# Could someone give me feedback on my algorithm?

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Hello. I am writing an algorithm that paints an 8 x 8 checkerboard red, and white. However the algorithm at this moment is 87 lines long. Sadly It would be only 64 lines to individually color all 64 squares. Please give me feedback.

Here is some more relevant information. At first I tried to paint every other square red. However it looked like this. (rough illustration)

Now after implementing my algorithm it looks like this.

Variable k represents the current square. j is the counter. So I use it to run a % on in order to determine even or odd.

 if(rects.size()<=8) { if(j%2 == 0 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=16) { if(j%2 == 1 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 0) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=24) { if(j%2 == 0 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=32) { if(j%2 == 1 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 0) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=40) { if(j%2 == 0 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=48) { if(j%2 == 1 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 0) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=56) { if(j%2 == 0 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } } else if(rects.size()<=64) { if(j%2 == 1 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 0) { (*rects[(k)]).set_fill_color(Color::white); } }  Edited by black_darkness

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I don't understand much of what you're doing or working with, but it looks like, without changing the logic of how you wanna paint it, you could use a for loops instead of repeating 8,16,24...64 you could make it 8 * i where i = 1,2,3...8

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I don't understand much of what you're doing or working with, but it looks like, without changing the logic of how you wanna paint it, you could use a for loops instead of repeating 8,16,24...64 you could make it 8 * i where i = 1,2,3...8

Here is an explanation of the goal.

I have this information for my algorithm to work with.

* current square (or k)
* j is a counter. So it changes from even to odd every turn, letting me know if current square is even or odd. (I could use k to do this and make simpler though I think)
* I have the rects.size() (This is the size of the vector)

I want to use this information to paint a checkerboard pattern. In order to do this the first 8 must have odds painted white, and evens painted red. The second 8 must have odds painted red and evens painted white. This will alternate line by line until I get to the bottom.

I couldn't think of how to do this without manually switching every time k increases by 8.

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I think you could immediately shorten that to:

 // This if statement might seem kind of magical if (((rects.size() / 8) % 2) == 0) // if <= 16, 32, 48, 64 { if(j%2 == 1 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 0) { (*rects[(k)]).set_fill_color(Color::white); } } else // <= 8, 24, 40, 56 { if(j%2 == 0 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } } 

However, I wouldn't actually recommend the above, and if I were you, I'd do something more like:
 for (int i = 0; i < rects.size(); ++i) { if (((i + i / 8) % 2) == 0) { rects->set_fill_color(Color::red); } else { rects->set_fill_color(Color::white); } } 

Edited: sorry, I messed up the above the first time. I think it's good now...

Edit again: Another alternative is the below double loop:

 bool white = false; for (int i = 0; i < rects.size(); i += 8) { for (int k = 0; k < 8; ++k) { if (white) { rects[i + k]->set_fill_color(Color::white); } else { rects[i + k]->set_fill_color(Color::red); } white = !white; } white = !white; }  Edited by Cornstalks

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First step is to obtain the 2-dimensional coordinate of a square. For example, if k is what you call the current square is a 1-dimensional sequential index (for example you number the squares sequentially along a row, and column by column), then you can obtain the (x,y) coordinate as x=k%8 and y=k/8.

Once you have a 2-dimensional coordinate into the checker board, you can color it according to:
 if(x%2 == y%2) red square else white square 

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I think you could immediately shorten that to:
 // This if statement might seem kind of magical if (((rects.size() / 8) % 2) == 0) // if <= 16, 32, 48, 64 { if(j%2 == 1 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 0) { (*rects[(k)]).set_fill_color(Color::white); } } else // <= 8, 24, 40, 56 { if(j%2 == 0 ) { (*rects[(k)]).set_fill_color(Color::red); } else if(j%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } } 

Yeah I kind of want to avoid another solution like this.

However, I wouldn't actually recommend the above, and if I were you, I'd do something more like:
 for (int i = 0; i < rects.size(); ++i) { if ((i % 2) == 0) { rects->set_fill_color(Color::red); } else { rects->set_fill_color(Color::white); } } 

I tried that at first but since the first square on the left is always even it made a red and white stripe pattern. So that is why I was forced to switch every time k increased by 8.

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First step is to obtain the 2-dimensional coordinate of a square. For example, if k is what you call the current square is a 1-dimensional sequential index (for example you number the squares sequentially along a row, and column by column), then you can obtain the (x,y) coordinate as x=k%8 and y=k/8.

Once you have a 2-dimensional coordinate into the checker board, you can color it according to:
 if(x%2 == y%2) red square else white square 

This seems like it will work thank you. I will try to implement a 2d system.

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*snip*
I tried that at first but since the first square on the left is always even it made a red and white stripe pattern. So that is why I was forced to switch every time k increased by 8.

Sorry, yeah, I forgot to watch out for that. I've edited that post just barely to fix that and add another alternative way of doing it.

Edit: Holy ninja'ing fest... We're all replying at once ha Edited by Cornstalks

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Can be reduced further:
 int sqIdx = 0; for ( row = 0; row < rowCount; ++row ) { // Each row in checkerboard for ( col = 0; col < colCount; ++col ) { // Each cell in column Color clr; if ( (row+col)%2 == 1 ) clr = Color::red; else clr = Color::white; rects[sqIdx++]->set_fill_color(clr); // Or "row*colCount+col" in place of "sqIdx++" } } 

And reduced further:
 Color lookup[2] = { Color::red, Color::white }; for ( row = 0; row < rowCount; ++row ) for ( col = 0; col < colCount; ++col ) rects[row*colCount+col] = lookup[(row+col)&1]; 

Also, ninja'd to the floor.

