Finding all combinations of well-formed brackets

Question

This came up while talking to a friend and I thought I\'d ask here since it\'s an interesting problem and would like to see other people\'s solutions.

The task is to

Answer 1

//C program to print all possible n pairs of balanced parentheses  


#include<stdio.h>

void fn(int p,int n,int o,int c);

void main()
{
    int n;
    printf("\nEnter n:");
    scanf("%d",&n);
    if(n>0)  
        fn(0,n,0,0);
}

void fn(int p,int n,into,int c)
{  
    static char str[100];
    if(c==n)
    {
        printf("%s\n",str);
        return;
    }
    else
    {
        if(o>c)
        {
            str[p]='}';
            fn(p+1,n,o,c+1);
        }
        if(o<n)
        {
            str[p]='{';
            fn(p+1,n;o+1,c);
        }
    }
}

Answer 2

F#:

UPDATE: this answer is wrong. My N=4 misses, for example "(())(())". (Do you see why?) I will post a correct (and more efficient) algorithm shortly.

(Shame on all you up-voters, for not catching me! :) )

---

Inefficient, but short and simple. (Note that it only prints the 'nth' line; call in a loop from 1..n to get the output asked for by the question.)

#light
let rec brackets n =
    if n = 1 then
        ["()"]
    else
        [for s in brackets (n-1) do
            yield "()" ^ s
            yield "(" ^ s ^ ")"
            yield s ^ "()"]

Example:

Set.of_list (brackets 4) |> Set.iter (printfn "%s")
(*
(((())))
((()()))
((())())
((()))()
(()(()))
(()()())
(()())()
(())()()
()((()))
()(()())
()(())()
()()(())
()()()()
*)

Answer 3

why cant this is as simple as this, this idea is quite simple

brackets(n) --> '()' + brackets(n-1) 0 '(' + brackets(n-1) + ')' 0 brackets(n-1) + '()'

where 0 is the concatenation operation above

public static Set<String> brackets(int n) {
    if(n == 1){
        Set<String> s = new HashSet<String>();
        s.add("()");
        return s;
    }else{
        Set<String> s1 = new HashSet<String>();
        Set<String> s2 = brackets(n - 1);
        for(Iterator<String> it = s2.iterator(); it.hasNext();){
            String s = it.next();
            s1.add("()" + s);
            s1.add("(" + s + ")");
            s1.add(s + "()");
        }
        s2.clear();
        s2 = null;
        return s1;
    }
}

Answer 4

Groovy version based on markt's elegant c# solution above. Dynamically checking open and close (information was repeated in output and args) as well as removing a couple of extraneous logic checks.

3.times{ 
    println bracks(it + 1)
}

def bracks(pairs, output=""){
    def open = output.count('(')
    def close = output.count(')')

    if (close == pairs) {
        print "$output "
    }
    else {
        if (open < pairs) bracks(pairs, "$output(")
        if (close < open) bracks(pairs, "$output)")
    }
    ""
}

Answer 5

In C#

    public static void CombiParentheses(int open, int close, StringBuilder str)
    {
        if (open == 0 && close == 0)
        {
            Console.WriteLine(str.ToString());
        }
        if (open > 0) //when you open a new parentheses, then you have to close one parentheses to balance it out.
        {                
            CombiParentheses(open - 1, close + 1, str.Append("{"));
        }
        if (close > 0)
        {                
            CombiParentheses(open , close - 1, str.Append("}"));
        }
    }

Answer 6

Simple solution in C++:

#include <iostream>
#include <string>

void brackets(string output, int open, int close, int pairs)
{
    if(open == pairs && close == pairs)
            cout << output << endl;
    else
    {
            if(open<pairs)
                    brackets(output+"(",open+1,close,pairs);
            if(close<open)
                    brackets(output+")",open,close+1,pairs);
    }
}

int main()
{
    for(int i=1;i<=3;i++)
    {
            cout << "Combination for i = " << i << endl;
            brackets("",0,0,i);
    }
}

Output:

Combination for i = 1
()
Combination for i = 2
(())
()()
Combination for i = 3
((()))
(()())
(())()
()(())
()()()