How to set a binary int? - arduino-uno

I am new to coding in C using a Arduino Uno. I would like to do the following:
int randomNumber;
int randomNumberBinairy = 0;
void setup() {
Serial.begin(9600);
randomSeed(analogRead(A0));
}
void loop() {
randomNumber = random(1, 16);
randomNumberBinairy = ((randomNumber, BIN));
Serial.println(randomNumberBinairy);
delay(5000);
}
This prints out:
2
2
2
2
etc..
However I would like it to print out the random number (between 1 and 16) in binary. So it should look something like:
101
1100
110
10
etc..
Any help on this please?

in arduino you can use the function bitRead(x, n)
int randomNumber;
void setup() {
Serial.begin(9600);
randomSeed(analogRead(A0));
}
void loop() {
randomNumber = random(1, 16);
Serial.print(bitRead(randomNumber, 0));
Serial.print(bitRead(randomNumber, 1));
Serial.print(bitRead(randomNumber, 2));
Serial.println(bitRead(randomNumber, 3));
delay(5000);
}

Just for your own reference, the documentation of Serial.print says:
Serial.print(78, BIN) gives "1001110"
Serial.print(78, OCT) gives "116"
Serial.print(78, DEC) gives "78"
Serial.print(78, HEX) gives "4E"
That is, if you want to print something in binary then you only need
to write
Serial.print(randomNumber, BIN);
or
Serial.println(randomNumber, BIN);
It's simple as that.

If you are sure it will be always 0 to 15 you can write a switch-case block that handles 16 different probability. That works faster than bitread(x, n) for each bit.

Related

How to code Arduino used countdown timer (buzzer), 4 pir sensors, and relay? They worked parallel but what I need is work sequentially

How do you program Arduino to do a countdown timer 10 times using a buzzer, then after that, the light will turn on (relay). In addition, 4 PIR sensors will detect movement, if there is movement, the light will turn off for 10 seconds after which the light will turn on again. I already try to code but something feels off because the light is on at the start and the sensor detects when the timer is still running.
Here is the code that I try to code:
#define relayPin 12
#define pirPin1 8
#define pirPin2 9
#define pirPin3 10
#define pirPin4 11
int val1 = 0,val2 = 0, val3 = 0, val4 = 0 ;
long buzzerFrequency = 1000;
float buzzerDuration = 100;
int startNumber = 9; //countdown starts with this number
int endNumber = 0; //countdown ends with this number
const int buzzerPin = 6;
void setup() {
// put your setup code here, to run once:
pinMode(buzzerPin,OUTPUT);
pinMode(relayPin,OUTPUT);
pinMode(pirPin1, INPUT);
pinMode(pirPin2, INPUT);
pinMode(pirPin3, INPUT);
pinMode(pirPin4, INPUT);
Serial.begin(115200);
}
void buzzer(){
if (startNumber >= endNumber) {
for (int i = 0; i <= 1; i++){
tone(buzzerPin,buzzerFrequency,buzzerDuration);
}
delay(1000);
startNumber--;
}
}
void allPir(){
val1 = digitalRead(pirPin1);
val2 = digitalRead(pirPin2);
val3 = digitalRead(pirPin3);
val4 = digitalRead(pirPin4);
if(val1 == HIGH || val2 == HIGH || val3 == HIGH || val4 == HIGH){
digitalWrite(relayPin, HIGH);
Serial.println("Movement! delay 10 second");
Serial.println("Light off");
delay(10000);
}else{
digitalWrite(relayPin, LOW);
Serial.println("No Movement");
Serial.println("Light on");
}
}
void loop() {
// put your main code here, to run repeatedly:
buzzer();
allPir();
}
The countdown should have finished first with a buzzer sound, then the light will turn on, then the PIR sensor will check the environment. How to fix it? and what do I need to add?

