Is it possible to handle relations between jobs (in this case services) in jsprit?
For instance one job has to start at the same time with one or more other jobs.
Or one job must not start before the end of another job (normal sequence).
If not, do you know other java libraries that can handle such kind of restrictions?
Thank you!
Yes you can handle such relations with jsprit.
It is a bit of coding, but - I think - easy to implement. To illustrate it, I prepared you five examples that are based on the Vehicle Routing Problems (VRP) developed by Christofides-Mingozzi-Toth's* (first benchmarking instance). Follow the links behind the individual headlines and you get to the code. Note that the first activity in a route is marked with a triangle. The labels denote jobIds.
noConstraints
capacityConstraints
job 13 and 21 in same route
job 13 and 21 in same route AND 21 before 13
job 13 and 21 in same route AND 13 right after 21
Visit http://jsprit.github.io/ and you get to a number of code examples and docs.
*N. Christofides, A. Mingozzi, and P. Toth. The vehicle routing problem. In N. Christofides, A. Mingozzi, P. Toth, and C. Sandi, editors, Combinatorial Optimization. John Wiley, Chichester, 1979.
Related
I am saving the best solution into the DB, and we display that on the web page. I am looking for some solution where a user can add more visits, but that should not change already published trips.
I have checked the documentation and found ProblemFactChange can be used, but only when the solver is already running.
In my case solver is already terminated and the solution is also published. Now I want to add more visits to the vehicle without modifying the existing visits of the Vehicle. Is this possible with Optaplanner? if yes any example of documentation would be very helpful.
You can use PlanningPin annotation for avoiding unwanted changes.
Optaplanner - Pinned planning entities
If you're not looking for pinning (see Ismail's excellent answer), take a look at the OptaPlanner School Timetabling example, which allows adding lessons between solver runs. The lessons simply get stored in the database and then get loaded when the solver starts.
The difficulty with VRP is the chained model complexity (we're working on an alternative): If you add a visit X between A and B, then make sure that afterwards A.next = X, B.previous = X, X.previous = A, X.next = B and X.vehicle = A.vehicle. Not the mention the arrival times etc.
My suggestion would be to resolve what is left after the changes have been introduced. Let's say you are you visited half of your destinations (A -> B -> C) but not yet (C - > D -> E) when two new possible destinations (D' and E') are introduced. Would not this be the same thing as you are starting in C and trying plan for D, D', E and E'? The solution needs to be updated on the progress though so the remainder + changes can be input to the next solution.
Just my two cent.
I have columns taken from excel as a dataframe, the columns are as follows:
HolidayTourProvider|Packages|Meals|Accommodation|LocalTravelVehicle|Cancellationfee
Holiday Tour Provider has a couple of company names
Packages, the features provided in each package are mostly the same like
Meals,Accommodation etc... even though one company may call it "Saver", others may call it "Budget". (each of column mostly follow Yes/No, except Local travel vehicle are again car names like Ford Taurus,jeep cherokee etc..
Cancellation amount is integers)
I need to write a function like
match(HolidayTP,Package)
where the user can give input like
match(AdventureLife, Luxury)
then I need to return all the packages that have similar features with Luxury by other Holiday Tour Providers, no matter what name they give the package like 'Semi Lux', 'Comfort' etc...
I want to give a counter for every match and display all the packages that exceed the counter by 3 or 4.
This is my first python code. I am stuck here.
fb is the total df I exported to
def mapHol(HTP, PACKAGE):
mfb = (fb['HTP']== HTP)&(fb['package']== package)
B = fb[mfb]
for i in fb[i]:
for j in B[j]:
if fb[i]==B[j]:
count+=1
I dont know how to proceed, please help me this is my first major project, I started on my own.
The project I am working on involves static offline GTFS data in a mobile app. All the GTFS data is available inside realm-objects (or SQLite if needed).
Now, I would like to establish all train- or bus-connections from A to B (starting after a certain departure-time).
How do I query the GTFS-data in order to get a connection from A to B ???
I reealized to get all trips leaving from A.
I realized to get all station-names along that trip including times.
But I find it very hard to get the connection information between two locations A and B. What SQL queries do i have to set up in order to get that information ?
Any help appreciated !
