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After the player has chosen some blue pencils (by means of a rubber rectangle).

Further, the program paints in red color some pencils and remove them from one heap.

## Introduction

### What is the Nim Game

Nim is a two-player mathematical game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.

Variants of Nim have been played since ancient times. The game is said to have originated in China (it closely resembles the Chinese game of "Jianshizi", or "picking stones"), but the origin is uncertain; the earliest European references to Nim are from the beginning of the 16^{th} century.

Its current name was coined by Charles L. Bouton of Harvard University who also developed the complete theory of the game in 1901, but the origins of the name were never fully explained. The name is probably derived from German nimm! meaning "take!", or the obsolete English verb nim of the same meaning. Some people have noted that turning the word NIM upside-down and backwards results in WIN.

Nim is usually played as a misere game, in which the player to take the last object loses. Nim can also be played as a normal play game, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because most games follow this convention, even though Nim usually does not.

## XOR Tricks for Win at Nim

Although it takes some high-level math to find the secret strategy of Nim, employing that strategy really only requires an understanding of binary numbers.

In order to win at Nim, you have to know how to use the binary operation XOR, which stands for "exclusive or". What XOR really means is "x XOR y = 1 if either x or y is 1, but not if they are both 1." So, 0 XOR 0 = 0, 1 XOR 0 = 1, 0 XOR 1 = 1, and 1 XOR 1 = 0. Or, more simply, the result of the XOR operation is 0 if both arguments are the same and 1 if the arguments are different.

For any given situation in Nim, there is a number which determines whether or not that situation is a losing one. To find that number, you have to perform the XOR operation on the number of objects in each row successively. For example, the position is:

So to see if this is a losing position, we have to XOR the number of objects in each row, as follows:

1 XOR 3 XOR 5

which when shown in binary is:

001 XOR 011 XOR 101

Now we have to XOR each digit in each number and put each result below the column for that digit. Let's start with the rightmost digits.

1 XOR 1 XOR 1 is the same as (1 XOR 1) XOR 1

1 XOR 1 = 0, and 0 XOR 1 = 1. So 1 XOR 1 XOR 1 = 1.

Now continue with the rest of the columns until we have a new binary number.

001 XOR 011 XOR 101 = 111

If all the digits of this final number are zero, the position is a losing position!!!!!!! If it's your turn and the position is a losing position, you're in trouble. However, in this example, the number is non-zero, so we can turn the position into a losing position for our opponent.

Let's take the number 111 that we got from XORing the rows and try to find a row which, when XORed with 111, gives us a lower number than the row previously had. Well, we know if we do 001 XOR 111 it will be greater than 1, and 011 XOR 111 will be greater than 3, so we must do 101 XOR 111, which is 010, or in decimal 2, and is less than 5. So in order to give your opponent a losing position, you just have to remove 3 objects from the row of 5, leaving 2.

### Summary

The steps needed to win are:

- When it is your turn, convert the number of objects in each row into binary numbers and XOR them.
- If the resulting number is
`0`

, there's not much you can do to win. If it isn't `0`

, XOR it with a row and make a move so as to leave that many objects in that row.

For more information about Mathematical theory of game, see the Wikipedia article.

## Using the Code

The following code demonstrates how to create a `Form`

. Let us draw something on a `Form`

's Client Area.

- The following code illustrates the process of dynamic creation and initialization of these heaps of pencils. For save and display images, we use Jagged array:
private Bitmap[][] images;
void CreateNew_Images()
{
images = new Bitmap[8][];
for(int i=0; i<8; i++)
{
images[i] = new Bitmap [30];
for(int j=0; j<30; j++)
{
images[i][j]= new Bitmap(tempImage);
images[i][j].MakeTransparent();
}
}
}
void Fill_Images()
{
int new_number_elements;
Random rdm;
for(int i=0; i < numberOfColumns; i++)
{
rdm =
newRandom(unchecked((int)DateTime.Now.Ticks));
new_number_elements = rdm.Next(1,30);
SumOfImages+ =
new_number_elements; for(int
j=
0;j < new_number_elements; j++)
{
images[i][j]= bitmapGray;
images[i][j].MakeTransparent();
}
numberLoadImages.SetValue(new_number_elements,i);
System.Threading.Thread.Sleep(35);
}
Invalidate();
}

**Overriding appropriate method**
The `OnPaint`

method for a form or a control is called by the operating system whenever part or all of that form or control is obscured by another form or control and needs to be redrawn. A `Graphics`

object is passed as part of the `PaintEventArgs`

parameter to the `OnPaint`

method. Using this `Graphics`

object, you redraw the portion of the display that has been obscured.

