I truely want to understand the functionalities of the highlighted words in the code for better appreciation of the code
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phh=zeros(10,10); pmm=zeros(10,10); pmh=zeros(10,10); phm=zeros(10,10); *M=zeros(10,10); H=zeros(10,10)*; source=zeros(10,200,100); *avgT=zeros(10,10); eval= zeros(1, 100); sim = zeros(10, 1)*; *se=zeros(10,1)*; *ce =zeros(10,1)*; *pc= zeros(10,100); c=zeros(10,100); T=zeros(100,100,10); actM=[1 1 1 1 1 0 0 0 0 0]; %cw2= [ 1 0 0 0 0 1 1 1 1 1] ; tq=zeros(10,T); cw3=zeros(1,100)*; q=10; while q <50 d=1; T=zeros(10,100,100); while d <5 %%%%%%%%%%%%%%%%%%%iterating over percentage of malicious packets
phh=zeros(10,10); pmm=zeros(10,10); pmh=zeros(10,10); phm=zeros(10,10); M=zeros(10,10); H=zeros(10,10); iter=1; while iter < 10 %%%%%%look into the following code, it does not run for iterations %%%%%%%%%%%%%%%%generating multiple runs for the given percentage %%%%%%%%%%%%%%%%of malicious packets
eval=zeros(1,100);
rp=randperm(100);
eval(rp(1:q)) = 1;
cw1=bsc(eval,0.04);
cw2=zeros(1,100);
%%%%%%%%%%%formulating malicious behavior
% for i = 1:100
%if i < q
% if eval(i)==1
% cw2(i)=0;
% else
% cw2(i)=1;
% end
% else
% cw2(i)=eval(i);
% end
% end
cw2=bsc(eval,(q/100));
source=zeros(100,200,100);
sim = zeros(10, 1); se=zeros(10,1); ce =zeros(10,1);
pc= zeros(10,100);
c=zeros(10,100);
for i=1:100
for j=1:10
source(j,i,:) =eval;% cw1;
if rem(i,10)<(d )&& j<((10/2) +1) &&i>1 %%%%%%%%%%%%%%%%%%%%malicious behavior %%%%%%%%%%%%%%
source(j,i,:)=cw2;%bsc(eval,q/100);
end
% if rem(i,10)<d &&j>3&&j<6 %%%%%%%%%%%%%%to have different amounts
% of misbehavior
% source(j,i,:)=cw3;
% end
for k=1:100
if (i ==1)
c(j,k)=1;
end
if (i>1&& source(j,i,k) == source(j,i-1,k))
c(j,k) =c(j,k)+1;
end
end
sim(j)=similar(source(j,i,:),eval);
se(j)= -sim(j)*log10(sim(j))/log10(100);
if isnan(se(j))
se(j)=1;
end
pc(j,:)=c(j,:)/i; pc(1,:)=c(1,:)/i;
sum =0;
for k=1:100
sum=(-pc(j,k)*log10(pc(j,k))/log10(100) )+ sum; % computing the entropy of each source for the whole code word
end
ce(j) =sum;
if isnan(ce(j))
ce(j)=1;
end
if i <2
T(d,i,j)=1-se(j); %%%%%%%%d is replaced by d
else
T(d,i,j)=(1-(se(j)+ce(j)))*0.5+ 0.5*T(d,i-1,j);
T(d,i,j)= T(d,i,j)/max(T(d,:,j));
end
if (isnan(T(d,i,j)) )
T(d,i,j)=0;
end
if (T(d,i,j)>1)
T(d,i,j)=1;
end
if (T(d,i,j)<0)
T(d,i,j)=0;
end
end
end
for i = 1: 10 % d*10 to 10 avgT(d,i)=mean(T(d,:,i)); %%%%%%%%%%replacing 'd 'by d if avgT(d,i)<0.75 M(d,i)=M(d,i)+1; if actM(i)==1 pmm (q/10,d)= pmm(q/10,d)+1; else pmh(q/10,d) =pmh(q/10,d)+1; end else H(d,i)=H(d,i)+1; if actM(i)==0 phh(q/10,d)=phh(q/10,d)+1; else phm(q/10,d) =phm(q/10,d)+1; end end end iter =iter+1; end % tp(q/10,d)=pmm(q/10,d)/(pmm(q/10,d)+phm(q/10,d)); tp(q/10,d)=pmm(q/10,d)/(pmm(q/10,d)+phm(q/10,d)); % fp(q/10,d)=pmh(q/10,d)/(phh(q/10,d)+pmh(q/10,d)); fp(q/10,d)=pmh(q/10,d)/(phh(q/10,d)+pmh(q/10,d)); d=d+1;
end
q =q+10; x=1:100; figure(q/10) plot(x,T(1,:,9),'-ob', x,T(1,:,2),'-dr', x, T(2,:,2),'-sg',x,T(3,:,2),'-m*',x,T(5,:,2),'-k^'); legend('Honest', 'Malicious, p =0.1, 10% recommendations are false', 'Malicious,p=0.2, 10% recommendations are false','Malicious,p=0.3, 10% recommendations are false','Malicious,p=0.5, 10% recommendations are false') xlabel('Iteration'); ylabel('RecommendationTrust'); title('Impact of size of recommendation vector');
1 Commento
Walter Roberson
il 5 Feb 2017
Which part should we consider to be highlighted?
Do we get a clue about what the code is intended to do?
Risposte (2)
KSSV
il 12 Nov 2016
It is called initialization.....when you are going to store the output/result inside the loop, we initialization the respective arrays/Matrices to zeros. This will increase the speed of the code as the memory is already allocated to the result. If there is no initialization code will take long time to run. You can see the difference by commenting those lines and running the code.
0 Commenti
elvin eziama
il 5 Feb 2017
Modificato: Walter Roberson
il 5 Feb 2017
1 Commento
Walter Roberson
il 5 Feb 2017
Run-length encoding of rows. As long as the current location keeps the same value, just increment a counter. When the current location is not the same as the one before, you are just past the end of a run.
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