# Algorithm To Compute The Gaps Between A Set of Intervals

Problem

Given a set of intervals with possibly non-distinct start and end points, find all maximal gaps. A gap is defined as an interval that does not overlap with any given interval. All endpoints are integers and inclusive.

For example, given the following set of intervals:

$$\{[2,6], [1,9], [12,19]\}$$

The set of all maximal gaps is:

$$\{[10,11]\}$$

For the following set of intervals:

$$\{[2,6], [1,9], [3,12], [18,20]\}$$

The set of all maximal gaps is:

$$\{[13,17]\}$$

because that produces the maximal gap.

Proposed Algorithm

My proposed algorithm (modified approach taken by John L.) to compute these gaps is:

1. Order the intervals by ascending start date.
2. Initialise an empty list gaps that will store gaps
3. Initialise a variable, last_covered_point, to the end point of the first interval.
4. Iterate through all intervals in the sorted order. For each interval [start, end], do the following.
1. If start > last_covered_point + 1, add the gap, [last_covered_point + 1, start - 1] to gaps.
2. Assign max(last_covered_point, end) to last_covered_point.
5. Return gaps

I have tested my algorithm on a few cases and it produces the correct results. But I cannot say with 100% guarantee that it works for every interval permutation and combination. Is there a way to prove this handles every permutation and combination?

• @JohnL. yes the question is correct now - thanks
– Mojo
Dec 18, 2020 at 21:25

The key to prove your algorithm is correct is to find enough invariants of the loop, step 4 so that we apply use mathematical induction.

Let $$I_1, I_2, \cdots, I_n$$ denote the sorted intervals. When the algorithm has just finished processing $$I_i$$, we record the values of gaps and last_covered_point as $$\text{gaps}_i$$ and $$\text{last_covered_point}_i$$ respectively.

Let us prove the following proposition, $$P(i)$$, for $$i=1, 2, \cdots, n$$.

$$\text{gaps}_i$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_i$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$.

When $$i=1$$, $$\text{gaps}_1$$ is the empty set and $$\text{last_covered_point}_1$$ is the right endpoint of $$I_1$$. So $$P(1)$$ is correct.

For the sake of mathematical induction, assume $$P(i)$$ is correct, where $$1\le i\lt n$$. Let $$I_{i+1}=[s, e]$$. There are two cases.

1. If $$s\gt\text{last_covered_point}_i+1$$, then $$\text{gaps}_i\cup[\text{last_covered_point}_i +1, s-1]=\text{gaps}_{i+1}.$$
Let $$m$$ be any point between the start point of $$I_1$$ and the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$. Suppose $$m$$ not covered by any of $$I_1, I_2, \cdots, I_{i+1}$$.

• If $$m\le\text{last_covered_point}_i$$, the induction hypothesis says that $$m$$ is covered by some interval in $$\text{gap}_i$$.
• Otherwise, $$m\gt\text{last_covered_point}_i$$. Since $$m$$ is not covered by $$I_{i+1}$$, we know $$m. So $$m$$ is covered by $$[\text{last_covered_point}_i +1, s-1]$$.

In both cases, $$m$$ is covered by some interval in $$\text{gaps}_{i+1}$$. Since $$\text{last_covered_point}_i$$ is the largest point covered by one of $$I_1, I_2, \cdots, I_i$$ and $$s$$ is the smallest point covered by $$I_{i+1}$$, $$[\text{last_covered_point}_i +1, s-1]$$ is a maximal gap.

2. Otherwise, we have $$s\le\text{last_covered_point}_i$$+1. We can also verify that $$\text{gaps}_{i+1}=\text{gaps}_{i}$$ is the set of all maximal gaps for $$I_1, I_2, \cdots, I_{i+1}$$.

Finally, since step 4.2 says $$\text{last_covered_point}_{i+1}=\max(\text{last_covered_point}_i, e)$$ and $$\text{last_covered_point}_i$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_i$$, $$\text{last_covered_point}_{i+1}$$ is the maximum of all right endpoints of $$I_1, I_2, \cdots, I_{i+1}$$.

So, $$P(i+1)$$ is correct. $$\quad\checkmark$$.

• I probably didn't make my post precise enough and so I think you may have misunderstood what I meant by a gap. I have cleaned up my original post to define the problem better but I also took your example and customised it to what I think the algorithm should be. The main issue I have is that I am not sure if I have covered all possible interval permutations.
– Mojo
Dec 17, 2020 at 15:19
• @Mojo, thanks for noticing that last_covered_point should be updated in step 4.2 as well. Dec 18, 2020 at 16:30
• @Mojo, I just wrote a proof. The proof is not completely formal, since "maximal gap" is given by a definition (although it is easy to define) and the case 2 is not proved in detail. However, the idea should be clear enough. Dec 20, 2020 at 5:24
• In my last comment, 'since "maximal gap" is given by a definition' should have been 'since "maximal gap" is not given by a definition'. Jan 4, 2021 at 1:35

I'd like to propose a quite simple algorithm as well. The idea is this: we're going to place open and close parentheses on the number line at the boundaries of each interval. For example, for the intervals $$(1,5), (2,7), (9, 10)$$, the number line would look like this:

1 2 3 4 5 6 7 8 9 10
( (     )   )   ( )


Then we'll just scan left to right, counting parentheses. When all the parentheses get closed, we start a gap. Note in particular that in the above diagram, we have lost the information about which close parenthesis is paired with which open parenthesis -- because it doesn't actually matter. So:

1. Convert each interval $$(a,b)$$ to pairs $$(a,\mathtt o),(b,\mathtt c)$$. ($$\mathtt o$$ and $$\mathtt c$$ are for open and close, respectively.)
2. Sort all the pairs you get from this process. (When sorting, use lexicographic ordering and $$\mathtt o < \mathtt c$$.)
3. Iterate through them, keeping a counter that starts from $$0$$.
• When you see a pair with $$\mathtt o$$ in the second part, increment the counter.
• When you see a pair with $$\mathtt c$$ in the second part, decrement the counter.
• When you decrement the counter, if that causes it to drop to $$0$$, then look at the next element of the list to decide what to do; empty list means you're done, otherwise if the next pair's first part is just one bigger than the current one's you do nothing, and in the last case you record a maximal gap between the current end point and the next open point.