getWidth()<((disk)green.peekTop()).getWidth())

    {

          carry.changePosition((carry.getXPosition()%100)+100,180-(stackHeight(green)*20));

                        green.push(carry); nextValid = true; break;

    }

          else

          {

        System.out.println("Cannot put bigger disk on smaller disk");

          }

            case 'y': if (yellow.isEmpty() || carry.getWidth()<((disk)yellow.peekTop()).getWidth())

              {

            carry.changePosition((carry.getXPosition()%100)+200,180-(stackHeight(yellow)*20));

                        yellow.push(carry); nextValid = true; break;

              }

              else

              {

            System.out.println("Cannot put bigger disk on smaller disk");

              }

            default : System.out.println("error, try again"); 

        }//switch

        }

        reDraw(blue,green,yellow); 

    }//while(main loop)

    System.out.println("Well Done");

}//play

4_12

I have NO idea how to implement the algorithm i found online (which I believe is correct) into a solve method. 

4_13

Everything worked perfectly, for (a/0) I just got an error, as expected.

4_14

Worked fine, it returned false.

4_15

I honestly don't know how to start this. I looked at my classmate's work for this and it's not that I don't understand the syntax.. this time im having trouble with the logic.

4_16

You would need to change every int in your code into into double, besides ofcourse for the int i =... in loops.

4_17

Easy.

switch (operator)

            {

                case '+': result=operand1+operand2; break;

                case '*': result=operand1*operand2; break;

                case '/': {if (operand1==0) {System.out.println("can't divide by 0");return 0;} result=operand2/operand1; break;}

                case '-': result=operand2-operand1; break;

            }//switch

4_18

I have a few ideas of how to approach this, however I am really running out of time. This is something for me to come back to hopefully sometime this week.

Finally. I caught up.

The CatchUp- Part 1

I went back to my previous post and attempted the last exercise and noticed how simple it was. I therefore managed to successfully modify LinkedListTest_Application to incorporate the new methods I made before.
-------------
Here's my recursiveFind method. Wasn't a problem at all.

public ListNode recursiveFind(ListNode list, Object goal){

 if (goal.toString().compareTo(list.getData().toString()) == 0){return list;}

 if (list.getData().toString().compareTo(goal.toString()) == 1){return null;}

 return recursiveFind(list.getNext(), goal);

}

Here is my recursive Length method.  Not a problem either.

public int recursiveListSize(ListNode list){

int count = 0;

if (list.getData()==null) 

{return count;}

{count++;}{return recursiveListSize(list.getNext());}

}

for 3_8 I couldn't think of a good way to approach this. The only thing i thought of doing was doing a count, putting the data on an array and then printing that backwards, but I felt like that was somewhat, cheap.

------------

CircularListNodes wasn't tough, it was pretty straight forward. I have it coded on my Eclipse workspace.

3_11 Having an end node complicates it for me and I get confused however it saves you a bunch of coding lines which you can ignore, but I would prefer writing those lines rather than to get confused.

------------

All I want to say at this stage is that a DoubleList is something beautiful and I wish I knew about it earlier.


3_12-3_14 I did, but I wish there were more specifications so I could know if my  method is right or wrong.


------------
LINKED LIST APPLICATION.


Mr Daly - I would just like to say that I spent about 10 minutes trying to fix your spelling mistakes in your code which gave me errors. It drove me mad.


I am now on 3_16 and can't keep my eyes open, I will continue my catchup mission tomorrow.

Linked Lists Student Tasks 1-6

Student task number one disappointingly confused me a lot.
I managed to get task number two; I know it is not a huge accomplishment because it is quite basic, but it's a start.

public ListNode findInOrderedList(Object goal)

{

  if(listEmpty()){return null;}

  ListNode temp = list;

  while (!(temp.getData().toString().compareTo(goal.toString()) == 1)) // condition that its still in the part that is smaller or equal to it

    {

        if (goal.toString().compareTo(temp.getData().toString()) == 0)

        return temp;

        else

        temp=temp.getNext();//move on to node 2

       }

  return null;

}

Task three took a while, but Amit helped me with it, and I finally understood it. This helped me a lot with the syntax of linked lists and their logic.

public void deleteFromList(Object goal)

 {

if (findInOrderedList(goal) == null) {return;}

else {

ListNode temp = list;

if (goal.toString().compareTo(temp.getNext().getData().toString()) == 0)

temp.setNext((temp.getNext()).getNext());

else

temp = temp.getNext();}

 }

I am not sure why, but the link for task 4 was blocked for me, I am not sure what I was supposed to do, but it didn't seem like such a necessity.

Task 5&6: I honestly was not very sure what was going on. I know realize I need help.

The following week's work is about halfway done, I will post it when I'm done, hopefully in the next day or two. I know I am behind, and it's a little bit frustrating, but I'm catching up.

Week 2 (Insertion and Selection Sort, Recursive Iterative, MergeSort QuickSort)

This week's student tasks were surprisingly not as difficult as I expected. I managed to complete the first student tasks in about 10 minutes which was surprising to me. The rest was more or less a review again, but it got me very confident in using Eclipse. The recursive and iterative methods were not a problem. When I arrived to the mergesort activity I was pretty lost. I know how it works yet I can't seem to find the right way to program it, hopefully we could brush over this in class. I am progressing with python, I have done plenty to give me confidence for the in class quizzes. Overall I am doing well, I might just need a little bit of help dealing with the mergesort and quicksorts.

You have probably heard of the group called Folding@Home, created in Stanford. http://folding.stanford.edu/
I heard about this when I was in London and saw a cool 3D visual representation of protein folding on my friends computer. What it does is it runs through a large amount of algorithms and records the data of the outcomes. When  proteins don't fold correctly they can cause different diseases or syndromes.

Week 1 (Searches, Big O, Visual Rep)

I have completed my first few coding problems using python. They were not difficult at all, the logic is the same and I feel like I am going to grasp the syntax quickly. I went through all the assignments of this week today and noticed that it was more or else of a review of what I already knew, however it really helped me get familiar with Eclipse. The student assignemtn at the end was tough for me I wasn't sure where to start, hopefully we could quickly talk about it in class, otherwise everything else was pretty smooth and self explanatory.