Determine Whether Two Date Ranges Overlap

(StartA <= EndB) and (EndA >= StartB)

Proof: Let ConditionA Mean that DateRange A Completely After DateRange B

_                        |---- DateRange A ------|
|---Date Range B -----|                          _

(True if StartA > EndB)

Let ConditionB Mean that DateRange A is Completely Before DateRange B

|---- DateRange A -----|                        _ 
_                          |---Date Range B ----|

(True if EndA < StartB)

Then Overlap exists if Neither A Nor B is true - (If one range is neither completely after the other, nor completely before the other, then they must overlap.)

Now one of De Morgan's laws says that:

Not (A Or B) <=> Not A And Not B

Which translates to: (StartA <= EndB) and (EndA >= StartB)

NOTE: This includes conditions where the edges overlap exactly. If you wish to exclude that,
change the >= operators to >, and <= to <

NOTE2. Thanks to @Baodad, see this blog, the actual overlap is least of:
{ endA-startA, endA - startB, endB-startA, endB - startB }

(StartA <= EndB) and (EndA >= StartB) (StartA <= EndB) and (StartB <= EndA)

NOTE3. Thanks to @tomosius, a shorter version reads:
DateRangesOverlap = max(start1, start2) < min(end1, end2)
This is actually a syntactical shortcut for what is a longer implementation, which includes extra checks to verify that the start dates are on or before the endDates. Deriving this from above:

If start and end dates can be out of order, i.e., if it is possible that startA > endA or startB > endB, then you also have to check that they are in order, so that means you have to add two additional validity rules:
(StartA <= EndB) and (StartB <= EndA) and (StartA <= EndA) and (StartB <= EndB) or:
(StartA <= EndB) and (StartA <= EndA) and (StartB <= EndA) and (StartB <= EndB) or,
(StartA <= Min(EndA, EndB) and (StartB <= Min(EndA, EndB)) or:
(Max(StartA, StartB) <= Min(EndA, EndB)

But to implement Min() and Max(), you have to code, (using C ternary for terseness),:
(StartA > StartB? Start A: StartB) <= (EndA < EndB? EndA: EndB)

NOTE4. Thanks to Carl for noticing this, but another answer shows an equivalent mathematical expression for this logical expression. Because the product of any two real numbers with opposite sign is negative and with the same sign it is positive, if you convert the datetimes to fractional numbers (and most DBMSs internally use numbers to represent datetimes), the above logical expression can also be evaluated using the following mathematical expression:

(EndA  - StartA) * (StartB - EndB) <= 0

---

I believe that it is sufficient to say that the two ranges overlap if:

(StartDate1 <= EndDate2) and (StartDate2 <= EndDate1)

---

This article Time Period Library for .NET describes the relation of two time periods by the enumeration PeriodRelation:

// ------------------------------------------------------------------------
public enum PeriodRelation
{
    After,
    StartTouching,
    StartInside,
    InsideStartTouching,
    EnclosingStartTouching,
    Enclosing,
    EnclosingEndTouching,
    ExactMatch,
    Inside,
    InsideEndTouching,
    EndInside,
    EndTouching,
    Before,
} // enum PeriodRelation

https://i.stack.imgur.com/0c6q0.png

---

For reasoning about temporal relations (or any other interval relations, come to that), consider Allen's Interval Algebra. It describes the 13 possible relations that two intervals can have with respect to each other. You can find other references — "Allen Interval" seems to be an operative search term. You can also find information about these operations in Snodgrass's Developing Time-Oriented Applications in SQL (PDF available online at URL), and in Date, Darwen and Lorentzos Temporal Data and the Relational Model (2002) or Time and Relational Theory: Temporal Databases in the Relational Model and SQL (2014; effectively the second edition of TD&RM).

The short(ish) answer is: given two date intervals A and B with components .start and .end and the constraint .start <= .end, then two intervals overlap if:

A.end >= B.start AND A.start <= B.end

You can tune the use of >= vs > and <= vs < to meet your requirements for degree of overlap.

ErikE comments:

You can only get 13 if you count things funny... I can get "15 possible relations that two intervals can have" when I go crazy with it. By sensible counting, I get only six, and if you throw out caring whether A or B comes first, I get only three (no intersect, partially intersect, one wholly within other). 15 goes like this: [before:before, start, within, end, after], [start:start, within, end, after], [within:within, end, after], [end:end, after], [after:after].

