Binary image operations pine spruce friday & monday GW 2.5, - PDF document

Binary image operations pine spruce friday & monday GW 2.5, 11.1.1, 11.1.5, 11.2.2, 11.4 different objects different shape different texture stina@cb.uu.se recognition and interpretation using computerized image analysis Friday 22/3 partitioning of 2D space • connectedness – neighbourhood relations – connected components • identification of connected components – labelling • measuring distances in an image square rectangular hexagonal – distance transforms (non-square) 1

neighbours neighbourhood pixel p with coordinates (x,y) 6-neighbours 4-neighbours 8-neighbours N 4 (p): (hexagonal) (square) N 8 (p): 4-neighbourhood • neighbourhood 8-neighbourhood • neighbours of a pixel ( , − 1 ) ( − 1 , − 1 ) ( , − 1 ) ( + 1 , − 1 ) x y x y x y x y ( 1 , ) ( 1 , ) x − y x + y ( x − 1 , ) y ( x + 1 , ) y – horizontal/vertical, edge, kant ( , + 1 ) ( − 1 , + 1 ) ( , + 1 ) ( + 1 , + 1 ) x y x y x y x y – diagonal, point, punkt, hörn, vertex adjacency similarity object O – all pixels with a certain ”similarity” • 4-adjacent • m-adjacent (mix) V – set of grey-levels – p & q belongs to O – q is in N 4 (p) example: and q is in N 4 (p) – q is a point neighbour grey-level in V={20,21,…,30} AND p and q have no • 8-adjacent edge neighbours in – p & q belongs to O O common and q are in N 8 (p) O binary ( bi-level! ) images: only 0s and 1s object/region (1) and background (0) 2

path path between two pixels p=(x,y) and q=(s,t) (x,y)=(x 0 ,y 0 ) adjacent (x 1 ,y 1 ) ... (x n-1 ,y n-1 ) adjacent 8-path 4-path (x n ,y n )=(s,t) p and q are connected if there is a path between p and q connected component connected component labelling • a pixel p belongs to the object O • every connected component is given an • connected component in O containing p: identity • represented by a grey-level image – all pixels that are connected with p – label image • every two pixels in a connected component are connected – grey-level for each pixel = identity • connected components are 4- or 8- connected if O consists of only one connected component � connected set 3

labelling using FloodFill example of FloodFill • initially all pixel in the label image = 0 • given the binary image – find a pixel in the first connected component – fill the first connected component with label 1 – look for the next connected component pixel in the binary image having 0 in the label image – fill the next connected component – ... labelling using a two scan example of algorithm two scan algorithm • initially • change all labels in the equivalence table – object pixels = 255 to the smallest – background pixels = 0 • if object pixel A<255 – if object pixel B or C<255 • A=min(B,C) – otherwise • A=new label • equivalence table updated 4-connectedness 8-connectedness scan 2 scan 1 4

label image in IMP connectivity – object and background • connected components of the object (O i ) • the object O is the union of all O i • the complement of O (O C ) consists of – background • subset of O C connected to the border of the image – holes • remaining part of O C using connected component connectivity paradox labelling • removal of connected components touching the a closed curve divides the plane into two (distinct) connected components image border – label – two scans 1. store label for object pixel at image border 2. if pixel=stored value � set to 0 • fill holes in the object – label the inverse image – remove connected components touching the image border – holes + original image how many background components?!? 5

using connected component filling holes labelling • count number of connected components • extract information for a specific connected component original image inverse image original image holes – size – compactness – ... labelled remove connected output inverse image components at image border distances boundary, border, contour, edge, ... consider an object O (connected set) boundary = the set of pixels with one or more neighbours not in O length, straightness, ... NOTE! connectivity used for object/background in case 8-connected object: thickness boundary = pixels with 4-neighbour in background 6

