Hey guys
I was wondering if you could give me a hand with something. If I have two
tuples that define a range, eg: (10, 20), (15, 30), I need to determine
whether the ranges overlap each other. The algo needs to catch:
(10, 20) (15, 25)
(15, 25) (10, 20)
(10, 25) (15, 20)
and
(15, 20) (10, 25)
I can think of lots of ways to do this but it's in a tight loop so I need
it to be as efficient as possible. Any help welcome :)
Cheers
Rob 6 18907
Robin Haswell wrote: Hey guys
I was wondering if you could give me a hand with something. If I have two tuples that define a range, eg: (10, 20), (15, 30), I need to determine whether the ranges overlap each other.
def overlap(a,b):
return a[0] <= b[0] <= a[1] or b[0] <= a[0] <= b[1]
 Paul
On Thu, 16 Feb 2006 15:22:15 +0000, Robin Haswell wrote: Hey guys
I was wondering if you could give me a hand with something. If I have two tuples that define a range, eg: (10, 20), (15, 30), I need to determine whether the ranges overlap each other. The algo needs to catch:
(10, 20) (15, 25) (15, 25) (10, 20) (10, 25) (15, 20) and (15, 20) (10, 25)
What do you mean "catch"? Do you expect the algorithm to recognise these
combinations as special and return something different?
I'm going to assume that there are no magic values that need to be caught,
because I don't know what that means, and just proceed as if the task is
to compare two tuples of the form (x1, x2) where x1 <= x2.
# warning: untested!
def isoverlap((x1,x2), (y1,y2)):
"""Given two numeric ranges, returns a flag True or False
indicating whether they overlap.
Assumes that the ranges are ordered (smallest,largest). If
that assumption is wrong, incorrect results may occur."""
# Fully overlapping cases:
# x1 <= y1 <= y2 <= x2
# y1 <= x1 <= x2 <= y2
# Partially overlapping cases:
# x1 <= y1 <= x2 <= y2
# y1 <= x1 <= y2 <= x2
# Nonoverlapping cases:
# x1 <= x2 < y1 <= y2
# y1 <= y2 < x1 <= x2
return not (x2 < y1 or y2 < x1)
I can think of lots of ways to do this but it's in a tight loop so I need it to be as efficient as possible.
"Efficient as possible" in what way? Least memory used? Smallest
number of bytes of source code? Fastest performance?
Assuming you need fastest performance, the general method to do that is to
write it in handtuned assembly language, taking care to use all the
tricks to optimize it for the particular CPU you are running on. If you
can find a way to push the processing into any highend graphics
processors you might have, that typically will speed it up even more. It
is still, sometimes, possible for the best assembly programmers to just
barely outperform optimizing C compilers, or so I'm told.
If you are willing to set your sights just a little lower, and rather than
aiming for the absolute fastest code possible, settle for code which is
fast enough, the usual technique is to stick to Python. Write different
functions implementing the various methods you can think of, and then time
them with the timeit module. Pick the fastest. Is it fast enough that
performance is satisfactory? If so, then you are done.
Here is a technique that may speed your code up a little: it is quicker to
find local variables than global. So, instead of this:
def main():
... code here ...
while condition:
flag = isoverlap(t1, t2)
... code ...
you can gain a little speed by making a local reference to the function:
def main():
... code here ...
iso = isoverlap # make a local reference for speed
while condition:
flag = iso(t1, t2)
... code ...
If your code is still too slow, you might speed it up with Psycho, or you
may need to write a C extension. Either way, you won't know if the code is
fast enough until you actually run it.

Steven.
Robin Haswell wrote: I was wondering if you could give me a hand with something. If I have two tuples that define a range, eg: (10, 20), (15, 30), I need to determine whether the ranges overlap each other. The algo needs to catch:
(10, 20) (15, 25) (15, 25) (10, 20) (10, 25) (15, 20) and (15, 20) (10, 25)
I can think of lots of ways to do this but it's in a tight loop so I need it to be as efficient as possible. Any help welcome :)
how about :
def overlap(a, b):
return a[1] > b[0] and a[0] < b[1]
</F>
Thanks guys, you've all been very helpful :)
Rob
Robin Haswell wrote: I can think of lots of ways to do this but it's in a tight loop so I need it to be as efficient as possible. Any help welcome :)
There are 24 possibilities of having 4 numbers in a row, and
the following 6 of them describe nonempty intervals (the remaining
18 are obtained by reversing the endpoints of one or both
intervals, thereby making them empty):
# separate
a0 a1 b0 b1
b0 b1 a0 a1
# inside
a0 b0 b1 a1
b0 a0 a1 b1
# cross
a0 b0 a1 b1
b0 a0 b1 a1
The simplest is to exclude the first pair, i.e.
not (a1<b0 or b1<a0)
which simplifies to
a1>=b0 and b1>=a0, given by Fredrik without proof ;)
You will have to decide whether you want ">" or ">="
(beware of the former when using floats).
If you cannot assume that the pairs are in order, this
will break.
Ralf
Fredrik Lundh wrote: def overlap(a, b): return a[1] > b[0] and a[0] < b[1]
Assuming that x[1] can't be smaller than x[0], this
is the right way to compare two ranges. Well, I guess
you need to figure out the borderline cases. Is it
a[1] > b[0] or a[1] >= b[0] which is relevant?
Anyway, if we have N ranges and N is big, we'll get
lots of comparisions: N*(N1)/2.
For large values of N, it might be better to somehow
cache/aggregate the intervals if this makes it possible
to make this into an N*M comparision where M is smaller
than (N1)/2....
E.g, with discrete values for the intervals such as with
(10, 15), one might consider something like (untested):
hits = sets.Set()
for interval in intervals:
for i in range(interval[0], interval[1]+1):
if i in hits:
raise ValueError('%s overlaps previous interval.'
% interval)
hits.add(i)
E.g. if we have 1000 intervals where the average size of
an interval is 10, you get 10000 "i in hits" and "hits.add(i)"
instead of 499500 'a[1] > b[0] and a[0] < b[1]'.
As always, it's good to measure...assuming you have fairly
realistic data to test with. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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