I have created a game, (4 in a row), in prolog. My heuristic function requires me to know how many Player's and Opponent's chips are in each possible 4-row combination on the board. The method I am using is as follows (in psuedocodish):
I have 1 list of all possible fours of the board (ComboList) =of the form==> [[A,B,C,D]|Rest].
I have 1 list of all the moves of the 1st player (List1) =of the form==> [[1],[7],[14]]
And 1 for opponent's moves (List2).
Step 1: obtain the first combo from ComboList, 2:
Check all of List1 to see how many are in this combo, 3:
Check all of List2 to see how many are in this combo,
Move onto the next combo from ComboList and start over...
This PROCESS takes waay too much runtime for what is required.
Please can someone suggest something better and more efficient! Much thanks in advance!
The following code uses member/3, which has also to become
known as nth1/3. See here:
http://storage.developerzen.com/fourrow.pro.txt
The predicate is nowadays found in library(lists) and has
possibly native support or a fast implementation:
http://www.swi-prolog.org/pldoc/man?predicate=nth1/3
But I guess asserting some facts and relying on argument
indexing might get you an even better result. See for
example here:
http://www.mxro.de/applications/four-in-a-row
Hope this helps.
Bye
Related
The requirements:
100k lines
One of the columns is not text - its custom painted with wxDC*.
The items addition is coming from another thread using wxThreadEvent.
Up until now I used wxDataViewListCtrl, but it takes too long to AppendItem 100 thousand time.
wxListCtrl (in virtual mode) does not have the ability to use wxDC* - please correct me if I am wrong.
The only thing I can think of is using wxDataViewCtrl + wxDataViewModel. But I can't understand how to add items.
I looked at the samples (https://github.com/wxWidgets/wxWidgets/tree/WX_3_0_BRANCH/samples/dataview), too complex for me.
I cant understand them.
I looked at the wiki (https://wiki.wxwidgets.org/WxDataViewCtrl), also too complex for me.
Can somebody please provide a very simple example of a wxDataViewCtrl + wxDataViewModel with one string column and one wxDC* column.
Thanks in advance.
P.S.
Per #HajoKirchhoff's request in the comments, I am posting some code:
// This is called from Rust 100k times.
extern "C" void Add_line_to_data_view_list_control(unsigned int index,
const char* date,
const char* sha1) {
wxThreadEvent evt(wxEVT_THREAD, 44);
evt.SetPayload(ViewListLine{index, std::string(date), std::string(sha1)});
wxQueueEvent(g_this, evt.Clone());
}
void TreeWidget::Add_line_to_data_view_list_control(wxThreadEvent& event) {
ViewListLine view_list_line = event.GetPayload<ViewListLine>();
wxVector<wxVariant> item;
item.push_back(wxVariant(static_cast<int>(view_list_line.index)));
item.push_back(wxVariant(view_list_line.date));
item.push_back(wxVariant(view_list_line.sha1));
AppendItem(item);
}
Appending 100k items to a control will always be slow. That's because it requires moving 100k items from your storage to the controls storage. A much better way for this amount of data is to have a "virtual" list control or wxGrid. In both cases the data is not actually transferred to the control. Instead when painting occurs, a callback function will transfer only the data required to paint. So for a 100k list you will only have "activity" for the 20-30 lines that are visible.
With wxListCtrl see https://docs.wxwidgets.org/3.0/classwx_list_ctrl.html, specify the wxLC_VIRTUAL flag, call SetItemCount and then provide/override
OnGetItemText
OnGetItemImage
OnGetItemColumnImage
Downside: You can only draw items contained in a wxImageList, since the OnGetItemImage return indizes into the list. So you cannot draw arbitrary items using a wxDC. Since the human eye will be overwhelmed with 100k different images anyway, this is usually acceptable. You may have to provide 20/30 different images beforehand, but you'll have a fast, flexible list.
That said, it is possible to override the OnPaint function and use that wxDC to draw anything in the list. But that'll get difficult pretty soon.
So an alternative would be to use wxGrid, create a wxGridTableBase derived class that acts as a bridge between the grid and your actual 100k data and create wxGridCellRenderer derived classes to render the actual data onscreen. The wxGridCellRenderer class will get a wxDC. This will give you more flexibility but is also much more complex than using a virtual wxListCtrl.
The full example of doing what you want will inevitably be relatively complex. But if you decompose in simple parts, it's really not that difficult: you do need to define a custom model, but if your list is flat, this basically just means returning the value of the item at the N-th position, as you can trivially implement all model methods related to the tree structure. An example of such a model, although with multiple columns can be found in the sample, so you just need to simplify it to a one (or two) column version.
