Tips for golfing in <all languages>

Question

The aim of this post is to gather all the golfing tips that can be easily applied to <all languages> rather than a specific one.

Only post answers that its logic can be applied to the majority of the languages

Please, one tip per answer

Answer 1

Merge Loops

You can usually merge two consequent loops, or two nested loops, into one.

Before:

for (i=0; i<a; i++) foo();
for (i=0; i<b; i++) bar();

After:

for (i=0; i<a+b; i++) i<a?foo():bar();

Answer 2

Just to mention the obvious:

Question your choice of algorithm and try something entirely new.

When golfing (especially harder problems that result in longer programs) all too often you might stick to the path you first chosen without trying other fundamental options. Of course, you may micro-golf one or a few lines at a time or a part of the overall idea, but often not try a totally different solution.

This was especially noticeable in Hitting 495 (Kaprekar) where deviating from the actual algorithm and looking for patterns you can apply to get to the same result was shorter in many languages (just not J).

The downside is that you possibly solve the same thing half a dozen times. But it works in really all languages except HQ9+ (where finding another way to output Hello World would be slightly futile).

Answer 3

Use Test-Driven Development

If the code must handle various inputs, then write comprehensive tests and make it easy to run them all very quickly. This allows you to try risky transforms one baby step at a time. Golfing then becomes like refactoring with perverse intent.

Answer 4

Try to reduce logical statements

For example, if A and B are booleans and your language treats booleans like numbers to some extent, A and (not B) and A>B are equivalent. For example in Python

if A and not B:
    foo()

is the same as:

if A>B:
    foo()

Answer 5

Initialize variables using values you already have.

Instead of x=1, try to look for something that already equals 1.
For example, a function's return value: printf("..");x=0; -> x=!printf("..");. It's easiest with 0, because you can always negate, or when all you need is the right truth value (and don't care if it's 1 or 19).

Answer 6

Use unary ~ for x+1 and x-1

This trick applies to languages that have a unary bitwise negation operator ~ and a unary regular negation operator -.

If your program, by chance, contains the expression -x-1, you can replace it with ~x to save bytes. This doesn't occur all too often, but watch what happens if we negate (-) both expressions: x+1 equals -~x! Similarly, x-1 equals ~-x. (Think of which way the tilde points: right is +, left is -.)

This is useful, because in all languages I can think of that have these operators, they have higher precedence than most operators. This allows you to save on parentheses. Watch how we save four bytes here:

(x+1)*(y-1)     ==>    -~x*~-y

Answer 7

Squeeze whitespace

Know the rules for whitespace in your language. Some punctuation marks, or other characters, might not need any surrounding whitespace. Consider this Bourne shell function:

f () { echo a; echo b; }

In Bourne shell, (); are metacharacters, and do not need surrounding whitespace. However, {} are words and need whitespace unless they are next to metacharacters. We can golf away 4 spaces next to ();, but must keep the space between { and echo.

f(){ echo a;echo b;}

In Common Lisp and PicoLisp, () are metacharacters. Consider this code to find the average of two numbers:

(/ (+ a b) 2)

We can golf away 2 spaces.

(/(+ a b)2)

---

Some languages have strange and subtle rules for whitespace. Consider this Ruby program, which prints the sum and product of a line of integers.

#!ruby -an
i=$F.map &:to_i
puts"#{i.reduce &:+} #{i.reduce &:*}"

Each & needs a space before itself. In Ruby, i=$F.map &:to_i means i=$F.map(&:to_i) where & passes a block parameter. But, i=$F.map&:to_i means i=$F.map.&(:to_i) where & is a binary operator.

This weirdness happens in languages, like Perl or Ruby, that use ambiguous punctuation. If in doubt, use a REPL or write short programs to test the whitespace rules.

Answer 8

Single Letter Variable Names

You have 52 of them; use them all! Don't be afraid to try different approaches and compare lengths. Know the language and the specific shortcuts/library functions available.

Answer 9

assign functions new names if used multiple times

x = SomeLongFunctionName
x(somedata)
x(somemoredata)
etc

Answer 10

Write an explanation of your code

Writing an explanation forces you to thoroughly look at each part of your code again and to make your thoughts and choices in writing a certain passage explicit. In doing so, you might find that different approaches are possible which may save some bytes, or that you subconsciously made assumptions which don't necessarily hold.

