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astra-53 [7]
3 years ago
12

Which of the following describes a hash algorithms ability to avoid the same output from two guessed inputs?A. Collision avoidan

ceB. Collision resistanceC. Collision strengthD. Collision metric
Computers and Technology
1 answer:
Minchanka [31]3 years ago
7 0

Answer: B) Collision Resistance

Explanation: A hash algorithm is the function that works by changing the string of data into the string of numeric values. The numeric values that are obtained are of the fixed length.

The collision is the attack in found when the produced outcome is same for the  two different string inputs. The hash algorithm has the ability to resist the collision by the cryptography technique . Cryptographic hash function are is for maintaining the authenticity of the input function.

Thus, the correct option is option(B). The other option are collision avoidance refers as the escaping the collision ,collision strength refers to the strength ability of collision and collision metric refers to pattern of collision, which are incorrect.

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2 years ago
Two character strings may have many common substrings. Substrings are required to be contiguous in the original string. For exam
ryzh [129]

Answer:

Explanation:

The following function is written in Java. It takes two strings as parameters and calculates the longest substring between both of them and returns that substring length.

import java.util.*;

class Test

{

   public static void main(String[] args)

   {

       String X = "photograph";

       String Y = "tomography";

       System.out.println(X + '\n' + Y);

       System.out.println("Longest common substring length: " + longestSub(X, Y));

   }

   static int longestSub(String X, String Y)

   {

       char[] word1 = X.toCharArray();

       char[] word2 = Y.toCharArray();

       

       int length1 = X.length();

       int length2 = Y.length();

       int substringArray[][] = new int[length1 + 1][length2 + 1];

       int longestSubstringLength = 0;

       

       for (int i = 0; i <= length1; i++)

       {

           for (int j = 0; j <= length2; j++)

           {

               if (i == 0 || j == 0)

                   substringArray[i][j] = 0;

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               {

                   substringArray[i][j]

                           = substringArray[i - 1][j - 1] + 1;

                   longestSubstringLength = Integer.max(longestSubstringLength,

                           substringArray[i][j]);

               }

               else

                   substringArray[i][j] = 0;

           }

       }

       return longestSubstringLength;

   }

}

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3 years ago
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