Some exercises with the "Caesar cipher" Pt 2.

Plaintext    |a |b |c  |d |e |f  |g |h |i   |j | k | l  | m| n| o | p | q  | r  | s | t  | u | v | w | x | y | z |
P-Number  |0 |1 |2 | 3 |4 |5 |6 |7 |8 |9 |10|11|12|13|14|15 |16 |17|18|19|20|21|22 |23|24|25|
Ciphertext |d |e | f | g |h |i  |j  |k |l |m | n | o | p | q | r  | s| t  | u | v | w | x | y | z  | a | b  | c |

In the Caesar cipher, the modulus operator was used to achieve a fast way of computing the result.
The use of an array from 0 to 25 is inherent to the discrete implementation methods widely used. Look-up tables also get this particularity many a times !
If it is not obvious why, wait until one of the next posts on the subject. Other examples will surely make obvious why .
Let's remember  what was said in a previous post :

// Caesar cipher ==> C = (P+s) Mod (Number_of_Characters)
// Number_of_Characters also known as Keyspace = 26 (0 to 25)
// P = numerical equivalent of character plaintext
// C = Numerical equivalent of cipher text character
// s = number of shifts,adds (n_shift)
// Ke = s = 3 ==> (P+s) mod (Number_of_Characters)
// Kd = -s = -3 ==> (C -s)mod (Number_of_Characters) (reverse operation)

From this, we can then adapt this idea into the way the ascii numerical code uses to represent printable characters.

// Keyspace = 96 ( Nchars; )

// P = retrieved

// s =  n_shift

// [ from "ascii'" character number 32-decimal; 040-OCT; 20-HEX; 00100000-BIN ==> "SPACE" 

// to "ascii'" character number 126 decimal; 176-OCT; 7E-HEX; 01111110-BIN; ==> "~" Equivalence sign - tilde

The way i found was to first subtract  an ascii_offset (retrieved -=32;) compute the algorithm ( uint16_t c =(retrieved + n_shift)% Nchars; ) and then re-add the ascii_offset (retrieved +=32;).

C = (P+s) Mod ( Keyspace )

I should probably note the fact that it should not be confused with the remainder.
Let us see why...

If in C = A modulus B ,  A=13 and B=5 , then C=3 ;
If    C = A remainder B, then it is also C=3.

BUT (and there are always but's) what if we give A=13 and B= -5 ?!
Then, we would get A Modulus B ==> C= -2; In turn, A Remainder B ==> C=3 !!

Oh, yeah !! The art of the Modulus operator in itself  ( specially with only the use of a hand calculator), and its relationship with cryptography, is an art in itself , and can easy allow for a series of posts elaborating on it !
As an example, check the table below (created with Gnumeric) .

Code here
https://docs.google.com/document/d/1CbK-v1OaQIhbr0ao5EY8CrGZ71M9-t4cav4bK4JBwo0/edit?usp=sharing