Math calculations in arduino

I wrote a code to do a basic calculations with arduino uno. but it returns me a wrong answer for filledPercentage. It is always -1. what is the problem with this?
//include the library code:
#include <LiquidCrystal.h>
// initialize the library by associating any needed LCD interface pin
// with the arduino pin number it is connected to
const int rs = 13, en = 11, d4 = 5, d5 = 4, d6 = 3, d7 = 2;
LiquidCrystal lcd(rs, en, d4, d5, d6, d7);
int lowLevelDistance =112;
int highLevelDistance = 47;
void setup() {
// set up the LCD's number of columns and rows:
lcd.begin(16, 2);
Serial.begin(9600);
}
void loop() {
// set the cursor to column 0, line 1
// (note: line 1 is the second row, since counting begins with 0):
lcd.setCursor(0, 1);
int currentLevel = 101;
int filledLevel = (lowLevelDistance - currentLevel);
int fullLevel = (lowLevelDistance - highLevelDistance);
int filledPercentage = filledLevel / (fullLevel / 100);
Serial.println(lowLevelDistance);
Serial.println(highLevelDistance);
Serial.println(currentLevel);
Serial.println(filledLevel);
Serial.println(fullLevel);
Serial.println(filledPercentage);
delay(1000);
}
You shouldn't use an integer for the division, your fulllevel / 100 returns 0.65 but then is converted into an integer (truncating it into a 0).
Dividing by 0 throws an error (-1).

Issues with Arduino Timing

I am working with an Arduino on a project for which timing is very important. I use TimerOne to trigger timer interrupts and use micros() for delays (delayMicroseconds() was causing problems worse than the one explained below). The program is sending a manual PWM signal to an LED and it is very important that the signal is sent with an error that is less than 8 microseconds (ideally, the signal is sent at the same time in each period). My test code is shown below:
#include <TimerOne.h>
#include <SPI.h>
const int LED_PIN = 3;
const int CHIP_SELECT = 12;
const int PERIOD = 4000;
const double DUTY_CYCLE = .5;
const int HIGH_TIME = PERIOD * DUTY_CYCLE;
const int LOW_TIME = PERIOD - HIGH_TIME;
const int INITIAL_SIGNAL_DELAY = LOW_TIME / 2;
const int HIGH_TIME_TOTAL_DELAY = INITIAL_SIGNAL_DELAY + HIGH_TIME;
const int RESISTOR_VALUE = 255;
boolean triggered = false;
boolean data = false;
unsigned long triggeredTime;
unsigned long s;
unsigned long e;
boolean found;
int i = 0;
void setup()
{
s = micros();
Timer1.initialize(PERIOD);
Timer1.attachInterrupt(trigger);
pinMode(LED_PIN, 3);
pinMode(CHIP_SELECT, OUTPUT);
SPI.begin();
digitalWrite(CHIP_SELECT, LOW);
SPI.transfer(B00010001);
SPI.transfer(RESISTOR_VALUE);
digitalWrite(CHIP_SELECT, HIGH);
e = micros();
Serial.begin(115200);
Serial.print("s: ");
Serial.println(s);
Serial.print("e: ");
Serial.println(e);
}
void loop()
{
if(triggered)
{
while(micros() - triggeredTime < INITIAL_SIGNAL_DELAY)
{ }
s = micros();
digitalWrite(LED_PIN, data);
while(micros() - triggeredTime < HIGH_TIME_TOTAL_DELAY)
{ }
digitalWrite(LED_PIN, LOW);
data = !data;
triggered = false;
e = micros();
//micros();
if(s % 100 > 28 || s % 100 < 12)
{
found = true;
}
if(!found)
{
Serial.print("s: ");
Serial.println(s);
}
else
{
Serial.print("ERROR: ");
Serial.println(s);
}
//Serial.print("e: ");
//Serial.println(e);
}
}
void trigger()
{
triggeredTime = micros();
triggered = true;
}
(it should be noted that the first signal sent is always xx20, usually 5020).
So, with this code, I eventually get an error. I am not sure why, but this error occurs at the same point every single time:
.
.
.
s: 1141020
s: 1145020
s: 1149020
ERROR: 1153032
ERROR: 1157020
ERROR: 1161020
.
.
.
Now, the really weird part is if I remove the comments before micros() (the micros() right after e = micros()), there is no error (or at least there is not an error within the first 30 seconds). I was wondering if anybody could provide an explanation for why this happens. I have dedicated many hours trying to get the timing working properly and everything was working well until I encountered this error. Any help would be very much appreciated. Thank you!

