Determine all consecutive subsets of the set {1,2,3,…,n}. The subsets should have at least 2 elements

Question

I need to partition a set S={1, 2, 3, … , n} consisting of consecutive numbers such that each subset has has at least 2 elements (rule 1) and it consists of consecutive numbers (rule 2).

The rules are:

1. Each subset has at least two elements.

2. All elements of all subsets are consecutive.

3. All elements of S are included in the partition.

Examples:

There is 1 subset for n = 2: 
1 2
There is 1 subset for n = 3:  
1 2 3   
There are 2 subset combinations for n = 4:
1 2 3 4
1 2 - 3 4
There are 3 subset combinations for n = 5:
1 2 3 4 5
1 2 - 3 4 5
1 2 3 - 4 5
There are 5 subset combinations for n = 6:
1 2 3 4 5 6
1 2 - 3 4 5 6
1 2 3 - 4 5 6
1 2 3 4 - 5 6
1 2 - 3 4 - 5 6
There are 8 subset combinations for n = 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
1 2 - 3 4 - 5 6 7
1 2 - 3 4 5 - 6 7
1 2 3 - 4 5 - 6 7
There are 13 subset combinations for n = 8:
1 2 3 4 5 6 7 8
1 2 - 3 4 5 6 7 8
1 2 3 - 4 5 6 7 8
1 2 3 4 - 5 6 7 8
1 2 3 4 5 - 6 7 8
1 2 3 4 5 6 - 7 8
1 2 - 3 4 - 5 6 7 8
1 2 - 3 4 5 - 6 7 8
1 2 - 3 4 5 6 - 7 8
1 2 3 - 4 5 - 6 7 8 
1 2 3 - 4 5 6 - 7 8
1 2 3 4 - 5 6 - 7 8
1 2 - 3 4 - 5 6 - 7 8
There are 21 subset combinations for n = 9:
1 2 3 4 5 6 7 8 9
1 2 - 3 4 5 6 7 8 9
1 2 3 - 4 5 6 7 8 9
1 2 3 4 - 5 6 7 8 9
1 2 3 4 5 - 6 7 8 9
1 2 3 4 5 6 - 7 8 9
1 2 3 4 5 6 7 - 8 9
1 2 - 3 4 - 5 6 7 8 9
1 2 - 3 4 5 - 6 7 8 9
1 2 - 3 4 5 6 - 6 7 9
1 2 - 3 4 5 6 7 - 8 9
1 2 3 - 4 5 - 6 7 8 9 
1 2 3 - 4 5 6 - 7 8 9
1 2 3 - 4 5 6 7 - 8 9
1 2 3 4 - 5 6 - 7 8 9
1 2 3 4 - 5 6 7 - 8 9
1 2 3 4 5 - 6 7 - 8 9
1 2 - 3 4 - 5 6 - 7 8 9
1 2 - 3 4 - 5 6 7 - 8 9
1 2 - 3 4 5 - 6 7 - 8 9
1 2 3 - 4 5 - 6 7 - 8 9
There are 34 subset combinations for n = 10:
1 2 3 4 5 6 7 8 9 10
1 2 - 3 4 5 6 7 8 9 10
1 2 3 - 4 5 6 7 8 9 10
1 2 3 4 - 5 6 7 8 9 10
1 2 3 4 5 - 6 7 8 9 10
1 2 3 4 5 6 - 7 8 9 10
1 2 3 4 5 6 7 - 8 9 10
1 2 3 4 5 6 7 8 - 9 10
1 2 - 3 4 - 5 6 7 8 9 10
1 2 - 3 4 5 - 6 7 8 9 10
1 2 - 3 4 5 6 - 6 7 9 10
1 2 - 3 4 5 6 7 - 8 9 10
1 2 - 3 4 5 6 7 8 - 9 10
1 2 3 - 4 5 - 6 7 8 9 10
1 2 3 - 4 5 6 - 7 8 9 10
1 2 3 - 4 5 6 7 - 8 9 10
1 2 3 - 4 5 6 7 8 - 9 10
1 2 3 4 - 5 6 - 7 8 9 10
1 2 3 4 - 5 6 7 - 8 9 10
1 2 3 4 - 5 6 7 8 - 9 10
1 2 3 4 5 - 6 7 - 8 9 10
1 2 3 4 5 - 6 7 8 - 9 10
1 2 3 4 5 6 - 7 8 - 9 10
1 2 - 3 4 - 5 6 - 7 8 9 10
1 2 - 3 4 - 5 6 7 - 8 9 10
1 2 - 3 4 - 5 6 7 8 - 9 10
1 2 - 3 4 5 - 6 7 - 8 9 10
1 2 - 3 4 5 - 6 7 8 - 9 10
1 2 - 3 4 5 6 - 7 8 - 9 10
1 2 3 - 4 5 - 6 7 - 8 9 10
1 2 3 - 4 5 - 6 7 8 - 9 10
1 2 3 - 4 5 6 - 7 8 - 9 10
1 2 3 4 - 5 6 - 7 8 - 9 10
1 2 - 3 4 - 5 6 - 7 8 - 9 10

