Total of all numbers from 1 to N will always be zero












4















The problem is I have to print all combinations of a sequence of
numbers from 1 to N that will always result to zero. It is allowed
to insert "+" (for adding) and "-" (for subtracting) between each
numbers so that the result will be zero.



//Output
N = 7

1 + 2 - 3 + 4 - 5 - 6 + 7 = 0
1 + 2 - 3 - 4 + 5 + 6 - 7 = 0
1 - 2 + 3 + 4 - 5 + 6 - 7 = 0
1 - 2 - 3 - 4 - 5 + 6 + 7 = 0


So how can I implement this? I am not asking for the actual
codes to do this, just a hint and ideas to solve this will
do. Thank you..










share|improve this question























  • I suppose that your constraint is that numbers cannot be repeated and always begin with the smallest and finish with the biggest, I would try to find any recursive solution you know that you have to stop when n is equal to N and only print when the result is 0.

    – vmrvictor
    19 hours ago











  • Can the numbers be repeated? Can you add the minus before the first number?

    – Karol Dowbecki
    19 hours ago











  • @KarolDowbecki no just conscecutive integer from 1 to N

    – robert
    19 hours ago











  • @robert If it can't, please update your question to fill in this extra constraint.

    – tryman
    19 hours ago






  • 1





    @Lino what it was exactly stated is the sum of consecutive integers that will always result to zero. It says that we can either insert + or - between each numbers. So the number 1 will always be positive...

    – robert
    19 hours ago
















4















The problem is I have to print all combinations of a sequence of
numbers from 1 to N that will always result to zero. It is allowed
to insert "+" (for adding) and "-" (for subtracting) between each
numbers so that the result will be zero.



//Output
N = 7

1 + 2 - 3 + 4 - 5 - 6 + 7 = 0
1 + 2 - 3 - 4 + 5 + 6 - 7 = 0
1 - 2 + 3 + 4 - 5 + 6 - 7 = 0
1 - 2 - 3 - 4 - 5 + 6 + 7 = 0


So how can I implement this? I am not asking for the actual
codes to do this, just a hint and ideas to solve this will
do. Thank you..










share|improve this question























  • I suppose that your constraint is that numbers cannot be repeated and always begin with the smallest and finish with the biggest, I would try to find any recursive solution you know that you have to stop when n is equal to N and only print when the result is 0.

    – vmrvictor
    19 hours ago











  • Can the numbers be repeated? Can you add the minus before the first number?

    – Karol Dowbecki
    19 hours ago











  • @KarolDowbecki no just conscecutive integer from 1 to N

    – robert
    19 hours ago











  • @robert If it can't, please update your question to fill in this extra constraint.

    – tryman
    19 hours ago






  • 1





    @Lino what it was exactly stated is the sum of consecutive integers that will always result to zero. It says that we can either insert + or - between each numbers. So the number 1 will always be positive...

    – robert
    19 hours ago














4












4








4


1






The problem is I have to print all combinations of a sequence of
numbers from 1 to N that will always result to zero. It is allowed
to insert "+" (for adding) and "-" (for subtracting) between each
numbers so that the result will be zero.



//Output
N = 7

1 + 2 - 3 + 4 - 5 - 6 + 7 = 0
1 + 2 - 3 - 4 + 5 + 6 - 7 = 0
1 - 2 + 3 + 4 - 5 + 6 - 7 = 0
1 - 2 - 3 - 4 - 5 + 6 + 7 = 0


So how can I implement this? I am not asking for the actual
codes to do this, just a hint and ideas to solve this will
do. Thank you..










share|improve this question














The problem is I have to print all combinations of a sequence of
numbers from 1 to N that will always result to zero. It is allowed
to insert "+" (for adding) and "-" (for subtracting) between each
numbers so that the result will be zero.



//Output
N = 7

1 + 2 - 3 + 4 - 5 - 6 + 7 = 0
1 + 2 - 3 - 4 + 5 + 6 - 7 = 0
1 - 2 + 3 + 4 - 5 + 6 - 7 = 0
1 - 2 - 3 - 4 - 5 + 6 + 7 = 0


So how can I implement this? I am not asking for the actual
codes to do this, just a hint and ideas to solve this will
do. Thank you..







java






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 19 hours ago









robertrobert

856




856













  • I suppose that your constraint is that numbers cannot be repeated and always begin with the smallest and finish with the biggest, I would try to find any recursive solution you know that you have to stop when n is equal to N and only print when the result is 0.

    – vmrvictor
    19 hours ago











  • Can the numbers be repeated? Can you add the minus before the first number?

    – Karol Dowbecki
    19 hours ago











  • @KarolDowbecki no just conscecutive integer from 1 to N

    – robert
    19 hours ago











  • @robert If it can't, please update your question to fill in this extra constraint.

    – tryman
    19 hours ago






  • 1





    @Lino what it was exactly stated is the sum of consecutive integers that will always result to zero. It says that we can either insert + or - between each numbers. So the number 1 will always be positive...

    – robert
    19 hours ago



















  • I suppose that your constraint is that numbers cannot be repeated and always begin with the smallest and finish with the biggest, I would try to find any recursive solution you know that you have to stop when n is equal to N and only print when the result is 0.

    – vmrvictor
    19 hours ago











  • Can the numbers be repeated? Can you add the minus before the first number?

    – Karol Dowbecki
    19 hours ago











  • @KarolDowbecki no just conscecutive integer from 1 to N

    – robert
    19 hours ago











  • @robert If it can't, please update your question to fill in this extra constraint.

    – tryman
    19 hours ago






  • 1





    @Lino what it was exactly stated is the sum of consecutive integers that will always result to zero. It says that we can either insert + or - between each numbers. So the number 1 will always be positive...

    – robert
    19 hours ago

















I suppose that your constraint is that numbers cannot be repeated and always begin with the smallest and finish with the biggest, I would try to find any recursive solution you know that you have to stop when n is equal to N and only print when the result is 0.

– vmrvictor
19 hours ago





I suppose that your constraint is that numbers cannot be repeated and always begin with the smallest and finish with the biggest, I would try to find any recursive solution you know that you have to stop when n is equal to N and only print when the result is 0.

– vmrvictor
19 hours ago













Can the numbers be repeated? Can you add the minus before the first number?

– Karol Dowbecki
19 hours ago





Can the numbers be repeated? Can you add the minus before the first number?

– Karol Dowbecki
19 hours ago













@KarolDowbecki no just conscecutive integer from 1 to N

– robert
19 hours ago





@KarolDowbecki no just conscecutive integer from 1 to N

– robert
19 hours ago













@robert If it can't, please update your question to fill in this extra constraint.

– tryman
19 hours ago





@robert If it can't, please update your question to fill in this extra constraint.

– tryman
19 hours ago




1




1





@Lino what it was exactly stated is the sum of consecutive integers that will always result to zero. It says that we can either insert + or - between each numbers. So the number 1 will always be positive...

– robert
19 hours ago





@Lino what it was exactly stated is the sum of consecutive integers that will always result to zero. It says that we can either insert + or - between each numbers. So the number 1 will always be positive...

– robert
19 hours ago












7 Answers
7






active

oldest

votes


















10














You could also use recursion here. Just remember your current integer, your max integer, your current sum and some kind of history of operations (could also be your final sequence).
In every level you proceed the path in two dirdctions: adding to your sum and substracting from it.



