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Game Theory in Balanced Ternary Numeral System | (Moving 3k steps at a time)

  • Difficulty Level : Medium
  • Last Updated : 07 Jun, 2021

Just like base 2 Binary numeral system having 0s and 1s as digits, Ternary(Trinary) Numeral System is a base 3 number system having 0s, 1s and -1 as digits. 
It’s better to use alphabet ‘Z’ in place of -1, since while denoting full ternary number -1 looks odd in between 1s and 0s.

Conversion of decimal into Balanced Ternary: 
As in binary conversion, first represent the decimal number into the normal ternary system having 0, 1, 2 as reminders. 
Now Iterating from the lowest digit safely skip any 0s and 1s, however turn 2 into Z and add 1 to the next digit. Turn 3 into 0 on the same terms( such digits are not present in the number initially but they can be encountered after increasing some 2s. )

 Examples: 

Decimal: 128 
Ternary: 11202 
Balanced Ternary: 1ZZZ1Z
Decimal: 1000 
Ternary: 1102101 
Balanced Ternary: 111Z101 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Numbers are in range of pow(3, 32)
int arr[32];
 
// Conversion of ternary into balanced ternary as
// start iterating from Least Significant Bit (i.e 0th),
// if encountered 0 or 1, safely skip and pass carry 0
// further 2, replace it to -1 and pass carry 1 further
// 3, replace it to 0 and pass carry 1 further
void balTernary(int ter)
{
    int carry = 0, base = 10;
    int i = 32;
    while (ter > 0) {
        int rem = ter % base;
        rem = rem + carry;
        if (rem == 0) {
            arr[i--] = 0;
            carry = 0;
        }
        else if (rem == 1) {
            arr[i--] = 1;
            carry = 0;
        }
        else if (rem == 2) {
            arr[i--] = -1;
            carry = 1;
        }
        else if (rem == 3) {
            arr[i--] = 0;
            carry = 1;
        }
        ter = ter / base;
    }
    if (carry == 1)
        arr[i] = 1;
}
 
// Similar to binary conversion
int ternary(int number)
{
    int ans = 0, rem = 1, base = 1;
    while (number > 0) {
        rem = number % 3;
        ans = ans + rem * base;
        number /= 3;
        base = base * 10;
    }
    return ans;
}
 
// Driver code
int main()
{
    int number = 3056;
 
    int ter = ternary(number);
    memset(arr, 0, sizeof(arr));
    balTernary(ter);
 
    int i = 0;
 
    // Moving on to first occupied bit
    while (arr[i] == 0) {
        i++;
    }
 
    // Printing
    for (int j = i; j <= 32; j++) {
 
        // Print 'Z' in place of -1
        if (arr[j] == -1)
            cout << 'Z';
        else
            cout << arr[j];
    }
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Numbers are in range of pow(3, 32)
static int []arr = new int[33];
 
// Conversion of ternary into balanced ternary as
// start iterating from Least Significant Bit (i.e 0th),
// if encountered 0 or 1, safely skip and pass carry 0
// further 2, replace it to -1 and pass carry 1 further
// 3, replace it to 0 and pass carry 1 further
static void balTernary(int ter)
{
    int carry = 0, base = 10;
    int i = 32;
    while (ter > 0)
    {
        int rem = ter % base;
        rem = rem + carry;
        if (rem == 0)
        {
            arr[i--] = 0;
            carry = 0;
        }
        else if (rem == 1)
        {
            arr[i--] = 1;
            carry = 0;
        }
        else if (rem == 2)
        {
            arr[i--] = -1;
            carry = 1;
        }
        else if (rem == 3)
        {
            arr[i--] = 0;
            carry = 1;
        }
        ter = (int)(ter / base);
    }
    if (carry == 1)
        arr[i] = 1;
}
 
// Similar to binary conversion
static int ternary(int number)
{
    int ans = 0, rem = 1, base = 1;
    while (number > 0)
    {
        rem = number % 3;
        ans = ans + rem * base;
        number = (int)(number/3);
        base = base * 10;
    }
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    int number = 3056;
 
    int ter = ternary(number);
    Arrays.fill(arr,0);
    balTernary(ter);
 
    int i = 0;
 
    // Moving on to first occupied bit
    while (arr[i] == 0)
    {
        i++;
    }
 
