How are min and max of cumulative variables assigned? - or-tools

I created a routing problem and added some dimension to it. A solution assignment is found and I want to know the cumulative value at each index. I noticed that the CumulVar of an assignment does not only have a Value method but also Min and Max methods. Apparently the cumulative variables are implemented in such a way that they can represent intervals. I can see how setting
slack_max>0
fix_start_cumul_to_zero=False
introduces an ambiguity for the cumulative variables as their is a choice in how to start and how much slack to add at each stop. But
Question: How are the Min and Max at each index computed?

You can get the Min and Max range of a given node index from solution.Min(dimension.Cumulvar(index))
Note you'll get Min and Max exactly the same when slack_max=0 unless you know something I don't ;)
Assuming you are using an output solution object solution and a time dimension time_dimension, this will store em as a dict with min-max tuples, you may wish to adapt the output format however you wish:
time_dict = {}
for vehicle_id in range(num_vehicles):
vehicle_time_dict={}
index = routing.Start(vehicle_id)
start_time = solution.Min(time_dimension.CumulVar(index))
vehicle_time_dict[index]=(index_min,index_max)
while not routing.isEnd(index):
previous_index = index
index = solution.Value(routing.NextVar(index))
index_min = solution.Min(time_dimension.CumulVar(index))
index_max = solution.Max(time_dimension.CumulVar(index))
vehicle_time_dict[index]=(index_min,index_max)
time_dict[vehicle_id]=vehicle_time_dict
routing.IsEnd(index) returns True if it's the last index of that vehicle's route (or anywhere after the last index, so if it's 10 nodes long:
routing.IsEnd(8) will return False,
routing.IsEnd(9) will return True,
routing.IsEnd(10) will also return True, etc)

Related

Mean of values before and after a specific element

I have an array of 1 x 400, where all element values are above 1500. However, I have some elements that have values<50 which are wrong measures and I would like to have the mean of the elements before and after the wrong measured data points and replace it in the main array.
For instance, element number 17 is below 50 so I want to take the mean of elements 16 and 18 and replace element 17 with the new mean.
Can someone help me, please? many thanks in advance.
No language is specified in the question, but for Python you could work with List Comprehension:
# array with 400 values, some of which are incorrect
arr = [...]
arr = [arr[i] if arr[i] >= 50 else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
That is, if arr[i] is less than 50, it'll be replaced by the average value of the element before and after it. There are two issues with this approach.
If i is the first or last element, then one of the two values will be undefined, and no mean can be obtained. This can be fixed by just using the value of the available neighbour, as specified below
If two values in a row are very low, the leftmost one will use the rightmost one to calculate its value, which will result in a very low value. This is a problem that may not occur for you in practice, but it is an inherent result of the way you wish to recalculate values, and you might want to keep it in mind.
Improved version, keeping in mind the edge cases:
# don't alter the first and last item, even if they're low
arr = [arr[i] if arr[i] >= 50 or i == 0 or i+1 == len(arr) else (arr[i-1]+arr[i+1])/2 for i in range(len(arr))]
# replace the first and last element if needed
if arr[0] < 50:
arr[0] = arr[1]
if arr[len(arr)-1] < 50:
arr[len(arr)-1] = arr[len(arr)-2]
I hope this answer was useful for you, even if you intend to use another language or framework than python.

Couldn't understand this SWIFT Fucntion [closed]

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Hope you are doing well. Would anyone please explain this code to me? I am still not getting how we got 120 here. When the parameters were passed to the function, where was it saved? How did it determine max and min before calculating?
Would be really appreciated if anyone could explain it for me please..
You're getting 120 because you're asking the tuple for the sum value twice. The tuple passed back is (min: Int, max: Int, sum: Int). In this case, sum is also index number 2 (min is index zero, max is index one). So in this case statistics.sum == statistics.2
If you want to access min or max you can access them using statistics.min or statistics.max respectively. You can also use statistics.0 or statistics.1 if you prefer (I recommend using the labels though).
Final thought: I would change your initial assignment of max and min inside the function to read as follows:
guard var max = scores.first, var min = scores.first else {
return (0, 0, 0)
}
This will protect against the user accidentally passing an empty array.
You could have also set min and max to 0 ,not to scores[0].
This func is very simple, I think you should read somewhere How functions are working.
For loop is used to go through every element, in your case every element in array scores. In every iteration, variable sum is increased by current element in array. So:
At the beginning of the function : sum = 0
1. iteration: sum = 0 + 5
2. iteration: sum = 5 + 3
3. iteration: sum = 8 + 100
4. iteration: sum = 108 + 3
5. iteration: sum = 111 + 9
For loop ends. Sum = 120.
Score in Scores means like every single element in score. So name Score can be whatever you want to name it. You could have also name it element , x, a or whatever you like.
The function returns a tuple of type (min: Int, max: Int, sum: Int).
max and min are not determined before being calculated. When they're first assigned at the top of the function, they just contain the first element of the passed in scores array. But by the time the loop below finishes, they indeed have the correct minimum and maximum values. Those values are then returned in a tuple matching the return type.

