## Image rotation with bilinear interpolation

In this article, I’ll show you how to rotate an image about its centre. 3 assignment methods will be shown,

• assign source pixels to destination pixels
• assign destination pixels from source pixels
• assign destination pixels from source pixels with bilinear interpolation

I’ll show you the code, then the results for comparison. So what’s bilinear interpolation?

### Bilinear interpolation

Read up on linear interpolation first if you haven’t done so. “Bilinear” means there are 2 directions to interpolate. Let me illustrate.

In our case, we’re interpolating between 4 pixels. Visualise each pixel as a single point. Linearly interpolate between the top 2 pixels. Linearly interpolate between the bottom 2 pixels. Then linearly interpolate between the calculated results of the previous two.

You can expand on this concept to get trilinear interpolation.

LERPs is a short form of linear interpolations. When would trilinear interpolation be useful? Voxels, which is out of scope in this article.

### Defining the centre of an image

I’m going to be fuzzy about this. I’m going to just take one pixel in the image and define it as the centre. This pixel is defined as having a horizontal index equal to half of its width (rounded down), and a vertical index equal to half its height (rounded down).

This means the image isn’t rotated about its “true” centre, but with a relatively large size, it won’t matter anyway. It’s not like you’re rotating an image of 5 pixel width and 3 pixel height, right?

### The preparation part

The actual code is quite long, so I’m separating it into 4 parts.

• Initialisation and variable declaration
• Assigning source pixels to destination pixels
• Assigning destination pixels from source pixels
• Assigning destination pixels from source pixels with bilinear interpolation

It’s hard-coded with -30 degrees as the angle of rotation, but you can easily write it into a function.

```// 30 deg = PI/6 rad
// rotating clockwise, so it's negative relative to Cartesian quadrants
const double cnAngle = -0.52359877559829887307710723054658;
// use whatever image you fancy
Bitmap bm = new Bitmap("rotationsource.jpg");
// general iterators
int i, j;
// calculated indices in Cartesian coordinates
int x, y;
double fDistance, fPolarAngle;
// for use in neighbouring indices in Cartesian coordinates
int iFloorX, iCeilingX, iFloorY, iCeilingY;
// calculated indices in Cartesian coordinates with trailing decimals
double fTrueX, fTrueY;
// for interpolation
double fDeltaX, fDeltaY;
// pixel colours
Color clrTopLeft, clrTopRight, clrBottomLeft, clrBottomRight;
// interpolated "top" pixels
double fTopRed, fTopGreen, fTopBlue;
// interpolated "bottom" pixels
double fBottomRed, fBottomGreen, fBottomBlue;
// final interpolated colour components
int iRed, iGreen, iBlue;
int iCentreX, iCentreY;
int iWidth, iHeight;
iWidth = bm.Width;
iHeight = bm.Height;

iCentreX = iWidth / 2;
iCentreY = iHeight / 2;

Bitmap bmSourceToDestination = new Bitmap(iWidth, iHeight);
Bitmap bmDestinationFromSource = new Bitmap(iWidth, iHeight);
Bitmap bmBilinearInterpolation = new Bitmap(iWidth, iHeight);

for (i = 0; i < iHeight; ++i)
{
for (j = 0; j < iWidth; ++j)
{
// initialise when "throwing" values
bmSourceToDestination.SetPixel(j, i, Color.Black);
// since we're looping, we might as well do for the others
bmDestinationFromSource.SetPixel(j, i, Color.Black);
bmBilinearInterpolation.SetPixel(j, i, Color.Black);
}
}
```

Some of it might not mean anything to you yet. Just wait for the rest of the code. You might want to read up on converting between raster, Cartesian and polar coordinates first before moving on.

### Throwing values from source to destination

```// assigning pixels from source image to destination image
for (i = 0; i < iHeight; ++i)
{
for (j = 0; j < iWidth; ++j)
{
// convert raster to Cartesian
x = j - iCentreX;
y = iCentreY - i;

// convert Cartesian to polar
fDistance = Math.Sqrt(x * x + y * y);
fPolarAngle = 0.0;
if (x == 0)
{
if (y == 0)
{
// centre of image, no rotation needed
bmSourceToDestination.SetPixel(j, i, bm.GetPixel(j, i));
continue;
}
else if (y < 0)
{
fPolarAngle = 1.5 * Math.PI;
}
else
{
fPolarAngle = 0.5 * Math.PI;
}
}
else
{
fPolarAngle = Math.Atan2((double)y, (double)x);
}

// the crucial rotation part
fPolarAngle += cnAngle;

// convert polar to Cartesian
x = (int)(Math.Round(fDistance * Math.Cos(fPolarAngle)));
y = (int)(Math.Round(fDistance * Math.Sin(fPolarAngle)));

// convert Cartesian to raster
x = x + iCentreX;
y = iCentreY - y;

// check bounds
if (x < 0 || x >= iWidth || y < 0 || y >= iHeight) continue;

bmSourceToDestination.SetPixel(x, y, bm.GetPixel(j, i));
}
}
bmSourceToDestination.Save("rotationsrctodest.jpg", System.Drawing.Imaging.ImageFormat.Jpeg);
```

It should be fairly easy to read. Note the part about checking for the central pixel of the image. No rotation calculation necessary, so we assign and move to the next pixel. Note also the part about checking boundaries.