If you just want to dynamically calculate shortest travel routes between static hubs, in an offline application, determined like the following image, you can use the following formula:
(Source)
Here's the pseudo code:
1 function Dijkstra(Graph, source):
2
3 create vertex set Q
4
5 for each vertex v in Graph: // Initialization
6 dist[v] ← INFINITY // Unknown distance from source to v
7 prev[v] ← UNDEFINED // Previous node in optimal path from source
8 add v to Q // All nodes initially in Q (unvisited nodes)
9
10 dist[source] ← 0 // Distance from source to source
11
12 while Q is not empty:
13 u ← vertex in Q with min dist[u] // Node with the least distance
14 // will be selected first
15 remove u from Q
16
17 for each neighbor v of u: // where v is still in Q.
18 alt ← dist[u] + length(u, v)
19 if alt < dist[v]: // A shorter path to v has been found
20 dist[v] ← alt
21 prev[v] ← u
22
23 return dist[], prev[]
(Source)
Alright, I'll admit so far my answer was a tad facetious...
I suspect you don't realize just how complicated it is to do what you're asking, especially in an offline environment. Companies like Google and Microsoft have spent millions on research with huge teams of data scientists.
If this is something you are serious about, I'd encourage you to start with a 10×10 grid and work on the logic of getting from "Point A → Point B" when you start adding barriers in random places (this simulating roads beginning & ending). Recently I was surprised how complicated a seemingly-simple, somewhat-related Stack Overflow "pipe sizes conversion" question that I answered had become.
If you didn't have the "offline" condition, I would've suggested looking into getting a [free] API Key for Google Web Services Directions API. Basically you tell it a where Points A & B are, and it gives you detailed route information, via transit or other methods.
Another of Google's many API's that could be helpful is the Snap to Roads API, which turns partial or error-ridden paths into driveable ones.
The study Route Logic and Shortest Path Logic is actually really fascinating stuff, and there are some amazing resources to learn about the related theories (see below), including a video from Google explaining how they went about it.
...but unfortunately this isn't something that's going to be accomplished with a simple SQL Query.
Actual Resources:
YouTube : Google Tech Talk: Fast Route Planning
Wikipedia : Shortest path problem
Wikipedia : Dijkstra's algorithm
...and some slightly Lighter Reading:
Wikipedia : Travelling Salesman Problem
Why UPS drivers don’t turn left and you probably shouldn’t either
Google Web Services : Directions API Developer's Guide
Stack Overflow : Shortest Route of converters between two different pipe sizes?
Wikipedia : Seven Bridges of Königsberg
I'm building a search engine for a game (path of exile). In this game, items have sockets.
items can have from 0 to 6 sockets
some sockets can be linked
there are 3 types of sockets (Red, Green, Blue)
when 3 sockets are linked (for instance), we consider that they're all linked to each other, not just to the closest one.
Example of an item with 5 linked sockets and 1 lonely sixth socket.
My goal is to allow users to search for items with specific socket colors and links with ElasticSearch.
A user would describe the socket combination like that:
"RRRBB" would be an item with 5 linked sockets, 3 red and 2 blue.
"GGG RB" would have 3 green socket linked and 1 red and 1 blue linked.
Still following?
When I index an item, I order the linked sockets alphabetically and by group size (group size might not be necessary): "RB GGB" would become "BGG RB".
And I would reorder the user input the same way before doing the search.
So, searching for an exact combination is super easy and fast(?). I wouldn't even need to analyze the data, would I?
But I would like these scenarios to work:
"ggg" would find "ggg rb" -> this is easy with just a whitespace analyzer?
But what I'm struggling with is:
"brr* r b" would match "bgrr r b" -> the way I'm doing things right now: as soon as a group has a *, I add * between every subgroup, so I would search for: bg* r b.
It works, but I think it's really inefficient, right? (even with small optimizations like remove the first and last * if the first and last letters are B or R).
Do you have an idea on how to index the data (analyzers, as a string or terms or...) to make this possible and fast?
The End.
Simple and common tree like data structures
Data Structure example
Animated Cartoons have 4 extremities (arm, leg,limb..)
Human have 4 ext.
Insects have 6 ext.
Arachnids have 6 ext.
Animated Cartoons have 4 by extremity
Human have 5 by ext.
Insects have 1 by ext.
Arachnids have 1 by ext.
