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I meant
4 is at location image1[0][1]
4 is the value.


so,

top = 1;
bottom = 2;
left = 1;

{1, 2, 3},
{4, 5, 6},
{7, 8, 9}

1 is (0,0)
2 is (1,0)
3 is (2, 0)
4 is (0, 1)
.
.

Sorry for the confusion.

The first 2d array is:
int[][] image1 = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9}, };
The second 2d array is:
top = 1;
bottom = 2;
left = 1;

(top, left) is the coordinates for the top left corner of the 2d array.
(bottom,right) is the coordinates for the bottom right corner of the 2d array.

So, we have 4 variables.
...

Given a 2d array
and

4 variables that creates another 2d array.
(top, left) and (bottom,right)

(Those are the coordinates to form another 2d array.)
These variables can partially cover the...

How do I calculate the mean?
The answer for this case is 7.
Why?

My assignment is here(if you want to know the question but I don't think you need it to answer "my" question):...

So, a "sub-region of the image" is a rectangle. So, how do I compute the average of a rectangle?

This is my assignment:
http://www.cse.yorku.ca/course/2011/assignments/assignment%201/Q2/Q2.html

I have no idea what is an image value. I tired looking it up but there's no help.

All the codes...

Sorry, I just posted a question under a section and I notice there is a better section for my type of question.
So, I posted a question twice...
eek...

I don't understand the palindrome word a(b^n)a.

Based on the pumping lemma definition:
The pumping states that: |word| >=n
and
The length of this word is (1 + n + 1) which is...

I think he is asking us to find a relation with those numbers and to print those numbers using a nested for loop.

=)

if t.hasALeft(v) then
{
left = t.left(v)
}

return weight(t, left) + weight(t, right)

if t.hasARight(v) then
{
right = t.Right(v)

I figured it out.


static boolean method (BinaryTree t)
{
...
return method2(t, t.root());
}

I'm not sure how to write a recursive method with a tree as a parameter.
I'm guessing you have to call its subtree every time.

In my book, I'm given BinaryTree t and Node v as my parameter. So, I can traverse the tree by doing like:



static boolean add(BinaryTree t, Node v )
return add(t, t.left(v)) + add(t,...

Ok. =)
I usually ask people if I'm on the right track. So, I guess it's ok to ask them to have a quick check on my work.
I'll try to post more detailed questions and posts.
Thanks for your tips.

I think you're right.

So, I changed my code so that the base case meets your requirement.
So my code right???



// returns the weight of node v
static int weight(BinaryTree t, Node v)
if...

I was about to do it your way, but I notice that in my book, there's no method t.hasRight(v) and t.right(v) . It only checks for the left children.
So, I guess I have to stick to my original code.
...

Hi, I'm new here.

I found the reply for my posted question to be slow. If I'm stuck on something, I have to wait a day until I get a reply.

My suggestion is to make a chatbox on this forum, so...

I'm suppose to add up all the number that are in a node of a binary tree.

Is this correct???

This is just the algorithm:::


static int weight(BinaryTree t, Node v)
if t.size == 0 then //if...

public static int f(int a, int b)
{
Z.h1(a, b); //// arbitrary code

if (Z.h2(a, b))
return Z.h3(a, b);

else
{
...