Next, you are going to need a custom renderer too, but this is not difficult neither and, again, there is an example of this in the sample too.
If you have any concrete questions, you should ask them, but it's going to be difficult to do much better than what the sample shows and it does already show exactly what you want to do.
Thank you every one who replied!
#Vz.'s words "If you have any concrete questions, you should ask them" got me thinking and I took another look at the samples of wxWidgets. The full code can be found here. Look at the following classes:
TreeDataViewModel
TreeWidget
TreeCustomRenderer
I am building a work-logging app which starts by showing a list of projects that I can select, and then when one is selected you get a collection of other buttons, to log data related to that selected project.
I decided to have a selected_project : Maybe Int in my model (projects are keyed off an integer id), which gets filled with Just 2 if you select project 2, for example.
The buttons that appear when a project is selected send messages like AddMinutes 10 (i.e. log 10 minutes of work to the selected project).
Obviously the update function will receive one of these types of messages only if a project has been selected but I still have to keep checking that selected_project is a Just p.
Is there any way to avoid this?
One idea I had was to have the buttons send a message which contains the project id, such as AddMinutes 2 10 (i.e. log 10 minutes of work to project 2). To some extent this works, but I now get a duplication -- the Just 2 in the model.selected_project and the AddMinutes 2 ... message that the button emits.
Update
As Simon notes, the repeated check that model.selected_project is a Just p has its upside: the model stays relatively more decoupled from the UI. For example, there might be other UI ways to update the projects and you might not need to have first selected a project.
To avoid having to check the Maybe each time you need a function which puts you into a context wherein the value "wrapped" by the Maybe is available. That function is Maybe.map.
In your case, to handle the AddMinutes Int message you can simply call: Maybe.map (functionWhichAddsMinutes minutes) model.selected_project.
Clearly, there's a little bit more to it since you have to produce a model, but the point is you can use Maybe.map to perform an operation if the value is available in the Maybe. And to handle the Maybe.Nothing case, you can use Maybe.withDefault.
At the end of the day is this any better than using a case expression? Maybe, maybe not (pun intended).
Personally, I have used the technique of providing the ID along with the message and I was satisfied with the result.
I am a novice at using TradingView's Pinescript and having a hard time finding an easy to understand example of a script. I am used to Java/C++ and Pinescript is very different. I am trying to build a script that will scan a stock chart and look for gaps of over 5%. Here is psuedocode for what I am trying to create:
if(difference between open of current day and previous day close > 5%) {
plot a green circle or red circle, depending on if gap was up or down
}
Thank you in advance!
You're best bet would be to go through their tutorial
There's some odds choices in this language if you have any programming background so it's probably a good idea to read it all (it's not that much). E.g.
open is the current bars open price, but open[1] is the previous bar open price (so should be read as open[current_index-1])
you can't use the plot calls inside function bodies
as for you question (not tested, but should be close enough to give the right idea):
study(title='gap detector', overlay=true)
//plotshape(<condition>, <options>) // condition must be true to plot something
is_percentage_increase = if (close-close[1])/close[1] > 0.05
true
plotshape(is_percentage_increase, style=shape.circle, color=green)
Pine scripting is easy to use; Initially it was bit hard to understand, Once started using it it becomes so useful to strategize the logic.
In your case you can use conditional operator as well to detect this.This will work in Version 2 .The version 3 is bit differrent
//version =2
study(title ="Experementing the code ",overlay =true ,shorttitle ="testing") //overlay=false to get this down of the chart as seperate layout
plotchar( (close-close[1])/close[1] >0.05 ? 1:na ,char =' ',text ="plot\nTest",textcolor=red,size.huge)
Instead of if the condition you can use ?: operator to do this job.
Please make sure plotchar(.....) coming in the same line, not in separate line.
Pine has lot of cool features to use and helped me to derive my own strategy. The tutorial is really good.
Note if you don't put char='' above it will print STAR as the default character. And in the character even if you put char='testtest' it will print the only t .
I'm developing a project what similarly with the example "meetingscheduling", when I have a MeetingAssignment, I want to get its "previous MeetingAssignment" in the same room.
In other words, when I have a MeetingAssignment - TA1, I want to find the MeetingAssignment which is in the same room with TA1, and it should be nearest one on the left of TA1.
My idea is: When I get TA1 :
1. Get all the MeetingAssignments that have the same room with TA1.
2. Get the MeetingAssignments from the list generate by previous step what have
the less startingTimeGrain than TA1.
3. Find the MeetingAssignment what have the largest startingTimeGrain value.
it's the "previous MeetingAssignment".
But when I get the MeetingAssignment list of the sulotion class during planning(WorkingSolution), all of the room is null, I got the wrong solution?
Any better idea for it? Many thanks.
