Chess 960 position generator

Projects General Guidance
1

Hi All,

I wanted to build a chess960 position generator, and I wanted to get some guidance on the way to store the data. The output would be a char array like: 'RNBQKBNR' that I generatte from the basic rules.

Here is my code, but I don't know how to handle the data. I thought I could shove it all into an array[8]. Like
{R, N, B, Q, K, B, N, R}.

int idn = random(960) ;         //Divide idn by 4, yielding quotient 'bishop_intermediate' and remainder 'bishop light'
int bishop_light = idn % 4;    // place a bishop on light square (b, d, f, h) 
int bishop_light_pos[] = {1, 3, 5, 7};
int a = bishop_light_pos[bishop_light];
int bishop_intermediate = idn/4; // Divide 'bishop_intermediate' by 4 again, yielding quotient N3 and remainder 'bishop_dark'.
int N3 = bishop_intermediate/4;
int bishop_dark = bishop_intermediate % 4; // place a bishop on dark square (a, c, e, g)
int bishop_dark_pos[] = {0, 2, 4, 6};
int b = bishop_dark_pos[bishop_dark];
int N4 = N3/6; 
int queen = N3 % 6;  //  Divide N3 by 6, yielding quotient N4 and remainder queen.
                                   //  Place the Queen according to Q, where 0 is the first free square starting from a, 1 is the second, etc.
                                 //  N4 will be a single digit, 0 ... 9. 
                                //  Place the Knights according to its value by consulting the following table:
int knight_table[] = { 11000, 10100, 10010, 10001, 01100, 01010, 01001, 00110, 00101, 00011 };

void setup() {
  Serial.begin(9600);
}


void loop() {
  Serial.println(idn);
  Serial.print("First Bishop on ");
  Serial.println(a);
  
  Serial.print("Second Bishop on ");
  Serial.println(b);
  
  
  delay(10000);
  
}

Here is the link to the rules of the scheme: Fischer random chess numbering scheme - Wikipedia

And the text description of what I am trying to do:

Direct derivation[edit]
White's Chess960 starting array can be derived from its number N (0 ... 959) as follows:

a) Divide N by 4, yielding quotient N2 and remainder B1. Place a Bishop upon the bright square corresponding to B1 (0=b, 1=d, 2=f, 3=h).

b) Divide N2 by 4 again, yielding quotient N3 and remainder B2. Place a second Bishop upon the dark square corresponding to B2 (0=a, 1=c, 2=e, 3=g).

c) Divide N3 by 6, yielding quotient N4 and remainder Q. Place the Queen according to Q, where 0 is the first free square starting from a, 1 is the second, etc.

d) N4 will be a single digit, 0 ... 9. Place the Knights according to its value by consulting the following table:

Digit Knight positioning
0 N N - - -
1 N - N - -
2 N - - N -
3 N - - - N
4 - N N - -
5 - N - N -
6 - N - - N
7 - - N N -
8 - - N - N
9 - - - N N
e) There are three blank squares remaining; place a Rook in each of the outer two and the King in the middle one.

Thanks for any ideas about the best way I should try to do this.

Jimmy

2

The best method of storage depends on what you do with the data afterwards. That being said, an array should be fine. What are you trying to do?

3

I'm just going to print it on a pair of DL2416 LED displays, or some kind of 8-digit intelligent display. I have a few HDSP 2115 displays that would work nice. I'll have it print the 8 digits, wait a little time and then go into sleep/power down. The reset button will start it back up and display a new position.

I started messing with trying to shove the results char by char into an array called back_rank, but I was getting 'array bound is not an integer constant.'

4

Looking closer at your code, you should probably define the length of all of your arrays:
int bishop_light_pos[4] = {1, 3, 5, 7};
int bishop_dark_pos[4] = {0, 2, 4, 6};

The biggest downside to arrays is that in C they are not adjustable length. Generally This just means you need to plan ahead and make sure it is long enough to handle anything you intend on putting into it.

5

Thanks for this info, I'll declare the size of those arrays.

I'm having trouble working out where to put the queen. I have a variable 'queen' that would be 0 to 6, and I have to put the Q into the index of the array that corresponds to the first open index.

So if the 'queen is 4', and the array is {B, B, 0, 0, 0, 0, 0, 0}, I need to put Q at index 6 like { B, B, 0, 0, 0, 0, Q, 0}.

So do I iterate over the array and count both the index count, and the 0-count?

This works for the 0th open index:

if (queen == 0){
    for ( int i = 0; i < 8; i++ ){
        if (back_rank[i] == '0') {
            current_index = i;
            break;
        }
   } 
back_rank[current_index] = 'Q';

But I'm having a hard time figuring out how to count the index, and the open squares (represented by '0'), when it is more than 0.

Thanks for any help.

Jimmy

6

I don't know anything about chess programming. If the columns in the array have specific meanings (e.g. the queen is always in column 4 or something similar) then it may make the code easier to follow if you give the index numbers a name. For example

byte queenCol = 4;

and then you could refer to it like this

if (array[queenCol] == 0) { ....

Sorry if the specifics are stupid but I hope it illustrates my point.

...R

7

I got this done. I still have a few bugs, and need to clean up the code a little. It currently looks like it was written by honey badgers.
I've attached a picture to show what it looks like. Thanks to all who offered advice.

960_generator.jpg
960_generator.jpg480×640 41.3 KB