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READ ME FIRST:
I have just found the official solutions online (have been looking for them for a while, but after posting this I quickly found it), and I'm currently trying to understand it. As I can tell, it uses a much simpler DP, which only uses an O(N) array.
EDIT 2
I kinda figured out what and how it does, but I still do not understand why it gives the correct answer. It uses a dp[index] -> weight sum if we select that interval and previous ones to cover [A, start_of_current]. It sorts the intervals by end time, than process each one in order. If it's start is A, we 'select' it and set it's dp[i] to it's weight. If not, it tries to find the previous selected interval whose end is closest to the current intervals start, but there is no gap in between. If found one, we 'select' the current and set it's dp[i]=it's weight + dp[j] where j is the found previous interval. If we can not find any previous, dp[i] = inf. We keep a list of the 'selected' intervals, which is, as we only push items to it, is sorted by the interval's end. When we selected one, we delete all the others from the end of this list whose cost (dp[i]) is bigger than the current one's.
What I do not get from this, is why this searching-the-closest-end yields the smallest overall weight sum, and also, why we delete all those intervals from the selection list?

Story:
This year I attended a programming competition, where we had a few interesting problems. I had trouble with some of them, and since the next round approaches, I want to clear them up.

The problem in a nutshell
We are given N weighted intervals, and a [A,B] interval. We have to cover the [A,B] interval with the given ones, in such a way, that we minimize the overall weight sum (and then the number of required intervals). We need to print the sum, the number of intervals, and then the intervals themselves. If we can not cover [A, B], we need to report that as a special value (-1).

First thoughts
If we sort the intervals by begin time, then we can do simple 0-1 Knapsack-like DP and solve the problem. Also, if the priorities were swapped (minimize count THAN sum), a simple greedy would do it.

The limits
Basically, all interval starts and ends are in the range 1-1 000 000, and N<=100 000. All intervals lie within [A,B].

My approach
I wrote a recursive algorithm like the 0-1 Knapsack one in python, that also stored the last selected interval - thus allowing the recovering of the selection list from the DP array later. It was a (current_interval, last_covered_day) -> (cost, last_selected_interval, last_covered_day')-like function. I used a dict as a DP array, as a regular array that big would have violated memory constraints and filling it fully would also increase runtime (at least that's what I thought - but a 1000000 * 100000 array would certainly would!). I wrote the function as a recursive one so it would not fill in the entire DP array and be faster & more memory-efficient.

The problem with this
Simply, I got RecursionError: maximum recursion depth exceededs on larger datasets - 100k deep recursion was simply too much. I read since on GeeksForGeeks that it should be possible to increase it, but I am still not confident that it would be safe. My recursive function is also not tail-call optimizatiable, so that would also not work.

So my questions
Is there a way of solving this problem without DP? If no, is filling in a full table an option with those high limits? Maybe we can come up with a different DP approach that does not use such big tables? Is it safe to just increase recursion depth limits with this kinds of problems?

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  • $\begingroup$ Could you link the solution you found? $\endgroup$ – Daniel Feb 27 at 8:07
  • $\begingroup$ @Daniel it's in Hungarian, along with the problems. I'll update my question with it instead, as I now understand WHAT & HOW it does, but I still do not see WHY it works... $\endgroup$ – Sasszem Feb 27 at 12:16
  • $\begingroup$ Are you still interested in an explanation to "why this searching-the-closest-end yields the smallest overall weight sum, and also, why we delete all those intervals from the selection list"? $\endgroup$ – John L. Apr 14 at 2:15
  • $\begingroup$ @JohnL. well, kind of yes. I've put this problem (and others) on a self since the competition, but I am still curious. $\endgroup$ – Sasszem Apr 15 at 4:06
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You can always convert any recursive algorithm (that uses the call stack) into a iterative algorithm where you explicitly push and pop onto a stack that you manage. This should eliminate recursion depth exceeded errors.

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  • $\begingroup$ Well, yes, I thought about that, but I still feel that I should not needed to do that - there must be a simpler solution. PS: There is, just I can not understand it yet, see my edited question. $\endgroup$ – Sasszem Feb 26 at 23:49

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