TA: Jessie Hsu yh2117@columbia.edu
1. Compandor Design:
(a) Idea of the compandor
Given a pdf of the random variable u , the goal is to determine decision levels and reconstruction levels in w domain, and then tranform them back to u domain.
Thus, what we are going to do here is to
(1) Determine the transform function both from u to w and from w to u
(2) Quantize w , obtaining decision and reconstruction levels
(3) Transform decision and reconstruction levels back to u
(b) pdf of u
The pdf of u is given by p u
( u )
 
1
0
| u |

1
 u

1 otherwise
 p u
( u )


1


1
 u
 u
0
0
 u

1

1
 u

0 otherwise
Determining the transform function from u to w
The tranformation w=f(u) , as given in the problem, is
(c) pdf of w
1

As 0
 u

1 , w
 f ( u )
  u x

0 p u
( x ) dx
 x

 u
0
( 1
 x ) dx
 u

1
2 u
2
As

1
 u

0 , w
  f (
 u )
 
[(
 u )

1
2
(
 u )
2
]
 u

1
2 u
2

0
 w

1
2
 
1
2
 w

0
 w
 f ( u )



 u

 u


1
2
1 u
2 u
2
2
0
 u

1
 u

1

0 as illustrated in (e).
Determining the inverse transform function from w to u
Accordingly, we can derive the transformation from u back to w ,
As 0
 w

1
2
, u
 f

1 ( w )

1

1

2 w pick u
 f

1 ( w )

1

1

2 w so that u falls inside the interval [0,1]

0
 u

1
As

1
2
 w

0 , u
 f

1
( w )
 
1

1

2 w
 
1
2
 w

0 pick u
 f

1 ( w )
 
1

1

2 w so that u falls inside the interval [-1,0]

0
 u

1
 u
 f

1
( w )




1

1


1

2 w
1

2 w 
0
1
2

 w

1
2 w

0 as illustrated in (f).
2

(e) Transformation from u to w (f) Transformation from w to u
(g) Quantizing w (h) Quantizing u
Quantizing w , obtaining decision and reconstruction levels w can be quantized with uniform quantization,
 t
1
, t
2
, t
3
, t
4
, t
5


[

1
2
,

1
4
, 0 ,
1
4
,
1
2
]
 r
1
, r
2
, r
3
, r
4


[

3
8
,

1
8
,
1
8
,
3
8
]
Transforming decision and reconstruction levels back to u
 u
 f

1
( w )
 t
1
, t
2
, t
3
, t
4
, t
5


[

1 ,

1

1
2
, 0 , 1

1
2
, 1 ]
 r
1
, r
2
, r
3
, r
4


[

1
2
,

1

2
3
, 1

2
3
,
1
2
]
3

Calculating Mean Square Error
MSE
 i
4 

1
E [( u
 r i
)
2
]
 i
4 

1 t i t i

1
( u
 r i
) 2 p u
( u ) du
 

1

1

1
2 ( u

(

1
2
))
2
( 1
 u ) du
 0 

1

1
2
( u

(

1

 
0
1

1
2 ( u

( 1

2
3
))
2
( 1
 u ) du
  1
1

2
3
))
2
( 1
 u ) du
1
2
( u

1
2
)
2
( 1
 u ) du

1

2
6

0 .
0178

2
3
3

6
6
[Bonus]
This quantizer is NOT optimal for u .
To show this claim, we provide a better quantizer in terms of mean square error.
Sample MATLAB programs doing Lloyd-Max quantization are provided below:
% DIP HW#2 Prob 1 Bonus
% Lloyd-Max Quantizer for Triangular distribution clear;
N=20; % Number of iterations
M=2001; % Number of samples along u axis
% pdf construction u=zeros(1,M); u=[-1:2/(M-1):1]; pu=zeros(1,M); pu(1:(M-1)/2+1)=1+u(1:(M-1)/2+1); pu((M-1)/2+1:M)=1-u((M-1)/2+1:M);
% Quantization data: thresholds and reconstruction levels t=zeros(N,5); t(:,1)=-1; t(:,3)=0; t(:,5)=1; r=zeros(N,4);
% Iteration: Start with the levels obtained from the compandor
MSE=zeros(1,N); t(1,2)=-1+1/sqrt(2); t(1,4)=1-1/sqrt(2); r(1,1)=-0.5; r(1,2)=-1+sqrt(3)/2; r(1,3)=1-sqrt(3)/2; r(1,4)=0.5;
MSE(1)=MSECompute(u,pu,squeeze(t(1,:)),squeeze(r(1,:)));
4