This tip is similar to Question your choice of algorithm and try something entirely new; however, I have found that the step of actually writing down how each part is supposed to work is sometimes crucial for becoming aware of alternatives.

As a bonus, answers including an explanation are more interesting for other users and are hence more likely to be upvoted.

Answer 11

Use the conditional operator.

A conditional operator

bool ? condition_true : condition_false

is more beneficial, character wise, than an IF statement.

if(a>b){r=a;}else{r=b;}

can be written as

r=a>b?a:b;

Answer 12

Double check your character count

Sounds like a no-brainer, but by being careful you might be able to "save" a few characters by not actually doing anything!

If you're using Windows, you may be inputting \r\n instead of just \r or \n when you hit Return - adding an extra byte per line! Turn control characters just to double check you're not doing this.

In Notepad++ you can convert all \r\n line endings to just \r by going to Edit > EOL Conversion > UNIX/OSX Format.

Also make sure you don't include any trailing whitespace in your character count! The line feed on the bottom line in your code is also inconsequential, so that won't need to be counted either.

Answer 13

Read the question carefully

Code golfing is as much about understanding the question (what is asked and what is not asked, even though it would be implied in any other setting) as producing code that (may) only satisfy what is asked.

Any input other than what is explicitly asked for need not be handled. If there are some test cases and no generic requirement, your code may only work in those cases. Etc.

Answer 14

Use bitwise operations for checking numbers between 0 and any 2ⁿ-1

Might be a bit of an edge case, but it could come in handy sometimes. It relies on the fact that all numbers to which m=2ⁿ-1 applies have the rightmost n bits set to 1.

So, 7₁₀ == 00000111₂, 15₁₀ == 00001111₂, 31₁₀ == 00011111₂ and so on.

The trick is x&~m. This will return true whenever x is not between 0 and m (inclusive), and false otherwise. It saves 6 bytes from the next shortest equivalent expression: x>=0&&x<=m, but obviously only works when m satisfies 2ⁿ-1.

Answer 15

Greater/Less than to save a digit:

//use:
if(n>9){A}else{B}
//instead of:
if(n<10){B}else{A}

Just remember to swap the code from if to the else and they will do exactly the same thing (or switch the sides of the inequality)!

Note: this can be applied with any power of 10 and their negatives: ...-100, -10, 10, 100...

(source link)

Answer 16

Reuse function parameters instead of new variables

Answer 17

Use > and < instead of >= and <=

When checking against hard-coded integer values, use > and < instead of >= and <= where possible. For example, using

if(x>24&&x<51)

Is 2 bytes shorter than using

if(x>=25&&x<=50)

Answer 18

use - instead of !=

for numeric comparisons:

If a equals b, a-b results in 0, which is falsy. Anything else than 0 is truthy; so
if used in a boolean context, a-b <=> a!=b

If you use it with if/else or with the ternary operator, this can also save you one byte for equality:
a==b?c:d <=> a-b?d:c

Answer 19

Avoid premature loop breaks

If running through a loop to check for 1 or more instances of a boolean check, it might make for a more efficient program to exit the loop on the first true value. However, removing the break and looping through all iterations allows for shorter code.

int main() {
bool m = false;
int n = 1000;
for (int i = 0; i < n; i++) {
if (i >= 100) {
m = true;
break; // remove this line
}
} 
return 0;
}

Answer 20

Split strings for long arrays

Most languages have a way to split a string into an array of strings around a token of some kind. This will inevitably be shorter than an array literal once the length reaches a language-dependent threshold, because the extra overhead per string will be one copy of a one-char token rather than (at least) two string delimiters.

E.g. in GolfScript

["Foo""Bar""Baz""Quux"]  # 23 chars

becomes

"Foo
Bar
Baz
Quux"n/  # 20 chars

For some languages, the threshold is as low as one string. E.g. in Java,

new String[]{"Foo"}  // 19 chars

becomes

"Foo".split("~")  // 16 chars

Answer 21

Understand what other people did

In addition to being fun, if you examine other people's code, you can sometimes discover a good algorithm that you didn't think about, or a trick (sometimes an obvious one) that you overlook.

Sometimes there is an existing answer that you can translate to another language, and benefit from the other language's goodies.

Answer 22

know your operator precedence

Whenever You combine several expressions, check the operator precedence table for your language to see if you can reorder stuff to save parentheses.

