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The International Obfuscated C Code Contest

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Information about this entry

Morse codes for letters and digits use sequences of 1 to 5 (inclusive) Morse symbols (denoted as . or -), e.g.

    E .
    Z --..
    3 ...--

(BTW: digits always have 5 symbols, letters 1 to 4. I didn’t use this fact.)

Of course we can encode these symbols with bits in a byte. I chose - to be a bit set to 0, . to be a bit set to 1. Characters in C are at least 8 bits, but they are either signed or unsigned. To be portable, I’ve used only values from 0 to 127 (inclusive) (both in the encoding of Morse codes, as well as in subexpressions in the program). Hence we can use type char (using unsigned char for an extra bit would have made the source code excessively large ;-)

The least significant bit contains the first symbol of a Morse code, so we have

    E > 0......1
    Z > 0...1100
    3 > 0..00111

The dots still have to be determined. But we also need to encode the length of the sequence. Since there are only 4 different kind of lengths, two bits would suffice, e.g.

    E > 000....1
    Z > 010.1100
    3 > 01100111

That means bits 6 and 5 set to 00 is length 1, set to 01 is length 2, etc. But this would mean that we would have to extract the length into a counter before shifting out each symbol, such as

    length = (encoding & 96) >> 5;
    do {
            putchar(encoding & 1 ? '-' : '.');
            encoding >>= 1;
    } while (length--);

We can also use the remaining bits to include the counter

    length 1: 00000.1x     `E > 00000.11
    length 2: 0000.1xx
    length 3: 000.1xxx
    length 4: 00.1xxxx     Z > 00.11100
    length 5: 0.1xxxxx     3 > 0.100111

with 1 bit still to be chosen freely (let’s assume it is 0). Now we can do

    do {
            putchar(encoding & 1 ? '-' : '.');
            encoding >>= 1;
    while (encoding > 1);

As you see, the encoded Morse sequence can be the counter itself. But this is not very pleasant, as some Morse sequences will end up having an ASCII value < 32, i.e. they will be control codes. This would make the initializing (encoding) string awkward. Hence I came up with the following

length 1: 0010010x     E > 00100101  (= '%')
length 2: 001010xx
length 3: 00110xxx
length 4: 0100xxxx     Z > 01001100  (= 'L')
length 5: 011xxxxx     3 > 01100111  (= 'g')

Notice these encodings in the initialization string. Only the coding for 5 needs an escape (5 = ..... > 01111111 = '\177'). The way the length is encoded still allows the encoding to be its own counter:

    do {
            putchar(encoding & 1 ? '-' : '.');
            encoding = (encoding + 32) >> 1;
    } while (encoding > 35);

The last shift will always result in 00100010, hence the test for >35. Note that the calculation for the next value of encoding is done at int precision, so that the actual value of the char remains <= 127.

The character '-' (ASCII 45) and '.' (ASCII 46) are conveniently in sequence, so the putchar(3) can simply be

    putchar(45 + encoding % 2);

I used the modulo operator this time, because + takes precedence over &. The assignment can be incorporated into the while control test, and a do-while loop with a body of a single expression statement can be written as a while loop with a comma operator in the test:

    while(putchar(45 + encoding % 2),
        encoding = encoding + 32 >> 1);

Voila.

Now, the decoding of Morse codes is just the reverse. We shift in 0s and 1s as we scan the line. But: we have to scan from back to front (the left most Morse symbol is in the least significant bit, and thus has to shifted in last). The test for the scanning while loop must stop when there are no more . or - on the left. Hence we need a terminating character in front of the line. The input line is read one past the encoding string (into the same character array) the terminating character for the reverse scan is therefore provided by the implicit '\0' of the array initializer ! The test also stops when at most 5 symbols have been shifted in.

The encoded character is then looked up in the encoding string with memchr(3) (why memchr(3) is explained later). Either a ? (when there are more than 5 consecutive Morse symbols, or if the sequence is unknown) or the decoded character is printed. Or a ' ' if there was an extra space in the source line. Note the use of “or” in these last sentences, as they explain the use of the conditional operator.

The strspn(3) calls need either " .-" or ".-" as argument. Of course they are overlapped. I put these strings between the encodings for digits and letters. This is OK, since neither of these characters, or the '\0', is a valid encoding ! We must, however, use memchr(3) instead of strchr(3) in the decoding lookup because of this '\0'. Since it is far more obscure this way, I hope you’ll forgive me.

The outline of the program is roughly as follows (pseudo C):

    while (more lines) {
        if (line is in morse) {
            while (more characters) {
                skip next sequence
                    if (' ')
                        putchar(' ');
                    else
                        decode_sequence; /* backwards */
            }
        } else {
            while (more characters) {
                if (isalnum)
                    encode_character;
                    putchar(' ');
            }
        }

Because of the two while(more characters) loops are the same, I’ve exchanged the if(line is in morse) and the two while loops (so there is single while loop). However, we must now store the if(line is in morse) test for efficiency (it would be stupid to test the entire line for every character). This is done in l[0] ! (the first encoding is thus in l[1]). The memchr(3) decoding lookup is not disturbed, because the test is always 0 for Morse lines, and '\0' is not a possible encoding. l[0] is also used as the temporary shift character while encoding a line of text. This does not disturb the if (line is in morse) test, as the temporary l[0] will remain != 0.

I’ve put the ifs in the above layout into a single switch. Hope you don’t mind ;-)

Summary of the use of l[999]:


Selected notes:


.. - …. .. -. -.- - …. .. … … .–. — .. .-.. . .-. .. … – — .-. . — -… ..-. ..- … -.-. .- - . -.. - …. .- -. - …. . .–. .-. — –. .-. .- –

. -..- . .-. -.-. .. … . -.-. …. .- -. –. . - …. . .–. .-. — –. .-. .- – … — .. - -.. — . … -. — - … …. — ..- -


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