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Hey Cornstalks I wrote a test and it works but It is switching 1 square too soon.

Here is the test I wrote.
 #include <iostream> int main() { using namespace std; int ooo_test = (1+1/8) % 2; cout<<"square 1 is: "<<ooo_test<<"\n"; ooo_test = (2+2/8) % 2; cout<<"square 2 is: "<<ooo_test<<"\n"; ooo_test = (3+3/8) % 2; cout<<"square 3 is: "<<ooo_test<<"\n"; ooo_test = (4+4/8) % 2; cout<<"square 4 is: "<<ooo_test<<"\n"; ooo_test = (5+6/8) % 2; cout<<"square 5 is: "<<ooo_test<<"\n"; ooo_test = (6+6/8) % 2; cout<<"square 6 is: "<<ooo_test<<"\n"; ooo_test = (7+7/8) % 2; cout<<"square 7 is: "<<ooo_test<<"\n"; ooo_test = (8+8/8) % 2; cout<<"square 8 is: "<<ooo_test<<"\n"; cin>>ooo_test; } 

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Hey Cornstalks I wrote a test and it works but It is switching 1 square too soon.

The test actually works, but you are misinterpreting the results. Square 1 is the second square of the first row, and square 8 is the first square of the second row. Both square 1 and square 8 should be red. Square 0, the first square of the first row, on the other hand, should be (and is) white.

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In C++, indices start at zero, not one, so your first one should be [font=courier new,courier,monospace]0 + 0 / 8[/font], your second [font=courier new,courier,monospace]1 + 1 / 8[/font], third [font=courier new,courier,monospace]2 + 2 / 8[/font], etc.

Thank you for the algorithm. I will have to keep this in mind. Is there a formal name for this type of algorithm?

anyways look how elegant you made it look. I am very pleased.

 if((k+k/8)%2 == 0) { (*rects[(k)]).set_fill_color(Color::red); } else if((k+k/8)%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); } k++; 

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[quote name='black_darkness' timestamp='1355441798' post='5010392']Hey Cornstalks I wrote a test and it works but It is switching 1 square too soon.

The test actually works, but you are misinterpreting the results. Square 1 is the second square of the first row, and square 8 is the first square of the second row. Both square 1 and square 8 should be red. Square 0, the first square of the first row, on the other hand, should be (and is) white.
[/quote]

I always thought that if you wrote a for loop it added one then ran what was inside the curly braces. oops. You were right. I wrote this test. I always did get a lot of 1 off errors.

 #include <iostream>; int main() { using namespace std; for(int i=0;i<5;i++) { cout<<i<<"\n"; } return 0; } 

and this was the output.

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Well then at least remove the "!" by switching the colors around

 rects[k]->set_fill_color((k+k/8)%2 ? Color::white : Color::red); 

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Thank you for the algorithm. I will have to keep this in mind. Is there a formal name for this type of algorithm?

Yes, the formal name is the Cornstalks Algorithm
Just kidding, there isn't a formal name for it. It might be possible to generalize it to a larger, more complex algorithm that has a formal name, but I just came up with this off the top of my head. Years of programming will do that to you...

I always thought that if you wrote a for loop it added one then ran what was inside the curly braces. oops. You were right. I wrote this test. I always did get a lot of 1 off errors.

I'll break down a for loop like this:
 for (A; B; C) D; 
Step 1: A is done (A is usually creating and setting a variable, like [font=courier new,courier,monospace]int [/font][font=courier new,courier,monospace]i = 0[/font]). Step 2: B is checked (B is the looping condition, and as long as it's true the loop is run). Step 3: D is run. Step 4: C is run (which is usually what updates the loop counter). Then it goes back to step 2 and repeats until B is false.

Couldn't resist bringing this thread to its logical conclusion by one-lining Álvaro's code.

One liner? (Added the loop )
 // This: for (int y = 0; y < 8; ++y) for (int x = 0; x < 8; ++x) rects[x + y * 8]->set_fill_color(((x + y) % 2) ? Color::white : Color::red); // Or: for (int i = 0; i < rects.size(); ++i) rects->set_fill_color(((i + i / 8) % 2) ? Color::white : Color::red); // But seriously, I hope no one ever does this in real life. Use *at least* two lines...  Edited by Cornstalks

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To test for odd vs even don't use modulation. Just binary-and with 1.

0 & 1 = 0
1 & 1 = 1
2 & 1 = 0
3 & 1 = 1
4 & 1 = 0
5 & 1 = 1 Edited by Khatharr

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This is true. Personally I always use parens to avoid ambiguity and in this case since you're feeding it to a ? operator you can just swap the result order and drop the explicit comparison to zero.

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For the record, this thread was a joy to read. So everyone got +1.

Carry on.

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 if((k+k/8)%2 == 0) { (*rects[(k)]).set_fill_color(Color::red); } else if((k+k/8)%2 == 1) { (*rects[(k)]).set_fill_color(Color::white); }

You need to read this article, and perhaps some of the comment as well:
http://thedailywtf.com/Articles/ButAnything-Can-Happen!.aspx
The thing you need to learn here is that you are mis-using else-if. There are only two possible outcomes of a number mod 2, ergo it is appropriate to just use an else for the second case.

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Lol @ the article.

[source lang="cpp"]for(int i = 0; i < rects.size(); ++i) rects->set_fill_color((i & 1) ? Color::white : Color::red);[/source] Edited by Khatharr

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To test for odd vs even don't use modulation. Just binary-and with 1.

Use whichever one you find more clear. I personally find %2 more clear than &1 in this situation. Edited by Álvaro