I did a quick implementation in Python, but it should be easy to transfer this to Java or whatever you are using.



def zero_sum(curr, n, seq, sum):
if curr == n and sum == 0:
print(seq)
elif curr < n:
zero_sum(curr + 1, n, seq + " - " + str(curr + 1), sum - (curr + 1))
zero_sum(curr + 1, n, seq + " + " + str(curr + 1), sum + (curr + 1))

zero_sum(1, 7, "1", 1)


Hopefully you get the idea.






share|improve this answer










New contributor




holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2





    You can not actually return anything using this approach, for eg: output String.

    – roundAbout
    16 hours ago











  • The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

    – holidayfun
    15 hours ago











  • What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

    – Solomon Ucko
    14 hours ago



















4














The first step is to turn the problem into an entirely regularly formed problem:



 n
∑ ±i = -1
i=2

n-2
∑ ±(i+2) = -1
i=0


The term 1 at the start has no prefix +/-. And the walking index better runs from 0 when using a Java array.



So one has n-1 coefficients -1 or +1 for the possible values.



A brute force approach would be to start with the highest values, i = n-2.



The upper/lower bounds for j = 0, ..., i would be ± (i + 1) * (2 + i + 2) / 2, so one can cut the evaluation there - when the till then calculated sum can no longer reach -1.



To represent the coefficients, one could make a new int[n - 1] or simply a new BitSet(n-1).



public void solve(int n) {
int i = n-2;
int sumDone = 0;
BigSet negates = new BitSet(n - 1);
solveRecursively(i, sumDone, negates);
}

private void solveRecursively(int i, int SumDone, BitSet negates) {
if (i < 0) {
if (sumDone == -1) {
System.out.println("Found: " + negates);
}
return;
}
...
}


The interesting, actual (home) work I leave to you. (With BitSet better i = n, ... , 2 by -1 seems simpler though.)






share|improve this answer































    3














    If I were you I would go for a graph implementation, and DFS algorithm. Imagine you have N nodes that are representing your numbers. Each number is connected to another via an "add" edge, or a "subtract" edge. So you have a fully connected graph. You can start from a node and compute all dfs paths that lead to zero.



    For more information about DFS algorithm, you can see the wikipage.



    Edit: In order to clarify my solution, the graph you will end up having will be a multigraph, which means that it has more than one edge between nodes. DFS in a multigraph is slightly more complicated, but it is not that hard.






    share|improve this answer


























    • is it possible to just use ordinary manipulations of array?

      – robert
      19 hours ago











    • @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

      – Farhood ET
      19 hours ago











    • @robert I have edited my post.

      – Farhood ET
      19 hours ago






    • 1





      Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

      – robert
      19 hours ago






    • 1





      @robert check holidayfun's answer.

      – Farhood ET
      19 hours ago



















    1














    Brute force approach:



    var input = 7; // change me

    var s = "1".repeat(input);
    IntStream.rangeClosed(0, Integer.parseInt(s, 2))
    .mapToObj(e -> {
    var str = Integer.toBinaryString(e);
    return "0".repeat(input - str.length()) + str; // padding zeros
    })
    .forEach(binStr -> {
    var total = 0;
    var plusMinus = binStr.split(""); // splitting binary string to String
    var sb = new StringBuilder();
    for(int i = 0; i < plusMinus.length; i++) {
    var pm = plusMinus[i].equals("1") ? "+": "-";
    var j = i + 1;
    total = pm.equals("+") ? (total + j): (total - j);
    sb.append(pm).append(j);
    }
    sb.append(" = ").append(total);
    if(total == 0) {
    System.out.println(sb.toString());
    }
    });


    The idea is to create binary string starting from 0000000 to 1111111 if the input is 7 where 0,1 represents -,+ and start looping. Now create another loop and apply the +/- operations for 1/0 to check if the total is 0 while creating a string from the operations performed. If the total is 0 then print the string that we got from performing the operations.






    share|improve this answer
























    • This approach takes same amount of time as a recursive method in java. STRANGE.

      – roundAbout
      16 hours ago






    • 1





      @roundAbout But recursion is a cleaner approach.

      – Aniket Sahrawat
      16 hours ago











    • @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

      – Baldrickk
      13 hours ago



















    1














    I would suggest a straight forward solution because as you mentioned you are dealing with conscecutive integer from 1 to N which are fixed. The only things that vary are the operators inbetween.



    Lets look at your example before we impliment a general solution:



    for n = 7 you need somehow to produce all possible combinations:



    1+2+3+4+5+6+7
    1+2+3+4+5+6-7
    1+2+3+4+5-6+7
    1+2+3+4+5-6-7
    ...
    1-2-3-4-5-6+7
    1-2-3-4-5-6-7


    If we remove the numbers from above strings/expressions then we'll have:



    ++++++
    +++++-
    ++++-+
    ++++--
    ...
    ----+-
    -----+
    ------


    Which reminds on binary numbers; if we interpret + as 0 and - as 1 the above can be mapped to the binary numbers from 000000 to 111111.



    For an input n you'll have n-1 operators inbetween, which means the count of all possible combinations will be 2^n-1.



    Putting all the above together something like below can be used to print those which sums are zero:



    public static void main(String args) throws IOException{
    permute(7);
    }
    public static void permute(int n){
    int combinations = (int)Math.pow(2, n-1);
    for(int i = 0; i < combinations; i++){
    String operators =String.format("%"+(n-1)+"s", Integer.toBinaryString(i)).replace(' ', '0');

    int totalSum = 1;
    StringBuilder sb = new StringBuilder();

    for(int x = 0; x< operators.length(); x++){
    sb.append(x+1);
    if(operators.charAt(x)=='0'){
    sb.append("+");
    totalSum = totalSum + (x+2);
    }
    else{
    sb.append("-");
    totalSum = totalSum-(x+2);
    }
    }
    sb.append(n);
    if(totalSum == 0){
    System.out.println(sb.toString() + " = " + totalSum);
    }
    }
    }


    Note/Example: String.format("%6s", Integer.toBinaryString(13)).replace(' ', '0') will produce a string with length = 6 from the binary representation of 13 with leading zeros, i.e 001101 instead of 1101 so that we get the required length of the operators.






    share|improve this answer































      1














      The question here is how much efficiency matters. If you're content to do a brute-force approach, a regression method like the one indicated by holidayfun is a fine way to go, though this will become unwieldy as n gets large.



      If performance speed matters, it may be worth doing a bit of math first. The easiest and most rewarding check is whether such a sum is even possible: since the sum of the first n natural numbers is n(n+1)/2, and since you want to divide this into two groups (a "positive" group and a "negative" group) of equal size, you must have that n(n+1)/4 is an integer. Therefore if neither n nor n+1 is divisible by four, stop. You cannot find such a sequence that adds to zero.



      This and a few other math tricks might speed up your application significantly, if speed is of the essence. For instance, finding one solution will often help you find others, for large n. For instance, if n=11, then {-11, -10, -7, -5} is one solution. But we could swap the -5 for any combination that adds to 5 that isn't in our set. Thus {-11, -10, -7, -3, -2} is also a solution, and similarly for -7, giving {-11, -10, -5, -4, -3} as a solution (we are not allowed to use -1 because the 1 must be positive). We could continue replacing the -10, the -11, and their components similarly to pick up six more solutions.



      This is probably how I'd approach this problem. Use a greedy algorithm to find the "largest" solution (the solution using the largest possible numbers), then keep splitting the components of that solution into successively smaller solutions. It is again fundamentally a recursion problem, but one whose running time decreases with the size of the component under consideration and which at each step generates another solution if a "smaller" solution exists. That being said, if you want every solution then you still have to check non-greedy combinations of your split (otherwise you'd miss solutions like {-7, -4, -3} in your n=7 example). If you only wanted a lot of solutions it would definitely be faster; but to get all of them it may be no better than a brute-force approach.






      share|improve this answer































        0














        It is a good question, but first you must have to try to solve it and show us what you tried so we can help you in the solution, this way you will improve more effectively.



