[3, 2, 1] has no valid modifications by the rules given.
I wonder how little wiggle-room there needs to be for a solution to be possible? Does [4, 2, 1] have a solution? You can get to with [4, -1, 3, 1] with one step, but I'm not sure where you can go from there.
To be pedantic, this isn't the operation he gave - "split an integer into 2 smaller integers" is the operation, but 2 into [3, -1] is splitting it into a larger and smaller integer.
Read literally, the numbers are constrained to not jump over the 0. It might depend on interpretation (absolute value or distance from positive infinity?); but negatives are tricky to handle with these rules and I think they might be illegal.
> You can get to with [4, -1, 3, 1] with one step, but I'm not sure where you can go from there.
Well, it's easy to show that you can't go anywhere from there:
1. You can't combine the 1 with the 3, because you'd have two 4s.
2. You can't combine the -1 with the 4, because you'd have two 3s.
3. You can't (usefully) combine the -1 with the 3, because that just means you never took your previous move of splitting that pair out of the 2.
Those are all the possible applications of the combining rule. You can't apply the splitting rule either:
1. Splitting the four into positive integers will cause a duplicate.
2. Splitting a negative number out of the four will violate the rule against numbers larger than 4.
3. Splitting the 3 has the same problems. You can't take out a negative number without leaving a residual of 4 or more, and you can't take out a positive number without leaving a residual of 1.
4. Taking a -1, -2, or -3 out of the 1 will leave a duplicate, and taking a positive value out of the 1 is impossible.
5. You can split the -1 into [-3, 2]. But at that point you need to get rid of the 1 in final position so that you can combine the -3 into the 4, and there isn't a way to do that. Once you've gotten here, your list already contains every legal positive value.
Huh. There's bound to be some weird constraint there.
E: Looking at the sibling comments, I'm disregarding negative numbers here. Probably that breaks the line of thought here.
For example, a sequence made up of all numbers from 1 to n is unsolvable. A sequence of [1, 2, ..., n-1, n+1] for n >= 3 also seems to be unsolvable, since no number up to n-1 can split into a smaller number (they are all in the list) and n+1 also cannot split, because 1 is in there.
So it seems that there would be an unsolvable class of lists along the lines of [1, 2, ..., k, ..., n-1, n+k] by that argument, because the n+k cannot split ever, because all numbers up to k are in that list.
And I guess for any unsolvable list with a max value of M, you can add any number below M into the unsolvable list, because that just removes moves.
The article says you start with a list of positive integers.
It also clearly shows right in the first example of the first rule that you can go smaller than the smallest initial value: [7, 5, 3] -> [6, 1, 5, 3] is the first legal move it shows, and you'll notice that 1 is smaller than 7, 5, or 3.
* You can never make an integer greater than the largest integer in the original list.
* You can never make a move that results in the same integer appearing in the list more than once.
You can make a negative number and you can go lower than the smallest but not greater than the largest, so you're wrong on both counts.
However, as a result of #1, you in fact cannot make negative integers. It’s not possible to split a nonnegative integer into two lesser-or-equal integers such that either of them is negative.
I believe Sharlin is correct here. Rule #1 states `Split an integer into two smaller integers.` 1 -> [2, -1] breaks this rule due to 2 not being smaller than 1.
I wonder how little wiggle-room there needs to be for a solution to be possible? Does [4, 2, 1] have a solution? You can get to with [4, -1, 3, 1] with one step, but I'm not sure where you can go from there.