I think that you cannot count the two entries 'before:before' and 'after:after'. I could see 7 entries if you equate some relations with their inverses (see the diagram in the referenced Wikipedia URL; it has 7 entries, 6 of which have a different inverse, with equals not having a distinct inverse). And whether three is sensible depends on your requirements.

----------------------|-------A-------|----------------------
    |----B1----|
           |----B2----|
               |----B3----|
               |----------B4----------|
               |----------------B5----------------|
                      |----B6----|
----------------------|-------A-------|----------------------
                      |------B7-------|
                      |----------B8-----------|
                         |----B9----|
                         |----B10-----|
                         |--------B11--------|
                                      |----B12----|
                                         |----B13----|
----------------------|-------A-------|----------------------

---

If the overlap itself should be calculated as well, you can use the following formula:

overlap = max(0, min(EndDate1, EndDate2) - max(StartDate1, StartDate2))
if (overlap > 0) { 
    ...
}

---

All the solutions that check a multitude of conditions based on where the ranges are in relation to one another can be greatly simplified by simply ensuring that one range starts before or at the same time as the other. You can do this by swapping the ranges if necessary up front.

Then, you can detect overlap if the second range start is either:

less than or equal to the first range end (if ranges are inclusive, containing both the start and end times); or

less than (if ranges are inclusive of start and exclusive of end).

For example (assuming inclusive at both ends), there's only four possibilities for range 2of which one is a non-overlap (the > at the end of the range means it doesn't matter where the range ends):

|-----|        range 1, lines below are all range 2.
|-->  :        overlap.
 |--> :        overlap.
      |--->    overlap (no overlap in exclusive-of-end case).
       |--->   no overlap.

The endpoint of the second range doesn't affect the result at all. So, in pseudo-code, you can do something like (assuming s <= e within a given range):

def hasOverlap (r1, r2):
    if r1.s > r2.s: swap r1, r2
    return r2.s <= r1.e

If the ranges are exclusive at the end, you just have to replace <= with < in the expression you return.

This greatly limits the number of checks you have to make because you remove half of the problem space early by ensuring the first range never starts after the second.

And, since "code talks", here is some Python code that shows this in action, with quite a few test cases. First, the InclusiveRange class:

class InclusiveRange:
    """InclusiveRange class to represent a lower and upper bound"""

    def __init__(self, start, end):
        """Initialisation ensures start <= end"""
        self.start = min(start, end)
        self.end = max(start, end)

    def __repr__(self):
        """ Return representation for f-string"""
        return f"({self.start}, {self.end})"

    def overlaps(self, other):
        """True if range overlaps with another"""

        # Very limited recursion to ensure start of first range
        # isn't after start of second.

        if self.start > other.start:
            return other.overlaps(self)

        # Greatly simplified check for overlap.

        return other.start <= self.end

Then a test case handler to allow us to nicely present the result of a single test case:

def test_case(range1, range2) -> None:
    """Single test case checker"""

    # Get low and high value for "graphic" output.

    low = min(range1.start, range2.start)
    high = max(range1.end, range2.end)

    # Output ranges and graphic.

    print(f"r1={range1} r2={range2}: ", end="")
    for val in range(low, high + 1):
        is_in_first = range1.start <= val <= range1.end
        is_in_second = range2.start <= val <= range2.end

        if is_in_first and is_in_second:
            print("|", end="")
        elif is_in_first:
            print("'", end="")
        elif is_in_second:
            print(",", end="")
        else:
            print(" ", end="")

    # Finally, output result of overlap check.

    print(f" - {range1.overlaps(range2)}\n")

Then finally, a decent chunk of test cases to which you can add your own if need be:

# Various test cases, add others if you doubt the correctness.

test_case(InclusiveRange(0, 1), InclusiveRange(8, 9))
test_case(InclusiveRange(0, 4), InclusiveRange(5, 9))
test_case(InclusiveRange(0, 4), InclusiveRange(4, 9))
test_case(InclusiveRange(0, 7), InclusiveRange(2, 9))
test_case(InclusiveRange(0, 4), InclusiveRange(0, 9))
test_case(InclusiveRange(0, 9), InclusiveRange(0, 9))
test_case(InclusiveRange(0, 9), InclusiveRange(4, 5))

test_case(InclusiveRange(8, 9), InclusiveRange(0, 1))
test_case(InclusiveRange(5, 9), InclusiveRange(0, 4))
test_case(InclusiveRange(4, 9), InclusiveRange(0, 4))
test_case(InclusiveRange(2, 9), InclusiveRange(0, 7))
test_case(InclusiveRange(0, 9), InclusiveRange(0, 4))
test_case(InclusiveRange(0, 9), InclusiveRange(0, 9))
test_case(InclusiveRange(4, 5), InclusiveRange(0, 9))

