Arduino Obstacle Avoidance Robot Car

Finally my MIDTERN exam is over~~ 
so i have time to share my arduino hw

This is our Arduino Obstacle Avoidance Robot Car.
It took us about 1hour to build it :O

 

This is part of our video with a lovely tempo~~

//超音波智能避障（ARDUINO）
//    L = 左
//    R = 右
//    F = 前
//    B = 後

#include  
int pinLB=6;     // 定義6腳位 左後
int pinLF=7;     // 定義9腳位 左前

int pinRB=8;    // 定義10腳位 右後
int pinRF=9;    // 定義11腳位 右前

int inputPin = A0;  // 定義超音波信號接收腳位
int outputPin =A1;  // 定義超音波信號發射腳位

int Fspeedd = 0;      // 前速
int Rspeedd = 0;      // 右速
int Lspeedd = 0;      // 左速
int directionn = 0;   // 前=8 後=2 左=4 右=6 
Servo myservo;        // 設 myservo
int delay_time = 50; // 伺服馬達轉向後的穩定時間

int Fgo = 8;         // 前進
int Rgo = 6;         // 右轉
int Lgo = 4;         // 左轉
int Bgo = 2;         // 倒車

void setup()
 {
  Serial.begin(9600);     // 定義馬達輸出腳位 
  pinMode(pinLB,OUTPUT); // 腳位 8 (PWM)
  pinMode(pinLF,OUTPUT); // 腳位 9 (PWM)
  pinMode(pinRB,OUTPUT); // 腳位 10 (PWM) 
  pinMode(pinRF,OUTPUT); // 腳位 11 (PWM)
  
  pinMode(inputPin, INPUT);    // 定義超音波輸入腳位
  pinMode(outputPin, OUTPUT);  // 定義超音波輸出腳位   

  myservo.attach(2);    // 定義伺服馬達輸出第5腳位(PWM)
 }
void advance(int a)     // 前進
    {
     digitalWrite(pinRB,LOW);  // 使馬達（右後）動作
     digitalWrite(pinRF,HIGH);
     digitalWrite(pinLB,LOW);  // 使馬達（左後）動作
     digitalWrite(pinLF,HIGH);
     delay(a * 100);     
    }

void right(int b)        //右轉(單輪)
    {
     digitalWrite(pinRB,LOW);   //使馬達（右後）動作
     digitalWrite(pinRF,HIGH);
     digitalWrite(pinLB,HIGH);
     digitalWrite(pinLF,HIGH);
     delay(b * 80);
    }
void left(int c)         //左轉(單輪)
    {
     digitalWrite(pinRB,HIGH);
     digitalWrite(pinRF,HIGH);
     digitalWrite(pinLB,LOW);   //使馬達（左後）動作
     digitalWrite(pinLF,HIGH);
     delay(c * 80);
    }
void turnR(int d)        //右轉(雙輪)
    {
     digitalWrite(pinRB,LOW);  //使馬達（右後）動作
     digitalWrite(pinRF,HIGH);
     digitalWrite(pinLB,HIGH);
     digitalWrite(pinLF,LOW);  //使馬達（左前）動作
     delay(d * 80);
    }
void turnL(int e)        //左轉(雙輪)
    {
     digitalWrite(pinRB,HIGH);
     digitalWrite(pinRF,LOW);   //使馬達（右前）動作
     digitalWrite(pinLB,LOW);   //使馬達（左後）動作
     digitalWrite(pinLF,HIGH);
     delay(e * 80);
    }    
void stopp(int f)         //停止
    {
     digitalWrite(pinRB,HIGH);
     digitalWrite(pinRF,HIGH);
     digitalWrite(pinLB,HIGH);
     digitalWrite(pinLF,HIGH);
     delay(f * 100);
    }
void back(int g)          //後退
    {

     digitalWrite(pinRB,HIGH);  //使馬達（右後）動作
     digitalWrite(pinRF,LOW);
     digitalWrite(pinLB,HIGH);  //使馬達（左後）動作
     digitalWrite(pinLF,LOW);
     delay(g * 100);     
    }
    
void detection()        //測量3個角度(0.90.179)
    {      
      int delay_time = 250;   // 伺服馬達轉向後的穩定時間
      ask_pin_F();            // 讀取前方距離
      
     if(Fspeedd < 10)         // 假如前方距離小於10公分
      {
      stopp(1);               // 清除輸出資料 
      back(2);                // 後退 0.2秒
      }
           