        However, the below code is a solution I have write before years, I think the code need improvement but it will help..



        public static void main(String args) {
        String plus = " + ", minus = " - ";
        Set<String> operations = new HashSet<>();
        operations.add("1" + plus);
        operations.add("1" + minus);

        // n >= 3
        int n = 7;

        for (int i = 1; i < n - 1; i++) {
        Set<String> newOperation = new HashSet<>();
        for (String opt : operations) {
        if ((i + 2) == n) {
        newOperation.add(opt + (i + 1) + plus + n);
        newOperation.add(opt + (i + 1) + minus + n);
        } else {
        newOperation.add(opt + (i + 1) + plus);
        newOperation.add(opt + (i + 1) + minus);
        }
        }
        operations.clear();
        operations.addAll(newOperation);
        }
        evalOperations(operations);
        }

        private static void evalOperations(Set<String> operations) {

        // from JDK1.6, you can use the built-in Javascript engine.
        ScriptEngineManager mgr = new ScriptEngineManager();
        ScriptEngine engine = mgr.getEngineByName("JavaScript");
        try {
        for (String opt : operations) {
        if ((int) engine.eval(opt) == 0) {
        System.out.println(opt + " = 0");
        }
        }
        } catch (ScriptException e) {
        e.printStackTrace();
        }
        }





        share|improve this answer

























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          7 Answers
          7






          active

          oldest

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          7 Answers
          7






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          10














          You could also use recursion here. Just remember your current integer, your max integer, your current sum and some kind of history of operations (could also be your final sequence).
          In every level you proceed the path in two dirdctions: adding to your sum and substracting from it.



          I did a quick implementation in Python, but it should be easy to transfer this to Java or whatever you are using.



          def zero_sum(curr, n, seq, sum):
          if curr == n and sum == 0:
          print(seq)
          elif curr < n:
          zero_sum(curr + 1, n, seq + " - " + str(curr + 1), sum - (curr + 1))
          zero_sum(curr + 1, n, seq + " + " + str(curr + 1), sum + (curr + 1))

          zero_sum(1, 7, "1", 1)


          Hopefully you get the idea.






          share|improve this answer










          New contributor




          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 2





            You can not actually return anything using this approach, for eg: output String.

            – roundAbout
            16 hours ago











          • The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

            – holidayfun
            15 hours ago











          • What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

            – Solomon Ucko
            14 hours ago
















          10














          You could also use recursion here. Just remember your current integer, your max integer, your current sum and some kind of history of operations (could also be your final sequence).
          In every level you proceed the path in two dirdctions: adding to your sum and substracting from it.



          I did a quick implementation in Python, but it should be easy to transfer this to Java or whatever you are using.



          def zero_sum(curr, n, seq, sum):
          if curr == n and sum == 0:
          print(seq)
          elif curr < n:
          zero_sum(curr + 1, n, seq + " - " + str(curr + 1), sum - (curr + 1))
          zero_sum(curr + 1, n, seq + " + " + str(curr + 1), sum + (curr + 1))

          zero_sum(1, 7, "1", 1)


          Hopefully you get the idea.






          share|improve this answer










          New contributor




          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 2





            You can not actually return anything using this approach, for eg: output String.

            – roundAbout
            16 hours ago











          • The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

            – holidayfun
            15 hours ago











          • What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

            – Solomon Ucko
            14 hours ago














          10












          10








          10







          You could also use recursion here. Just remember your current integer, your max integer, your current sum and some kind of history of operations (could also be your final sequence).
          In every level you proceed the path in two dirdctions: adding to your sum and substracting from it.



          I did a quick implementation in Python, but it should be easy to transfer this to Java or whatever you are using.



          def zero_sum(curr, n, seq, sum):
          if curr == n and sum == 0:
          print(seq)
          elif curr < n:
          zero_sum(curr + 1, n, seq + " - " + str(curr + 1), sum - (curr + 1))
          zero_sum(curr + 1, n, seq + " + " + str(curr + 1), sum + (curr + 1))

          zero_sum(1, 7, "1", 1)


          Hopefully you get the idea.






          share|improve this answer










          New contributor




          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.










          You could also use recursion here. Just remember your current integer, your max integer, your current sum and some kind of history of operations (could also be your final sequence).
          In every level you proceed the path in two dirdctions: adding to your sum and substracting from it.



          I did a quick implementation in Python, but it should be easy to transfer this to Java or whatever you are using.



          def zero_sum(curr, n, seq, sum):
          if curr == n and sum == 0:
          print(seq)
          elif curr < n:
          zero_sum(curr + 1, n, seq + " - " + str(curr + 1), sum - (curr + 1))
          zero_sum(curr + 1, n, seq + " + " + str(curr + 1), sum + (curr + 1))

          zero_sum(1, 7, "1", 1)


          Hopefully you get the idea.







          share|improve this answer










          New contributor




          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|improve this answer



          share|improve this answer








          edited 19 hours ago





















          New contributor




          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 19 hours ago









          holidayfunholidayfun

          1015




          1015




          New contributor




          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          holidayfun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.








          • 2





            You can not actually return anything using this approach, for eg: output String.

            – roundAbout
            16 hours ago











          • The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

            – holidayfun
            15 hours ago











          • What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

            – Solomon Ucko
            14 hours ago














          • 2





            You can not actually return anything using this approach, for eg: output String.

            – roundAbout
            16 hours ago











          • The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

            – holidayfun
            15 hours ago











          • What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

            – Solomon Ucko
            14 hours ago








          2




          2





          You can not actually return anything using this approach, for eg: output String.

          – roundAbout
          16 hours ago





          You can not actually return anything using this approach, for eg: output String.

          – roundAbout
          16 hours ago













          The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

          – holidayfun
          15 hours ago





          The OP asked for printing the valid combinations. If it was necessary to return a long string of valid combinations you could return a single valid combination when ending the recursion and the levels above could concatenate the results of the two recursive calls.

          – holidayfun
          15 hours ago













          What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

          – Solomon Ucko
          14 hours ago





          What about using yield seq instead of print(seq) and yield from zero_sum(...) instead of zero_sum(...) to make it more reusable?

          – Solomon Ucko
          14 hours ago













          4














          The first step is to turn the problem into an entirely regularly formed problem:



           n
          ∑ ±i = -1
          i=2

          n-2
          ∑ ±(i+2) = -1
          i=0


          The term 1 at the start has no prefix +/-. And the walking index better runs from 0 when using a Java array.



          So one has n-1 coefficients -1 or +1 for the possible values.



          A brute force approach would be to start with the highest values, i = n-2.



          The upper/lower bounds for j = 0, ..., i would be ± (i + 1) * (2 + i + 2) / 2, so one can cut the evaluation there - when the till then calculated sum can no longer reach -1.



          To represent the coefficients, one could make a new int[n - 1] or simply a new BitSet(n-1).



          public void solve(int n) {
          int i = n-2;
          int sumDone = 0;
          BigSet negates = new BitSet(n - 1);
          solveRecursively(i, sumDone, negates);
          }

          private void solveRecursively(int i, int SumDone, BitSet negates) {
          if (i < 0) {
          if (sumDone == -1) {
          System.out.println("Found: " + negates);
          }
          return;
          }
          ...
          }


          The interesting, actual (home) work I leave to you. (With BitSet better i = n, ... , 2 by -1 seems simpler though.)






          share|improve this answer




























            4














            The first step is to turn the problem into an entirely regularly formed problem:



             n
            ∑ ±i = -1
            i=2

            n-2
            ∑ ±(i+2) = -1
            i=0


            The term 1 at the start has no prefix +/-. And the walking index better runs from 0 when using a Java array.



