I have to create a lot of turtles forming a compact group of any shape, a simple 10x100 rectangle is enough. The important thing is that they must be near each others.
In c i would do something like this:
for(x = 1; x <= rows; x++)
{
for(y = 1; y <= columns; y++)
{
create_turtle(x,y);
}
}
And the equivalent in netlogo would be:
crt 1000
let n 0
let x 1
let y 1
while[y <= 10]
[
set x 1
while[x <= 100]
[
ask turtle n
[move-to patch x y]
set x x + 1
set n n + 1
]
set y y + 1
]
But it's not an elegant solution. Any suggestion?
Edit: More precisely I have to reproduce what has been done in this article: http://science.sciencemag.org/content/345/6198/795.full
Every turtle is a little robot.
And here you can see one way turtles could be positioned turtles schema
I'm using circle turtles like the robots of the article.
One of the trickiest things for programmers from other languages to do when learning NetLogo is getting rid of all the loops. Iterating through the agents or patches is embedded in the ask primitive, you don't need to code the iteration. ask also iterates in a random order so that repeated processes don't lead to any advantage to whichever agent happens to be first in the loop.
Also, when you create turtles, you can immediately give them instructions. You can also place them initially in an arbitrary position rather than move them there. Here is one solution that places them all in a rectangle that is 5 patches to the left/right of centre (0,0) and occupies half the height of the world.
create-turtles 1000 [ setxy random-float 10 - 5 random-ycor * 0.5 ]
From the edit, I think you are wanting them to be created at gridpoints rather than randomly within the space. If that is true, then select the patches you want and ask them to sprout a turtle.
let in-shape patches with [ pxcor >= -10 and pxcor <= 10 and pycor >= -10 and pycor <= 10 ]
ask in-shape [ sprout 1 ]
You will need to work out your own values and make sure they are within the world dimension.
Related
I have limited programming experience (mechanical engineering student, so a bit of matlab and labview experience) and am very new to NetLogo, so I apologize in advance if this question is pretty basic or my code is of poor quality.
I need to have my turtles move to 1 of 2 possible neighboring patches based on a given probability function. The two patches that I need to input to the probability function are the two neighboring patches with the lowest nest-scent value. I have been able to pull the two lowest nest-scent values, but I cannot figure out how to actually figure out which patches those are, and how to put those coordinates into an ifelse statement to move the turtle to one of them based on the aformentioned probability function. I have the following code that is obviously not working:
to move
set farthest-patch sort-by < [nest-scent] of neighbors
let a1x pxcor of item 0 farthest-patch
let a1y pycor of item 0 farthest-patch
let a2x pxcor of item 1 farthest-patch
let a2y pycor of item 1 farthest-patch
let a1 item 0 farthest-patch
let a2 item 1 farthest-patch
let x (((a1 + a2) / 100 ) - 1)
let probability-move 0.5 * (1 + ((exp(x) - exp( - x)) / (exp(x) + exp( - x))))
ifelse random-float 1 < probability-move
[set to-move 1]
[set to-move 0]
let a1-probability (a1 / (a1 + a2))
ifelse random-float 1 < a1-probability
[set destination [a1x a1y]]
[set destination [a2x a2y]]
ifelse count turtles-here >= 20
[set full 1]
[set full 0]
if [a1x a21] = full
[set destination [a2x a2y]]
if [a2x a2y] = full
[set destination [a1x a1y]]
if [a2x a2y] and [a1x a1y] = full
[set to-move 0]
ifelse to-move = 1
[move-to destination]
[stop]
end
Basically what I have (tried) to do here is sort a farthest-patches list by increasing nest-scent, and I have pulled the two lowest nest-scent values in order to input those values into my probability functions (both for whether or not to move, and if they are to move which of the two patches to select). I am not sure how to properly pull the patch coordinates of the patches that the a1 and a2 values were taken from.
Thanks for any help,
Brad
okay, you are making life way more complicated than it needs to be. You can select the two patches (or turtles) with the smallest values of a variable with min-n-of. Look it up in the dictionary to get the details.
Having found the two candidates, the best option is to use the rnd extension for choosing the destination because it has a primitive for random selection by weight. Finally, since you are using a function of your variable as the weight (rather than the variable value itself), you need a way to construct that weight. The best option is to separate it out - you could also have a second variable with the weight value, but that just proliferates variables.
