Checking For Time Range Overlap, The Watchman Problem [sql]

April 14, 2024 Post a Comment

I am running into a road block on a larger problem. As part of a large query I need to solve a 'night watchman' problem. I have a table with schedule shifts as such: ID | Start

Solution 1:

Here is a way to flatten date range like this

Start|End2009-1-106:00|2009-1-118:002009-2-120:00|2009-2-204:002009-2-206:00|2009-2-214:00

You have to compare previousandnext dates in each row and see whether

Current row's Start date falls between previous row's date range.
Current row's End date falls between next row's date range.

alt text

Using above code, implementing UDF is as simple as followed.

createfunction fnThereIsWatchmenBetween(@from datetime, @to datetime)
returns bit
asbegindeclare @_Result bit

    declare@FlattenedDateRangetable (
        Start   datetime,
        [End]   datetime
    )

    insert@FlattenedDateRange(Start, [End])
    selectdistinctStart=casewhen Pv.Start isnullthen Curr.Start 
                    when Curr.Start between Pv.Start and Pv.[End] then Pv.Start
                    else Curr.Start 
                end,
            [End] =casewhen Curr.[End] between Nx.Start and Nx.[End] then Nx.[End] 
                    else Curr.[End] 
                endfrom    shift Curr
            leftjoin shift Pv on Pv.ID = Curr.ID -1--; prevleftjoin shift Nx on Nx.ID = Curr.ID +1--; next

    if exists(  select1from    FlattenedDateRange R
                where@frombetween R.Start and R.[End]
                        and@tobetween R.Start and R.[End]) beginset @_Result =1--; There is/are watchman/men during specified date rangeendelsebeginset @_Result =0--; There is NO watchmanendreturn @_Result
end

Solution 2:

An unguarded interval obviously starts either at the end of a watched period or at the beginning of the whole time range that you are checking. So you need a query that selects all elements from this set that don't have an overlapping shift. The query would look like:

select1from shifts s1 wherenot exists
    (select1from shifts s2
     where s2.start<=s1.endand s2.end > s1.end
    )
    and s1.end>=start_of_range and s1.end<  end_of_range
union
select1wherenot exists
    (select1from shifts s2 
      where s2.start<=start_of_range and s2.end > start_of_range
    )

If this is non-empty, then you have an unguarded interval. I suspect it will run in quadratic time, so it might be slower than "sort, fetch and loop".

Solution 3:

One way is to create a temp table with a row for each time value requiring to be checked (which is a function of the resolution of your shifts).

If it were minutes it would have 60 * 24 = 1440 rows for a day; about 10K rows for a week.

Then the SQL is relatively simple:

SELECT COUNT(1) FROM #minutes m LEFT JOIN shifts s ON m.checktime BETWEEN s.start_time AND s.end_time HAVING COUNT(1) = 0

This has the benefit of also being able to show how many shifts are covering the same time.

The execution time should be negligible given the scales you've described.

Solution 4:

I was looking at date ranges and thought I would re-visit this question. I may fall flat on my face here, but it seems these two conditions would be enough

(1) Shift isnotat beginning ofrangeand has noleft neighbour

OR

(2) Shift isnotatendofrangeand has noright neighbour.

Appreciate this may not be the most efficient.

CREATETABLE times
(
TimeID int,
StartTime Time,
EndTime Time
)

INSERTINTO times
VALUES
(1,'10:00:00','11:00:00'),
(2,'11:00:00','12:00:00'),
(3,'13:00:00','14:00:00'),
(4,'14:30:00','15:00:00'),
(5,'15:00:00','16:00:00'),
(6,'16:00:00','17:00:00')

declare@start_of_range time='09:30:00'declare@end_of_range time='17:30:00'select timeID,StartTime,EndTime 
from times s1 where-- No left neighbour and not at beginning of rangenotexists
    (select1from times s2
     where s2.startTime < s1.startTime and s2.endTime >= s1.startTime
    )
    and s1.StartTime>@start_of_range
  or-- No right neighbour and not at end of rangenotexists
    (select1from times s2
     where s2.startTime <= s1.endTime and s2.endTime > s1.endTime
    )
    and s1.EndTime<@end_of_range

Result set

timeID  StartTime   EndTime
1   10:00:00.0000000    11:00:00.0000000
2   11:00:00.0000000    12:00:00.0000000
3   13:00:00.0000000    14:00:00.0000000
4   14:30:00.0000000    15:00:00.0000000
6   16:00:00.0000000    17:00:00.0000000

Actually it's only necessary to check either the right neighbours or the left neighbours, as long as you make sure that the start and end of range is checked, so you could introduce the start of range as a dummy interval and just check the right neighbours as follows:-

select*from
(
select timeID,StartTime,EndTime 
from times unionselect0,@start_of_range,@start_of_range) s1
wherenotexists
    (select1from times s2
     where s2.startTime<=s1.endTime and s2.endTime > s1.endTime
    )
    and s1.EndTime<@end_of_range