      if(Fspeedd < 15)         // 假如前方距離小於25公分
      {
        stopp(1);               // 清除輸出資料 
        ask_pin_L();            // 讀取左方距離
        delay(delay_time);      // 等待伺服馬達穩定
        ask_pin_R();            // 讀取右方距離  
        delay(delay_time);      // 等待伺服馬達穩定  
        
        if(Lspeedd > Rspeedd)   //假如 左邊距離大於右邊距離
        {
         directionn = Rgo;      //向右走
        }
        
        if(Lspeedd <= Rspeedd)   //假如 左邊距離小於或等於右邊距離
        {
         directionn = Lgo;      //向左走
        } 
        
        if (Lspeedd < 10 && Rspeedd < 10)   //假如 左邊距離和右邊距離皆小於10公分
        {
         directionn = Bgo;      //向後走        
        }          
      }
      else                      //加如前方不小於(大於)25公分     
      {
        directionn = Fgo;        //向前走     
      }
     
    }    
void ask_pin_F()   // 量出前方距離 
    {
      myservo.write(90);
      digitalWrite(outputPin, LOW);   // 讓超聲波發射低電壓2μs
      delayMicroseconds(2);
      digitalWrite(outputPin, HIGH);  // 讓超聲波發射高電壓10μs，這裡至少是10μs
      delayMicroseconds(10);
      digitalWrite(outputPin, LOW);    // 維持超聲波發射低電壓
      float Fdistance = pulseIn(inputPin, HIGH);  // 讀差相差時間
      Fdistance= Fdistance/5.8/10;       // 將時間轉為距離距离（單位：公分）
      Serial.print("F distance:");      //輸出距離（單位：公分）
      Serial.println(Fdistance);         //顯示距離
      Fspeedd = Fdistance;              // 將距離 讀入Fspeedd(前速)
    }  
 void ask_pin_L()   // 量出左邊距離 
    {
      myservo.write(5);
      delay(delay_time);
      digitalWrite(outputPin, LOW);   // 讓超聲波發射低電壓2μs
      delayMicroseconds(2);
      digitalWrite(outputPin, HIGH);  // 讓超聲波發射高電壓10μs，這裡至少是10μs
      delayMicroseconds(10);
      digitalWrite(outputPin, LOW);    // 維持超聲波發射低電壓
      float Ldistance = pulseIn(inputPin, HIGH);  // 讀差相差時間
      Ldistance= Ldistance/5.8/10;       // 將時間轉為距離距离（單位：公分）
      Serial.print("L distance:");       //輸出距離（單位：公分）
      Serial.println(Ldistance);         //顯示距離
      Lspeedd = Ldistance;              // 將距離 讀入Lspeedd(左速)
    }  
void ask_pin_R()   // 量出右邊距離 
    {
      myservo.write(177);
      delay(delay_time);
      digitalWrite(outputPin, LOW);   // 讓超聲波發射低電壓2μs
      delayMicroseconds(2);
      digitalWrite(outputPin, HIGH);  // 讓超聲波發射高電壓10μs，這裡至少是10μs
      delayMicroseconds(10);
      digitalWrite(outputPin, LOW);    // 維持超聲波發射低電壓
      float Rdistance = pulseIn(inputPin, HIGH);  // 讀差相差時間
      Rdistance= Rdistance/5.8/10;       // 將時間轉為距離距离（單位：公分）
      Serial.print("R distance:");       //輸出距離（單位：公分）
      Serial.println(Rdistance);         //顯示距離
      Rspeedd = Rdistance;              // 將距離 讀入Rspeedd(右速)
    }  
    
void loop()
 {
    myservo.write(90);  //讓伺服馬達回歸 預備位置 準備下一次的測量
    detection();        //測量角度 並且判斷要往哪一方向移動
      
   if(directionn == 2)  //假如directionn(方向) = 2(倒車)             
   {
     back(8);                    //  倒退(車)
     turnL(2);                   //些微向左方移動(防止卡在死巷裡)
     Serial.print(" Reverse ");   //顯示方向(倒退)
   }
   if(directionn == 6)           //假如directionn(方向) = 6(右轉)    
   {
     back(1); 
     turnR(6);                   // 右轉
     Serial.print(" Right ");    //顯示方向(左轉)
   }
   if(directionn == 4)          //假如directionn(方向) = 4(左轉)    
   {  
     back(1);      
     turnL(6);                  // 左轉
     Serial.print(" Left ");     //顯示方向(右轉)   
   }  
   if(directionn == 8)          //假如directionn(方向) = 8(前進)      
   { 
    advance(1);                 // 正常前進  
    Serial.print(" Advance ");   //顯示方向(前進)
    Serial.print("   ");    
   }
 }