Some Kind of Implementation
Level/Table0
Quantity, Item
Level/Table1
ItemName, Kingdom
Level/Table2
Kingdom, NumberOfExtremities
Level/Table3
ExtremityName, NumberOfFingers
Example Dataset
1 Homer Simpson, 1 Ralph Wiggum, 2 jon
skeet, 3 Atomic ant, 2 Shelob (spider)
Querying.. "Number of fingers"
Number = 1*4*4 + 1*4*4 + 1*4*5 + 3*6*1 + 2*6*1 = 82 fingers (Let Jon be a Human)
I wonder if there is any tool for define it parseable for automatic create the inherited data, and drawing this kind of trees, (with the plus of making this kind of data access, if where posible..)
It could be drawn manually with for example FreeMind, but AFAIK it dont let you define datatype or structures to automatically create inherited branch of items, so it's really annoying to have to repeat and repeat a structure by copying (and with the risk of mistake). Repeated Work over Repeated Data, (an human running repeated code), it's a buggy feature.
So I would like to write the data in the correct language that let me reuse it
for queries and visualization, if all data is in XML, or Java Classes, or in a Database File, etc.. there is some tool for viewing the tree and making the query?
PD : Creating nested folders in a filesystem and using Norton Commander in tree view, is not an option, I hope (just because It have to be builded manually)
Your answer is mostly going to depend on what programming skills you already have and what skills you are willing to acquire. I can tell you what I would do with what I know.
I think for drawing trees you want a LaTeX package like qtree. If you don't like this one, there are a bunch of others out there. You'd have to write a script in whatever your favorite scripting language is to parse your input into the LaTeX code to generate the trees, but this could easily be done with less than 100 lines in most languages, if I properly understand your intentions. I would definitely recommend storing your data in an XML format using a library like Ruby's REXML, or whatever your favorite scripting language has.
If you are looking to generate more interactive trees, check out the Adobe Flex Framework. Again, if you don't like this specific framework, there are bunches of others out there (I recommend the blog FlowingData).
Hope this helps and I didn't miserably misunderstand your question.
Data structure that You are describing looks like it can fit in xml format. Take a look at Exist XML database, and if I can say so it is the most complete xml database. It comes with many tools to get you started fast ! like XQuery Sandbox option in admin http interface.
Example Dataset
1 Homer Simpson, 1 Ralph Wiggum, 2 jon skeet, 3 Atomic ant, 2 Shelob (spider)
I am assuming that there are 2 instances of jon skeet, 3 instances of Atomic ant and 2 instances of Shelob
Here is a XQuery example:
let $doc :=
<root>
<definition>
<AnimatedCartoons>
<extremities>4</extremities>
<fingers_per_ext>4</fingers_per_ext>
</AnimatedCartoons>
<Human>
<extremities>4</extremities>
<fingers_per_ext>5</fingers_per_ext>
</Human>
<Insects>
<extremities>6</extremities>
<fingers_per_ext>1</fingers_per_ext>
</Insects>
<Arachnids>
<extremities>6</extremities>
<fingers_per_ext>1</fingers_per_ext>
</Arachnids>
</definition>
<subject><name>Homer Simpson</name><kind>AnimatedCartoons</kind></subject>
<subject><name>Ralph Wiggum</name><kind>AnimatedCartoons</kind></subject>
<subject><name>jon skeet</name><kind>Human</kind></subject>
<subject><name>jon skeet</name><kind>Human</kind></subject>
<subject><name>Atomic ant</name><kind>Insects</kind></subject>
<subject><name>Atomic ant</name><kind>Insects</kind></subject>
<subject><name>Atomic ant</name><kind>Insects</kind></subject>
<subject><name>Shelob</name><kind>Arachnids</kind></subject>
<subject><name>Shelob</name><kind>Arachnids</kind></subject>
</root>
let $definitions := $doc/definition/*
let $subjects := $doc/subject
(: here goes some query logic :)
let $fingers := fn:sum(
for $subject in $subjects
return (
for $x in $definitions
where fn:name($x) = $subject/kind
return $x/extremities * $x/fingers_per_ext
)
)
return $fingers
XML Schema Editor with visualization is perhaps what I am searching for
http://en.wikipedia.org/wiki/XML_Schema_Editor
checking it..