Examples:

• In all languages that I know, bitwise operators have a higher precedence than boolean operators: (a&b)&&c needs no parentheses: a&b&&c just as (a*b)+c does not.
• a+(b<<c) can be rewritten as a+b*2**c.
  That doesn´t save anything for this example, but it will if c is a small integer literal (<14).
• Bitwise operations have a lower precedence than most arithmetic operations, so if your language implicitly casts boolean to int, you can save a byte on a<b&&c<d with a<b&c<d (unless you need the short circuit evaluation)

Answer 23

Use unary ~ for a-b-1 and a+b+1

In addition to @Lynn's suggestions regarding x+1 → -~x; and x-1 → ~-x, you can also golf a-b-1 and a+b+1.

a-b-1    // 5 bytes
a+~b     // 4 bytes

a+b+1    // 5 bytes
a-~b     // 4 bytes

It might look like a tip you won't use all that often, kinda like using ~x instead of -x-1 doesn't happen often, but I've used it enough times to see it as a useful tip here. Especially with array-indexing you might use these above in some cases.

Answer 24

Shorter for-loops

If you have X statements {inside} your for-loop, you can move X-1 statements (inside) the for-loop after the second semicolon for(blah;blah;HERE) to save 3 bytes. (separate the statements by using a comma ,)

Instead of

for(int i=0;i<9;){s+=s.length();println(i++);}

you can move one of the statements into the for-loop's ( braces) while leaving the other out

for(int i=0;i<9;println(i++))s+=s.length();

and save 3 bytes (saved 1 more byte thanks to @ETHProductions)

---

Put simply,

instead of

for(blah;blah;){blah 1;blah 2;...;blah X}

move the statements around so you end up with this

for(blah;blah;blah 2,...,blah X)blah 1;

and save 3 bytes

Answer 25

Compress or/and streaks

Simple trick I've come up with when trying to squeeze a long streak of conditions chained by ands (or ors, in this case just substitute 'all' with 'any').

Eg:

if a>0 and a<10 and a+b==4 and a+3<1:

Becomes

if all([a>0,a<10,a+b==4,a+3<1]):

Answer 26

Rely on the compiler to provide the required performance.

Be sure to know which optimisations are guaranteed by the compiler and at which optimisation levels, and use them liberally. And even if performance isn't a concern requirement, you can still test with optimisations on, and then only discount one character because your code is still technically valid without the compiler flag.

Consider the following Haskell function to compute 2^n (ignoring the fact that Haskell already has a built-in exponentiation operator or three) (23 characters):

p 0=1;p x=p(x-1)+p(x-1)

The problem is - it's horrendously slow, it runs in exponential time. This might make your code untestable or to fail any performance constraints given by the question. You might be tempted to use a temporary variable or an immediately invoked function literal to avoid repeated function calls (25 characters):

p 0=1;p x=(\y->y+y)$p$x-1

But the compiler can already do that for you, you just need set -O as a compiler flag! Instead of spending few extra characters per site to eliminate common subexpressions manually, just tell the compiler to do basic optimisations for you for a grand total of one character or two across the entire program.

Answer 27

Utilize language version / compiler / environment quirks / new features

This is especially useful for polyglots, but can be applied to other challenges. Sometimes, a compiler bug can golf off a byte, a implementation bug can allow you to save a few chars, or a really bleeding-edge feature can improve your score.

Answer 28

Maybe somewhat obvious but...

Make use of operator return values

Keep in mind that the assignment operator returns a value!

For example, if you want to add y to x and then check if x is greater than something, you can do

if(25<x+=y)

instead of

x+=y;if(x>25)

Or maybe you want to find the length of a string after trimming it:

strlen(s=trim(s))

Rather than

s=trim(s);strlen(s)

Answer 29

Use bitwise operators in your if statements

In an if statement, you can (usually) use the bitwise operator "and" (usually & , if it is present) rather than && or and and save 1-2 bytes. Sames goes with other keywords (not is the same as a bitwise ! or ~ in many languages (saves 2 bytes), or is bitwise | in many languages (saves 1 byte), etc.).