            So one has n-1 coefficients -1 or +1 for the possible values.



            A brute force approach would be to start with the highest values, i = n-2.



            The upper/lower bounds for j = 0, ..., i would be ± (i + 1) * (2 + i + 2) / 2, so one can cut the evaluation there - when the till then calculated sum can no longer reach -1.



            To represent the coefficients, one could make a new int[n - 1] or simply a new BitSet(n-1).



            public void solve(int n) {
            int i = n-2;
            int sumDone = 0;
            BigSet negates = new BitSet(n - 1);
            solveRecursively(i, sumDone, negates);
            }

            private void solveRecursively(int i, int SumDone, BitSet negates) {
            if (i < 0) {
            if (sumDone == -1) {
            System.out.println("Found: " + negates);
            }
            return;
            }
            ...
            }


            The interesting, actual (home) work I leave to you. (With BitSet better i = n, ... , 2 by -1 seems simpler though.)






            share|improve this answer


























              4












              4








              4







              The first step is to turn the problem into an entirely regularly formed problem:



               n
              ∑ ±i = -1
              i=2

              n-2
              ∑ ±(i+2) = -1
              i=0


              The term 1 at the start has no prefix +/-. And the walking index better runs from 0 when using a Java array.



              So one has n-1 coefficients -1 or +1 for the possible values.



              A brute force approach would be to start with the highest values, i = n-2.



              The upper/lower bounds for j = 0, ..., i would be ± (i + 1) * (2 + i + 2) / 2, so one can cut the evaluation there - when the till then calculated sum can no longer reach -1.



              To represent the coefficients, one could make a new int[n - 1] or simply a new BitSet(n-1).



              public void solve(int n) {
              int i = n-2;
              int sumDone = 0;
              BigSet negates = new BitSet(n - 1);
              solveRecursively(i, sumDone, negates);
              }

              private void solveRecursively(int i, int SumDone, BitSet negates) {
              if (i < 0) {
              if (sumDone == -1) {
              System.out.println("Found: " + negates);
              }
              return;
              }
              ...
              }


              The interesting, actual (home) work I leave to you. (With BitSet better i = n, ... , 2 by -1 seems simpler though.)






              share|improve this answer













              The first step is to turn the problem into an entirely regularly formed problem:



               n
              ∑ ±i = -1
              i=2

              n-2
              ∑ ±(i+2) = -1
              i=0


              The term 1 at the start has no prefix +/-. And the walking index better runs from 0 when using a Java array.



              So one has n-1 coefficients -1 or +1 for the possible values.



              A brute force approach would be to start with the highest values, i = n-2.



              The upper/lower bounds for j = 0, ..., i would be ± (i + 1) * (2 + i + 2) / 2, so one can cut the evaluation there - when the till then calculated sum can no longer reach -1.



              To represent the coefficients, one could make a new int[n - 1] or simply a new BitSet(n-1).



              public void solve(int n) {
              int i = n-2;
              int sumDone = 0;
              BigSet negates = new BitSet(n - 1);
              solveRecursively(i, sumDone, negates);
              }

              private void solveRecursively(int i, int SumDone, BitSet negates) {
              if (i < 0) {
              if (sumDone == -1) {
              System.out.println("Found: " + negates);
              }
              return;
              }
              ...
              }


              The interesting, actual (home) work I leave to you. (With BitSet better i = n, ... , 2 by -1 seems simpler though.)







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 18 hours ago









              Joop EggenJoop Eggen

              78.5k755105




              78.5k755105























                  3














                  If I were you I would go for a graph implementation, and DFS algorithm. Imagine you have N nodes that are representing your numbers. Each number is connected to another via an "add" edge, or a "subtract" edge. So you have a fully connected graph. You can start from a node and compute all dfs paths that lead to zero.



                  For more information about DFS algorithm, you can see the wikipage.



                  Edit: In order to clarify my solution, the graph you will end up having will be a multigraph, which means that it has more than one edge between nodes. DFS in a multigraph is slightly more complicated, but it is not that hard.






                  share|improve this answer


























                  • is it possible to just use ordinary manipulations of array?

                    – robert
                    19 hours ago











                  • @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

                    – Farhood ET
                    19 hours ago











                  • @robert I have edited my post.

                    – Farhood ET
                    19 hours ago






                  • 1





                    Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

                    – robert
                    19 hours ago






                  • 1





                    @robert check holidayfun's answer.

                    – Farhood ET
                    19 hours ago
















                  3














                  If I were you I would go for a graph implementation, and DFS algorithm. Imagine you have N nodes that are representing your numbers. Each number is connected to another via an "add" edge, or a "subtract" edge. So you have a fully connected graph. You can start from a node and compute all dfs paths that lead to zero.



                  For more information about DFS algorithm, you can see the wikipage.



                  Edit: In order to clarify my solution, the graph you will end up having will be a multigraph, which means that it has more than one edge between nodes. DFS in a multigraph is slightly more complicated, but it is not that hard.






                  share|improve this answer


























                  • is it possible to just use ordinary manipulations of array?

                    – robert
                    19 hours ago











                  • @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

                    – Farhood ET
                    19 hours ago











                  • @robert I have edited my post.

                    – Farhood ET
                    19 hours ago






                  • 1





                    Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

                    – robert
                    19 hours ago






                  • 1





                    @robert check holidayfun's answer.

                    – Farhood ET
                    19 hours ago














                  3












                  3








                  3







                  If I were you I would go for a graph implementation, and DFS algorithm. Imagine you have N nodes that are representing your numbers. Each number is connected to another via an "add" edge, or a "subtract" edge. So you have a fully connected graph. You can start from a node and compute all dfs paths that lead to zero.



                  For more information about DFS algorithm, you can see the wikipage.



                  Edit: In order to clarify my solution, the graph you will end up having will be a multigraph, which means that it has more than one edge between nodes. DFS in a multigraph is slightly more complicated, but it is not that hard.






                  share|improve this answer















                  If I were you I would go for a graph implementation, and DFS algorithm. Imagine you have N nodes that are representing your numbers. Each number is connected to another via an "add" edge, or a "subtract" edge. So you have a fully connected graph. You can start from a node and compute all dfs paths that lead to zero.



                  For more information about DFS algorithm, you can see the wikipage.



                  Edit: In order to clarify my solution, the graph you will end up having will be a multigraph, which means that it has more than one edge between nodes. DFS in a multigraph is slightly more complicated, but it is not that hard.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 19 hours ago

























                  answered 19 hours ago









                  Farhood ETFarhood ET

                  1709




                  1709













                  • is it possible to just use ordinary manipulations of array?

                    – robert
                    19 hours ago











                  • @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

                    – Farhood ET
                    19 hours ago











                  • @robert I have edited my post.

                    – Farhood ET
                    19 hours ago






                  • 1





                    Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

                    – robert
                    19 hours ago






                  • 1





                    @robert check holidayfun's answer.

                    – Farhood ET
                    19 hours ago



















                  • is it possible to just use ordinary manipulations of array?

                    – robert
                    19 hours ago











                  • @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

                    – Farhood ET
                    19 hours ago











                  • @robert I have edited my post.

                    – Farhood ET
                    19 hours ago






                  • 1





                    Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

                    – robert
                    19 hours ago






                  • 1





                    @robert check holidayfun's answer.

