I am very new to labview and recently I had been trying to make this sequence loop.
E.g.
myRio's starting position is ((x <0.05) && (y <=0.05) &&( z>0.9))
next while detecting myRio's position changes to ((x<0.2) && (y <= -0.9) &&( z<=0.3)) and then back again to the starting position, it will turn validate this as one correct cycle and change the counter from 0 to 1. and loop this whole sequence again.
Would really appreciate if you could highlight how can I do these kind of sequence looping. Thank you very much.
It sounds as if what you are trying to do is to count transitions between states. A great design pattern for anything involving transitions between states - which covers a large part of what people generally use LabVIEW for - is a state machine.
The link gives an explanation but essentially what you need is:
a While loop with a shift register
a case structure inside the loop whose case selector is wired to the left shift register terminal
some logic inside each case that decides what value to output to the right shift register terminal, i.e. what state to go to next.
In your case you could implement this with just two states:
State 1: Check position and see if we have reached the target position.
If so go to State 2
If not go to State 1 again
State 2: Check position and see if we have got back to the start position.
If so, increment the count and go to State 1
If not, go to State 2 again
However it would be slightly more elegant to use three states:
State 2: Check position and see if we have got back to the start position.
If so, go to State 3
If not, go to State 2 again
State 3: Increment the count and go to State 1.
You can use a second shift register for the counter: initialise it to 0, wire the left terminal across to the right one in States 1 and 2 (leaving the count unchanged) and increment it in State 3.
You can use an integer value or a string for the case selector, but the best practice is to use an enum which you save as a typedef. This allows you to reorder, rename, add or remove states later on without breaking existing code.
Related
I've followed "Procedural Generation in Godot: Dungeon Generation" by KidsCanCode #https://www.youtube.com/watch?v=o3fwlk1NI-w and find myself unable to debug the current problem.
This specific commit has the code, but I'll try to explain in more detail bellow.
My main scene has a Camera2D node, a generic Node2D calles Rooms and a TileMap, everything is empty.
When the script starts, it runs a
func make_room(_pos, _size):
position = _pos
size = _size
var s = RectangleShape2D.new()
s.custom_solver_bias = 0.5
s.extents = size
$CollisionShape2D.shape = s
A few times and it fills $Rooms using .add_child(r) where r is a instance of the node that has the make_room() function. It will then iterate over $Rooms.get_children() a few times to create a AStar node to link all the rooms:
The magic comes when make_map() is called after afterwards, it fills the map with non-walkable blocks and then it carves the empty spaces, which works fine too:
There is a find_start_room() that is called to find the initial room, it also sets a global variable to the Main script start_room, which is used to write 'Start' on the map using draw_string(font, start_room.position - Vector2(125,0),"start",Color(3,4,8))
When I hit 'esc' it runs this simple code to instance the player:
player = Player.instance()
add_child(player)
player.position = start_room.position + Vector2(start_room.size.x/2, start_room.size.y/2)
play_mode = true
The problem comes when spawning the player. I tried doing some 'blind' fixing, such as adding or subtracting a Vector2(start_room.size.x/2, start_room.size.y/2) to player.position to see if I could make it fall within the room, to no avail.
Turning to the debugger didn't help, as the positions expressed by the variable inspectors don't seem to mean anything.
I tried implementing a simple 'mouse click print location':
print("Mouse Click/Unclick at: ", event.position)
print("Node thing",get_node("/root/Main/TileMap").world_to_map(event.position))
And also a 'start_room' print location:
print(get_node("/root/Main/TileMap").world_to_map(start_room.position))
And a when player moves print location, written directly into the Character script:
print(get_node("/root/Main/TileMap").world_to_map(self.position))
Getting results like the ones bellow:
Mouse Click/Unclick at: (518, 293)
Node thing(16, 9)
(-142, 0)
(-147, -3)
So, the player doesn't spawn on the same position as the start_room and the mouse position information is not the same as anything else.
Why is the player now spawning correctly? How can I debug this situation?
EDIT1: User Theraot mentioned about how the RigidBody2D is doing some weird collisions, and from what I understood, changing their collision behavior should fix the whole thing.
There's a section on the code that -after generating the random rooms- it removes some of the rooms like this:
for room in $Rooms.get_children():
if randf() < cull:
room.queue_free()
else:
room.mode = RigidBody2D.MODE_STATIC
room_positions.append(Vector3(room.position.x, room.position.y, 0))
From what I understand, if the room is randomly selected it will be deleted using queue_free() OR it will be appended to a room_positions for further processing. This means if I shift all the rooms to a different collision layer, the player/character instance would be alone with the TileMap on the same collision layer.
So I just added a simple room.collision_layer = 3 changing this section of the code to
for room in $Rooms.get_children():
if randf() < cull:
room.queue_free()
else:
room.mode = RigidBody2D.MODE_STATIC
room.collision_layer = 3
room_positions.append(Vector3(room.position.x, room.position.y, 0))
It doesn't seem to have changed anything, the player still spawns outside the room.
Do you see the rooms spread outwards?
You didn't write code to move the rooms. Sure, the code gives them a random position. But even if you set their position to Vector2.ZERO they move outwards, avoiding overlaps.
Why? Because these rooms are RigidBody2D, and they do push other physics objects. Such as other rooms or the player character.
That's the problem: These rooms are RigidBody2D, and you put your KinematicBody2D player character on top of one of them. The RigidBody2D pushes it out.
The tutorial you followed is exploiting this behavior of RigidBody2Ds to spread the rooms. However you don't need these RigidBody2D after you are done populating your TileMap.