Running that produces the output:

r1=(0, 1) r2=(8, 9): ''      ,, - False
r1=(0, 4) r2=(5, 9): ''''',,,,, - False
r1=(0, 4) r2=(4, 9): ''''|,,,,, - True
r1=(0, 7) r2=(2, 9): ''||||||,, - True
r1=(0, 4) r2=(0, 9): |||||,,,,, - True
r1=(0, 9) r2=(0, 9): |||||||||| - True
r1=(0, 9) r2=(4, 5): ''''||'''' - True
r1=(8, 9) r2=(0, 1): ,,      '' - False
r1=(5, 9) r2=(0, 4): ,,,,,''''' - False
r1=(4, 9) r2=(0, 4): ,,,,|''''' - True
r1=(2, 9) r2=(0, 7): ,,||||||'' - True
r1=(0, 9) r2=(0, 4): |||||''''' - True
r1=(0, 9) r2=(0, 9): |||||||||| - True
r1=(4, 5) r2=(0, 9): ,,,,||,,,, - True

where each line has:

the two ranges being evaluated;

a graphical representation of the "range space" (from lowest start to highest end) where each character is a value in that "range space": ' indicates a value in the first range only, , indicates a value in the second range only, | indicates a value in both ranges, and indicates a value in neither range.

the result of the overlap check.

You can see quite clearly that you only get true in the overlap check when there is at least one value in both ranges (i.e., a | character). Every other case gives false.

Feel free to use any other values if you want to add more test cases.

---

Here is yet another solution using JavaScript. Specialities of my solution:

Handles null values as infinity

Assumes that the lower bound is inclusive and the upper bound exclusive.

Comes with a bunch of tests

The tests are based on integers but since date objects in JavaScript are comparable you can just throw in two date objects as well. Or you could throw in the millisecond timestamp.

Code:

/**
 * Compares to comparable objects to find out whether they overlap.
 * It is assumed that the interval is in the format [from,to) (read: from is inclusive, to is exclusive).
 * A null value is interpreted as infinity
 */
function intervalsOverlap(from1, to1, from2, to2) {
    return (to2 === null || from1 < to2) && (to1 === null || to1 > from2);
}

Tests:

describe('', function() {
    function generateTest(firstRange, secondRange, expected) {
        it(JSON.stringify(firstRange) + ' and ' + JSON.stringify(secondRange), function() {
            expect(intervalsOverlap(firstRange[0], firstRange[1], secondRange[0], secondRange[1])).toBe(expected);
        });
    }

    describe('no overlap (touching ends)', function() {
        generateTest([10,20], [20,30], false);
        generateTest([20,30], [10,20], false);

        generateTest([10,20], [20,null], false);
        generateTest([20,null], [10,20], false);

        generateTest([null,20], [20,30], false);
        generateTest([20,30], [null,20], false);
    });

    describe('do overlap (one end overlaps)', function() {
        generateTest([10,20], [19,30], true);
        generateTest([19,30], [10,20], true);

        generateTest([10,20], [null,30], true);
        generateTest([10,20], [19,null], true);
        generateTest([null,30], [10,20], true);
        generateTest([19,null], [10,20], true);
    });

    describe('do overlap (one range included in other range)', function() {
        generateTest([10,40], [20,30], true);
        generateTest([20,30], [10,40], true);

        generateTest([10,40], [null,null], true);
        generateTest([null,null], [10,40], true);
    });

    describe('do overlap (both ranges equal)', function() {
        generateTest([10,20], [10,20], true);

        generateTest([null,20], [null,20], true);
        generateTest([10,null], [10,null], true);
        generateTest([null,null], [null,null], true);
    });
});

Result when run with karma&jasmine&PhantomJS:

PhantomJS 1.9.8 (Linux): Executed 20 of 20 SUCCESS (0.003 secs / 0.004 secs)

---

https://i.stack.imgur.com/bDVe1.png

Here is the code that does the magic:

 var isOverlapping =  ((A == null || D == null || A <= D) 
            && (C == null || B == null || C <= B)
            && (A == null || B == null || A <= B)
            && (C == null || D == null || C <= D));

Where..