CodeFight02

it's my midterm exam week!!!!!!  just so terrible~~

Year is leap if its number is divisible by 4 and isn't divisible by 100 or if its number is divisible by 400. Determine if a given year is leap or not.
[input] integer year number of the year
[output] boolean true if the given year is leap, false otherwise

bool leapYear(int year) {
    
  if (year % 4 == 0 && year % 100 !=0 || year % 400 == 0) {
    return true;
  }
  return false;
}

CodeFight01

1.Given array of integers, find the sum of absolute differences of consecutive numbers.
 Example [4, 7, 1, 2] -> 3 + 6 + 1 = 10
 [input] array.integer inputArray
 [output] integer sum of absolute differences of consecutive numbers from inputArray

int arraySumAdjacentDifference(std::vector inputArray) {

  int answer = 0;
  for (int i = 1; i < inputArray.size(); i++) {
    answer += std::abs(inputArray[i] - inputArray[i - 1]);
  }
  return answer;
}

2.Find the area of a rectangle with given sides.
 Example rectangleArea(2, 3) = 6
 [input] integer a positive integer
 [input] integer b positive integer [output] integer

int rectangleArea(int a, int b) {
   int result = a*b;
   for(int i=0; i < a; i++){
   result = a*b;
   }
   return result;

3.Determine if the given point is inside the given circle. [input] array.integer point array of length 2 representing a point [input] array.integer center array of length 2 representing the center of a circle [input] integer radius radius of the circle [output] boolean true if point is inside the circle centered at center with radius equal to radius or on its border, false otherwise

boolean insidecircle(std::vectorpoint ,std:vectorcenter,int radius){

  struct Helper{
    int sqr(int value){
      return value*value;
    }
  };

   Helper h;

   if(h.sqr(point[0]-center[0])+h.sqr(point[1]-center[1])<=h.sqr(radius)){
     return true;
   }
   return false;

}

4.Given integer n, find n! (factorial of n).
 [input] integer n non-negative integer
[output] integer

int factorial(int n) {
  if(n==0){
    return 1;
  }
  return n*factorial(n-1);
}

CodeFighter

1.For a given triangle determine if it is a right triangle.
[input] array.integer sides array of three integers representing triangle sides
[output] boolean true if the triangle with sides from the sides array is a right triangle, false otherwise

bool rightTringle(std::vector sides){
    struct Helper{
     int sqr(int value){
       return value*value;
     }
   };
   Helper h;
   std::sort(sides.gegin(),sides.end());
   if(h.sqr(sides[0]+h.sqr(sides[1]==h.sqr(sides[2])){
      return true;
   }
   return false
}

2.If the given integer is greater than 12 return this number, otherwise return 12.
 [input] integer number positive integer
 [output] integer

int returnTwelve(int number) {
  while(number < 12){
   number = number + 1;
  }
  return number
}

3.Given two integers a and b, find the remainder of a when divided by b.  
     Example for a=5, b=3 output should be 2

 [input] integer a a positive integer
 [input] integer b a positive integer
 [output] integer

int findTheRemainder(int a, int b) {
   while(a > b){
   a = a -b;
  }
  return a;
}

4.Given a string, find the number of different characters in it. for "cabca" output should be 3
[input] string s a string of lowercase latin letters
[output] integer

int differentSymbolsNaive(std::string s) {

  int result = 0;

  for (int i = 0; i < 26; i++) {
    bool found = false;
    for (int j = 0; j < s.size(); j++) {       
      if (s[j] = 'a' + i) {
        found = true;
        break;
      }
    }
    if (found) {
      result++;      
    }
  }
  return result;
}

5.Given an integer n, find the volume of the cube with edges equal to n.
     Example for n=5 output should be 125
[input] integer n positive integer
[output] integer

int cube(int n){
   int result =1;
   for(int i = 0; i < 3; i++){
     result = result * n; 
   }
   return result; 
}

#JAVA change line in txt

I tried to write string to txt file in java .
When i checked my txt file in my windows,i encountered some problems.
If i use \n to change a new line ,i saw some black block appear the position that should change line and change line doesn’t work at all.

But when i use my android device to check the result ,it’s ok!!!

It’s wired,but i found a way to solve it.

answer:
use \r\n instead of using \n and the txt file you open on you pc will be ok :")

如果單純用\n來換行，用記事本打開檔案時

會看到原本應該換行的地方變成很多黑色小方塊

而且每行會接在一起，看起來很奇怪
不過用Java讀取這個文字檔時卻又是正常的

解決法很簡單，只要在\n前面加上\r
變成"\r\n"，在記事本上看就不會亂掉了