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Sgn value of a polynomial

Given a polynomial function f(x) = 1+ a1*x + a2*(x^2) + … an(x^n). Find the Sgn value of these function, when x is given and all the coefficients also.

If value of polynomial greater than 0
   Sign = 1
Else If value of polynomial less than 0
   Sign = -1
Else if value of polynomial is 0
   Sign = 0

Examples:

Input: poly[] = [1, 2, 3] 
       x = 1 
Output:  1 
Explanation: f(1) = 6 which is > 0 
hence 1.

Input: poly[] = [1, -1, 2, 3] 
       x = -2 
Output: -1 
Explanation: f(-2)=-11 which is less 
then 0, hence -1.



A naive approach will be to calculate every power of x and then add it to the answer by multiplying it with its coefficient. Calculating power of x will take O(n) time and for n coefficients. Hence taking the total complexity to O(n * n)

An efficient approach is to use Horner’s method. We evaluate value of polynomial using Horner’s method. Then we return value according to sing of the value.

Below is the implementation of the above approach

C++

// CPP program to find sign value of a 
// polynomial
#include <iostream>
using namespace std;
   
// returns value of poly[0]x(n-1) + poly[1]x(n-2) 
// + .. + poly[n-1]
int horner(int poly[], int n, int x)
{
    int result = poly[0];  // Initialize result
   
    // Evaluate value of polynomial
    // using Horner's method
    for (int i=1; i<n; i++)
        result = result*x + poly[i];
   
    return result;
}
  
// Returns sign value of polynomial
int findSign(int poly[], int n, int x)
{
   int result = horner(poly, n, x);
   if (result > 0)
      return 1;
   else if (result < 0)
      return -1; 
   return 0;
}
   
// Driver program to test above function.
int main()
{
    // Let us evaluate value of 2x3 - 6x2
    // + 2x - 1 for x = 3
    int poly[] = {2, -6, 2, -1};
    int x = 3;
    int n = sizeof(poly)/sizeof(poly[0]);
    cout << "Sign of polynomial is " 
         << findSign(poly, n, x);
    return 0;
}

Java

// Java program to find sign value of a 
// polynomial
  
class GFG
{
    // returns value of poly[0]x(n-1) + poly[1]x(n-2) 
    // + .. + poly[n-1]
    static int horner(int poly[], int n, int x)
    {
        // Initialize result
        int result = poly[0]; 
      
        // Evaluate value of polynomial
        // using Horner's method
        for (int i = 1; i < n; i++)
            result = result * x + poly[i];
      
        return result;
    }
      
    // Returns sign value of polynomial
    static int findSign(int poly[], int n, int x)
    {
        int result = horner(poly, n, x);
        if (result > 0)
            return 1;
        else if (result < 0)
            return -1
        return 0;
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        // Let us evaluate value of 2x3 - 6x2
        // + 2x - 1 for x = 3
        int poly[] = {2, -6, 2, -1};
        int x = 3;
        int n = poly.length;
        System.out.print("Sign of polynomial is "+
                              findSign(poly, n, x));
    }
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find 
# sign value of a 
# polynomial
  
# returns value of poly[0]x(n-1) + 
# poly[1]x(n-2) + .. + poly[n-1]
def horner( poly, n, x):
      
    # Initialize result
    result = poly[0];
      
    # Evaluate value of 
    # polynomial using 
    # Horner's method
    for i in range(1,n):
        result = (result * x + 
                     poly[i]);
    return result;
  
# Returns sign value 
# of polynomial
def findSign(poly, n, x):
    result = horner(poly, n, x);
    if (result > 0):
        return 1;
    elif (result < 0):
        return -1;
    return 0;
  
# Driver Code
  
# Let us evaluate value 
# of 2x3 - 6x2
# + 2x - 1 for x = 3
poly = [2, -6, 2, -1];
x = 3;
n = len(poly);
  
print("Sign of polynomial is "
         findSign(poly, n, x));
  
# This code is contributed by mits

C#

// C# program to find sign value of a 
// polynomial
using System;
  
class GFG {
      
    // returns value of poly[0]x(n-1)
    // + poly[1]x(n-2) + .. + poly[n-1]
    static int horner(int []poly, int n, int x)
    {
          
        // Initialize result
        int result = poly[0]; 
      
        // Evaluate value of polynomial
        // using Horner's method
        for (int i = 1; i < n; i++)
            result = result * x + poly[i];
      
        return result;
    }
      
    // Returns sign value of polynomial
    static int findSign(int []poly, int n, int x)
    {
          
        int result = horner(poly, n, x);
          
        if (result > 0)
            return 1;
        else if (result < 0)
            return -1; 
              
        return 0;
    }
      
    // Driver code
    public static void Main () 
    {
          
        // Let us evaluate value of 2x3 - 6x2
        // + 2x - 1 for x = 3
        int []poly = {2, -6, 2, -1};
        int x = 3;
        int n = poly.Length;
          
        Console.Write("Sign of polynomial is "
                      + findSign(poly, n, x));
    }
}
  
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find 
// sign value of a 
// polynomial
  
// returns value of poly[0]x(n-1) + 
// poly[1]x(n-2) + .. + poly[n-1]
function horner( $poly, $n, $x)
{
      
    // Initialize result
    $result = $poly[0]; 
  
    // Evaluate value of polynomial
    // using Horner's method
    for ($i = 1; $i < $n; $i++)
        $result = $result * $x + $poly[$i];
  
    return $result;
}
  
// Returns sign value 
// of polynomial
function findSign($poly, $n, $x)
{
    $result = horner($poly, $n, $x);
    if ($result > 0)
        return 1;
    else if ($result < 0)
        return -1; 
    return 0;
}
  
    // Driver Code
    // Let us evaluate value 
    // of 2x3 - 6x2
    // + 2x - 1 for x = 3
    $poly = array(2, -6, 2, -1);
    $x = 3;
    $n = count($poly);
    echo "Sign of polynomial is "
        , findSign($poly, $n, $x);
  
// This code is contributed by anuj_67.
?>


Output:

Sign of polynomial is 1

Time complexity : O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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