### Finding values from the source

```// assigning pixels of destination image from source image
for (i = 0; i < iHeight; ++i)
{
for (j = 0; j < iWidth; ++j)
{
// convert raster to Cartesian
x = j - iCentreX;
y = iCentreY - i;

// convert Cartesian to polar
fDistance = Math.Sqrt(x * x + y * y);
fPolarAngle = 0.0;
if (x == 0)
{
if (y == 0)
{
// centre of image, no rotation needed
bmDestinationFromSource.SetPixel(j, i, bm.GetPixel(j, i));
continue;
}
else if (y < 0)
{
fPolarAngle = 1.5 * Math.PI;
}
else
{
fPolarAngle = 0.5 * Math.PI;
}
}
else
{
fPolarAngle = Math.Atan2((double)y, (double)x);
}

// the crucial rotation part
// "reverse" rotate, so minus instead of plus
fPolarAngle -= cnAngle;

// convert polar to Cartesian
x = (int)(Math.Round(fDistance * Math.Cos(fPolarAngle)));
y = (int)(Math.Round(fDistance * Math.Sin(fPolarAngle)));

// convert Cartesian to raster
x = x + iCentreX;
y = iCentreY - y;

// check bounds
if (x < 0 || x >= iWidth || y < 0 || y >= iHeight) continue;

bmDestinationFromSource.SetPixel(j, i, bm.GetPixel(x, y));
}
}
bmDestinationFromSource.Save("rotationdestfromsrc.jpg", System.Drawing.Imaging.ImageFormat.Jpeg);
```

The key difference here is the use of the rotation angle. Instead of adding it, we subtract it. The reason is, we rotate source pixels 30 degrees clockwise and assign it to destination pixels. But from destination pixels, we get source pixels which are rotated 30 degrees anticlockwise. Either way, we get a destination image that's the source image rotated 30 degrees clockwise.

Also compare the assignment, noting the indices:

```bmSourceToDestination.SetPixel(x, y, bm.GetPixel(j, i));
bmDestinationFromSource.SetPixel(j, i, bm.GetPixel(x, y));
```

x and y variables are calculated and thus "messy". I prefer my messy indices on the right. There's a practical reason for it too, which will be evident when I show you the rotation results.