A -> 1Start

B -> 1End

C -> 2Start

D -> 2End

Proof? Check out this test console code gist.

---

An easy way to remember the solution would be
min(ends)>max(starts)

---

Here's my solution in Java, which works on unbounded intervals too

private Boolean overlap (Timestamp startA, Timestamp endA,
                         Timestamp startB, Timestamp endB)
{
    return (endB == null || startA == null || !startA.after(endB))
        && (endA == null || startB == null || !endA.before(startB));
}

---

I would do

StartDate1.IsBetween(StartDate2, EndDate2) || EndDate1.IsBetween(StartDate2, EndDate2)

Where IsBetween is something like

    public static bool IsBetween(this DateTime value, DateTime left, DateTime right) {
        return (value > left && value < right) || (value < left && value > right);
    }

---

The solution posted here did not work for all overlapping ranges...

----------------------|-------A-------|----------------------
    |----B1----|
           |----B2----|
               |----B3----|
               |----------B4----------|
               |----------------B5----------------|
                      |----B6----|
----------------------|-------A-------|----------------------
                      |------B7-------|
                      |----------B8-----------|
                         |----B9----|
                         |----B10-----|
                         |--------B11--------|
                                      |----B12----|
                                         |----B13----|
----------------------|-------A-------|----------------------

my working solution was:

AND (
  ('start_date' BETWEEN STARTDATE AND ENDDATE) -- caters for inner and end date outer
  OR
  ('end_date' BETWEEN STARTDATE AND ENDDATE) -- caters for inner and start date outer
  OR
  (STARTDATE BETWEEN 'start_date' AND 'end_date') -- only one needed for outer range where dates are inside.
)

---

As there have been several answers for different languages and environments, here is one for standard ANSI SQL.

In standard SQL it is as as simple as

(StartDate1, EndDate1) overlaps (StartDate2, EndDate2)

assuming all four columns are DATE or TIMESTAMP columns. It returns true if both ranges have at least one day in common (assuming DATE values)

(However not all DBMS products support that)

In PostgreSQL it's also easy to test for inclusion by using date ranges

daterange(StartDate1, EndDate1) @> daterange(StartDate2, EndDate2)

the above returns true if the second range is completely included in the first (which is different to "overlaps")

---

This was my javascript solution with moment.js:

// Current row dates
var dateStart = moment("2014-08-01", "YYYY-MM-DD");
var dateEnd = moment("2014-08-30", "YYYY-MM-DD");

// Check with dates above
var rangeUsedStart = moment("2014-08-02", "YYYY-MM-DD");
var rangeUsedEnd = moment("2014-08-015", "YYYY-MM-DD");

// Range covers other ?
if((dateStart <= rangeUsedStart) && (rangeUsedEnd <= dateEnd)) {
    return false;
}
// Range intersects with other start ?
if((dateStart <= rangeUsedStart) && (rangeUsedStart <= dateEnd)) {
    return false;
}
// Range intersects with other end ?
if((dateStart <= rangeUsedEnd) && (rangeUsedEnd <= dateEnd)) {
    return false;
}

// All good
return true;

---

Short answer using momentjs:

function isOverlapping(startDate1, endDate1, startDate2, endDate2){ 
    return moment(startDate1).isSameOrBefore(endDate2) && 
    moment(startDate2).isSameOrBefore(endDate1);
}

the answer is based on above answers, but its shortened.

---

In Microsoft SQL SERVER - SQL Function

CREATE FUNCTION IsOverlapDates 
(
    @startDate1 as datetime,
    @endDate1 as datetime,
    @startDate2 as datetime,
    @endDate2 as datetime
)
RETURNS int
AS
BEGIN
DECLARE @Overlap as int
SET @Overlap = (SELECT CASE WHEN  (
        (@startDate1 BETWEEN @startDate2 AND @endDate2) -- caters for inner and end date outer
        OR
        (@endDate1 BETWEEN @startDate2 AND @endDate2) -- caters for inner and start date outer
        OR
        (@startDate2 BETWEEN @startDate1 AND @endDate1) -- only one needed for outer range where dates are inside.
        ) THEN 1 ELSE 0 END
    )
    RETURN @Overlap