At the end of the Construction Heuristic (if it has a chance to complete, see DEBUG logging), all planning variables will be non-null. If the CH takes too, see the "scaling CH" chapter in the docs.
I was trying to solve Google Code Jam problems and there is one of them that I don't understand. Here is the question (World Finals 2013 - problem C): https://code.google.com/codejam/contest/2437491/dashboard#s=p2&a=2
And here follows the problem analysis: https://code.google.com/codejam/contest/2437491/dashboard#s=a&a=2
I don't understand why we can use binary search. In order to use binary search the elements have to be sorted. In order words: for a given element e, we can't have any element less than e at its right side. But that is not the case in this problem. Let me give you an example:
Suppose we do what the analysis tells us to do: we start with a left bound angle of 90° and a right bound angle of 0°. Our first search will be at angle of 45°. Suppose we find that, for this angle, X < N. In this case, the analysis tells us to make our left bound 45°. At this point, we can have discarded a viable solution (at, let's say, 75°) and at the same time there can be no more solutions between 0° and 45°, leading us to say that there's no solution (wrongly).
I don't think Google's solution is wrong =P. But I can't figure out why we can use a binary search in this case. Anyone knows?
I don't understand why we can use binary search. In order to use
binary search the elements have to be sorted. In order words: for a
given element e, we can't have any element less than e at its right
side. But that is not the case in this problem.
A binary search works in this case because:
the values vary by at most 1
we only need to find one solution, not all of them
the first and last value straddle the desired value (X .. N .. 2N-X)
I don't quite follow your counter-example, but here's an example of a binary search on a sequence with the above constraints. Looking for 3:
1 2 1 1 2 3 2 3 4 5 4 4 3 3 4 5 4 4
[ ]
[ ]
[ ]
[ ]
*
I have read the problem and in the meantime thought about the solution. When I read the solution I have seen that they have mostly done the same as I would have, however, I did not thought about some minor optimizations they were using, as I was still digesting the task.
Solution:
Step1: They choose a median so that each of the line splits the set into half, therefore there will be two provinces having x mines, while the other two provinces will have N - x mines, respectively, because the two lines each split the set into half and
2 * x + 2 * (2 * N - x) = 2 * x + 4 * N - 2 * x = 4 * N.
If x = N, then we were lucky and accidentally found a solution.
Step2: They are taking advantage of the "fact" that no three lines are collinear. I believe they are wrong, as the task did not tell us this is the case and they have taken advantage of this "fact", because they assumed that the task is solvable, however, in the task they were clearly asking us to tell them if the task is impossible with the current input. I believe this part is smelly. However, the task is not necessarily solvable, not to mention the fact that there might be a solution even for the case when three mines are collinear.
Thus, somewhere in between X had to be exactly equal to N!
Not true either, as they have stated in the task that
You should output IMPOSSIBLE instead if there is no good placement of
borders.
Step 3: They are still using the "fact" described as un-true in the previous step.
So let us close the book and think ourselves. Their solution is not bad, but they assume something which is not necessarily true. I believe them that all their inputs contained mines corresponding to their assumption, but this is not necessarily the case, as the task did not clearly state this and I can easily create a solvable input having three collinear mines.
Their idea for median choice is correct, so we must follow this procedure, the problem gets more complicated if we do not do this step. Now, we could search for a solution by modifying the angle until we find a solution or reach the border of the period (this was my idea initially). However, we know which provinces have too much mines and which provinces do not have enough mines. Also, we know that the period is pi/2 or, in other terms 90 degrees, because if we move alpha by pi/2 into either positive (counter-clockwise) or negative (clockwise) direction, then we have the same problem, but each child gets a different province, which is irrelevant from our point of view, they will still be rivals, I guess, but this does not concern us.
Now, we try and see what happens if we rotate the lines by pi/4. We will see that some mines might have changed borders. We have either not reached a solution yet, or have gone too far and poor provinces became rich and rich provinces became poor. In either case we know in which half the solution should be, so we rotate back/forward by pi/8. Then, with the same logic, by pi/16, until we have found a solution or there is no solution.
Back to the question, we cannot arrive into the situation described by you, because if there was a valid solution at 75 degrees, then we would see that we have not rotated the lines enough by rotating only 45 degrees, because then based on the number of mines which have changed borders we would be able to determine the right angle-interval. Remember, that we have two rich provinces and two poor provinces. Each rich provinces have two poor bordering provinces and vice-versa. So, the poor provinces should gain mines and the rich provinces should lose mines. If, when rotating by 45 degrees we see that the poor provinces did not get enough mines, then we will choose to rotate more until we see they have gained enough mines. If they have gained too many mines, then we change direction.