Special numbers challenge in programming

First, sorry for my bad English.
Special numbers are numbers that the sum of the digits is divisible to the number of the digit.
Example: 135 is a special number because the sum of the digits is 1+3+5 = 9, the number of the digit is 3, and 9 is divisible to 3 because 9 % 3 == 0. 2,3,9,13,17,15,225, 14825 are also special numbers.
Requirement:
Write a program that read the number n (n <= 10^6) from a file named SNUMS.INP (SNUMS.INP can contain up to 10^6 numbers) and print the result out into the file SNUMS.OUT. Number n is the order of the special number and the result will be that special number in n order (sorry I don't know how to express it).
Example: n = 3 means you have to print out the 3rd special number which is 3, n = 10 you have to print out 10th special number which is 11, n = 13 you have to print out 13th special number which is 17, n = 15 you have to print out 15th special number which is 20.
The example bellow will demonstrate the file SNUMS.INP and SNUMS.OUT (Remember: SNUMS.INP can contain up to 10^6 numbers)
SNUMS.INP:
2
14
17
22
SNUMS.OUT:
2
19
24
35
I have my own alogrithm but the the running time exceeds 1 second (my SNUMS.INP has 10^6 numbers). So I need the optimal alogrithm so that the running time will be less than or equal 1s.
Guys I decide to post my own code which is written in Java, it always take more than 4 seconds to run. Could you guys please suggest some ideas to improve or how to make it run faster
import java.util.Scanner;
import java.io.*;
public class Test
{
public static void main(String[]args) throws IOException
{
File file = new File("SNUMS.INP");
Scanner inputFile = new Scanner(file);
int order = 1;
int i = 1;
int[] special = new int[1000000+1];
// Write all 10^6 special numbers into an array named "special"
while (order <= 1000000)
{
if (specialNumber(i) == true)
{
special[order] = i;
order++;
}
i++;
}
// Write the result to file
PrintWriter outputFile = new PrintWriter("SNUMS.OUT");
outputFile.println(special[inputFile.nextInt()]);
while (inputFile.hasNext())
outputFile.println(special[inputFile.nextInt()]);
outputFile.close();
}
public static boolean specialNumber(int i)
{
// This method check whether the number is a special number
boolean specialNumber = false;
byte count=0;
long sum=0;
while (i != 0)
{
sum = sum + (i % 10);
count++;
i = i / 10;
}
if (sum % count == 0) return true;
else return false;
}
}
This is file SNUMS.INP (sample) contains 10^6 numbers if you guys want to test.
https://drive.google.com/file/d/0BwOJpa2dAZlUNkE3YmMwZmlBOTg/view?usp=sharing
I've managed to solve it in 0.6 seconds on C# 6.0 (.Net 4.6 IA-64) at Core i7 3.2 GHz with HDD 7200 rpc; hope that precompution will be fast enough at your workstation:
// Precompute beautiful numbers
private static int[] BeautifulNumbers(int length) {
int[] result = new int[length];
int index = 0;
for (int i = 1; ; ++i) {
int sum = 0;
int count = 0;
for (int v = i; v > 0; sum += v % 10, ++count, v /= 10)
;
if (sum % count == 0) {
result[index] = i;
if (++index >= result.Length)
return result;
}
}
}
...
// Test file with 1e6 items
File.WriteAllLines(#"D:\SNUMS.INP", Enumerable
.Range(1, 1000000)
.Select(index => index.ToString()));
...
Stopwatch sw = new Stopwatch();
sw.Start();
// Precomputed numbers (about 0.3 seconds to be created)
int[] data = BeautifulNumbers(1000000);
// File (about 0.3 seconds for both reading and writing)
var result = File
.ReadLines(#"D:\SNUMS.INP")
.Select(line => data[int.Parse(line) - 1].ToString());
File.WriteAllLines(#"D:\SNUMS.OUT", result);
sw.Stop();
Console.Write("Elapsed time {0}", sw.ElapsedMilliseconds);
The output vary from
Elapsed time 516
to
Elapsed time 660
with average elapsed time at about 580 milliseconds
Now that you have the metaphor of abacus implemented below, here are some hints
instead of just incrementing with 1 inside a cycle, can we incremente more aggressively? Indeed we can, but with an extra bit of care.
first, how much aggressive we can be? Looking to 11 (first special with 2 digits), it doesn't pay to just increment by 1, we can increment it by 2. Looking to 102 (special with 3 digits), we can increment it by 3. Is it natural to think we should use increments equal with the number of digits?
now the "extra bit of care" - whenever the "increment by the number of digits" causes a "carry", the naive increment breaks. Because the carry will add 1 to the sum of digits, so that we may need to subtract that one from something to keep the sum of digits well behaved.
one of the issues in the above is that we jumped quite happily at "first special with N digits", but the computer is not us to see it at a glance. Fortunately, the "first special with N digits" is easy to compute: it is 10^(N-1)+(N-1) - 10^(N-1) brings an 1 and the rest is zero, and N-1 brings the rest to make the sum of digits be the first divisible with N. Of course, this will break down if N > 10, but fortunately the problem is limited to 10^6 special numbers, which will require at most 7 digits (the millionth specual number is 6806035 - 7 digits);
so, we can detect the "first special number with N digits" and we know we should try with care to increment it by N. Can we look now better into that "extra care"?.
The code - twice as speedy as the previous one and totally "orthodox" in obtaining the data (via getters instead of direct access to data members).