    // Printing
    for (int j = i; j <= 32; j++)
    {
 
        // Print 'Z' in place of -1
        if (arr[j] == -1)
            System.out.print('Z');
        else
        System.out.print(arr[j]);
    }
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Python3




# Python3 implementation of the approach
 
# Numbers are in range of pow(3, 32)
arr = [0] * 32
 
# Conversion of ternary into balanced ternary as
# start iterating from Least Significant Bit (i.e 0th),
# if encountered 0 or 1, safely skip and pass carry 0
# further 2, replace it to -1 and pass carry 1 further
# 3, replace it to 0 and pass carry 1 further
def balTernary(ter):
  
    carry, base, i = 0, 10, 31
    while ter > 0:
        rem = (ter % base) + carry
         
        if rem == 0
            arr[i] = 0
            carry, i = 0, i-1
          
        elif rem == 1
            arr[i] = 1
            carry, i = 0, i-1
          
        elif rem == 2
            arr[i] = -1
            carry, i = 1, i-1
          
        elif rem == 3
            arr[i] = 0
            carry, i = 1, i-1
          
        ter = ter // base
      
    if carry == 1:
        arr[i] = 1
  
# Similar to binary conversion
def ternary(number):
  
    ans, rem, base = 0, 1, 1
    while number > 0:
        rem = number % 3
        ans = ans + rem * base
        number //= 3
        base = base * 10
      
    return ans
  
# Driver code
if __name__ == "__main__":
  
    number = 3056
    ter = ternary(number)
    balTernary(ter)
 
    i = 0
 
    # Moving on to first occupied bit
    while arr[i] == 0
        i += 1
      
    # Printing
    for j in range(i, 32): 
 
        # Print 'Z' in place of -1
        if arr[j] == -1:
            print('Z', end = "")
        else:
            print(arr[j], end = "")
      
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
     
class GFG
{
 
// Numbers are in range of pow(3, 32)
static int []arr = new int[33];
 
// Conversion of ternary into balanced ternary as
// start iterating from Least Significant Bit (i.e 0th),
// if encountered 0 or 1, safely skip and pass carry 0
// further 2, replace it to -1 and pass carry 1 further
// 3, replace it to 0 and pass carry 1 further
static void balTernary(int ter)
{
    int carry = 0, b = 10;
    int i = 32;
    while (ter > 0)
    {
        int rem = ter % b;
        rem = rem + carry;
        if (rem == 0)
        {
            arr[i--] = 0;
            carry = 0;
        }
        else if (rem == 1)
        {
            arr[i--] = 1;
            carry = 0;
        }
        else if (rem == 2)
        {
            arr[i--] = -1;
            carry = 1;
        }
        else if (rem == 3)
        {
            arr[i--] = 0;
            carry = 1;
        }
        ter = (int)(ter / b);
    }
    if (carry == 1)
        arr[i] = 1;
}
 
// Similar to binary conversion
static int ternary(int number)
{
    int ans = 0, rem = 1, b = 1;
    while (number > 0)
    {
        rem = number % 3;
        ans = ans + rem * b;
        number = (int)(number / 3);
        b = b * 10;
    }
    return ans;
}
 
// Driver code
public static void Main(String []args)
{
    int number = 3056;
 
    int ter = ternary(number);
    balTernary(ter);
 
    int i = 0;
 
    // Moving on to first occupied bit
    while (arr[i] == 0)
    {
        i++;
    }
 
    // Printing
    for (int j = i; j <= 32; j++)
    {
 
        // Print 'Z' in place of -1
        if (arr[j] == -1)
            Console.Write('Z');
        else
            Console.Write(arr[j]);
    }
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Numbers are in range of pow(3, 32)
let arr = new Array(33);
 
// Conversion of ternary into balanced ternary as
// start iterating from Least Significant Bit (i.e 0th),
// if encountered 0 or 1, safely skip and pass carry 0
// further 2, replace it to -1 and pass carry 1 further
// 3, replace it to 0 and pass carry 1 further
function balTernary(ter)
{
    let carry = 0, base = 10;
    let i = 32;
    while (ter > 0)
    {
        let rem = ter % base;
        rem = rem + carry;
        if (rem == 0)
        {
            arr[i--] = 0;
            carry = 0;
        }
        else if (rem == 1)
        {
            arr[i--] = 1;
            carry = 0;
        }
        else if (rem == 2)
        {
            arr[i--] = -1;
            carry = 1;
        }
        else if (rem == 3)
        {
            arr[i--] = 0;
            carry = 1;
        }
        ter = Math.floor(ter / base);
    }
    if (carry == 1)
        arr[i] = 1;
}
 
