Comments by Auyik (Top 56 by date)

Thank you so much, no worries.

I applied your suggestion, but it seems that my code now, never counts the number of particles #1, #2 and #3. Also the counter for the fullcells remains zero :(

I also changed the new functions as "int" instead of "void", but that neither works.

I have another question ,why grid[y][x] inside "void absorb_particle1" hasn't the expression "psi ->" before?

Do I need to define psi, inside of the main function as "SimulInfo psi"?

Thanks! I hqve a question: why did you defined particle1, particle2, etc as integer arrays? Can I leave them as integers only?

Thank you!

Thanks. I have another question, what are the methods to automate the task of save the file before compiling again?

Interesting. Could clarify a bit more what you mean by "the product of the compilation process", please?

I think I have found a way to fix the issue. Thank you very much for your answer.

Well, I'm trying to reproduce the results of a paper from scratch and there, Y is defined as the partial pressure of particle #1, and it's in the interval [0,1]. It is used to decide which molecule will be selected to be trapped on the surface. So, we draw a random number "r" (also in [0,1]). Then, if it is less than Y, we choose particle #1, otherwise, we select particle #2.

In the mentioned paper the authors simulate the interactions between both particles. The idea is to plot Y versus both coverages and obtain the same results they obtained before. But first I'm trying to "fill" that surface with atoms, before adding interactions between them. However, I have this problem while trying to choose between particles #1 and #2.

I can't think of any other way to incorporate "Y" in the simulation to be able to choose between these two atoms. I don't have a lot of programming experience but this simulation is not supposed to be difficult, and I don't know what to do.

Sorry, I edited the original code of the post

Y is defined as the partial pressure of the particle #1, and it's in the interval [0,1]. It is used to decide which molecule will be selected to adsorb. So, we draw a random number "r" (also in [0,1]) so that if it is less than Y, we choose the particle #1. Otherwise, we choose the particle #2.

This is part of the algorithm the paper I'm following. In this paper the authors simulate the interactions between both particles. The idea, is to plot Y versus both coverages, to reproduce the results of a model developed some years ago.

Thanks. In my simulation "y" represents the "partial pressure" of an atom or the probability of such atom to be chosen. It's greater than zero and less than 1. Later on I will need to reproduce the dependence of the coverages with this parameter (after adding more interactions between particle #1 and particle #2).

The problem is that it's the best way I found to compare "r" with "y" so far

Thank you very much. I edited my code to include a function that calculates the average. Is it ok the way I did it?

Sorry, I fix my code. Now it's working

Thank you, I edited this post and paste a shorter version of the original code. I assigned to arrays the number of CO's and O's to storage the number of molecules trapped in the surface, then the blocks that calculate the mean and standard deviation where place outside the while loop.

As results, for the mean I get exactly the number of CO trapped. The same for O. Also, I think now the code gives me reasonable results for the corresponding standard deviations.

I'm gonna add again the interactions between the molecules to see how those results change.

Thanks, I'll do that.

Basically, I get "inf" outputs for the standard deviation

Yes, when I run tests, sometimes I get infinite values for the standard deviation

Oh sorry, I deleted that part while editing. The issue is that I am not completely sure if I placed the blocks that calculate the mean and standard deviation in the right part of the code and that's affecting the output

Ok, I'm not sure If I placed the code chunks that calculated the mean and standard dev in the right place and it is afecting the output. It compiles but that's my doubt.

HI, I just edited my question. Thank you

Thank you very much for your advice! I think I got it :)

Thank you, I tried with a shorter code (I don't include all the interactions between molecules mentioned above yet). I think it's fixed...Still checking

Thanks for responding. I have a question: in the definition of the function do I need to write it as

gstate set_rightneigbor(gstate state, int x, int y, gstate grid[MAX_X][MAX_Y]);

?

Thank you very much! I never thought about defining a new datatype to prevent the program from wrong assignments. Indeed, I think it will be useful since I would like to add more than one type of molecule in the code (for example CO, and O) and make interactions between them, but first I needed a good base like this. I wanna modify this and come back if new questions arise :)

Thank you so much. If I have more questions, I'll come back :)

Thank you!

