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"Caesar ciphers are among the simplest devised, and rely solely on remapping characters to others in the alphabet using a constant shift modulo the size of the alphabet. The amount shifted is the key used to encipher, or decipher, the message. This remapping is usually restricted to letters, so that with a key of 2, `A' is replaced by `C', `B' by `D', ..., `X' by `Z', `Y' by `A', and `Z' by `B'. Lower case letters are mapped in an identical way to give their lower case replacements. Thus here the key is an integer in the range 1 to 25 (not 0 to 25, as there is no point replacing every `A' by an `A' etc.).
Straightforward Caesar ciphers are rather too easy to recognise however. In an attempt to maintain the basic idea, but complicate the result, what we will do is systemmatically jumble up the letters. Treating `I' and `J' as one letter from the start (so now there are only 24 different keys, assuming we still exclude mapping `A' to `A'), the 25 letters will be written to a 5 by 5 grid, by rows, and read back by columns. So, for example, with a key of 10, `A' is mapped into `L', and the grid produced is:
L M N O P
Q R S T U
V W X Y Z
A B C D E
F G H I K
and the uppercase alphabet upon encryption becomes: LQVAFMRWBBGNSXCHOTYDIPUZEK Note that here the `B' is doubled up because it represents both `I' and `J'. Thus, with this key of 10, the following line is enciphered to the one after it.
Caesar ciphers are child's play.
Vlfylt vbhwfty ltf vwbna'y hnle.
An acceptable drawback of the treatment of `I' and `J' in the grid is that a decoded message will have all `J's mapped to `I's, and all `j's to `i's"
this is a part of requirement in the whole program,I almost finished it, but I do not know thata how to make" a decoded message will have all `J's mapped to `I's, and all `j's to `i's". "Treating `I' and `J' as one letter from the start"?what does that mean?
this is my code about the part:
public static char encode(char ch,int n)  {
  while (!Character.isLetter(ch))
   return ch;
  if(ch>='A'&&ch<='Z')
  return (char)('A'+(ch-'A'+n)%26);
if(ch>='a'&&ch<='z')
  return (char)('a'+(ch-'a'+n)%26);
return ch;
public static char decode(char ch,int n)  {
  while (!Character.isLetter(ch))
   return ch;
  if(ch>='A'&&ch<='Z')
  return (char)('A'+(ch-'A'-n)%26);
if(ch>='a'&&ch<='z')
  return (char)('a'+(ch-'a'-n)%26);
return ch;
}but
how to make" a decoded message will have all `J's mapped to `I's, and all `j's to `i's". "Treating `I' and `J' as one letter from the start"?

Answer 1

Your teacher/instructor/professor would obviously be the final authority on this, but I think all it's saying is that when you encode, treat each 'j' you come across as an 'i', and therefore, when you decode, because you've lost information that it was a 'j' to start with, it will be mapped back to an 'i'.
One easy way to accomplish this (that would save special cases later in your for-loop) would simply be a:
String toEncode = "Djibouti"
toEncode = toEncode.replaceAll("j","i"); //obviously handle both cases instead
String encoded = encode(toEncode);
String decoded = decode(encoded);
System.out.println(decoded); //this would print "Diibouti"The other way is to simply add a special case to your encode/decode (char) methods.
Edited by: endasil on 28-Sep-2009 9:35 AM