END
GO

--Execution of the above code
DECLARE @startDate1 as datetime
DECLARE @endDate1 as datetime
DECLARE @startDate2 as datetime
DECLARE @endDate2 as datetime
DECLARE @Overlap as int
SET @startDate1 = '2014-06-01 01:00:00' 
SET @endDate1 =   '2014-06-01 02:00:00'
SET @startDate2 = '2014-06-01 01:00:00' 
SET @endDate2 =   '2014-06-01 01:30:00'

SET @Overlap = [dbo].[IsOverlapDates]  (@startDate1, @endDate1, @startDate2, @endDate2)

SELECT Overlap = @Overlap

---

the simplest

The simplest way is to use a well-engineered dedicated library for date-time work.

someInterval.overlaps( anotherInterval )

java.time & ThreeTen-Extra

The best in the business is the java.time framework built into Java 8 and later. Add to that the ThreeTen-Extra project that supplements java.time with additional classes, specifically the Interval class we need here.

As for the language-agnostic tag on this Question, the source code for both projects is available for use in other languages (mind their licenses).

Interval

The org.threeten.extra.Interval class is handy, but requires date-time moments (java.time.Instant objects) rather than date-only values. So we proceed by using the first moment of the day in UTC to represent the date.

Instant start = Instant.parse( "2016-01-01T00:00:00Z" );
Instant stop = Instant.parse( "2016-02-01T00:00:00Z" );

Create an Interval to represent that span of time.

Interval interval_A = Interval.of( start , stop );

We can also define an Interval with a starting moment plus a Duration.

Instant start_B = Instant.parse( "2016-01-03T00:00:00Z" );
Interval interval_B = Interval.of( start_B , Duration.of( 3 , ChronoUnit.DAYS ) );

Comparing to test for overlaps is easy.

Boolean overlaps = interval_A.overlaps( interval_B );

You can compare an Interval against another Interval or Instant:

abuts

contains

encloses

equals

isAfter

isBefore

overlaps

All of these use the Half-Open approach to defining a span of time where the beginning is inclusive and the ending is exclusive.

---

I had a situation where we had dates instead of datetimes, and the dates could overlap only on start/end. Example below:

https://i.stack.imgur.com/qzRBx.png

(Green is the current interval, blue blocks are valid intervals, red ones are overlapping intervals).

I adapted Ian Nelson's answer to the following solution:

   (startB <= startA && endB > startA)
|| (startB >= startA && startB < endA)

This matches all overlap cases but ignores the allowed overlap ones.

---

The mathematical solution given by @Bretana is good but neglects two specific details:

aspect of closed or half-open intervals empty intervals

About the closed or open state of interval boundaries, the solution of @Bretana valid for closed intervals

(StartA <= EndB) and (EndA >= StartB)

can be rewritten for half-open intervals to:

(StartA < EndB) and (EndA > StartB)

This correction is necessary because an open interval boundary does not belong to the value range of an interval by definition.

And about empty intervals, well, here the relationship shown above does NOT hold. Empty intervals which do not contain any valid value by definition must be handled as special case. I demonstrate it by my Java time library Time4J via this example:

MomentInterval a = MomentInterval.between(Instant.now(), Instant.now().plusSeconds(2));
MomentInterval b = a.collapse(); // make b an empty interval out of a

System.out.println(a); // [2017-04-10T05:28:11,909000000Z/2017-04-10T05:28:13,909000000Z)
System.out.println(b); // [2017-04-10T05:28:11,909000000Z/2017-04-10T05:28:11,909000000Z)

The leading square bracket "[" indicates a closed start while the last bracket ")" indicates an open end.

System.out.println(
      "startA < endB: " + a.getStartAsInstant().isBefore(b.getEndAsInstant())); // false
System.out.println(
      "endA > startB: " + a.getEndAsInstant().isAfter(b.getStartAsInstant())); // true

System.out.println("a overlaps b: " + a.intersects(b)); // a overlaps b: false

As shown above, empty intervals violate the overlap condition above (especially startA < endB), so Time4J (and other libraries, too) has to handle it as special edge case in order to guarantee that the overlap of any arbitrary interval with an empty interval does not exist. Of course, date intervals (which are closed by default in Time4J but can be half-open, too, like empty date intervals) are handled in a similar way.

---

This is an extension to the excellent answer by @charles-bretana.

The answer however does not make a distinction among open, closed, and half-open (or half-closed) intervals.