Here is a complete working model. Please copy the whole thing into a new instance of NetLogo and try and understand how it works, rather than just copy the relevant bits into your code because min-n-of, using agentsets and passing variables to procedures are important aspects of NetLogo that you need to know about. I have also set up colouring etc so you can see the choices it makes.
extensions [rnd]
patches-own [ nest-scent ]
to setup
clear-all
create-turtles 1 [ set color red ]
ask patches
[ set nest-scent random 100
set plabel nest-scent
]
reset-ticks
end
to go
ask one-of turtles [ move ]
tick
end
to move
set pcolor blue
let targets min-n-of 2 neighbors [ nest-scent ]
let destination rnd:weighted-one-of targets [ calc-weight nest-scent ]
move-to destination
end
to-report calc-weight [ XX ]
let weight 0.5 * (1 + ((exp(XX) - exp( - XX)) / (exp(XX) + exp( - XX))))
report weight
end
In my model I have some agents collecting from another agent who they bump into at random after which they move back to their base. As they move back they drop off some material as defined by the random function. Here's some sample code
to go
ask searchers
[ set energy energy - 1
fd 0.0125
if random-float 1 < (1 / 50)
[ ifelse random 2 = 0
[ rt 45 ]
[ lt 45 ]
]
search
]
end
to search
if any? depots in-radius vision with [color = yellow]
[spread
set energy 0] ;; makes them to back to base
end
to spread
if random 10000 = 1 [hatch-rubbish 1 [ set color white
set shape "circle"
set size 0.5]]
end
If I set the visual radius to something enormous so they can always see the depot the number of bits of rubbish works out.
However if I allow them to move around with a radius of 1, the count of rubbish is much higher than 1 in 10,000.
Why would that make a difference?
Thanks
Using the below code I get agents are like this:
to setup
ask breadth-patches [sprout-walls wall-agents[set color 2]]
ask length-patches [sprout-walls wall-agents[set color 2]]
ask gap-patches [sprout-walls wall-agents[set color 2]]
ask length-patches[align-inside-at-top]
ask breadth-patches [align-inside-at-right-left]
ask gap-patches[align-inside-at-top]
end
to align-inside-at-top ;; patch procedure
let counter count walls-here ;; we will use this as a count-down, after using it in some calculations
if counter > 0 ;; could assume there are turtles, but we are not.
[ let gap1 1 / counter ;; size of turtles, gap between turtles
let half-gap gap1 / 2 ;; half-size of turtles
let ytop 0
if-else(pycor < 0)[set ytop pycor - .5 - half-gap]
[set ytop pycor + .5 - half-gap]
let xleft pxcor - .5 - half-gap
ask walls-here
[ set size gap1
set ycor ytop
set xcor xleft + gap1 * counter
set counter counter - 1 ;; so we're placing them from right to left
; set ycor ycor + 0.125
]
]
end
to align-inside-at-right-left ;; patch procedure
let counter count turtles-here ;; we will use this as a count-down, after using it in some calculations
if counter > 0 ;; could assume there are turtles, but we are not.
[ let gap1 1 / counter ;; size of turtles, gap between turtles
let half-gap gap1 / 2 ;; half-size of turtles
let ytop pycor + .5 + half-gap
let xleft 0
if-else (pxcor < 0)[
set xleft pxcor + .5 - half-gap]
[ set xleft pxcor - .5 + half-gap
]
ask turtles-here
[ set size gap1
set ycor ytop - gap1 * counter
set xcor xleft ;+ gap * counter
set counter counter - 1 ;; so we're placing them from right to left
]
]
end
Note: The gap in the rectangle is due to the following code
ask patches with [pxcor > (gap * (-1)) and pxcor < gap and pycor =(breadthrec - 1)][ask walls-here[die]]
Here, gap = 1 ,i.e, a width of 1 patch.
So the input parameter is the wall-agents which specifies the number of agents to created per patch along the length and breadth patches.
I wish to change to create overlapping agents as in the figure below(Sorry the figure is not so perfect, but I hope it explains it). Please help on how to achieve this.
This is a lot of code to ask anybody to debug for you.
I would suggest solving a simpler version of the problem first. Do you have code that can create wall-agents turtles in a single patch, evenly spaced along a line? Once you had a working code that did that, then you could attempt to generalize it to your more complex problem.