measuring distances in an distance image Z 2 • distance functions: x=(x 1 ,x 2 ),y=(y 1 ,y 2 ) ∈ Z – d E (x,y)= || x-y || 2 = � {(x 1 -y 1 ) 2 +(x 2 -y 2 ) 2 } l 2 – d city (x,y)= || x-y || 1 = | x 1 -y 1 | + | x 2 -y 2 | l 1 – d chess (x,y)= || x-y || � =max( | x 1 -y 1 | , | x 2 -y 2 | ) l � • intuitively: Euclidean distance • often represented by a distance transform – each object pixel is labelled with distance to 7 2 3 2 + Euclidean closest voxel in background 7 3 + city block, number of steps is 4-path 3 4 + chessboard, number of steps in 8-path metric – desired property local vs. global l 1 DEFINITION. A function d:Z 2 � Z 2 is a metric if and only if for any points (=pixels) x,y,z ∈ Z 2 the following holds: l 2 l � • d(x,y) � 0 and d(x,y)=0 only when x=y positive definite • d(x,y)=d(y,x) symmetric A known distance from A • d(x,z) � d(x,y)+d(x,z) fulfils � -inequality unknown distance from A 7

simple distance distance transform (DT) • number of steps in a minimal path • representation of distances in an image • city block: 4-connectedness • grey-level image – d city (x,y)= || x-y || 1 = | x 1 -y 1 | + | x 2 -y 2 | l 1 – replica of binary image – each object pixel is labelled with the distance • chessboard: 8-connectedness to its closest pixel in the background – d chess (x,y)= || x-y || � =max( | x 1 -y 1 | , | x 2 -y 2 | ) l � • from binary image to DT by a distance • mix � octagonal distance transformation • DT gives important information introduced by Rosenfeld and Pfaltz in 1966 & 1968 examples of distance DTs with numbers transforms 0 0 0 0 0 0 0 1 2 3 0 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 1 2 3 4 0 0 0 0 0 0 1 1 2 2 0 0 0 0 0 1 2 3 4 5 0 0 0 0 0 1 1 2 2 3 0 0 0 0 1 2 3 4 5 6 0 0 0 0 1 1 2 2 3 3 0 0 0 1 2 3 4 5 6 7 0 0 0 1 1 2 2 3 3 4 0 0 1 2 3 4 5 6 7 8 0 0 1 1 2 2 3 3 4 4 0 1 2 3 4 5 6 7 8 9 0 1 1 2 2 3 3 4 4 5 0 1 2 3 4 5 6 7 8 9 0 1 2 2 3 3 4 4 5 5 0 1 2 3 4 5 6 7 8 9 0 1 2 3 3 4 4 5 5 6 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 4 5 5 6 6 city block chess board city block chess board 8

computing DT city block forward backward • initially: object = � , background = 0 • one forward and one backward scan • current distance label of a pixel is compared to the distance labels for its already visited neighbours • update if shorter distance p p city block chess board chess board comparison: discs forward backward a disc is defined by centre pixel (p 0 ) and radius (r): all pixels closer than r to p 0 { } ( ) = ∈ n ; , ≤ B p Z d p p r 0 city block chess board Euclidean 9

simple distance transforms weighted distance transforms chamfering • use length of shortest path • assign weights to steps in path corresponding to neighbourhood relation – a for steps in edge directions city-block chess board octagonal – b for steps in point directions dist=11 dist=7 (diagonal only b a b a , b every even) a p a dist=8 b a b more or less: by Rutovitz & Hilditch 1969 city block 1 , 2 distance=number of steps 1 , 1 later Borgefors 1984/86 chess board 3 , 4 algorithm for weighted DTs for y=1 to n forward backward +b +a +b for x=1 to n ( ) ( ) ( ) ( ) I x y , = min I x + i y , + j + w i j , +a ( ) i j , ∈ fwd for y=n downto 1 for x=n downto 1 ( ) ( ) ( ) ( ) +0 +a , = min + , + + , I x y I x i y j w i j ( ) i j , ∈ bwd +b +a +b ( ) { } w i j i j , , , ∈ − 1 0 1 , , weights ( ) i , j pixel used in forward ( fwd ) and backward ( bwd ) scan ( ) I , i j image of size n x n 10

using a larger neighbourhood DTs in IMP even better approximation using knight step (c) original 2 2 2 b c a c b c b a b a 2 2 5 , 7 , 11 a a p a a region of interst c b a b c gives maximum difference 2 2 2 a c a c b with Euclidean distance of 2% DT (<3,4>) discs maximal difference – Euclidean 3 , 4 5 , 7 , 11 a b diff a b c diff 1 1.4142 9.0% 1 1.414 2.236 2.8% 3 , 4 5 , 7 , 11 1 2 58.6% 2 1 city block Euclidean 5 7 11 2.0% 1 1 41.4% a opt b opt c opt 1.4% a opt b opt 4.5% 2 3 13.4% 3 4 8.1% 8 11 7.3% 11