How do I recreate the Pong game with two pots and the map function?

Question

I am working on a project for my brothers and anyone else who is interested in playing some LED Pong. I was just stuck on the software part. I don't really know what to do. But, I have an idea: I want to use the map() function to make my life easier, however, I am not sure how to use the map function with the Dot Matrix. All help would be appreciated. All of the stuff came in this kit.

Here is the test code:

#include <LedControl.h>
int DIN = 2;
int CS = 3;
int CLK = 4;
const int playerOnePin = A0;
const int playerTwoPin = A1;
int playerOneValue;
int playerOneValue2;
int playerTwoValue;
int playerTwoValue2;
LedControl lc = LedControl(DIN, CLK, CS, 0);
void setup() {
 lc.shutdown(0, false); //The MAX72XX is in power-saving mode on startup
 lc.setIntensity(0, 15); // Set the brightness to maximum value
 lc.clearDisplay(0); // and clear the display
 pinMode(playerOnePin, INPUT);
 pinMode(playerTwoPin, INPUT);
}
void loop() {
 byte a[8] = {0xC0, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00};
 printByte(a);
 
}
void printByte(byte character []) {
 int i = 0;
 for (i = 0; i < 8; i++)
 {
 lc.setRow(0, i, character[i]);
 }
}
void readPot() {
 
 playerOneValue = analogRead(A0);
 playerOneValue2 = map(playerOneValue, 0, 1023, 0, 180);
}

Here is the other code:

#include <LedControl.h>
int DIN = 2;
int CS = 3;
int CLK = 4;
byte displayImage[8] = {0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00};
// considering a 2d co-ordinate system with origin (0,0) at bottom left corner
int ballX = 3; //X position
int ballY = 7; //Y position - top
int speedX = 0; // no X movement, can be -1, 0 or 1
int speedY = -1; // downward Y movement, can be -1, 0 or 1
int paddleX = 4; // X position of center of paddle - can be 1 to 6. Paddle width is 3
int score = 0;
LedControl lc = LedControl(DIN, CLK, CS, 0);
void setup() {
 pinMode (A0, INPUT);
 pinMode (A1, INPUT);
 lc.shutdown(0, false); // Keep MAX72XX awake
 lc.setIntensity(0, 15); // Set the brightness to maximum value
 lc.clearDisplay(0); // and clear the display
}
void loop() {
 // update ball position
 ballX = ballX + speedX;
 ballY = ballY + speedY;
 // check for ball interaction with walls
 if (ballX == 0 || ballX == 7) {
 speedX = speedX * -1; // bouncing off walls in horizontal direction
 }
 // bouncing off ceiling
 if (ballY == 7) {
 speedY = speedY * -1; // bouncing off the ceiling
 }
 // bouncing off the paddle
 if (ballY == 0 && ballX >= (paddleX - 1) && ballX <= (paddleX + 1)) {
 speedY = speedY * -1;
 score++; // player earns a point
 }
 // going past the paddle
 if (ballY == 0 && ballX < (paddleX - 1) && ballX > (paddleX + 1)) {
 // going past the paddle - player is out
 Serial.println(); Serial.print("Score: "); Serial.println(score);
 for (int i = 0; i < 8; i++) {
 displayImage[i] = 0x00;
 }
 displayImage[3] = 0xFF; displayImage[4] = 0xFF; // show a line
 renderByte(displayImage);
 while (1); // Freeze
 }
 // clearing the image variable
 for (int i = 0; i < 8; i++) {
 displayImage[i] = 0x00;
 }
 // generating new image
 addPixel(ballX, ballY); // adding the ball position to image
 addPixel(paddleX - 1, 0); addPixel(paddleX, 0); addPixel(paddleX + 1, 0);
 // adding paddle position to image
 renderByte(displayImage); // show the generated image
 // handling paddle control
 if (analogRead(A0) && paddleX > 0) {
 paddleX = paddleX - 1; // move paddle left
 }
 if (analogRead(A1) && paddleX < 7) {
 paddleX = paddleX + 1; // move paddle right
 }
 delay(200);
}
void addPixel(int xVal, int yVal) {
 int newRowval = 2 ^ (7 - xVal);
 int rowVal = displayImage[7 - yVal];
 displayImage[7 - yVal] = rowVal || newRowval; // inserting a 1 at the required pixel
}
void renderByte(byte image [])
{
 int i = 0;
 for (i = 0; i < 8; i++)
 {
 lc.setRow(0, i, image[i]);
 }
}

Here is the picture of the wiring and the schematic:

This is the wiring picture

Wiring Pic

Wiring Pic

Wiring Pic

Schematic

• DataFiddler: Your ball will either go straight or diagonally at 45° ?
• Austin: Yes, the ball will go straight, I don't know about the diagonally part, but it won't move the paddle.

Answer 1

First figure out the min and max values that paddleX can have. I assume if the paddle is two dots wide then its min,max will be 0,6 Then figure out the min and max of your pot, typically 0,1023 for a a 10 bit adc. Now to map your pot value to paddleX value you'd use the command :

paddleX = map(analogRead(A0),0,1023,0,6);

here is the map documentation : https://www.arduino.cc/reference/en/language/functions/math/map/

• DataFiddler: For an evenly spaced mapping paddleX = map(analogRead(A0),0,1024,0,7); would provide better results. ( Integer division truncates )
• Austin: Yes, the paddle is two dots wide.
• Austin: I don't really know what to do from there.