for k=1:N-1 % kth iteration
[tnew,rnew]=LloydMax( u,pu,squeeze(t(k,:)),squeeze(r(k,:)) );
t(k+1,:)=tnew;
r(k+1,:)=rnew;
clear tnew rnew;
MSE(k+1)=MSECompute(u,pu,squeeze(t(k+1,:)),squeeze(r(k+1,:))); end
Function for computing MSE: function y=MSECompute(u,pu,t,r)
% Computes quantization error (MSE)
% Takes input (u,pu,t,r)
M=length(r); % Number of quantization levels y=0; tceil=0;
MSE=zeros(1,M); for k=1:M
% MSE in the kth interval
tfloor=tceil+1;
tceil=min( max(find( u<=t(k+1) )),length(u) );
currentu=u(tfloor:tceil);
currentpu=pu(tfloor:tceil);
MSE(k)=(sum( currentpu.*((currentu-r(k)).^2) ));
clear currentu currentpu currentnum; end y=(sum(MSE))/(length(u)/range(u));
Function for Lloyd-Max iteration: function [tnew,rnew]=LloydMax(u,pu,t,r)
% Lloyd-Max quantizer subroutine
% Computes new quantization thresholds given old ones
% Takes input (u,pu,t,r)
M=length(r); % Number of quantization levels tceil=0; tnew=zeros(1,M+1); rnew=zeros(1,M);
5

tnew(1)=t(1); tnew(M+1)=t(M+1); for k=1:M
tfloor=tceil+1;
tceil=min( max(find( u<=t(k+1) )),length(u) );
currentu=u(tfloor:tceil);
currentpu=pu(tfloor:tceil);
currentnum=tceil-tfloor+1;
rnew(k)= sum(currentu.*currentpu) / sum(currentpu);
clear currentu currentpu currentnum; end for p=2:M
tnew(p)= (r(p-1)+r(p)) / 2 ; end
Starting with the decision levels and reconstruction levels from the compandor,

 t
1 r
1
,
, t
2 r
2
,
, t
3 r
3
,
, t
4 r
4
,
 t
5


[

[

1 ,

0 .
2929 , 0 , 0 .
2929 , 1 ]

0 .
5 ,

0 .
1340 , 0 .
1340 , 0 .
5 ] with MSE = 0.0178, in the end (after 20 iterations), we get
 r
1
, r
2
 t
1
,
, r
3 t
,
2
, r
4 t

3
,
 t
4
[
, t
5


[

1 ,

0 .
3816 , 0 , 0 .
3819 , 1 ]

0 .
5877 ,

0 .
1755 , 0 .
1760 , 0 .
5877 ] with MSE = 0.0155, less than the MSE yielded by the compandor.
Therefore the compandor is not the optimal quantizer for u .
6

2. Sample MATLAB code:
(1) Histogram calculation
% HW 2, Prob 2 (1)
% Histogram computation
% 8-bits (0~255) function [hind,h]=histcomp(img) hmax=255;
[M,N]=size(img); y=double(img(:)); h=zeros(hmax+1); hind=[0:1:hmax]; for k=1:M*N
h( y(k)+1 )=h( y(k)+1 )+1; end
(a) coin.bmp (b) Histogram of coin.bmp
7

(2) Cumulative histogram calculation
% HW 2, Prob 2 (2)
% Cumulative Histogram computation
% 8-bits (0~255) function [chind,ch]=histcomp(img) intmax=255;
[M,N]=size(img); y=double(img(:)); h=zeros(intmax+1); ch=zeros(intmax+1); chind=[0:1:intmax]; for k=1:M*N
h( y(k)+1 )=h( y(k)+1 )+1; end for k=1:length(h)
ch(k)=sum( h(1:k) ); end
% Plot the cumulative histogram chmax=max(max(ch)); figure set(gcf,'color',[1,1,1]); plot(chind,ch); grid; axis([0 intmax 0 chmax+1]); title('Cumulative Histogram'); xlabel('Gray Level'); ylabel('Cumulative Histogram');
(a) coin.bmp
8
(b) Cumulative histogram of coin.bmp