Answer 30

Combine multiple/nested if checks using And/Or when possible.

i.e.:

if (a && (b || c)) {

}

instead of:

if (a) {
    if (b) {
        //Do Stuff
    } elseif (c) {
        //Do same stuff
    }
}

Answer 31

Find better ways to initialise your variables

Some other answers came close to mentioning this already, but in many (strictly typed?) languages, it's shorter to initialise x as empty string like:

x:=""

or x as empty rune (char) like:

x:=''

than

var x string

and

var x rune

Using preexisting values is obviously preferred, but not so easy.

Answer 32

Don't declare variables in for-loops

If for-loops have one statement inside of them, you can move the iterating variable's initialisation with the other variables that you already have declared.

int a=1;for(int i=1;i<n;)a*=++i;  // 32 bytes
int a=1,i=1;for(;i<n;)a*=++i;     // 29 bytes

Also, if you have multiple un-nested for-loops, it would be better if you reuse the same variable. This sort of ties in with this answer.

int a=1;for(int i=1;i<n;)a*=++i;for(int j=2;j<n;)a-=j++; // 56 bytes
int a=1,i=1;for(;i<n;)a*=++i;for(i=2;i<n;)a-=i++;        // 49 bytes

(I know you can do a=i=1, but that is not the point)

---

Now if you don't have any statement before the for-loop (and you only have 1 for-loop), it would be best to initialise the variable inside the for-loop

for(int i=0;i<99;)i*=2;       // 23 bytes
int i=0;for(;i<99;)i*=2;      // 24 bytes

Answer 33

Use truth tables

An example condition can be (((A AND B) OR (A OR B)) AND C) OR (A OR C) with A, B and C being boolean variables. But it's so long! Is there any way to shorten it, and prove it acts the same after shortening? Well, truth tables to the rescue!

+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    A    |    B    |    C    |D=A AND B|E=A OR B |F=D OR E |G=F AND C|H=A OR C |I=G OR H |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    0    |    0    |    0    |    0    |    0    |    0    |    0    |    0    |    0    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    0    |    0    |    1    |    0    |    0    |    0    |    0    |    1    |    1    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    0    |    1    |    0    |    0    |    1    |    1    |    0    |    0    |    0    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    0    |    1    |    1    |    0    |    1    |    1    |    1    |    1    |    1    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    1    |    0    |    0    |    0    |    1    |    1    |    0    |    1    |    1    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    1    |    0    |    1    |    0    |    1    |    1    |    1    |    1    |    1    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    1    |    1    |    0    |    1    |    1    |    1    |    0    |    1    |    1    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+
|    1    |    1    |    1    |    1    |    1    |    1    |    1    |    1    |    1    |
+---------+---------+---------+---------+---------+---------+---------+---------+---------+

The truth table above represents the possible values and outcomes of the example condition we specified earlier. D, E, F, G and H are variables which make our life a lot easier. You can see what they're assigned to at the top row of the table. The final variable, I, is the outcome of the condition. So, now that we have the outcomes handy, we can simplify the table to just the base variables and the final outcome:

+-+-+-+-+
|A|B|C|I|
+-+-+-+-+
|0|0|0|0|
+-+-+-+-+
|0|0|1|1|
+-+-+-+-+
|0|1|0|0|
+-+-+-+-+
|0|1|1|1|
+-+-+-+-+
|1|0|0|1|
+-+-+-+-+
|1|0|1|1|
+-+-+-+-+
|1|1|0|1|
+-+-+-+-+
|1|1|1|1|
+-+-+-+-+

We can now relatively easily see that the condition above really reduces to just A OR C.

Answer 34

Using float -> integer conversion automatic truncates for flooring (positive values) in most languages that don't strongly enforce types:

f.floor     # Ruby
f.to_i      # -1 byte
int $f      # Perl
0|$f        # -2 bytes
parseInt(f) # Javascript (Never use Math.floor)
0|f

This will truncate downwards (for positive values), and to get the ceil instead, it can be achieved with the same trick by reflecting the value across 0 (not true in all languages / implementations, not always shorter):

(i/5.0).ceil # Ruby
-(i/-5)

Answer 35

Overloading for function

I find useful for golfing to use as C++ templates and give to the sys the chance of create function for the type (it would reduce the code written by one human).

Too if there are + function that use + or - the same algorithm, write all in one function only, that depends from one parameter for say what result one want. In this way recently condense 3 function in one with near 50% written code less; reference palpn() function in codegolf.stackexchange.com/a/148745/58988 .