Why `OnPaint`

method? The main reason is we can get `Graphics`

object easily in `OnPaint`

method. In this method, I draw objects rotation after mouse click.

protected override void OnPaint(PaintEventArgs e)
{
bool flag=false;
e.Graphics.DrawImageUnscaled(PictureBitmap, 0, 0);
if(flag_colors==ColorsPen.Blue)
{
bitmapBlueSharper.MakeTransparent(Color.White);
e.Graphics.DrawImage(bitmapBlueSharper,((int)Xposition.GetValue(
current_column)),0);
}
else if(flag_colors==ColorsPen.Red)
{
bitmapRedSharper.MakeTransparent(Color.White);
e.Graphics.DrawImage(bitmapRedSharper,((int)Xposition.GetValue(
current_column)),0);
}
int x=-160;
for(int i= 0;i<8; i++,x+=120)
{
int y=-18;
for(int j= 0;j<((int)numberLoadImages.GetValue(i)); j++)
{
if(i==current_column && j==current_element_in_column)
{
flag=true;
}
else
{
if(i== current_column &&j>current_element_in_column)
{
images[i][j]=this.bitmapGray;
e.Graphics.DrawImageUnscaled(images[i][j], xCurrentTop+x,
yCurrentTop+y+posDown);
y-=18;
e.Graphics.ResetTransform();
}
else
{
e.Graphics.DrawImageUnscaled(images[i][j],
xCurrentTop+x, yCurrentTop+y);
y-=18;
e.Graphics.ResetTransform();
}
}
}
yCurrentTop = yPrevTop;
}
if(flag)
{
positionY=yCurrentTop+posY+
images[current_column][current_element_in_column].Height/2;
e.Graphics.TranslateTransform( xCurrentTop+posX+
images[current_column][current_element_in_column].Width/2,positionY);
e.Graphics.RotateTransform(angle);
e.Graphics.ScaleTransform(1,(float) delta);
images[current_column][current_element_in_column].MakeTransparent(
Color.White);
e.Graphics.DrawImage(
images[current_column][current_element_in_column],
new Rectangle(-images[current_column]
[current_element_in_column].Width,
-images[current_column][current_element_in_column].Height,
images[current_column][current_element_in_column].Width,
images[current_column][current_element_in_column].Height));
e.Graphics.ResetTransform();
bitmapBlueSharper.MakeTransparent(Color.White);
if(flag_colors==ColorsPen.Blue)
e.Graphics.DrawImage(this.bitmapBlueSharper,
((int)Xposition.GetValue(current_column)),0);
else if(flag_colors==ColorsPen.Red )
e.Graphics.DrawImage(this.bitmapRedSharper,
((int)Xposition.GetValue(current_column)),0);
}
base.OnPaint(e);
}

- Also, I want to explain how to draw a rubber band rectangle.
Rubber rectangle is a Grouping object.

Group uses rubber band to collect a set of objects (pencils).

In other words, the `pencil`

objects are completely independent of one another (rubber rectangle can always be ungrouped - each member maintains almost complete autonomy from the other members of its group).

Figure on the left includes a set of grouped objects (pencils);

A **rubber** rectangle technique is commonly used to delimit a selection in response to user mouse-pointer input. This term is used to describe the situation where:

- the left mouse button is held down, defining one corner of the rectangle
- the mouse is dragged and released at the point defining the opposite corner

of the rectangle - the rectangle is drawn while the mouse is being dragged, so that it looks like the

rectangle is being stretched and contracted, like a rubber band

For more information about Rubber rectangle, visit MSDN at Microsoft here.

The following sample code demonstrates how to use the rubber rectangle:

private void MyDrawReversibleRectangle( Point p1, Point p2 )
{
p1 = PointToScreen( p1 );
p2 = PointToScreen( p2 );
Rectangle rc = new Rectangle(p1.X, p1.Y, p2.X - p1.X, p2.Y - p1.Y);
ControlPaint.DrawReversibleFrame( rc, Color.Black, FrameStyle.Thick );
}
private void Form1_MouseMove(object sender,
System.Windows.Forms.MouseEventArgs e)
{
if(this.timer1.Enabled==true || this.timer2.Enabled==true)
return;
if( rubber_rect_flag )
{
MyDrawReversibleRectangle( FirstPoint, SecondPoint );
SecondPoint = new Point(e.X,e.Y);
MyDrawReversibleRectangle( FirstPoint, SecondPoint );
}
}