(3) This is an open problem, with criteria, combinations and parameter adjustments determined by you. Here are sample codes of repective methods:
Contrast stretching (G&W p.85)
% HW 2, Prob 2 (3)-1
% Contrast stretching clear; intmax=255; img=double(imread('d:\courses\dip\hw\hw2\coin.bmp'));
[M,N]=size(img); imgcs=zeros(M,N);
% Two cut points r1=100; s1=100; r2=200; s2=200;
% Map intensities for k=1:M
for p=1:N
x=img(k,p);
% Section 1
if x>=0 & x<r1
y=x*s1/r1;
% Section 2
elseif x>=r1 & x<r2
y=(x-r1)*(s2-s1)/(r2-r1)+s1;
% Section 3
else
y=(x-r2)*(intmax-s2)/(intmax-r2)+s2;
end
imgcs(k,p)=round(y);
% Clip gray levels to 0~255
if imgcs(k,p)<0
imgcs(k,p)=0;
elseif imgcs(k,p)>intmax
imgcs(k,p)=intmax;
end
y=0;
end end
The parameters to be adjusted are (r
1
,s
1
) and (r
2
,s
2
)
9

Histogram equalization (G&W p.91)
% HW 2, Prob 2 (3)-2
% Histogram equalization clear; intmax=255; img=imread('d:\dip\hw\hw2\coin.bmp');
[M,N]=size(img); imgeq=histeq(img);
[hinind,hin]=histcomp(double(img)); imgeqy=double(imgeq(:));
[houtind,hout]=histcomp(imgeqy); hmax=max([ max(hin),max(hout) ]);
% Plot the input histogram figure set(gcf,'color',[1,1,1]); plot(hinind,hin); grid; axis([0 intmax 0 hmax+1]); title('Input Histogram'); xlabel('Gray Level'); ylabel('Histogram');
% Plot the output histogram figure set(gcf,'color',[1,1,1]); plot(houtind,hout); grid; axis([0 intmax 0 hmax+1]); title('Equalized Histogram'); xlabel('Gray Level'); ylabel('Histogram');
% Display the input and output image figure set(gcf,'color',[1,1,1]); subplot(2,1,1),imshow(uint8(img)); title('Image Before Histogram Equalization'); subplot(2,1,2),imshow(uint8(imgeq)); title('Image After Histogram Equalization'); imwrite(imgeq,gray(256),'d:\dip\hw\hw2\coineq.bmp','bmp');
10

(a) coin.bmp (b) coin.bmp after histogram equalization
(c) Histogram of coin.bmp (d) Histogram of coin.bmp after equalization
11

Histogram matching (G&W p.91)
% HW 2, Prob 2 (3)-3
% Histogram matching clear; intmax=255; img=imread('d:\courses\dip\hw\hw2\coin.bmp');
[M,N]=size(img);
% Specify target histogram hind=[0:1:intmax]; htgt=zeros(1,intmax+1); a=round( M*N/(intmax+1) ); htgt(1:(intmax+1))=a; % Flat histogram b=M*N-a*(intmax+1); for k=1:b
htgt(b)=htgt(b)+1; end imgeq=histeq(img,htgt);
[hinind,hin]=histcomp(double(img)); imgeqy=double(imgeq(:));
[houtind,hout]=histcomp(imgeqy);
The parameter to be adjusted is your target histogram. A flat histogram is used in the sample code, which yields equivalent results to histogram equalization.
12

Power-law transform (G&W p.81)
% HW 2, Prob 2 (3)-4
% Power transform clear; intmax=255; img=double(imread('d:\courses\dip\hw\hw2\coin.bmp'));
[M,N]=size(img); imgp=zeros(M,N);
% Power transform c=1; gamma=1; for k=1:M
for p=1:N
r=img(k,p);
y=c*r^(gamma);
imgp(k,p)=round(y);
% Clip gray levels to 0~255
if imgp(k,p)<0
imgp(k,p)=0;
elseif imgp(k,p)>intmax
imgp(k,p)=intmax;
end
end end
The parameter to be adjusted is γ .
13

Log transform (G&W p.79)
% HW 2, Prob 2 (3)-5
% Log transform clear; intmax=255; img=double(imread('d:\courses\dip\hw\hw2\coin.bmp'));
[M,N]=size(img); imglog=zeros(M,N);
% Log transform c=40; for k=1:M
for p=1:N
r=img(k,p);
y=c*log(1+r);
imglog(k,p)=round(y);
% Clip gray levels to 0~255
if imglog(k,p)<0
imglog(k,p)=0;
elseif imglog(k,p)>intmax
imglog(k,p)=intmax;
end
y=0;
end end
The parameter to be adjusted is c.
14