Case 1: A, B are closed intervals

A = [StartA, EndA]
B = [StartB, EndB]

                         [---- DateRange A ------]   (True if StartA > EndB)
[--- Date Range B -----]                           


[---- DateRange A -----]                             (True if EndA < StartB)
                         [--- Date Range B ----]

Overlap iff: (StartA <= EndB) and (EndA >= StartB)

Case 2: A, B are open intervals

A = (StartA, EndA)
B = (StartB, EndB)

                         (---- DateRange A ------)   (True if StartA >= EndB)
(--- Date Range B -----)                           

(---- DateRange A -----)                             (True if EndA <= StartB)
                         (--- Date Range B ----)

Overlap iff: (StartA < EndB) and (EndA > StartB)

Case 3: A, B right open

A = [StartA, EndA)
B = [StartB, EndB)

                         [---- DateRange A ------)   (True if StartA >= EndB) 
[--- Date Range B -----)                           

[---- DateRange A -----)                             (True if EndA <= StartB)
                         [--- Date Range B ----)

Overlap condition: (StartA < EndB) and (EndA > StartB)

Case 4: A, B left open

A = (StartA, EndA]
B = (StartB, EndB]

                         (---- DateRange A ------]   (True if StartA >= EndB)
(--- Date Range B -----]                           

(---- DateRange A -----]                             (True if EndA <= StartB)
                         (--- Date Range B ----]

Overlap condition: (StartA < EndB) and (EndA > StartB)

Case 5: A right open, B closed

A = [StartA, EndA)
B = [StartB, EndB]

                         [---- DateRange A ------)    (True if StartA > EndB)
[--- Date Range B -----]                           


[---- DateRange A -----)                              (True if EndA <= StartB)  
                         [--- Date Range B ----]

Overlap condition: (StartA <= EndB) and (EndA > StartB)

etc...

Finally, the general condition for two intervals to overlap is

(StartA <🞐 EndB) and (EndA >🞐 StartB)

where 🞐 turns a strict inequality into a non-strict one whenever the comparison is made between two included endpoint.

---

In case you're using a date range that has not ended yet (still on going) e.g. not set endDate = '0000-00-00' you can not use BETWEEN because 0000-00-00 is not a valid date!

I used this solution:

(Startdate BETWEEN '".$startdate2."' AND '".$enddate2."')  //overlap: starts between start2/end2
OR (Startdate < '".$startdate2."' 
  AND (enddate = '0000-00-00' OR enddate >= '".$startdate2."')
) //overlap: starts before start2 and enddate not set 0000-00-00 (still on going) or if enddate is set but higher then startdate2

If startdate2 is higher then enddate there is no overlap!

---

The answer is too simple for me so I have created a more generic dynamic SQL statement which checks to see if a person has any overlapping dates.

SELECT DISTINCT T1.EmpID
FROM Table1 T1
INNER JOIN Table2 T2 ON T1.EmpID = T2.EmpID 
    AND T1.JobID <> T2.JobID
    AND (
        (T1.DateFrom >= T2.DateFrom AND T1.dateFrom <= T2.DateTo) 
        OR (T1.DateTo >= T2.DateFrom AND T1.DateTo <= T2.DateTo)
        OR (T1.DateFrom < T2.DateFrom AND T1.DateTo IS NULL)
    )
    AND NOT (T1.DateFrom = T2.DateFrom)

---

Using Java util.Date, here what I did.

    public static boolean checkTimeOverlaps(Date startDate1, Date endDate1, Date startDate2, Date endDate2)
    {
        if (startDate1 == null || endDate1 == null || startDate2 == null || endDate2 == null)
           return false;

        if ((startDate1.getTime() <= endDate2.getTime()) && (startDate2.getTime() <= endDate1.getTime()))
           return true;

        return false;
    }

---

For ruby I also found this:

class Interval < ActiveRecord::Base

  validates_presence_of :start_date, :end_date

  # Check if a given interval overlaps this interval    
  def overlaps?(other)
    (start_date - other.end_date) * (other.start_date - end_date) >= 0
  end

  # Return a scope for all interval overlapping the given interval, including the given interval itself
  named_scope :overlapping, lambda { |interval| {
    :conditions => ["id <> ? AND (DATEDIFF(start_date, ?) * DATEDIFF(?, end_date)) >= 0", interval.id, interval.end_date, interval.start_date]
  }}

end

Found it here with nice explaination -> http://makandracards.com/makandra/984-test-if-two-date-ranges-overlap-in-ruby-or-rails

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The easiest way to do it in my opinion would be to compare if either EndDate1 is before StartDate2 and EndDate2 is before StartDate1.