                    – Farhood ET
                    19 hours ago

















                  is it possible to just use ordinary manipulations of array?

                  – robert
                  19 hours ago





                  is it possible to just use ordinary manipulations of array?

                  – robert
                  19 hours ago













                  @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

                  – Farhood ET
                  19 hours ago





                  @robert You can implement your graph with an Adjacency List. geeksforgeeks.org/graph-and-its-representations

                  – Farhood ET
                  19 hours ago













                  @robert I have edited my post.

                  – Farhood ET
                  19 hours ago





                  @robert I have edited my post.

                  – Farhood ET
                  19 hours ago




                  1




                  1





                  Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

                  – robert
                  19 hours ago





                  Is there a little bit more simpler than that? To be honest I don't know anything about a graph in java yet. This problem that I am asking is just included in the recursion section of our java class specifically about permutations, so I am expecting any ideas that relates in implementing permutations and combinations or something like that....

                  – robert
                  19 hours ago




                  1




                  1





                  @robert check holidayfun's answer.

                  – Farhood ET
                  19 hours ago





                  @robert check holidayfun's answer.

                  – Farhood ET
                  19 hours ago











                  1














                  Brute force approach:



                  var input = 7; // change me

                  var s = "1".repeat(input);
                  IntStream.rangeClosed(0, Integer.parseInt(s, 2))
                  .mapToObj(e -> {
                  var str = Integer.toBinaryString(e);
                  return "0".repeat(input - str.length()) + str; // padding zeros
                  })
                  .forEach(binStr -> {
                  var total = 0;
                  var plusMinus = binStr.split(""); // splitting binary string to String
                  var sb = new StringBuilder();
                  for(int i = 0; i < plusMinus.length; i++) {
                  var pm = plusMinus[i].equals("1") ? "+": "-";
                  var j = i + 1;
                  total = pm.equals("+") ? (total + j): (total - j);
                  sb.append(pm).append(j);
                  }
                  sb.append(" = ").append(total);
                  if(total == 0) {
                  System.out.println(sb.toString());
                  }
                  });


                  The idea is to create binary string starting from 0000000 to 1111111 if the input is 7 where 0,1 represents -,+ and start looping. Now create another loop and apply the +/- operations for 1/0 to check if the total is 0 while creating a string from the operations performed. If the total is 0 then print the string that we got from performing the operations.






                  share|improve this answer
























                  • This approach takes same amount of time as a recursive method in java. STRANGE.

                    – roundAbout
                    16 hours ago






                  • 1





                    @roundAbout But recursion is a cleaner approach.

                    – Aniket Sahrawat
                    16 hours ago











                  • @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

                    – Baldrickk
                    13 hours ago
















                  1














                  Brute force approach:



                  var input = 7; // change me

                  var s = "1".repeat(input);
                  IntStream.rangeClosed(0, Integer.parseInt(s, 2))
                  .mapToObj(e -> {
                  var str = Integer.toBinaryString(e);
                  return "0".repeat(input - str.length()) + str; // padding zeros
                  })
                  .forEach(binStr -> {
                  var total = 0;
                  var plusMinus = binStr.split(""); // splitting binary string to String
                  var sb = new StringBuilder();
                  for(int i = 0; i < plusMinus.length; i++) {
                  var pm = plusMinus[i].equals("1") ? "+": "-";
                  var j = i + 1;
                  total = pm.equals("+") ? (total + j): (total - j);
                  sb.append(pm).append(j);
                  }
                  sb.append(" = ").append(total);
                  if(total == 0) {
                  System.out.println(sb.toString());
                  }
                  });


                  The idea is to create binary string starting from 0000000 to 1111111 if the input is 7 where 0,1 represents -,+ and start looping. Now create another loop and apply the +/- operations for 1/0 to check if the total is 0 while creating a string from the operations performed. If the total is 0 then print the string that we got from performing the operations.






                  share|improve this answer
























                  • This approach takes same amount of time as a recursive method in java. STRANGE.

                    – roundAbout
                    16 hours ago






                  • 1





                    @roundAbout But recursion is a cleaner approach.

                    – Aniket Sahrawat
                    16 hours ago











                  • @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

                    – Baldrickk
                    13 hours ago














                  1












                  1








                  1







                  Brute force approach:



                  var input = 7; // change me

                  var s = "1".repeat(input);
                  IntStream.rangeClosed(0, Integer.parseInt(s, 2))
                  .mapToObj(e -> {
                  var str = Integer.toBinaryString(e);
                  return "0".repeat(input - str.length()) + str; // padding zeros
                  })
                  .forEach(binStr -> {
                  var total = 0;
                  var plusMinus = binStr.split(""); // splitting binary string to String
                  var sb = new StringBuilder();
                  for(int i = 0; i < plusMinus.length; i++) {
                  var pm = plusMinus[i].equals("1") ? "+": "-";
                  var j = i + 1;
                  total = pm.equals("+") ? (total + j): (total - j);
                  sb.append(pm).append(j);
                  }
                  sb.append(" = ").append(total);
                  if(total == 0) {
                  System.out.println(sb.toString());
                  }
                  });


                  The idea is to create binary string starting from 0000000 to 1111111 if the input is 7 where 0,1 represents -,+ and start looping. Now create another loop and apply the +/- operations for 1/0 to check if the total is 0 while creating a string from the operations performed. If the total is 0 then print the string that we got from performing the operations.






                  share|improve this answer













                  Brute force approach:



                  var input = 7; // change me

                  var s = "1".repeat(input);
                  IntStream.rangeClosed(0, Integer.parseInt(s, 2))
                  .mapToObj(e -> {
                  var str = Integer.toBinaryString(e);
                  return "0".repeat(input - str.length()) + str; // padding zeros
                  })
                  .forEach(binStr -> {
                  var total = 0;
                  var plusMinus = binStr.split(""); // splitting binary string to String
                  var sb = new StringBuilder();
                  for(int i = 0; i < plusMinus.length; i++) {
                  var pm = plusMinus[i].equals("1") ? "+": "-";
                  var j = i + 1;
                  total = pm.equals("+") ? (total + j): (total - j);
                  sb.append(pm).append(j);
                  }
                  sb.append(" = ").append(total);
                  if(total == 0) {
                  System.out.println(sb.toString());
                  }
                  });


                  The idea is to create binary string starting from 0000000 to 1111111 if the input is 7 where 0,1 represents -,+ and start looping. Now create another loop and apply the +/- operations for 1/0 to check if the total is 0 while creating a string from the operations performed. If the total is 0 then print the string that we got from performing the operations.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 17 hours ago









                  Aniket SahrawatAniket Sahrawat

                  6,14521339




                  6,14521339













                  • This approach takes same amount of time as a recursive method in java. STRANGE.

                    – roundAbout
                    16 hours ago






                  • 1





                    @roundAbout But recursion is a cleaner approach.

                    – Aniket Sahrawat
                    16 hours ago











                  • @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

                    – Baldrickk
                    13 hours ago



















                  • This approach takes same amount of time as a recursive method in java. STRANGE.

                    – roundAbout
                    16 hours ago






                  • 1





                    @roundAbout But recursion is a cleaner approach.

                    – Aniket Sahrawat
                    16 hours ago











                  • @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

                    – Baldrickk
                    13 hours ago

















                  This approach takes same amount of time as a recursive method in java. STRANGE.

                  – roundAbout
                  16 hours ago





                  This approach takes same amount of time as a recursive method in java. STRANGE.

                  – roundAbout
                  16 hours ago




                  1




                  1





                  @roundAbout But recursion is a cleaner approach.

