Caesarts Cipher encryption algorithm assistance
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Caesar's cypher is the simplest encryption algorithm. It adds a fixed value to the ASCII (unicode) value of each character of a text. In other words, it shifts the characters. Decrypting a text is simply shifting it back by the same amount, that is, it substract the same value from the characters. Write a function called caesar that accepts two arguments: the first is the character vector to be encrypted, while the second is the shift amount. The function returns the output argument coded, the encrypted text. The function needs to work with all the visible ASCII characters from space to ~. The ASCII codes of these are 32 through 126. If the shifted code goes outside of this range, it should wrap around. For example, if we shift ~ by 1, the result should be space. If we shift space by -1, the result should be ~.
Here are a few things you may want to try with MATLAB before starting on this assignment:
double(' ')
ans =
32
double('~')
ans =
126
char([65 66 67])
ans =
'ABC'
' ' : '~'
ans =
' !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~'
And here are a few example runs:
caesar('ABCD',1)
ans =
'BCDE'
caesar('xyz ~',1)
ans =
'yz{! '
caesar('xyz ~',-1)
ans =
'wxy~}'
Code to Call your Function:
coded = caesar('ABCD', 3)
decoded = caesar(coded, -3)
wrap = caesar('1234', 96)
back = caesar(wrap, -96)
Hi there, I am having trouble with the caesars cypher question for week 8 homework assignments and I was wondering if I could get some help with the code that I have so far. Here is what I have, any help would be appreciated.
function coded = caesar(v,n)
v = char(32:126)
secret = v + 2
mod(v,126)
coded = char(mod(v,126))
end
ps. I feel embarrased posting this I am trying to figure it out but it is very difficult for me.
20 comentarios
KALYAN ACHARJYA
el 14 de Jul. de 2019
May be your teacher, forgot to ask the question.
Andrew Marttini
el 14 de Jul. de 2019
dpb
el 14 de Jul. de 2019
"I'm sorry I not sure what you want me to do with this text?"
He's pointing out you tried to post the full question in the Question title instead of it in the question text box for that purpose and so it got truncated and the full question is not shown...
Andrew Marttini
el 14 de Jul. de 2019
Andrew Marttini
el 14 de Jul. de 2019
So, now finish the cleanup you started...Take the question text out of the title and make a short, appropriate title for the question (write a headline, iow)
Then, format the code to be legible...(select the code then use the "CODE" button or press CTRL-e)
Andrew Marttini
el 14 de Jul. de 2019
Walter Roberson
el 14 de Jul. de 2019
Your code ignores the two inputs.
Andrew Marttini
el 15 de Jul. de 2019
dpb
el 15 de Jul. de 2019
What Walter's pointing out is that you pass in v but immediately overwrite it and never use n
PS. Thanks for cleaning up the question--much easier to read...
Andrew Marttini
el 15 de Jul. de 2019
Walter Roberson
el 15 de Jul. de 2019
Editada: Walter Roberson
el 16 de Jul. de 2019
That code still ignores the input v.
Walter Roberson
el 15 de Jul. de 2019
Suppose you were restricting to the range 5, 6, 7, 8 . Then if you were to take the input minus 5, the result would be 0, 1, 2, 3 . If you were to make an arithmetic change to that, and take the result mod 4 then the result would still be in the range 0, 1, 2, 3. If you were then to add 5 to that, the result would be 5, 6, 7, or 8.
It should be trivial for you to extend this logic to the range you are using.
Andrew Marttini
el 17 de Jul. de 2019
Andrew Marttini
el 17 de Jul. de 2019
Walter Roberson
el 17 de Jul. de 2019
mod(32:126,32) does a calculation on some constants and then displays the results. The line does not make any change to any variables.
Andrew Marttini
el 17 de Jul. de 2019
Walter Roberson
el 17 de Jul. de 2019
Yes you need to include v new in the mod function.
The only place 32:126 should occur is in a comment.
evan muas
el 29 de Dic. de 2019
A link to my answer on stackexchange
https://stackoverflow.com/a/59522032/10810614
Mayank Joshi
el 5 de Ag. de 2021
coded=char(mod((double(v)+n-32),95)+32); Use this
Respuestas (4)
Akhil Thomas
el 16 de Mayo de 2020
function
y = caesar2(ch, key)
v = ' ' : '~';
[~, loc] = ismember(ch, v);
v2 = circshift(v, -key);
y = v2(loc);
end
Akhil Thomas
el 16 de Mayo de 2020
function txt = caesar(txt,key)
txt = double(txt) + key;
first = double(' ');
last = double('~');
% use mod to shift the characters - notice the + 1
% this is a common error and results in shifts
% being off by 1
txt = char(mod(txt - first,last - first + 1) + first);
end
Shaun Olisagu
el 27 de Jul. de 2020
function coded = caesar(text, amount)
new_text = double(text)
text_shift = new_text + amount
coded = char(mod(text_shift,(32:126,32)))
end
Pls help. I've been on this for quite a long time now
Yifan He
el 28 de Jul. de 2022
Editada: Walter Roberson
el 28 de Jul. de 2022
function coded = caesar(v,s)
x = double(v);
y = x + s;
if y >= 32
n = fix((y-32)/95);
else
n = fix((y-126)/95);
end
z = y - 95*n;
coded = char(z);
2 comentarios
Walter Roberson
el 28 de Jul. de 2022
What if the requested shift is more than +/- 95 ?
Yifan He
el 28 de Jul. de 2022
It still make sense. This problem is a step function in mathematics, actually. z=f(y)=y-95n (32+95n ≤ y ≤ 126+95n,n∈Z)
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