I didn't write them down here but there are 55 subset combinations for n = 11 and 89 subset combinations for n = 12.

I need to write a Visual Basic code listing all possible subset groups for n. I have been thinking on the solution for days but it seems that the solution of the problem is beyond my capacity. The number of required nested loops increases with n and I could not figure out how to program the nested loops with increasing number. Any help will be greatly appreciated.

After some research, I found out this is the problem of "compositions of n with all parts >1" and the total number of possible compositions are Fibonacci numbers (Fn-1 for n).

Answer 1

My answer to you is to try and come up with a recurrence relation of the given pattern. Think recursively. How can I break this problem down into smaller subproblems until reaching the smallest problem. Solve that smallest problem. After solving that smallest problem, think induction. Hypothesize on the what the nth step will be and how you will reach the (n+1)th step. Try to solve that (n+1)th step. Once you have come up with a recurrence relation on the pattern being given, it should not be too difficult to think about how to solve this pattern recursively. Instead of trying to use nested loops, this approach may be more intuitive.

Answer 2

We already know the answer for these cases (as you wrote in your examples):

1. n=2
2. n=3
3. n=4

For n=5:

• You can partition from 2: 1 2 - 3 4 5. This is like dividing the 5 member set into two sets, first one n=2, and second one n=3. We can now continue dividing each half, but we already know the solutions when n=2 and n=3!
• You can partition from 3: 1 2 3 - 4 5. This is like dividing the 5 member set into two sets, first one n=3, and second one n=2. We can now continue dividing each half, but we already know the solution when n=2 and n=3!

For n=6:

• You can partition into two sets from 2: 1 2 - 3 4 5 6. This is like dividing the 6 member set into two sets, first one n=2, and second one n=4. We can now continue dividing each half, but we already know the solution when n=2. Solve the second half by assuming n=4!
• You can partition into two sets from 3: 1 2 3 - 4 5 6. This is like dividing the 6 member set into two sets, first one n=3, and second one n=3, We can now continue dividing each half, but we already know the solution when n=3 and n=3!
• You can partition into two sets from 3: 1 2 3 4 - 5 6. This is like dividing the 6 member set into two sets, first one n=4, and second one n=2, We can now continue dividing each half. Solve the first half by setting n=4. For the second half, we already know the solution when n=2!

This is a simple recursion relationship. The general case:

Partition (S): (where |S|>4)
- For i from 2 to |S|-2, partition the given set into two halves: s1 and s2 from i (s1={1,...,i}, s2={i+1,...,n}), and print the two subsets as a solution.
- Recursively continue for each half by calling Partition(s1) and Partition(s2)

---

Another perhaps harder solution is to assume we are dividing the numbers 1 to n into n sections, where the length of each section can be either 0, 2, or a number greater than 2. In other words let xi be the length of each section:

x1 + x2 + ... xn = n, where the range of xi is: {0} + [2,n]

This is a system of linear non-equalities can can be solved by methods described here.