Thanks!

Thank you for your help! I just deleted this part of my code. Yes! I'm trying to understand "calloc()" and "malloc()", but it's hard for me right now.

I'll go to work with a 40x40 lattice indeed. Do you think I need to get memory with this lattice size?

Yes, you are right. I will fix it. Thank you

Oh, I totally forgot about the part of the code that makes my matrix always "occupied". It makes sense. But, how can I define the grid without this problem happens?

Thanks. I got rid of the gotos and now the matrix renains the same as expected. Now I'm trying to fix the issue that the program doesn't remember if a site is or not occupied

Thank you for your remarks. I already got rid of the gotos and currently finding a better way to write the loops

Thank you. I will check it right now:)

Also, I see that the program doesn't even remember if a "site" of the matrix is occupied or not by NO, O, CO or N.

To explain myself better, the following is the output of my code

1 2 
3 4 
(x,y)=(0,0)

The coordinates are (0,0) and the solicited matrix entry is: 1

IS IT OCCUPIED???


It's occupied, the trial ends. Choose another site again

5 6 
7 8 
(x,y)=(0,1)

The coordinates are (0,1) and the solicited matrix entry is: 6

IS IT OCCUPIED???


It's empty. Let's go to generate a random number to choose the adsorbates

random q= 0.663232

q = 0.663232 is greater than xco = 0.400000. We choose NO

r= 0.045448

r= 0.045448 is less than rno= 0.600000

 We seek another neighbor to decide if NO can disociate

FIND THE NEIGHBORS SO THAT NO CAN BE DISOCIATED

Right: 5
Left: 5
Up: 8
Down: 8

****WE DIDN'T FIND AN EMPTY NEIGHBOR SO,THE TRIAL ENDS, START AGAIN****

9 10 
11 12 
(x,y)=(1,0)

The coordinates are (1,0) and the solicited matrix entry is: 11

IS IT OCCUPIED???


It's empty. Let's go to generate a random number to choose the adsorbates

random q= 0.896695

q = 0.896695 is greater than xco = 0.400000. We choose NO

r= 0.358901

r= 0.358901 is less than rno= 0.600000

 We seek another neighbor to decide if NO can disociate

FIND THE NEIGHBORS SO THAT NO CAN BE DISOCIATED

Right: 12
Left: 12
Up: 9
Down: 9

****WE DIDN'T FIND AN EMPTY NEIGHBOR SO,THE TRIAL ENDS, START AGAIN****

13 14 
15 16 
(x,y)=(0,0)

The coordinates are (0,0) and the solicited matrix entry is: 13

IS IT OCCUPIED???


It's empty. Let's go to generate a random number to choose the adsorbates

random q= 0.540815

q = 0.540815 is greater than xco = 0.400000. We choose NO

r= 0.274174

r= 0.274174 is less than rno= 0.600000

 We seek another neighbor to decide if NO can disociate

FIND THE NEIGHBORS SO THAT NO CAN BE DISOCIATED

Right: 14
Left: 14
Up: 15
Down: 15

****WE DIDN'T FIND AN EMPTY NEIGHBOR SO,THE TRIAL ENDS, START AGAIN****

17 18 
19 20 
(x,y)=(1,1)

The coordinates are (1,1) and the solicited matrix entry is: 20

IS IT OCCUPIED???


It's empty. Let's go to generate a random number to choose the adsorbates

random q= 0.474475

q = 0.474475 is greater than xco = 0.400000. We choose NO

r= 0.114339

r= 0.114339 is less than rno= 0.600000

 We seek another neighbor to decide if NO can disociate

FIND THE NEIGHBORS SO THAT NO CAN BE DISOCIATED

Right: 19
Left: 19
Up: 18
Down: 18

****WE DIDN'T FIND AN EMPTY NEIGHBOR SO,THE TRIAL ENDS, START AGAIN****

21 22 
23 24 
(x,y)=(1,1)

The coordinates are (1,1) and the solicited matrix entry is: 24

IS IT OCCUPIED???


