# 2.1 Implementation of scale-invariant feature transform

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## Introduction

The Scale- Invariant Feature Transform is an algorithm in computer vision to detect and describe points of interest in an image.

## Approach

According to Prof. Dowe’s paper on Distinctive Image Features from Scale-Invariant Keypoints(2004), there are four stages to SIFT:

1. “Scale-space extrema detection”- Searches the entire image for candidate interest points
2. “Keypoint localization”- Calculate the location and scale for each candidate, remove candidates that are not stable
3. “Orientation assignment”- Assign each key point with one or more orientations that are calculated based on the gradient direction at that key point location in the image
4. “Keypoint descriptor”- For each key point, calculate the gradient in its surrounding area. This allows the transform to be distortion resistant. The above approach “transforms image data into scale-invariant coordinates relative to local features”.

Our project seeks to achieve scale, rotation and translation resistance. However due to time-constraint, we did not implement stage 4 “Keypoint descriptor” of SIFT.

## Stage 1

Apply Gaussian filters of different scales to the image. By using different scales the Gaussian filters would have different variances. Due to the inherent properties of Gaussian filters, this would “smooth” out the images, removing finer details of the image. At different scales, the details of the image that are insignificant compared to the standard deviation of the Gaussian filter applied would be removed. The Gaussians are generated using the following formula:

Then the image, represented as an array of digits, is convolved with the Gaussian.

L(x,y,sigma) is the value of the resulting image at location (x,y) under the Gaussian filter with standard deviation sigma. I stands for the original image.

We applied Gaussians with scale 0, 1, and 2 to the image. At scale 0, we are essentially preserving the original image, at scales 1 and 2 we are “smoothing out” the image to an increasing extend. We have 3 octaves of resulting images, each octave consists of images resulting from repeated applying the gaussian filter of the same scale to the original image. After each octave, the image is down-sampled by two.

## Stage 2

Now we have the image smoothed to different extends, with variant amount of fine detailed preserved in the resulting images. Within each octave, we use Difference of Gaussian, which is basically subtracting neighboring images from each other. Difference of Gaussian is proven to be a close approximation of scale-normalized Laplacian of Gaussian, which is shown to "produce the most stable image features compared to a range of other possible image functions, such as the gradient, Hessian, or Harris corner function”. Moreover, Difference of Gaussian is efficient to compute since it’s just subtracting images.

## Code

Then for each pixel in a resulting image, we compare it to its eight neighboring pixels in the same image and nine neighboring pixels in the images processed by adjacent scales. It’s selected if it’s greater or smaller than all its neighbors. The result is a candidate key point.

## Stage 3

To calculate the magnitude and orientation of each key point, we look at all it’s neighboring pixels in the image that is processed with the same scale.

m(x,y) stands for the gradient magnitude of the point and theta(x,y) stands for the orientation of the point.

## Code

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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Sherica
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Sherica
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Tamia
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a perfect square v²+2v+_
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Asali
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Samantha
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Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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what's the easiest and fastest way to the synthesize AgNP?
China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
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Azam
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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