How I can solve, index exceeds matrix dimensions

Question

Mohammed alfath Ahmed 2019-9-9

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https://ww2.mathworks.cn/matlabcentral/answers/479616-how-i-can-solve-index-exceeds-matrix-dimensions

编辑： Adam Danz 2019-9-10

采纳的回答： Adam Danz

26224.pdf

The error appears in this line(fprintf('\t%d',c1(o)))

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Answer 1

Adam Danz 2019-9-9

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https://ww2.mathworks.cn/matlabcentral/answers/479616-how-i-can-solve-index-exceeds-matrix-dimensions#answer_391116

编辑：Adam Danz 2019-9-10

在 MATLAB Online 中打开

The code that contains the problem is copied below from your pdf attachment. The error happens when 'o' becomes larger than the number of elements in 'c1'.

while(o<=over)
    fprintf('\t%d',c1(o));
    o=o+1;
end

The while-loop that comes before this only has 1 iteration because of the break at the end! Without the break, the while loop will never end because 'o' will always be 1 since you are not increasing that value within the loop.

Solution

This is how your last 2 while-loops should appear. The arrows show where I corrected your code.

while(o<=over);
    m=m1(o);
    diff=0;
    if(m>n);
        diff=m-n+1;
    end
    m=m-diff;
    
    qm=dec2bin(e);
    len=length(qm);
    c=1;
    xz=1;
    while(xz<=len)
        if(qm(xz)=='1')
            c=mod(mod((c^2),n)*m,n);
        elseif(qm(xz)=='0')
            c=(mod(c^2,n));
        end
        xz=xz+1;
    end
    c1(o)=c;
	o = o+1;     % <----------------- here (remove the break)
end
o=1;
fprintf('\nThe encrypted message is \n');
while(o<=over)
    fprintf('\t%d',c1(o));
    o=o+1;
end
fprintf('\n')    % <-------------------- here (add this)

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Mohammed alfath Ahmed 2019-9-10

Yes

Adam Danz 2019-9-10

编辑：Adam Danz 2019-9-10

在 MATLAB Online 中打开

randint() was part of the Communication System Toolbox and is now replaced with randi(). You'll see the use of randi() in places of these two lines in KALYAN 's answer.

n1=randint(1,1,n);        
e=randint(1,1,[2 n1]);

These function produce randome integers. Random number generation cannot be reverse-engineered. You'll need to save the random numbers somewhere.

I haven't looked too deeply into your algrithm so I can't say whether the rest is deterministic or not. However, before you get into dycription, I'd check that your encryption is doing what it's supposed to do.

My answer gets rid of your errors but I doubt the the algorithm is correct. For example, when I enter the inputs 7, 3, and "secret message" (without quotes), the encryption is just all 20s (I confirmed with c1==20). You can be certain that this encyption cannot be decrypted and carries no meaning.

The encrypted message is 
	20	20	20	20	20	20	20	20	20	20	20	20	20	20

KALYAN 's answer produces a different result but again, I doubt that's the correct algorithm since the 'break' only permits 1 iteration of the while(o<=over)-loop.

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Answer 2

KALYAN ACHARJYA 2019-9-9

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编辑：KALYAN ACHARJYA 2019-9-9

在 MATLAB Online 中打开

In your example PDF C1 is

>> whos c1
  Name      Size            Bytes  Class     Attributes
  c1        1x1                 8  double      

you can do c1(1) only

>> c1(1)
ans =
   107
>> c1(2)
Index exceeds matrix dimensions.

You are trying to access the vector elements more than its length, hence Index exceeds matrix dimensions.while loop continue o=1,2,3... In c1(o) but you have c1(1) only, not c1(2),c1(3)...hence code shows the Index exceeds matrix dimensions

Is It:

p=input('Enter the prime no. for p: ');
q=input('Enter the prime no. for q: ');
n=p*q;
fprintf('\nn=%d',n);
phi=(p-1)*(q-1);
fprintf('\nphi(%d) is %d',n,phi);
val=0;
cd=0;
while(cd~=1||val==0)
n1=randi(n);
e=randi([2,n],1,1)
 val=isprime(e);
 cd=gcd(e,phi);
 end
val1=0;
d=0;
while(val1~=1);
d=d+1;
val1=mod(d*e,phi);
end
fprintf('\nd=%d',d);
fprintf('\nPublic key is (%d,%d)',e,n);
fprintf('\nPrivate key is (%d,%d)',d,n);
m=input('\nEnter the message: ','s');
m1=m-0;
disp('ASCII equivalent of message ');
disp(m1);
over=length(m1);
 o=1;
 while(o<=over);
 m=m1(o);
 diff=0;
 if(m>n);
 diff=m-n+1;
 end
 m=m-diff;
qm=dec2bin(e);
len=length(qm);
c=1;
xz=1;
while(xz<=len)
 if(qm(xz)=='1')
 c=mod(mod((c^2),n)*m,n);
 elseif(qm(xz)=='0')
 c=(mod(c^2,n));
 end
 xz=xz+1;
end
c1(o)=c;
break
 end
fprintf('\nThe encrypted message is \n');
fprintf('\t%d',c1);

Sample Results: % I have removed the while loop, as c1 having length 1 only, hence no role to access the c1(2), c(3)..invalid in the case.

Enter the prime no. for p: 9
Enter the prime no. for q: 13
n=117
phi(117) is 96
e =
    12
e =
    89
d=41
Public key is (89,117)
Private key is (41,117)
Enter the message: hello
ASCII equivalent of message 
   104   101   108   108   111
The encrypted message is 
	65>> 