Instead, you can store the start position in a variable for later placing the player character (you don't need offsets - by the way - the position of the room is the center of the room), and then remove the RigidBody2Ds. If you want to keep the code that writes the text, you would also have to modify it, so it does not fail when the room no longer exists.
Alternatively, you can edit their collision layer and mask so they don't collide with the player character (or anything for that matter, but why would you want these RigidBody2Ds that collide with nothing?).
Addendum post edit: Collision layers and mask don't work as you expect.
First of all, the collision layer and mask are flags. The values of the layers are powers of two (1, 2, 4, 8...). So, when you set it to 3, it is the layer 1 plus the layer 2. So it still collides with a collision mask of 1.
And second, even if you changed the collision layer of the rooms to 2 (so it does not match the collision mask of 1 that the player character has). The player character still has a layer 1 which match the collision mask of the rooms.
See also the proposal Make physics layers and masks logic simple and consistent.
Thus, you would need to change the layer and mask. Both. in such way that they don't collide. For example, you can set layer and mask to 0 (which disable all collisions). The algorithm that populates the TileMap does not use the layer and mask.
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 still having a little bit of trouble understanding the sensitivity list and it activating a process.
most textbooks say that a process is activated every time an event occurs on a signal inside the sensitivity list.
process(in)
begin
x <= in;
end process;
Now looking at this example, "in" is an input declared in the entity. Now if "in" starts off at 0 and changes to 1 then the process would activate and the value of x would take in value "in". Now suppose after in changed from 0 to 1 that it now stays at constant value of 1. Does this mean the process will not get activated? Will x still give output of '1'? I want to say that it wont get activated and will only activate once in changes back from 1 back to 0. Can someone please confirm?
Within the sensitivity list (I assume this is hardware language, VHDL has the same exact syntax and format), whenever there is some type of signal change (L-> H, 0 ->1, 1-> 0... any change in the variable you listed within the sensitivity list), it will activate the process and the process will execute until completion, which then the process will end. When the process end, signals/outputs (depends on how you interpret them) will be stored on a driver, which will update given signals after some propagation delay.
So from your second statement, yes. If it changes to 0 -> 1, the process activates, if its 1 -> 0, the process activates, and if in remains 1, the process will not be activated. So x's value remains.
While converting multitape turing machine into an equivalent singletape turing machine we have to shift the data and insert a blank to it.
e.g :
Multitape = [1,2,3,4] [5,6,7,8] [9,10,11,12]
Equivalent singletape = [1,2,3,4,#,5,6,7,8,#,9,10,11,12]
consider this transition function in multitape turing machine :
[(q1,4) = (q2,4,R) and similar for others]
after this transition next element of tape1 is Blank
But in single tape
[(q1,4) = (q2,4,R) and for others]
After this transition next element of tape1 is # so we have to shift remaining data to insert a blank at this position.
How to do that? please give answers with respect to transition function.
Your transition function only reads the first tape. A transition for the three.tape machien should look something like:
(q1,4,5,10) = (q2,4,R,6,L,11,R)
because you need instructions for all three tapes.
For the shifting: there is no better way than to shift all the data on one side of the current position one position by one. For example:
Mark current position
Run to the right-most symbol
Move right-most symbol one to the right
Move next symbol on the left one position to the right and so on until you reach the marked symbol
I have a task, where I have to calculate optimal policy
(Reinforcement Learning - Markov decision process) in the grid world (agent movies left,right,up,down).
In left table, there are Optimal values (V*).
In right table, there is sollution (directions) which I don't know how to get by using that "Optimal policy" formula.
Y=0.9 (discount factor)
And here is formula:
So if anyone knows how to use that formula, to get solution (those arrows), please help.
Edit: there is whole problem description on this page:
http://webdocs.cs.ualberta.ca/~sutton/book/ebook/node35.html
Rewards: state A(column 2, row 1) is followed by a reward of +10 and transition to state A', while state B(column 4, row 1) is followed by a reward of +5 and transition to state B'.
You can move: up,down,left,right. You cannot move outside the grid or stay in same place.
Break the math down, piece by piece:
The arg max (....) is telling you to find the argument, a, which maximizes everything in the parentheses. The variables a, s, and s' are an action, a state you're in, and a state that results from that action, respectively. So the arg max (...) is telling you to find an action that maximizes that term.
You know gamma, and someone did the hard work of calculating V*(s'), which is the value of that resulting state. So you know what to plug in there, right?
So what is p(s,a,s')? That is the probability that, starting from s and doing a, you end in some s'. This is meant to represent some kind of faulty actuator-- you say "go forward!" and it foolishly decides to go left (or two squares forward, or remain still, or whatever.) For this problem, I'd expect it to be given to you, but you haven't shared it with us. And the summation over s' is telling you that when you start in s, and you pick an action a, you need to sum over all possible resulting s' states. Again, you need the details of that p(s,a,s') function to know what those are.
And last, there is r(s,a) which is the reward for doing action a in state s, regardless of where you end up. In this problem it might be slightly negative, to represent a fuel cost. If there is a series of rewards in the grid and a grab action, it might be positive. You need that, too.
So, what to do? Pick a state s, and calculate your policy for it. For each s, you're going have the possibility of (s,a1), and (s,a2), and (s,a3), etc. You have to find the a that gives you the biggest result. And of course for each pair (s,a) you may (in fact, almost certainly will) have multiple values of s' to stick in the summation.
If this sounds like a lot of work, well, that's why we have computers.
PS - Read the problem description very carefully to find out what happens if you run into a wall.