It's empty. Let's go to generate a random number to choose the adsorbates

random q= 0.141615

CO is adsorbed

adsorbed CO ---> co = 1

FIND THE NEIGHBORS OF THE ADSORBED CO

Right: 23
Left: 23
Up: 22
Down: 22

It wasn't encountered a site with O :(

CO2 IS NOT PRODUCED

It shows that the original matrix is not preserved, and even when it said the coordinate (0,0) is occupied (originally it has the label "1" in the original matrix), later on, it says that the same coordinate (now labeled by "13") is empty.

That's the problem I'm having. It seems that the matrix doesn't remember its original state

Oh, thank you very much. You're right. Let me check my code one more time.

OK, thanks. I made some modifications to my code, but it never passes through the following part of the code (where I'm trying to verify the occupation of a site):

//EVALUATE IF THE CHOSEN SITE IS EITHER EMPTY OR OCCUPIED
    printf("IS IT OCCUPIED???\n\n\n");

               //random = (double) rand () / RAND_MAX;
               //printf("random = %lf\n\n",random);
               // Include particle based on the probability
   check:          if (grid[y][x] == occupied){

                   printf("It's occupied, the trial ends. Choose another site again\n\n");
                   goto step1;

               }
               else{
                   if (grid[y][x] == empty){

                       printf("It's empty. Let's go to generate a random number to choose the adsorbates\n\n");
                       goto step2;
                   }

               }

Why it would happen?

Hi! thank you for your answer. I was thinking about using "==" but doesn't that action replace the value of the chosen site? In my code grid[x][y] is the chosen entry in the surface I'm working with.

I solved it!

Thank you very much for your response. I fixed my code 20 minutes after I posted this question and now is working as I expected! But I'm happy I posted the question anyways :)

Also, thank you for your advice. I will check it out right now. Yes, I think I need to improve my debugging skills and this definitely will help me in my current project :D

Okay, I set aside the approach of applying transformations to a square grid to obtain a triangular one, and instead, I wrote code directly considering a triangular lattice. For that, I considered if the row of the chosen site had an even or odd x coordinate. Additionally, I included the corresponding periodic boundary conditions... The code works as expected for most of the chosen sites, but for others it gives me incorrect nearest neighbors. I'm not quite sure why this is happening :(

Should I open a new thread considering this new approach?

Hi again! I have left this aside a bit due to other things I have to do but, I reformulated the code corresponding to a square lattice by using 2 coordinates. I'll try to use the above transformations in this new code. If new questions arise, I will come back without a doubt. Thank you for checking :)

Okay, no worries. Let me think about this a bit more.

Btw, thank you very much for your help, it was very useful to clarify things! I will let you know if I come to any conclusions :)

Yes, I just noticed that depending on the site I need to consider if it's in an even or odd row, for example. I think there's when this part of the suggested code become useful:

//(X,Y) --> (U,V)
inline void XY2UV( int X, int Y, int &U, int &V )
{
U = X*2 + ( Y%2 );
V = Y;
}

But I've been working with 1 coordinate, not two. Here is where I'm stuck now :(

Thanks! Below there is a similar suggestion than yours. Let me see if I'm correct by shifting coordinates :)

Hmm, based on that diagram, could I say that the transformation gives me the following changes:

L--->1
U--->2
UR-->3
R--->4
D--->5
DL--->6?

Am I correct?

I think that the orientation it's what they are considering in the formulation for the triangular lattice. For that reason, they included an ordered pair in that case (?). But I don't understand how to apply the transformation described there since effectively, I have been working with 1 orientation. I tried to ask on stack overflow in fact, but this is the part that I don't get yet (the link you provided, was my question)

I've included some links showing diagrams for both the square and triangular lattices I was talking about :)

Yes, you are right. But for now, I only considered 4 nearest neighbors and ignore the diagonals when I was talking about the square grid:)

@Graeme_Grant, yes you are correct. It's like the diagram showed in this link: https://i.stack.imgur.com/1N4pq.png

I had problems including links in my post, sorry.