Is there a range data structure in Scala?

I'm looking for a way to handle ranges in Scala.
What I need to do is:
given a set of ranges and a range(A) return the range(B) where range(A) intersect range (B) is not empty
given a set of ranges and a range(A) remove/add range(A) from/to the set of ranges.
given range(A) and range(B) create a range(C) = [min(A,B), max(A,B)]
I saw something similar in java - http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/collect/RangeSet.html
Though subRangeSet returns only the intersect values and not the range in the set (or list of ranges) that it intersects with.
RangeSet rangeSet = TreeRangeSet.create();
rangeSet.add(Range.closed(0, 10));
rangeSet.add(Range.closed(30, 40));
Range range = Range.closed(12, 32);
System.out.println(rangeSet.subRangeSet(range)); //[30,32] (I need [30,40])
System.out.println(range.span(Range.closed(30, 40))); //[12,40]
There is an Interval[A] type in the spire math library. This allows working with ranges of arbitrary types that define an Order. Boundaries can be inclusive, exclusive or omitted. So e.g. (-∞, 0.0] or [0.0, 1.0) would be possible intervals of doubles.
Here is a library intervalset for working with sets of non-overlapping intervals (IntervalSeq or IntervalTrie) as well as maps of intervals to arbitrary values (IntervalMap).
Here is a related question that describes how to use IntervalSeq with DateTime.
Note that if the type you want to use is 64bit or less (basically any primitive), IntervalTrie is extremely fast. See the Benchmarks.
As Tzach Zohar has mentioned in the comment, if all you need is range of Int - go for scala.collection.immutable.Range:
val rangeSet = Set(0 to 10, 30 to 40)
val r = 12 to 32
rangeSet.filter(range => range.contains(r.start) || range.contains(r.end))
If you need it for another underlying type - implement it by yourself, it's easy for your usecase.

Assigning times to events

include "globals.mzn";
%Data
time_ID = [11,12,13,14,15];
eventId = [0011, 0012, 0013, 0021, 0022, 0031, 0041, 0051, 0061, 0071];
int:ntime = 5;
int:nevent = 10;
set of int: events =1..nevent;
set of int: time = 1..ntime;
array[1..nevent] of int:eventId;
array[1..nevent] of var time:event_time;
array[1..ntime] of int:time_ID;
solve satisfy;
constraint
forall(event in eventId)(
exists(t in time_ID)(
event_time[event] = t ));
output[ show(event_time) ];
I'm trying to assign times to an event using the code above.
But rather than randomly assign times to the events, it returns an error " array access out of bounds"
How can I make it select randomly from the time array?
Thank you
The error was because you tried to assign the index 11 (the first element in eventId array) in "event_time" array.
The assigment of just 1's is correct since you haven't done any other constraints on the "event_time" array. If you set the number of solutions to - say - 3 you will see other solutions. And, in fact, the constraint as it stand now is not really meaningful since it just ensures that there is some assignment to the elements in "event_time", but this constraint is handled by the domain of "event_time" (i.e. that all indices are in the range 1..ntime).

Find value in vector "p" that corresponds to maximum value in vector "r = f(p)"

As simple as in title. I have nx1 sized vector p. I'm interested in the maximum value of r = p/foo - floor(p/foo), with foo being a scalar, so I just call:
max_value = max(p/foo-floor(p/foo))
How can I get which value of p gave out max_value?
I thought about calling:
[max_value, max_index] = max(p/foo-floor(p/foo))
but soon I realised that max_index is pretty useless. I'm sorry asking this, real beginner here.
Having dropped the issue to pieces, I realized there's no unique corrispondence between values p and values in my related vector p/foo-floor(p/foo), so there's a logical issue rather than a language one.
However, given my input data, I know that the solution is unique. How can I fix this?
I ended up doing:
result = p(p/foo-floor(p/foo) == max(p/foo-floor(p/foo)))
Looks terrible, so if you know any other way...
Once you have the index, use it:
result = p(max_index)
You can create a new vector with your lets say "transformed" values:
p2 = (p/foo-floor(p/foo))
and then just use find to find the max values on p2:
max_index = find(p2 == max(p2))
that will return the index or indices of p2 with the max value of that operation, and finally just lookup the original value in p
p(max_index)
in 1 line, this is:
p(find((p/foo-floor(p/foo) == max((p/foo-floor(p/foo))))))
which is basically the same thing you did in the end :)