Thanks to @Crimsonfox, the complete answer is:
Christmas Comes But Once a Year
The corresponding ASCII code points we get by running the commands are:
67 104 114 105 115 116 109 97 115 32 67 111 109 101 115 32 66 117 116 32 79 110 99 101 32 97 32 89 101 97 114
To solve the puzzle, I created the tiny JavaScript program below. The code iterates through all the characters of the string (the asterisk and spaces are removed), assigning a specific command to each symbol.
var input =
"\n" +
">>>\n" +
">>>O>\n" +
">O<O>o@\n" +
"O>o>o>>O@\n" +
">>@<O>o<@>>\n" +
"@O>o@O<o<@<<@\n" +
"O<<<o>>>@O<<<o<\n" +
"o<O<@>>>O>>>o>@>>\n" +
">>>O>o@O<<o@<O>o@>>\n" +
"O<<<<o@O<<<<<o>o>>@O<\n" +
"o<<O@>>>>>O<<<<o@O<o@><\n" +
"<<<<<<<@>O<o>>>@>O>>>>>>o\n" +
">@O>>o<@O>>o@<O>>>o<o>>@O<<\n" +
"o>>>>>@<<O>o<o>>@>>>>O>o@<O>>\n" +
"o>>@O<<<<o@O>>o><><<>><<>>><><@";
var output = [],
x = 0,
line = 0;
for(var i = 0; i < input.length; i++) {
if(input[i] === ">") {
// increment value
x += line;
} else if(input[i] === "<") {
// decrement value
x -= line;
} else if(input[i] === "O") {
// multiply by line number
x *= line;
} else if(input[i] === "o") {
// divide by line number and round off
x = Math.floor(x / line);
} else if(input[i] === "@") {
// print ASCII character
output.push(x);
} else if(input[i] === "\n") {
// increment line number
line++;
}
}
console.log(String.fromCharCode(...output));
As we can see, we need to do four simple arithmetic operations in order to decode the Christmas tree: addition (>), subtraction (<), multiplication (O) and division (o), using the line number of the corresponding command. @ prints the value of x which represents an ASCII code point.
Original answer
Note: Since this used to be a partial answer, some of the information below might be incorrect.
The hints provided in the question (and one of OP’s comments) let us conclude the following:
As stated in the “large hint,” < and > are used to increment and decrement a variable by a certain value, while @ means “print ASCII character.”
O/o are not bracketed groups. Between two @s, you will often find a structure of the form O{x times > or <}o. This is a “trick” that does a very simple operation.
→ O/o do not represent brackets or parenthesis. However, O{…}o seems to imply that O and o do come in pairs and thus probably have a similar purpose. Since each of the five different symbols represent a “very simple operation,” we should probably focus on simple additions (or subtractions) rather than multiplications and divisions.
It's employed to compensate for what \n does. The difference between the number of O and o should turn out to be small, if not 0.
→ This suggests that O and o represent a particular number and have something to do with line numbers. Since O adds 1 to the line variable and o subtracts the same number in the code above, we can say there is only little difference between the two commands.
Which line a command is on influences its result (except if the command is @).
→ The variable line in the code must somehow be involved in incrementing and decrementing the value x. In the example above, a > on line 4 increments x by 4, while a < decrements it by 4. @ simply outputs the value of the counter x.
Maybe you should focus on the start, which gets us to a printable ASCII letter in (14+3) commands.
→ There are exactly 14 symbols or commands before the first print command (@), excluding line breaks, as well as 3 Os: >>>>>>O>>O<O>o. We have make sure x is between 65 (A) and 90 (Z) (or 97 and 122 for the lowercase letters), so that the commands yield a valid letter of the alphabet.
I had to be very careful with my wording in the original note. (The asterisk (and the leading spaces) are for decoration only.)
→ If OP had written “whitespaces” instead of “spaces,” that would mean we could also ignore line breaks and line numbers, which is not the case, according to the hints.
Now, the question arises whether >/< and O/o really constitute two separate counters that can be incremented and decremented.
>s, 65<s, 32Os, 32os, and 31@s. $\endgroup$Oandoform balanced "brackets". $\endgroup$