Question

How to represent the ABCs as a 5-letter binary code using A for 0 and B for 1 in C

Answer 1

#include <stdio.h>

// a = AAAAA -> 00000 -> 0
// b = AAAAB -> 00001 -> 1
// c = AAABA -> 00010 -> 2
// d = AAABB -> 00011 -> 3
// e = AABAA -> 00100 -> 4
// f = AABAB -> 00101 -> 5 
// g = AABBA -> 00110 -> 6

// Convert integer n (0–31) into 5-letter A/B code
void toAB(int n, char out[6]) {
    for (int i = 4; i >= 0; --i) {
        int bit = (n >> i) & 1;     // extract bit
        out[4 - i] = bit ? 'B' : 'A';
    }
    out[5] = '\0';                  // null terminator
}

int main() {
    char code[6] = "";

    for (char ch = 'a'; ch <= 'z'; ++ch) {
        int value = ch - 'a';      // a=0, b=1, c=2...
        toAB(value, code);

        printf("%c -> %d -> %s\n", ch, value, code);
    }

    return 0;
}


/*
run:

a -> 0 -> AAAAA
b -> 1 -> AAAAB
c -> 2 -> AAABA
d -> 3 -> AAABB
e -> 4 -> AABAA
f -> 5 -> AABAB
g -> 6 -> AABBA
h -> 7 -> AABBB
i -> 8 -> ABAAA
j -> 9 -> ABAAB
k -> 10 -> ABABA
l -> 11 -> ABABB
m -> 12 -> ABBAA
n -> 13 -> ABBAB
o -> 14 -> ABBBA
p -> 15 -> ABBBB
q -> 16 -> BAAAA
r -> 17 -> BAAAB
s -> 18 -> BAABA
t -> 19 -> BAABB
u -> 20 -> BABAA
v -> 21 -> BABAB
w -> 22 -> BABBA
x -> 23 -> BABBB
y -> 24 -> BBAAA
z -> 25 -> BBAAB

*/

 

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