That of course if you are considering intervals where StartDate is always before EndDate.

---

If you provide a date range as input and want to find out if it overlaps with the existing date range in database, the following conditions can successfully meet your demand

Assume you provide a @StartDate and @EndDate from your form input.

conditions are :

If @StartDate is ahead of existingStartDate and behind existingEndDate then we can say @StartDate is in the middle of a existing date range, thus we can conclude it will overlap

@StartDate >=existing.StartDate And @StartDate <= existing.EndDate) 

If @StartDate is behind existingStartDate but @EndDate is ahead of existingStartDate we can say that it will overlap

 (@StartDate <= existing.StartDate And @EndDate >= existing.StartDate)

If @StartDate is behind existingStartDate And @EndDate is ahead of existingEndDate we can conclude that the provided date range devours a existing date range , thus overlaps

 (@StartDate <= existing.StartDate And @EndDate >= existing.EndDate))

If any of the condition stands true, your provided date range overlaps with existing ones in the database.

---

Split the problem into cases then handle each case.

The situation 'two date ranges intersect' is covered by two cases - the first date range starts within the second, or the second date range starts within the first.

---

You can try this:

//custom date for example
$d1 = new DateTime("2012-07-08");
$d2 = new DateTime("2012-07-11");
$d3 = new DateTime("2012-07-08");
$d4 = new DateTime("2012-07-15");

//create a date period object
$interval = new DateInterval('P1D');
$daterange = iterator_to_array(new DatePeriod($d1, $interval, $d2));
$daterange1 = iterator_to_array(new DatePeriod($d3, $interval, $d4));
array_map(function($v) use ($daterange1) { if(in_array($v, $daterange1)) print "Bingo!";}, $daterange);

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public static class NumberExtensionMethods
    {
        public static Boolean IsBetween(this Int64 value, Int64 Min, Int64 Max)
        {
            if (value >= Min && value <= Max) return true;
            else return false;
        }

        public static Boolean IsBetween(this DateTime value, DateTime Min, DateTime Max)
        {
            Int64 numricValue = value.Ticks;
            Int64 numericStartDate = Min.Ticks;
            Int64 numericEndDate = Max.Ticks;

            if (numricValue.IsBetween(numericStartDate, numericEndDate) )
            {
                return true;
            }

            return false;
        }
    }

public static Boolean IsOverlap(DateTime startDate1, DateTime endDate1, DateTime startDate2, DateTime endDate2)
        {
            Int64 numericStartDate1 = startDate1.Ticks;
            Int64 numericEndDate1 = endDate1.Ticks;
            Int64 numericStartDate2 = startDate2.Ticks;
            Int64 numericEndDate2 = endDate2.Ticks;

            if (numericStartDate2.IsBetween(numericStartDate1, numericEndDate1) ||
                numericEndDate2.IsBetween(numericStartDate1, numericEndDate1) ||
                numericStartDate1.IsBetween(numericStartDate2, numericEndDate2) ||
                numericEndDate1.IsBetween(numericStartDate2, numericEndDate2))
            {
                return true;
            }

            return false;
        } 


if (IsOverlap(startdate1, enddate1, startdate2, enddate2))
            {
                Console.WriteLine("IsOverlap");
            }

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This was my solution, it returns true when the values don't overlap:

X START 1 Y END 1

A START 2 B END 2

TEST1: (X <= A || X >= B)
        &&
TEST2: (Y >= B || Y <= A) 
        && 
TEST3: (X >= B || Y <= A)


X-------------Y
    A-----B

TEST1:  TRUE
TEST2:  TRUE
TEST3:  FALSE
RESULT: FALSE

---------------------------------------

X---Y
      A---B

TEST1:  TRUE
TEST2:  TRUE
TEST3:  TRUE
RESULT: TRUE

---------------------------------------

      X---Y
A---B

TEST1:  TRUE
TEST2:  TRUE
TEST3:  TRUE
RESULT: TRUE

---------------------------------------

     X----Y
A---------------B

TEST1:  FALSE
TEST2:  FALSE
TEST3:  FALSE
RESULT: FALSE

---