### Image rotation code with bilinear interpolation

```// assigning pixels of destination image from source image
// with bilinear interpolation
for (i = 0; i < iHeight; ++i)
{
for (j = 0; j < iWidth; ++j)
{
// convert raster to Cartesian
x = j - iCentreX;
y = iCentreY - i;

// convert Cartesian to polar
fDistance = Math.Sqrt(x * x + y * y);
fPolarAngle = 0.0;
if (x == 0)
{
if (y == 0)
{
// centre of image, no rotation needed
bmBilinearInterpolation.SetPixel(j, i, bm.GetPixel(j, i));
continue;
}
else if (y < 0)
{
fPolarAngle = 1.5 * Math.PI;
}
else
{
fPolarAngle = 0.5 * Math.PI;
}
}
else
{
fPolarAngle = Math.Atan2((double)y, (double)x);
}

// the crucial rotation part
// "reverse" rotate, so minus instead of plus
fPolarAngle -= cnAngle;

// convert polar to Cartesian
fTrueX = fDistance * Math.Cos(fPolarAngle);
fTrueY = fDistance * Math.Sin(fPolarAngle);

// convert Cartesian to raster
fTrueX = fTrueX + (double)iCentreX;
fTrueY = (double)iCentreY - fTrueY;

iFloorX = (int)(Math.Floor(fTrueX));
iFloorY = (int)(Math.Floor(fTrueY));
iCeilingX = (int)(Math.Ceiling(fTrueX));
iCeilingY = (int)(Math.Ceiling(fTrueY));

// check bounds
if (iFloorX < 0 || iCeilingX < 0 || iFloorX >= iWidth || iCeilingX >= iWidth || iFloorY < 0 || iCeilingY < 0 || iFloorY >= iHeight || iCeilingY >= iHeight) continue;

fDeltaX = fTrueX - (double)iFloorX;
fDeltaY = fTrueY - (double)iFloorY;

clrTopLeft = bm.GetPixel(iFloorX, iFloorY);
clrTopRight = bm.GetPixel(iCeilingX, iFloorY);
clrBottomLeft = bm.GetPixel(iFloorX, iCeilingY);
clrBottomRight = bm.GetPixel(iCeilingX, iCeilingY);

// linearly interpolate horizontally between top neighbours
fTopRed = (1 - fDeltaX) * clrTopLeft.R + fDeltaX * clrTopRight.R;
fTopGreen = (1 - fDeltaX) * clrTopLeft.G + fDeltaX * clrTopRight.G;
fTopBlue = (1 - fDeltaX) * clrTopLeft.B + fDeltaX * clrTopRight.B;

// linearly interpolate horizontally between bottom neighbours
fBottomRed = (1 - fDeltaX) * clrBottomLeft.R + fDeltaX * clrBottomRight.R;
fBottomGreen = (1 - fDeltaX) * clrBottomLeft.G + fDeltaX * clrBottomRight.G;
fBottomBlue = (1 - fDeltaX) * clrBottomLeft.B + fDeltaX * clrBottomRight.B;

// linearly interpolate vertically between top and bottom interpolated results
iRed = (int)(Math.Round((1 - fDeltaY) * fTopRed + fDeltaY * fBottomRed));
iGreen = (int)(Math.Round((1 - fDeltaY) * fTopGreen + fDeltaY * fBottomGreen));
iBlue = (int)(Math.Round((1 - fDeltaY) * fTopBlue + fDeltaY * fBottomBlue));

// make sure colour values are valid
if (iRed < 0) iRed = 0;
if (iRed > 255) iRed = 255;
if (iGreen < 0) iGreen = 0;
if (iGreen > 255) iGreen = 255;
if (iBlue < 0) iBlue = 0;
if (iBlue > 255) iBlue = 255;

bmBilinearInterpolation.SetPixel(j, i, Color.FromArgb(iRed, iGreen, iBlue));
}
}
bmBilinearInterpolation.Save("rotationbilinearinterpolation.jpg", System.Drawing.Imaging.ImageFormat.Jpeg);
```

This part is similar to the destination-from-source part, with a lot more calculations. We have to find the 4 pixels that surrounds our "true" position-calculated pixel. Then we perform linear interpolation on the 4 neighbouring pixels.

We need to interpolate for the red, green and blue components individually. Refer to my article on colour theory for a refresher.

### Pictures, pictures!

After doing all that, we're finally done. Let me show you my source image first.

I added the marble cylinder for emphasising image quality. I needed something that's straight (vertically or horizontally) in the source image.

Here's what we get after rotating with the source-to-destination method:

Note the speckled black pixels dotted all over. This is because some of the destination pixels (which are within the image bounds) were unassigned.

Note also that the specks even form patterns. This is due to the sine and cosine functions, and the regularity of pixel width and height. Sine and cosine are periodic functions. Since pixel indices are regular, therefore sine and cosine results are regular too. Hence, calculations regularly fail to assign pixel values.

There might be source pixels that have the same calculated destination pixel (due to sine and cosine and rounding). This also implies that there might be anti-gravity destination pixels that no source pixel can ever matched to! I haven't verified this, but it seems a possibility.

Still think you should iterate over the source (image/array) instead of over the destination?

Next, we have the image result of the destination-from-source method:

Compare the quality with the source-to-destination part. No missing pixels. It's still sort of grainy though. This is because some of the destination pixels get their values from the same source pixel, so there might be 2 side-by-side destination pixels with the same colour. This gives mini blocks of identical colour in the result, which on the whole, gives an unpolished look.

Now, we have the bilinear interpolation incorporated version.

It looks smoother, right? Note the straight edge of the marble cylinder. Compare with the image result without bilinear interpolation.

I might do a version with cubic interpolation for even smoother results, but I feel the bilinear version is good enough for now. Have fun!

[UPDATE: Now, there's a version without the resulting image being clipped.]

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## Linear and cubic interpolation

Interpolation is a method of calculating a value from a set of given values. We’ll be looking at interpolation with a bias towards image processing, but the theory can be generalised for other purposes. You’ve probably already solved some interpolation problems without knowing it. Let me give you an example.

### A distance problem

Suppose there are 3 towns A, B, C and they happen to lie on a straight line, in that order. B is 5 kilometres away from A, and C is 15 kilometres away from A. If you travel one quarter of the way from town B to town C, how far are you from town A?

