How to cyclically increment and decrement 26 latin characters in a loop

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I want to increment or decrement characters, but have them cycle back to a when going beyond z and to z when going before a.

For example incrementing 'w' by 2 gives 'y' and decrementing 'w' by 2 gives 'u'. Another example decrementing 'w' by 28 gives 'u' and decrementing 'a' by 256 gives 'e'.

I've figure out how to increment: char(int(A[i]+B-97)%26 +97) where B is the shift amount and A[i] is current character.

Don't overcomplicate. Use modulo to keep the increment or decrement amount in a range of 26 characters, then simply do a range check:

char cyclicIncrementDecrement(char ch, int amount)
{
    int newValue = int(ch) + (amount % 26);
    if (newValue < 'a') newValue += 26;
    if (newValue > 'z') newValue -= 26;
    return char(newValue);
}

This method of course assumes ch already is in range of 'a' to 'z'. If not, you need to handle that (put it in range or throw an exception or whatever is appropriate for your application).

Running this:

int main()
{
    std::cout << cyclicIncrementDecrement('w', -2) << std::endl;
    std::cout << cyclicIncrementDecrement('w', 2) << std::endl;
    std::cout << cyclicIncrementDecrement('w', -28) << std::endl;
    std::cout << cyclicIncrementDecrement('a', -256) << std::endl;
    std::cout << cyclicIncrementDecrement('z', -256) << std::endl;
    std::cout << cyclicIncrementDecrement('z', -51) << std::endl;
    std::cout << cyclicIncrementDecrement('z', -52) << std::endl;
}

gives:

u y u e d a z

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Using modular arithmetic, calculate your answer as modulo 26 and then add 'a' (ASCII 97) to your result.

char cyclic_increment(char ch, int n) {
  int tmp = ((ch - 97) + n) % 26;
  if (tmp < 0 )
    tmp += 26;
  return (char)(tmp + 97);
}

Alternatively, you could write the above (without an if) as:

char cyclic_increment(char ch, int n) {
  return (((ch - 'a') + n) % 26 + 26) % 26 + 'a';
}

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This handles both positive and negative offsets:

unsigned char az_cyclic(int in_ch)
{
    constexpr int mod = 26; // There are 26 letters in the English alphabet

    int offset = (in_ch - 'a') % mod; // (ASCII To zero-based offset) Mod_26 remainder

    if (offset < 0) // If negative offset,
        offset += mod; // normalize to positive. For example: -1 to 25

    return 'a' + offset; // Normalize to ASCII
}

Use-cases:

int main()
{
    unsigned char out_ch = '\0';

    out_ch = az_cyclic('a' - 1); // 'z'
    out_ch = az_cyclic('a' - 1 - 26); // 'z'
    out_ch = az_cyclic('a' - 2); // 'y'

    out_ch = az_cyclic('a' + 4); // 'e'
    out_ch = az_cyclic('a' + 4 + 26); // 'e'
    out_ch = az_cyclic('a' + 2); // 'c'

    return 0;
}

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Comments
  • "decrementing 'a' by 256 gives 'e'." not if you use ascii code nor if a char is on 8 bits ...
  • Increment/decrement happen in cyclic fashion. a to z and z to a
  • @PrakharShukla only modulo 26, with a taken as the lowest value. You need to be clear on that. There are a lot more char values than those in C and C++.
  • more examples to make it clear: decrementing z by 256 gives d, decrementing z by 51 gives a, decrementing z by 52 gives z.
  • You were close: char(int((A[i]+B-97)%26 + 26) % 26 + 97) gives the correct result as mentioned in my answer below. In case of negative remainder, one can always add the divisor (26) without worrying about the result.
  • decrementing w by 28 will give 'y' in this case, but we need 'u'
  • I got the problem here. The point to note is that mod function in c++ does not work like normal mod function. -5 mod 4 is -1 in C++ and not 3 !
  • @PrakharShukla -1 (mod 4) in modular arithmetic is 3.
  • Alternatively, you can get rid of if and do tmp = (tmp + 26) % 26; instead. I'm not sure which option is faster.
  • I suspect that should be slower without compiler optimizations since modulo is slower than checking the sign bit, but, you're right.