If you run into trouble writing that simpler version, you'll have a smaller question to ask here on Stack Overflow that will be much easier for someone to answer than your current, very large question.
If you are able to write the simpler version, don't throw it away — keep it, so you can go back to it if you need to. Then tackle the bigger problem.
You might even be able to take the simpler version, put it into a procedure, and then call that procedure from your larger solution. Making small procedures that work, and then calling those smaller procedures from other ones, is often a good way to break a problem down into manageable parts.
I wish to find out whether in a given turtle's heading there is another agent present upto a given distance.
Here the Distance is "D".
Note:
Any agent present before D in the given direction should be also considered.
Even the direction doesn't coincide with the other's agent centre but just touches it ,even then that agent should be considered.
Problem:
No turtle-ahead procedure available. Combination of patch-ahead and turtles-on not applicable due to patch-size>> turtle-size.
Possible approach:
1.Represent the turtle's heading by the equation of a line.
to-report calculate-line[x y angle]
let m tan angle
let A m
let B -1
let C (- m * x + y)
report (list A B C)
end
to-report heading-to-angle [ h ]
report (90 - h) mod 360
end
let line-equ calculate-line (xcor) (ycor) (heading-to-angle heading)
2.Calculate the perpendicular distance from other turtles here, Check if there are within a range that the size of other turtles.
to-report value[A X1 Y1];;A is list of coefficents of line, x1 and y1 are coordinates of red turtle
if-else(abs((item 0 A * X1 + item 1 A * Y1 + item 2 A) / (sqrt((item 0 A ^ 2) + (item 1 A ^ 2) ))) < [size] of wall )
[ report "true"][report "false"]
end
3.To check if the red turtle is within D. One could obtain a line perpendicular to black one and compute the red turtle distance from it to check if it is less than or equal to D. But then that adds more complication.(Though one can simplify rotate the turtle by 90 left or right and get the line equation.)
This is what I meant by my comment. Run this code (as its own model). What it does is turn all the turtles on a few 'ahead' patches a different colour. I know this is not what you are trying to do, but the agentset candidates is a relatively small number of turtles. These are the only ones that you have to check whether they are on the correct path. So instead of turning them a different colour you could check the direction that they are from your initial turtle.
to setup
clear-all
set-patch-size 25
resize-world -10 10 -10 10
create-turtles 1000
[ setxy random-xcor random-ycor
set color yellow
set size 0.4
]
ask one-of turtles
[ set color red
set size 1
check-from-me 5
]
end
to check-from-me [howfar]
let counter 0
let candidates turtles-here
while [counter < howfar]
[ set counter counter + 1
set candidates (turtle-set candidates turtles-on patch-ahead counter)
]
ask candidates [set color red]
end
to-report check-wall
let return false
hatch 1[
set color black
set size ([size] of one-of walls) / 2
show (2.5 * ([size] of myself))
while [distance myself < (2.5 * ([size] of myself))]
[
fd ([size] of one-of walls) / 64
if any? walls in-radius size
[
set return true
]
show distance myself
]
]
report return
end
The above works. But still is approx. I am looking for better solution with probably less maths as one elucidated in the question.
I was wondering if anyone has a good idea to filter turtles who have the caller turtle in their vision and in their cone of vision with the least computational cost?
Now I am using something like this:
Member? Caller-Agent agents in-cone 5 100
or just who are agents who can see me?
I was wondering if anyone knows how can I check if heading of other agents in some radius X is toward the caller agent ?
So , this is the way I did it:
I used abs( towards myself - heading) < 50
I thought it might include cone with angle of 100 since it uses absolute value.
This is just a test program to see if it works:
ask turtles[set label "" set color green]
ask turtle 7 [
set color red
set label "Caller"
ask other (turtles with [distance myself <= 3 and abs( towards myself - heading) < 50 ])
[
set color yellow
]
ask other turtles [
set label (Member? turtle 7 Other turtles in-cone 3 100)
]
]
and this one cone 80 and abs( towards myself - heading) < 40 :
There is a problem with this approach:
if an agent is as same location of the caller then I will get the error that there is no heading defined to same point or something like that!
Error is : No heading is defined from a point (7,7) to that same point. error while turtle 28 running TOWARDS
Update:
[(distance myself <= 3 and distance myself > 0 and abs( subtract-headings towards myself heading ) < 60) or distance myself = 0 ]