To solve it, you can figure out the distance between B and C, which is 15 – 5 = 10 km. One quarter of the way means 1/4 * 10 = 2.5 km. Then add the distance between A and B to this and you have 5 + 2.5 = 7.5 km.

### Linear interpolation

If you visualise the problem as interpolating between 2 points, then B becomes the point p0 with a value of 5 (km) and C becomes the point p1 with a value of 15 (km). The usual variable used is t, so the generic formula is:
f(t) = (1 – t) * p0 + t * p1, where t lies between 0 and 1 inclusive.

Using this, we have
f(1/4) = (1 – 1/4) * 5 + 1/4 * 15
= 3/4 * 5 + 1/4 * 15
= 7.5

This is linear interpolation. Linearity refers to the power of the variable t, which is 1. Note that there’s no stopping you from using negative values of t or values greater than 1.

Suppose you travelled from B to A one quarter of the distance between B and C. How far are you from town A?
f(-1/4) = (1 – (-1/4)) * 5 + (-1/4) * 15
= 5/4 * 5 – 1/4 * 15
= 2.5

Suppose you travelled from B to C and went past C by a quarter of the way. How far are you from town A?
f(5/4) = (1 – 5/4) * 5 + 5/4 * 15
= -1/4 * 5 + 5/4 * 15
= 17.5

What happens if you get a negative result?
f(-1) = (1 – (-1)) * 5 + (-1) * 15
= 2 * 5 – 15
= -5

It means you’re 5 kilometres away from town A. You’re just in the opposite direction from towns B and C. The calculation result is correct. It’s how you interpret the value.

### Applications in image processing

A common operation in image processing is manipulating height maps. Height maps are usually greyscale bitmap files where a white pixel (RGB values are 255 for all 3) is the highest point, and a black pixel (RGB values are 0 for all 3) is the lowest point.

You know enlarging photographs can give you some weird results. What happens is you’re trying to fill in the blanks in a larger image using values from the original image. Where do you think the image editing software comes up with values? Interpolation.

If you think of the red, green and blue values of image pixels as 3 different “height maps”, then you’re just performing interpolation on 3 values. Suppose we’re talking about linear interpolation between two pixels. You’ll interpolate between the red component of the 2 pixels and get a value. Similarly you do it for the green and blue components. The calculated results of the red, green and blue become the interpolated colour.

### Cubic Bezier interpolation

There are all kinds of cubic curves available. The Catmullâ€“Rom spline, the non-uniform rational B-spline (NURBS) and I didn’t really want to write anything on the subject after I remember my Hermite splines… I love Bezier curves though, so I thought maybe I can write something with that.

Instead of 2 points used in linear interpolation, cubic interpolation uses 4 points. To illustrate, suppose you’re on an undulating plain with small hills undulating in their usual carefree manner. You’re in between two such (undulating) hills and you want to find out how high you are.

Instead of linear interpolating your way through these two (undulating) hills, the better way will be to interpolate with another 2 (undulating) hills! Ok, I’m stopping with the undulating thing…

The Bezier curve equation looks like this:
B(t) = (1-t)^3 * p0 + 3*(1-t)^2 * t * p1 + 3*(1-t)* t^2 * p2 + t^3 * p3
where p0, p1, p2, p3 are the (height) values, and t lies between 0 and 1 inclusive.

You will be between p1 and p2. Let’s also assume that the hills are equidistant from each other. Like the pixels on an image, the hills shall be of equal distance from its neighbour.

Because of this equidistant property, p1 is 0.33 (roughly 1/3) units away from p0, p2 is 0.67 (roughly 2/3) units away from p0 and p3 is 1 unit away from p0.

How do you know what’s the value of t to use? You might be able to calculate the t if you do linear interpolation between p1 and p2. But that t value is different from the t value in the Bezier curve.

Ahhh… once you get the t-linear value, you interpolate with 0.33 and 0.67 to get the t-Bezier value. Confused? Suppose you’re one quarter way from p1 to p2. Your t-linear value is 1/4. Interpolate that with 0.33 and 0.67 to get
f(1/4) = (1 – 1/4) * 0.33 + 1/4 * 0.67
= 0.415

And 0.415 is your t-Bezier value. Voila!

### You skipped the quadratic power!

I know. It’s logical to think that there’s a power 2 somewhere. But there isn’t. There is one fundamental flaw with quadratic interpolation. Which segment do you use?

### In closing

Interpolation is just a method of creating data values using a set of existing data. What those created values mean is up to you to interpret.

In image processing, interpolation can be used to fill in blanks when enlarging an image. It doesn’t guarantee that the enlarged image looks good. Image processing is very much an aesthetic-based operation. I’ll talk a bit more on this when I get to writing code to rotate images.