// Similar to binary conversion   
function ternary(number)
{
    let ans = 0, rem = 1, base = 1;
    while (number > 0)
    {
        rem = number % 3;
        ans = ans + rem * base;
        number = Math.floor(number/3);
        base = base * 10;
    }
    return ans;
}
 
// Driver code
let number = 3056;
   
    let ter = ternary(number);
    for(let i=0;i<arr.length;i++)
    {
        arr[i]=0;
    }
    balTernary(ter);
   
    let i = 0;
   
    // Moving on to first occupied bit
    while (arr[i] == 0)
    {
        i++;
    }
   
    // Printing
    for (let j = i; j <= 32; j++)
    {
   
        // Print 'Z' in place of -1
        if (arr[j] == -1)
            document.write('Z');
        else
            document.write(arr[j]);
    }
 
// This code is contributed by patel2127
</script>
Output: 



111ZZ1ZZ

 

Recovering original decimal number from a balanced ternary number:- 
Procedure:- Similarly as it’s done in binary to decimal conversion 
Example:- 111ZZ1ZZ 
3^7*(1) + 3^6*(1) + 3^5*(1) + 3^4*(-1) + 3^3*(-1) + 3^2*(1) + 3^1*(-1) + 3^0*(-1)
= 2187 + 729 + 243 - 81 - 27 + 9 - 3 - 1 = 3056

Game Rules: 
There are two robots allowed to move in steps on x-axis starting from 0. 
They can make several steps starting from 0 but there are some limitations on their movement. 
In k_t_h    step robot will move exact 3^k    units of distance. 
In each step robot must choose one of the two directions left (x- coordinate decreases) or right (x-coordinate increases), in a particular step only one robot will move and another will wait. 
It is not allowed to skip any step.

Statement: 
Given two integers x1 and x2. Robot 1 and 2 are separately required to cover their respective distances x1 and x2. Is it possible?? 
If it is possible you won otherwise you lose.

Approach: 
There is only one balanced ternary representation of each Decimal number (distance here), this means there is only one way to cover a particular distance satisfying above rules. 
So, if it is possible to cover distances x1 and x2 such that when one robot moves other remains still and both can’t remain still at the same time then it’s a victory.

Logic: 
First represent x1 and x2 as balanced ternary number using above procedure. 
Iterate from LSB check:- 
At a time(step) only one value should be 1 or Z. 
Both can’t be 0 at the same time(step).
If rule breaks at any step it’s your lose otherwise you won.

Example:  

Input: x1 = 6890, x2 = 18252 
Output: 
Balanced ternary representation of x1 = 01001101ZZ 
Balanced ternary representation of x2 = 10Z1001000 
Victory

Input: x1 = 18, x2 = 45 
Output: 
Balanced ternary representation of x1 = 01Z00 
Balanced ternary representation of x2 = 1ZZ00 
Defeat 
 

Iterate bitwise over both the arrays and break wherever rule breaks. 
To do first make length of both arrays equal by adding 0s at beginning of the shortest one, such that length becomes same.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true
// if the game cannot be won
bool isDefeat(string s1, string s2, int n)
{
    for (int i = 0; i < n; i++) {
        if ((s1[i] == '0' && s2[i] == '1')
            || (s1[i] == '1' && s2[i] == '0'))
            continue;
        else if ((s1[i] == '0' && s2[i] == 'Z')
                 || (s1[i] == 'Z' && s2[i] == '0'))
            continue;
        else {
            return true;
        }
    }
    return false;
}
 
// Driver code
int main()
{
    string s1 = { "01001101ZZ" };
    string s2 = { "10Z1001000" };
 