                  – Aniket Sahrawat
                  16 hours ago





                  @roundAbout But recursion is a cleaner approach.

                  – Aniket Sahrawat
                  16 hours ago













                  @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

                  – Baldrickk
                  13 hours ago





                  @roundAbout it does the same amount of work, it's just an iterative implementation of the same algorithm.

                  – Baldrickk
                  13 hours ago











                  1














                  I would suggest a straight forward solution because as you mentioned you are dealing with conscecutive integer from 1 to N which are fixed. The only things that vary are the operators inbetween.



                  Lets look at your example before we impliment a general solution:



                  for n = 7 you need somehow to produce all possible combinations:



                  1+2+3+4+5+6+7
                  1+2+3+4+5+6-7
                  1+2+3+4+5-6+7
                  1+2+3+4+5-6-7
                  ...
                  1-2-3-4-5-6+7
                  1-2-3-4-5-6-7


                  If we remove the numbers from above strings/expressions then we'll have:



                  ++++++
                  +++++-
                  ++++-+
                  ++++--
                  ...
                  ----+-
                  -----+
                  ------


                  Which reminds on binary numbers; if we interpret + as 0 and - as 1 the above can be mapped to the binary numbers from 000000 to 111111.



                  For an input n you'll have n-1 operators inbetween, which means the count of all possible combinations will be 2^n-1.



                  Putting all the above together something like below can be used to print those which sums are zero:



                  public static void main(String args) throws IOException{
                  permute(7);
                  }
                  public static void permute(int n){
                  int combinations = (int)Math.pow(2, n-1);
                  for(int i = 0; i < combinations; i++){
                  String operators =String.format("%"+(n-1)+"s", Integer.toBinaryString(i)).replace(' ', '0');

                  int totalSum = 1;
                  StringBuilder sb = new StringBuilder();

                  for(int x = 0; x< operators.length(); x++){
                  sb.append(x+1);
                  if(operators.charAt(x)=='0'){
                  sb.append("+");
                  totalSum = totalSum + (x+2);
                  }
                  else{
                  sb.append("-");
                  totalSum = totalSum-(x+2);
                  }
                  }
                  sb.append(n);
                  if(totalSum == 0){
                  System.out.println(sb.toString() + " = " + totalSum);
                  }
                  }
                  }


                  Note/Example: String.format("%6s", Integer.toBinaryString(13)).replace(' ', '0') will produce a string with length = 6 from the binary representation of 13 with leading zeros, i.e 001101 instead of 1101 so that we get the required length of the operators.






                  share|improve this answer




























                    1














                    I would suggest a straight forward solution because as you mentioned you are dealing with conscecutive integer from 1 to N which are fixed. The only things that vary are the operators inbetween.



                    Lets look at your example before we impliment a general solution:



                    for n = 7 you need somehow to produce all possible combinations:



                    1+2+3+4+5+6+7
                    1+2+3+4+5+6-7
                    1+2+3+4+5-6+7
                    1+2+3+4+5-6-7
                    ...
                    1-2-3-4-5-6+7
                    1-2-3-4-5-6-7


                    If we remove the numbers from above strings/expressions then we'll have:



                    ++++++
                    +++++-
                    ++++-+
                    ++++--
                    ...
                    ----+-
                    -----+
                    ------


                    Which reminds on binary numbers; if we interpret + as 0 and - as 1 the above can be mapped to the binary numbers from 000000 to 111111.



                    For an input n you'll have n-1 operators inbetween, which means the count of all possible combinations will be 2^n-1.



                    Putting all the above together something like below can be used to print those which sums are zero:



                    public static void main(String args) throws IOException{
                    permute(7);
                    }
                    public static void permute(int n){
                    int combinations = (int)Math.pow(2, n-1);
                    for(int i = 0; i < combinations; i++){
                    String operators =String.format("%"+(n-1)+"s", Integer.toBinaryString(i)).replace(' ', '0');

                    int totalSum = 1;
                    StringBuilder sb = new StringBuilder();

                    for(int x = 0; x< operators.length(); x++){
                    sb.append(x+1);
                    if(operators.charAt(x)=='0'){
                    sb.append("+");
                    totalSum = totalSum + (x+2);
                    }
                    else{
                    sb.append("-");
                    totalSum = totalSum-(x+2);
                    }
                    }
                    sb.append(n);
                    if(totalSum == 0){
                    System.out.println(sb.toString() + " = " + totalSum);
                    }
                    }
                    }


                    Note/Example: String.format("%6s", Integer.toBinaryString(13)).replace(' ', '0') will produce a string with length = 6 from the binary representation of 13 with leading zeros, i.e 001101 instead of 1101 so that we get the required length of the operators.






                    share|improve this answer


























                      1












                      1








                      1







                      I would suggest a straight forward solution because as you mentioned you are dealing with conscecutive integer from 1 to N which are fixed. The only things that vary are the operators inbetween.



                      Lets look at your example before we impliment a general solution:



                      for n = 7 you need somehow to produce all possible combinations:



                      1+2+3+4+5+6+7
                      1+2+3+4+5+6-7
                      1+2+3+4+5-6+7
                      1+2+3+4+5-6-7
                      ...
                      1-2-3-4-5-6+7
                      1-2-3-4-5-6-7


                      If we remove the numbers from above strings/expressions then we'll have:



                      ++++++
                      +++++-
                      ++++-+
                      ++++--
                      ...
                      ----+-
                      -----+
                      ------


                      Which reminds on binary numbers; if we interpret + as 0 and - as 1 the above can be mapped to the binary numbers from 000000 to 111111.



                      For an input n you'll have n-1 operators inbetween, which means the count of all possible combinations will be 2^n-1.



                      Putting all the above together something like below can be used to print those which sums are zero:



                      public static void main(String args) throws IOException{
                      permute(7);
                      }
                      public static void permute(int n){
                      int combinations = (int)Math.pow(2, n-1);
                      for(int i = 0; i < combinations; i++){
                      String operators =String.format("%"+(n-1)+"s", Integer.toBinaryString(i)).replace(' ', '0');

                      int totalSum = 1;
                      StringBuilder sb = new StringBuilder();

                      for(int x = 0; x< operators.length(); x++){
                      sb.append(x+1);
                      if(operators.charAt(x)=='0'){
                      sb.append("+");
                      totalSum = totalSum + (x+2);
                      }
                      else{
                      sb.append("-");
                      totalSum = totalSum-(x+2);
                      }
                      }
                      sb.append(n);
                      if(totalSum == 0){
                      System.out.println(sb.toString() + " = " + totalSum);
                      }
                      }
                      }


                      Note/Example: String.format("%6s", Integer.toBinaryString(13)).replace(' ', '0') will produce a string with length = 6 from the binary representation of 13 with leading zeros, i.e 001101 instead of 1101 so that we get the required length of the operators.






                      share|improve this answer













                      I would suggest a straight forward solution because as you mentioned you are dealing with conscecutive integer from 1 to N which are fixed. The only things that vary are the operators inbetween.



                      Lets look at your example before we impliment a general solution:



                      for n = 7 you need somehow to produce all possible combinations:



                      1+2+3+4+5+6+7
                      1+2+3+4+5+6-7
                      1+2+3+4+5-6+7
                      1+2+3+4+5-6-7
                      ...
                      1-2-3-4-5-6+7
                      1-2-3-4-5-6-7


                      If we remove the numbers from above strings/expressions then we'll have:



                      ++++++
                      +++++-
                      ++++-+
                      ++++--
                      ...
                      ----+-
                      -----+
                      ------


                      Which reminds on binary numbers; if we interpret + as 0 and - as 1 the above can be mapped to the binary numbers from 000000 to 111111.



