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## Data Structure Questions and Answers-Wagner-Fischer Algorithm

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Question 1 |

Wagner-Fischer is a ..... algorithm.

Brute force | |

Greedy | |

Dynamic programming | |

Recursive |

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Question 1 Explanation:

Wagner-Fischer belongs to the dynamic programming type of algorithms.

Question 2 |

Wagner-Fischer algorithm is used to find .....

Longest common subsequence | |

Longest increasing subsequence | |

Edit distance between two strings | |

All of the mentioned |

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Question 2 Explanation:

Wagner-Fischer algorithm is used to find the edit distance between two strings.

Question 3 |

What is the edit distance between the strings "abcd" and "acbd" when the allowed operations are insertion, deletion and substitution?

1 | |

2 | |

3 | |

4 |

**Reasoning Questions answers**

Question 3 Explanation:

The string "abcd" can be changed to "acbd" by substituting "b" with "c" and "c" with "b". Thus, the edit distance is 2.

Question 4 |

For which of the following pairs of strings is the edit distance maximum?

sunday & monday | |

monday & tuesday | |

tuesday & wednesday | |

wednesday & thursday |

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Question 4 Explanation:

The edit distances are 2, 4, 4 and 5 respectively. Hence, the maximum edit distance is between the strings wednesday and thursday.

Question 5 |

Consider the following implementation of the Wagner-Fischer algorithm:

int get....min(int a, int b) { if(a < b) return a; return b; } int edit....distance(char *s1, char *s2) { int len1, len2, i, j, min; len1 = strlen(s1); len2 = strlen(s2); int arr[len1 + 1][len2 + 1]; for(i = 0;i <= len1; i++) arr[i][0] = i; for(i = 0; i <= len2; i++) arr[0][i] = i; for(i = 1; i <= len1; i++) { for(j = 1; j <= len2; j++) { min = get....min(arr[i-1][j], arr[i][j-1]) + 1; if(s1[i - 1] == s2[j - 1]) { if(arr[i-1][j-1] < min) .....; } else { if(arr[i-1][j-1] + 1 < min) min = arr[i-1][j-1] + 1; } arr[i][j] = min; } } return arr[len1][len2]; }

Which of the following lines should be inserted to complete the above code?

arr[i][j] = min | |

min = arr[i-1][j-1] - 1; | |

min = arr[i-1][j-1]. | |

min = arr[i-1][j-1] + 1; |

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Question 5 Explanation:

The line min = arr[i-1][j-1] completes the above code.

There are 5 questions to complete.