Feel free to inline:
import java.util.ArrayList;
import java.util.Arrays;
public class Abacus {
static protected int pow10[]=
{1,10,100,1000, 10000, 100000, 1000000, 10000000, 100000000}
;
// the value stored for line[i] corresponds to digit[i]*pow10[i]
protected int lineValues[];
protected int sumDigits;
protected int representedNumber;
public Abacus() {
this.lineValues=new int[0];
this.sumDigits=0;
this.representedNumber=0;
}
public int getLineValue(int line) {
return this.lineValues[line];
}
public void clearUnitLine() {
this.sumDigits-=this.lineValues[0];
this.representedNumber-=this.lineValues[0];
this.lineValues[0]=0;
}
// This is how you operate the abacus in real life being asked
// to add a number of units to the line presenting powers of 10
public boolean addWithCarry(int units, int line) {
if(line-1==pow10.length) {
// don't have enough pow10 stored
pow10=Arrays.copyOf(pow10, pow10.length+1);
pow10[line]=pow10[line-1]*10;
}
if(line>=this.lineValues.length) {
// don't have enough lines for the carry
this.lineValues=Arrays.copyOf(this.lineValues, line+1);
}
int digitOnTheLine=this.lineValues[line]/pow10[line];
int carryOnTheNextLine=0;
while(digitOnTheLine+units>=10) {
carryOnTheNextLine++;
units-=10;
}
if(carryOnTheNextLine>0) {
// we have a carry, the sumDigits will be affected
// 1. the next two statememts are equiv with "set a value of zero on the line"
this.sumDigits-=digitOnTheLine;
this.representedNumber-=this.lineValues[line];
// this is the new value of the digit to set on the line
digitOnTheLine+=units;
// 3. set that value and keep all the values synchronized
this.sumDigits+=digitOnTheLine;
this.lineValues[line]=digitOnTheLine*pow10[line];
this.representedNumber+=this.lineValues[line];
// 4. as we had a carry, the next line will be affected as well.
this.addWithCarry(carryOnTheNextLine, line+1);
}
else { // we an simply add the provided value without carry
int delta=units*pow10[line];
this.lineValues[line]+=delta;
this.representedNumber+=delta;
this.sumDigits+=units;
}
return carryOnTheNextLine>0;
}
public int getSumDigits() {
return this.sumDigits;
}
public int getRepresentedNumber() {
return this.representedNumber;
}
public int getLinesCount() {
return this.lineValues.length;
}
static public ArrayList<Integer> specials(int N) {
ArrayList<Integer> ret=new ArrayList<>(N);
Abacus abacus=new Abacus();
ret.add(1);
abacus.addWithCarry(1, 0); // to have something to add to
int increment=abacus.getLinesCount();
while(ret.size()<N) {
boolean hadCarry=abacus.addWithCarry(increment, 0);
if(hadCarry) {
// need to resynch the sum for a perfect number
int newIncrement=abacus.getLinesCount();
abacus.clearUnitLine();
if(newIncrement!=increment) {
// we switched powers of 10
abacus.addWithCarry(newIncrement-1, 0);
increment=newIncrement;
}
else { // simple carry
int digitsSum=abacus.getSumDigits();
// how much we should add to the last digit to make the sumDigits
// divisible again with the increment?
int units=increment-digitsSum % increment;
if(units<increment) {
abacus.addWithCarry(units, 0);
}
}
}
ret.add(abacus.getRepresentedNumber());
}
return ret;
}
// to understand how the addWithCarry works, try the following code
static void add13To90() {
Abacus abacus; // starts with a represented number of 0
// line==1 means units of 10^1
abacus.addWithCary(9, 1); // so this should make the abacus store 90
System.out.println(abacus.getRepresentedNumber());
// line==0 means units of 10^0
abacus.addWithCarry(13, 0);
System.out.println(abacus.getRepresentedNumber()); // 103
}
static public void main(String[] args) {
int count=1000000;
long t1=System.nanoTime();
ArrayList<Integer> s1=Abacus.specials(count);
long t2=System.nanoTime();
System.out.println("t:"+(t2-t1));
}
}
Constructing the numbers from their digits is bound to be faster.
Remember the abacus? Ever used one?
import java.util.ArrayList;
public class Specials {
static public ArrayList<Integer> computeNSpecials(int N) {
ArrayList<Integer> specials = new ArrayList<>();
int abacus[] = new int[0]; // at index i we have the digit for 10^i
// This way, when we don't have enough specials,
// we simply reallocate the array and continue
while (specials.size() < N) {
// see if a carry operation is necessary
int currDigit = 0;
for (; currDigit < abacus.length && abacus[currDigit] == 9; currDigit++) {
abacus[currDigit] = 0; // a carry occurs when adding 1
}
if (currDigit == abacus.length) {
// a carry, but we don't have enough lines on the abacus
abacus = new int[abacus.length + 1];
abacus[currDigit] = 1; // we resolved the carry, all the digits below
// are 0
} else {
abacus[currDigit]++; // we resolve the carry (if there was one),
currDigit = 0; // now it's safe to continue incrementing at 10^0
}
// let's obtain the current number and the sum of the digits
int sumDigits = 0;
for (int i = 0; i<abacus.length; i++) {
sumDigits += abacus[i];
}
// is it special?
if (sumDigits % abacus.length == 0) {
// only now compute the number and collect it as special
int number = 0;
for (int i = abacus.length - 1; i >= 0; i--) {
number = 10 * number + abacus[i];
}
specials.add(number);
}
}
return specials;
}
static public void main(String[] args) {
ArrayList<Integer> specials=Specials.computeNSpecials(100);
for(int i=0; i<specials.size(); i++) {
System.out.println(specials.get(i));
}
}
}