                      For an input n you'll have n-1 operators inbetween, which means the count of all possible combinations will be 2^n-1.



                      Putting all the above together something like below can be used to print those which sums are zero:



                      public static void main(String args) throws IOException{
                      permute(7);
                      }
                      public static void permute(int n){
                      int combinations = (int)Math.pow(2, n-1);
                      for(int i = 0; i < combinations; i++){
                      String operators =String.format("%"+(n-1)+"s", Integer.toBinaryString(i)).replace(' ', '0');

                      int totalSum = 1;
                      StringBuilder sb = new StringBuilder();

                      for(int x = 0; x< operators.length(); x++){
                      sb.append(x+1);
                      if(operators.charAt(x)=='0'){
                      sb.append("+");
                      totalSum = totalSum + (x+2);
                      }
                      else{
                      sb.append("-");
                      totalSum = totalSum-(x+2);
                      }
                      }
                      sb.append(n);
                      if(totalSum == 0){
                      System.out.println(sb.toString() + " = " + totalSum);
                      }
                      }
                      }


                      Note/Example: String.format("%6s", Integer.toBinaryString(13)).replace(' ', '0') will produce a string with length = 6 from the binary representation of 13 with leading zeros, i.e 001101 instead of 1101 so that we get the required length of the operators.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 17 hours ago









                      EritreanEritrean

                      3,81211015




                      3,81211015























                          1














                          The question here is how much efficiency matters. If you're content to do a brute-force approach, a regression method like the one indicated by holidayfun is a fine way to go, though this will become unwieldy as n gets large.



                          If performance speed matters, it may be worth doing a bit of math first. The easiest and most rewarding check is whether such a sum is even possible: since the sum of the first n natural numbers is n(n+1)/2, and since you want to divide this into two groups (a "positive" group and a "negative" group) of equal size, you must have that n(n+1)/4 is an integer. Therefore if neither n nor n+1 is divisible by four, stop. You cannot find such a sequence that adds to zero.



                          This and a few other math tricks might speed up your application significantly, if speed is of the essence. For instance, finding one solution will often help you find others, for large n. For instance, if n=11, then {-11, -10, -7, -5} is one solution. But we could swap the -5 for any combination that adds to 5 that isn't in our set. Thus {-11, -10, -7, -3, -2} is also a solution, and similarly for -7, giving {-11, -10, -5, -4, -3} as a solution (we are not allowed to use -1 because the 1 must be positive). We could continue replacing the -10, the -11, and their components similarly to pick up six more solutions.



                          This is probably how I'd approach this problem. Use a greedy algorithm to find the "largest" solution (the solution using the largest possible numbers), then keep splitting the components of that solution into successively smaller solutions. It is again fundamentally a recursion problem, but one whose running time decreases with the size of the component under consideration and which at each step generates another solution if a "smaller" solution exists. That being said, if you want every solution then you still have to check non-greedy combinations of your split (otherwise you'd miss solutions like {-7, -4, -3} in your n=7 example). If you only wanted a lot of solutions it would definitely be faster; but to get all of them it may be no better than a brute-force approach.






                          share|improve this answer




























                            1














                            The question here is how much efficiency matters. If you're content to do a brute-force approach, a regression method like the one indicated by holidayfun is a fine way to go, though this will become unwieldy as n gets large.



                            If performance speed matters, it may be worth doing a bit of math first. The easiest and most rewarding check is whether such a sum is even possible: since the sum of the first n natural numbers is n(n+1)/2, and since you want to divide this into two groups (a "positive" group and a "negative" group) of equal size, you must have that n(n+1)/4 is an integer. Therefore if neither n nor n+1 is divisible by four, stop. You cannot find such a sequence that adds to zero.



                            This and a few other math tricks might speed up your application significantly, if speed is of the essence. For instance, finding one solution will often help you find others, for large n. For instance, if n=11, then {-11, -10, -7, -5} is one solution. But we could swap the -5 for any combination that adds to 5 that isn't in our set. Thus {-11, -10, -7, -3, -2} is also a solution, and similarly for -7, giving {-11, -10, -5, -4, -3} as a solution (we are not allowed to use -1 because the 1 must be positive). We could continue replacing the -10, the -11, and their components similarly to pick up six more solutions.



                            This is probably how I'd approach this problem. Use a greedy algorithm to find the "largest" solution (the solution using the largest possible numbers), then keep splitting the components of that solution into successively smaller solutions. It is again fundamentally a recursion problem, but one whose running time decreases with the size of the component under consideration and which at each step generates another solution if a "smaller" solution exists. That being said, if you want every solution then you still have to check non-greedy combinations of your split (otherwise you'd miss solutions like {-7, -4, -3} in your n=7 example). If you only wanted a lot of solutions it would definitely be faster; but to get all of them it may be no better than a brute-force approach.






                            share|improve this answer


























                              1












                              1








                              1







                              The question here is how much efficiency matters. If you're content to do a brute-force approach, a regression method like the one indicated by holidayfun is a fine way to go, though this will become unwieldy as n gets large.



                              If performance speed matters, it may be worth doing a bit of math first. The easiest and most rewarding check is whether such a sum is even possible: since the sum of the first n natural numbers is n(n+1)/2, and since you want to divide this into two groups (a "positive" group and a "negative" group) of equal size, you must have that n(n+1)/4 is an integer. Therefore if neither n nor n+1 is divisible by four, stop. You cannot find such a sequence that adds to zero.



                              This and a few other math tricks might speed up your application significantly, if speed is of the essence. For instance, finding one solution will often help you find others, for large n. For instance, if n=11, then {-11, -10, -7, -5} is one solution. But we could swap the -5 for any combination that adds to 5 that isn't in our set. Thus {-11, -10, -7, -3, -2} is also a solution, and similarly for -7, giving {-11, -10, -5, -4, -3} as a solution (we are not allowed to use -1 because the 1 must be positive). We could continue replacing the -10, the -11, and their components similarly to pick up six more solutions.



                              This is probably how I'd approach this problem. Use a greedy algorithm to find the "largest" solution (the solution using the largest possible numbers), then keep splitting the components of that solution into successively smaller solutions. It is again fundamentally a recursion problem, but one whose running time decreases with the size of the component under consideration and which at each step generates another solution if a "smaller" solution exists. That being said, if you want every solution then you still have to check non-greedy combinations of your split (otherwise you'd miss solutions like {-7, -4, -3} in your n=7 example). If you only wanted a lot of solutions it would definitely be faster; but to get all of them it may be no better than a brute-force approach.






                              share|improve this answer













                              The question here is how much efficiency matters. If you're content to do a brute-force approach, a regression method like the one indicated by holidayfun is a fine way to go, though this will become unwieldy as n gets large.



                              If performance speed matters, it may be worth doing a bit of math first. The easiest and most rewarding check is whether such a sum is even possible: since the sum of the first n natural numbers is n(n+1)/2, and since you want to divide this into two groups (a "positive" group and a "negative" group) of equal size, you must have that n(n+1)/4 is an integer. Therefore if neither n nor n+1 is divisible by four, stop. You cannot find such a sequence that adds to zero.



                              This and a few other math tricks might speed up your application significantly, if speed is of the essence. For instance, finding one solution will often help you find others, for large n. For instance, if n=11, then {-11, -10, -7, -5} is one solution. But we could swap the -5 for any combination that adds to 5 that isn't in our set. Thus {-11, -10, -7, -3, -2} is also a solution, and similarly for -7, giving {-11, -10, -5, -4, -3} as a solution (we are not allowed to use -1 because the 1 must be positive). We could continue replacing the -10, the -11, and their components similarly to pick up six more solutions.



