EE 355 OCR Explanation Mark Redekopp 2 Make Your Own Image Note: - - PowerPoint PPT Presentation

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EE 355 OCR Explanation

Mark Redekopp

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Make Your Own Image

• Note: You can make your own images by:

– Using the "Verdana" font and writing some text in a word processor – Using the Windows "snipping" tool to capture a section of the screen – Paste it into an image editor (Windows Paint) – Resize the image to 256x256 – Save it as 8-bit Grayscale

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Note about BFS

Scan top to bottom, left to right Stop when you hit a black, unfound pixel 1 Now start a BFS using 8 connected neighbors (N, NW, W, SW, S, SE, E, NE) 2

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Note about BFS

Row r Row r+1

Note: Due to slight image rendering discrepancies, characters further right might be found first since one of their pixels is in a higher row. In this image, 2 and 3 might be found before 1

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Note about BFS

Scan top to bottom, left to right Stop when you hit a black, unfound pixel 1 Now start a BFS using 8 connected neighbors (N, NW, W, SW, S, SE, E, NE) 2

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Euler Number

• Euler number = # of components - # of holes
• Component is connected black region

– Each character is just 1 component

• Holes are enclosed white regions

– '8' has 2 holes, '0' has 1, '2' has none

• Thus, Euler number, E = C-H = 1-H

– E=1 => {1, 2, 3, 5, 7} – E=0 => {0, 4, 6, 9} – E=-1 => {8}

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Center of Mass (Centroids)

• Imagine black pixels represents mass in the image

below

• Where would the center of mass be?

– Somewhere in the upper right

• This can be useful for identifying certain characters

('7' = more mass in the upper half, '3' = more mass in the right half, etc.)

Vertical center

• f mass is on

row 1.1

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Vertical Centroid

• Equation

– isblack(i,j) = 1 if pixel (i,j) is black, 0 otherwise

𝑗,𝑘=0,0 𝐼,𝑋

𝑗 ∗ 𝑗𝑡𝑐𝑚𝑏𝑑𝑙(𝑗, 𝑘) [

𝑗,𝑘=0,0 𝐼,𝑋

𝑗𝑡𝑐𝑚𝑏𝑑𝑙(𝑗, 𝑘)]

– Multiple row index * 1 if the pixel is black and sum

• ver all pixels

– 0*1 + 0*1 + 0*1 + 0*1 + 0*1 + 1*1 + 2*1 + 3*1 + 4*1 = 10 – Total black pixels = 9 – Vertical centroid = 10/9 = 1.11

1 2 3 4 1 2 3 4

Vertical center

• f mass is on

row 1.1

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Horizontal Centroid

• Equation

– isblack(i,j) = 1 if pixel (i,j) is black, 0 otherwise

𝑗,𝑘=0,0 𝐼,𝑋

𝑘 ∗ 𝑗𝑡𝑐𝑚𝑏𝑑𝑙(𝑗, 𝑘) [

𝑗,𝑘=0,0 𝐼,𝑋

𝑗𝑡𝑐𝑚𝑏𝑑𝑙(𝑗, 𝑘)]

– Multiple column index * 1 if the pixel is black and sum over all pixels – 0*1 + 1*1 + 2*1 + 3*1 + 4*1 + 4*1 + 4*1 + 4*1 + 4*1 = 26 – Total black pixels = 9 – Vertical centroid = 26/9 = 2.89

1 2 3 4 1 2 3 4

Horizontal center

• f mass is on

column 2.89

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Center of Mass (Centroids)

• Putting the VC (Vertical centroid) and HC (Horizontal centroid)

together we arrive at a center of mass of (1.1, 2.89)

• It may be more useful to translate the coordinate system to

have 0,0 be right in the middle which can easily be accomplished by:

– (VC – H/2 , HC – W/2 ) => (1.1-2, 2.89-2) = (-0.9, 0.89) – This tells us the center of mass is in the upper (negative), left (positive) quadrant of the image

Center of Mass

• 2
• 1

1 2

• 2
• 1

1 2

Center of Mass

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Symmetry

• Another useful characteristic might be vertical

and horizontal symmetry

– Mirror images top to bottom (vertical symmetry)

• r left to right (horizontal symmetry)

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

Vertical Symmetry Horizontal Symmetry Both Vertical & Horizontal Symmetry

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Computing Vertical Symmetry

• Scan down the top row and compare pixel values to

the bottom row, then compare second row to second to last row, etc.

– Count how many pixels you compared and how many matched – Likely helpful to normalize to a percentage

1 2 3 4 M1 M2 M3 M4 M5 1 M6 M7 M8 M9 MA 2 3 M6 M7 M8 M9 MA 4 M1 M2 M3 M4 M5 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

Vertical Symmetry Horizontal Symmetry Both Vertical & Horizontal Symmetry

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Computing Bit Quads

• To compute bit quads remember to start at one row

and column above and to the left and continue to

• ne row and column below and to the right (i.e.

leave a frame of one row/column of white around the black pattern)

1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4

Vertical Symmetry Horizontal Symmetry Both Vertical & Horizontal Symmetry

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