How to send integers via popen?

I need to be able to send an integer from cocoa to an arduino.
It is easy to send characters, i.e. single digit integers, but I can't seem to find a way of sending two and three digit integers.
The purpose of this is to control the brightness of an LED continuously from 0 to 255.
So far, I can either turn it on and off using the following code:
int ledPin = 9; // LED connected to digital pin 9
int incomingByte = 0; // for incoming serial data
void setup() {
// initialize the digital pin as an output:
pinMode(ledPin, OUTPUT);
Serial.begin(9600);
}
void loop()
{
if (Serial.available() > 0) {
// read the incoming byte:
incomingByte = Serial.read();
if(incomingByte == 105){ //105 corresponds to i and is programmed in cocoa to turn the LED on
digitalWrite(ledPin, HIGH);
}
else if(incomingByte == 111){ //111 corresponds to o and is programmed in cocoa to turn the LED on
digitalWrite(ledPin, LOW);
}
}
}
However, I can't work out how to set values between 0 and 255. Instead of 'digitalWrite', I would use 'AnalogWrite', however, I don't know how to send the incomingByte to be a value between 0 and 255.
This is the cocoa code:
#import "MainController.h"
#implementation MainController
-(IBAction)ledOn:(id)sender{
popen("echo i > /dev/cu.usbmodem1411", "r");
}
-(IBAction)ledOff:(id)sender{
popen("echo o > /dev/cu.usbmodem1411", "r");
}
#end
Thanks.
You can treat the integers with more than one digits as cstring in your arduino code then convert them to integer via atoi().
The following code will capture the bytes from the serial buffer as cstring:
char buffer[MAX_BUFFER_SIZE]; // global c character array variable
boolean inputReady()
{
byte index = 0;
byte avail = Serial.available();
if(avail > 0)
{
while(index < avail)
{
byte val;
do
{
val = Serial.read();
}
while (val == -1); // if value is no longer -1, bytes are captured
buffer[index] = val;
index++;
}
buffer[index] = 0; //terminate the character array
return true;
}
return false;
}
Note: I prefer cstrings than the String class built in the arduino IDE since it uses more memory, but be careful using cstrings as they are prone to memory leaks if not managed well.
This is how your loop() block should look like:
void loop()
{
if(inputReady())
{
int pwmValue = atoi(buffer); // convert ascii buffer to integer.
if(pwmValue >= 0 && pwmValue <= 255) // filter values from 0-255
analogWrite(ledPin, pwmValue);
else
digitalWrite(ledPin, LOW); // turn off if pwmValue is more than 255
}
delay(100);
}
You can then send pwm values as string from your cocoa code via popen().
#import "MainController.h"
#implementation MainController
-(IBAction)ledPWMValue:(id)sender{
popen("echo 254 > /dev/cu.usbmodem1411", "r");
}
#end
I've tested this on my Arduino Uno and will probably work on other variants. I hope this helps you in your project and best of luck!

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