    // Common length
    int n = 10;
 
    if (isDefeat(s1, s2, n))
        cout << "Defeat";
    else
        cout << "Victory";
 
    return 0;
}

Java




// Java implementation of the approach
class GfG
{
     
// Function that returns true
// if the game cannot be won
static boolean isDefeat(String s1, String s2, int n)
{
    for (int i = 0; i < n; i++)
    {
        if ((s1.charAt(i) == '0' && s2.charAt(i) == '1')
            || (s1.charAt(i) == '1' && s2.charAt(i) == '0'))
            continue;
        else if ((s1.charAt(i) == '0' && s2.charAt(i) == 'Z')
                || (s1.charAt(i) == 'Z' && s2.charAt(i) == '0'))
            continue;
        else
        {
            return true;
        }
    }
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    String s1 = ("01001101ZZ" );
    String s2 = ("10Z1001000" );
 
    // Common length
    int n = 10;
 
    if (isDefeat(s1, s2, n))
        System.out.println("Defeat");
    else
        System.out.println("Victory");
 
}
}
 
// This code is contributed by Code_Mech

Python3




# Python3 implementation of the approach
 
# Function that returns true
# if the game cannot be won
def isDefeat(s1, s2, n):
 
    for i in range(n):
        if ((s1[i] == '0' and s2[i] == '1') or
            (s1[i] == '1' and s2[i] == '0')):
            continue
        elif ((s1[i] == '0' and s2[i] == 'Z') or
              (s1[i] == 'Z' and s2[i] == '0')):
            continue
        else:
            return True
         
    return False
 
# Driver code
s1 = "01001101ZZ"
s2 = "10Z1001000"
 
# Common length
n = 10
 
if (isDefeat(s1, s2, n)):
    print("Defeat")
else:
    print("Victory")
 
# This code is contributed by mohit kumar

C#




// C# implementation of the approach
using System;
 
class GfG
{
     
// Function that returns true
// if the game cannot be won
static bool isDefeat(string s1, string s2, int n)
{
    for (int i = 0; i < n; i++)
    {
        if ((s1[i] == '0' && s2[i] == '1')
            || (s1[i] == '1' && s2[i] == '0'))
            continue;
        else if ((s1[i] == '0' && s2[i] == 'Z')
                || (s1[i] == 'Z' && s2[i]== '0'))
            continue;
        else
        {
            return true;
        }
    }
    return false;
}
 
// Driver code
public static void Main()
{
    string s1 = ("01001101ZZ" );
    string s2 = ("10Z1001000" );
 
    // Common length
    int n = 10;
 
    if (isDefeat(s1, s2, n))
        Console.WriteLine("Defeat");
    else
        Console.WriteLine("Victory");
 
}
}
 
// This code is contributed by Code_Mech

PHP




<?php
// PHP implementation of the approach
// Function that returns true
// if the game cannot be won
function isDefeat($s1, $s2, $n)
{
    for ($i = 0; $i < $n; $i++)
    {
        if( ($s1[$i] == '0' && $s2[$i] == '1')
            || ($s1[$i] == '1' && $s2[$i] == '0'))
            continue;
        else if (($s1[$i] == '0' && $s2[$i] == 'Z')
                || ($s1[$i] == 'Z' && $s2[$i] == '0'))
            continue;
        else
        {
            return true;
        }
    }
    return false;
}
 
// Driver code
    $s1 = ("01001101ZZ" );
    $s2 = ("10Z1001000" );
 
    // Common length
    $n = 10;
 
    if (isDefeat($s1, $s2, $n))
        echo("Defeat");
    else
        echo("Victory");
 
// This code is contributed by Code_Mech

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function that returns true
// if the game cannot be won
function isDefeat(s1, s2, n)
{
    for(let i = 0; i < n; i++)
    {
        if ((s1[i] == '0' && s2[i] == '1') ||
            (s1[i] == '1' && s2[i] == '0'))
            continue;
        else if ((s1[i] == '0' && s2[i] == 'Z') ||
                 (s1[i] == 'Z' && s2[i]== '0'))
            continue;
        else
        {
            return true;
        }
    }
    return false;
}
 
// Driver Code
let s1 = ("01001101ZZ");
let s2 = ("10Z1001000");
 
// Common length
let n = 10;
 
if (isDefeat(s1, s2, n))
    document.write("Defeat");
else
    document.write("Victory");
     
// This code is contributed by code_hunt
 
</script>
Output: 
Victory

 




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