                              This is probably how I'd approach this problem. Use a greedy algorithm to find the "largest" solution (the solution using the largest possible numbers), then keep splitting the components of that solution into successively smaller solutions. It is again fundamentally a recursion problem, but one whose running time decreases with the size of the component under consideration and which at each step generates another solution if a "smaller" solution exists. That being said, if you want every solution then you still have to check non-greedy combinations of your split (otherwise you'd miss solutions like {-7, -4, -3} in your n=7 example). If you only wanted a lot of solutions it would definitely be faster; but to get all of them it may be no better than a brute-force approach.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 11 hours ago









                              GalendoGalendo

                              1185




                              1185























                                  0














                                  It is a good question, but first you must have to try to solve it and show us what you tried so we can help you in the solution, this way you will improve more effectively.



                                  However, the below code is a solution I have write before years, I think the code need improvement but it will help..



                                  public static void main(String args) {
                                  String plus = " + ", minus = " - ";
                                  Set<String> operations = new HashSet<>();
                                  operations.add("1" + plus);
                                  operations.add("1" + minus);

                                  // n >= 3
                                  int n = 7;

                                  for (int i = 1; i < n - 1; i++) {
                                  Set<String> newOperation = new HashSet<>();
                                  for (String opt : operations) {
                                  if ((i + 2) == n) {
                                  newOperation.add(opt + (i + 1) + plus + n);
                                  newOperation.add(opt + (i + 1) + minus + n);
                                  } else {
                                  newOperation.add(opt + (i + 1) + plus);
                                  newOperation.add(opt + (i + 1) + minus);
                                  }
                                  }
                                  operations.clear();
                                  operations.addAll(newOperation);
                                  }
                                  evalOperations(operations);
                                  }

                                  private static void evalOperations(Set<String> operations) {

                                  // from JDK1.6, you can use the built-in Javascript engine.
                                  ScriptEngineManager mgr = new ScriptEngineManager();
                                  ScriptEngine engine = mgr.getEngineByName("JavaScript");
                                  try {
                                  for (String opt : operations) {
                                  if ((int) engine.eval(opt) == 0) {
                                  System.out.println(opt + " = 0");
                                  }
                                  }
                                  } catch (ScriptException e) {
                                  e.printStackTrace();
                                  }
                                  }





                                  share|improve this answer






























                                    0














                                    It is a good question, but first you must have to try to solve it and show us what you tried so we can help you in the solution, this way you will improve more effectively.



                                    However, the below code is a solution I have write before years, I think the code need improvement but it will help..



                                    public static void main(String args) {
                                    String plus = " + ", minus = " - ";
                                    Set<String> operations = new HashSet<>();
                                    operations.add("1" + plus);
                                    operations.add("1" + minus);

                                    // n >= 3
                                    int n = 7;

                                    for (int i = 1; i < n - 1; i++) {
                                    Set<String> newOperation = new HashSet<>();
                                    for (String opt : operations) {
                                    if ((i + 2) == n) {
                                    newOperation.add(opt + (i + 1) + plus + n);
                                    newOperation.add(opt + (i + 1) + minus + n);
                                    } else {
                                    newOperation.add(opt + (i + 1) + plus);
                                    newOperation.add(opt + (i + 1) + minus);
                                    }
                                    }
                                    operations.clear();
                                    operations.addAll(newOperation);
                                    }
                                    evalOperations(operations);
                                    }

                                    private static void evalOperations(Set<String> operations) {

                                    // from JDK1.6, you can use the built-in Javascript engine.
                                    ScriptEngineManager mgr = new ScriptEngineManager();
                                    ScriptEngine engine = mgr.getEngineByName("JavaScript");
                                    try {
                                    for (String opt : operations) {
                                    if ((int) engine.eval(opt) == 0) {
                                    System.out.println(opt + " = 0");
                                    }
                                    }
                                    } catch (ScriptException e) {
                                    e.printStackTrace();
                                    }
                                    }





                                    share|improve this answer




























                                      0












                                      0








                                      0







                                      It is a good question, but first you must have to try to solve it and show us what you tried so we can help you in the solution, this way you will improve more effectively.



                                      However, the below code is a solution I have write before years, I think the code need improvement but it will help..



                                      public static void main(String args) {
                                      String plus = " + ", minus = " - ";
                                      Set<String> operations = new HashSet<>();
                                      operations.add("1" + plus);
                                      operations.add("1" + minus);

                                      // n >= 3
                                      int n = 7;

                                      for (int i = 1; i < n - 1; i++) {
                                      Set<String> newOperation = new HashSet<>();
                                      for (String opt : operations) {
                                      if ((i + 2) == n) {
                                      newOperation.add(opt + (i + 1) + plus + n);
                                      newOperation.add(opt + (i + 1) + minus + n);
                                      } else {
                                      newOperation.add(opt + (i + 1) + plus);
                                      newOperation.add(opt + (i + 1) + minus);
                                      }
                                      }
                                      operations.clear();
                                      operations.addAll(newOperation);
                                      }
                                      evalOperations(operations);
                                      }

                                      private static void evalOperations(Set<String> operations) {

                                      // from JDK1.6, you can use the built-in Javascript engine.
                                      ScriptEngineManager mgr = new ScriptEngineManager();
                                      ScriptEngine engine = mgr.getEngineByName("JavaScript");
                                      try {
                                      for (String opt : operations) {
                                      if ((int) engine.eval(opt) == 0) {
                                      System.out.println(opt + " = 0");
                                      }
                                      }
                                      } catch (ScriptException e) {
                                      e.printStackTrace();
                                      }
                                      }





                                      share|improve this answer















                                      It is a good question, but first you must have to try to solve it and show us what you tried so we can help you in the solution, this way you will improve more effectively.



                                      However, the below code is a solution I have write before years, I think the code need improvement but it will help..



                                      public static void main(String args) {
                                      String plus = " + ", minus = " - ";
                                      Set<String> operations = new HashSet<>();
                                      operations.add("1" + plus);
                                      operations.add("1" + minus);

                                      // n >= 3
                                      int n = 7;

                                      for (int i = 1; i < n - 1; i++) {
                                      Set<String> newOperation = new HashSet<>();
                                      for (String opt : operations) {
                                      if ((i + 2) == n) {
                                      newOperation.add(opt + (i + 1) + plus + n);
                                      newOperation.add(opt + (i + 1) + minus + n);
                                      } else {
                                      newOperation.add(opt + (i + 1) + plus);
                                      newOperation.add(opt + (i + 1) + minus);
                                      }
                                      }
                                      operations.clear();
                                      operations.addAll(newOperation);
                                      }
                                      evalOperations(operations);
                                      }

                                      private static void evalOperations(Set<String> operations) {

                                      // from JDK1.6, you can use the built-in Javascript engine.
                                      ScriptEngineManager mgr = new ScriptEngineManager();
                                      ScriptEngine engine = mgr.getEngineByName("JavaScript");
                                      try {
                                      for (String opt : operations) {
                                      if ((int) engine.eval(opt) == 0) {
                                      System.out.println(opt + " = 0");
                                      }
                                      }
                                      } catch (ScriptException e) {
                                      e.printStackTrace();
                                      }
                                      }






                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited 17 hours ago

























                                      answered 17 hours ago









                                      Ebraheem Alrabee'Ebraheem Alrabee'

                                      627924




                                      627924






























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