For two vectors , the Fixed Length Levenshtein distance between and is defined as the smallest number such that can be transformed to through deletions and insertions. \nFind the maximum size of the balls of radius in centered at .
[THOUGHT] To determine the maximum size of a ball of radius 1 in using the Fixed Length Levenshtein distance, we need to identify all vectors that can be transformed from a center vector through exactly one deletion and one insertion.
Given a vector , we need to count the number of ways to:
- Delete one element from , resulting in a vector of length 99.
- Insert one element into the resulting vector of length 99 to get back to a vector of length 100.
Let's break this down:
- Deletion: We can delete any one of the 100 elements of . This gives us 100 choices.
- Insertion: After deleting an element, we have a vector of length 99. We can insert either a 0 or a 1 at any one of the 100 positions (including the beginning and the end). This gives us 100 positions and 2 choices for the value to insert.
Thus, for each of the 100 deletions, we have 200 possible insertions (100 positions * 2 values). However, we must be careful not to double-count the original vector , as it should not be included in the ball of radius 1 around itself.
Therefore, the total number of vectors that can be transformed from through one deletion and one insertion is:
However, this count includes the original vector multiple times, since inserting the same value back into the same position after deletion results in . Each deletion-insertion pair that results in corresponds to:
- Deleting and inserting back at the -th position (100 such cases).
Thus, we need to subtract these 100 cases from our total count:
So, the maximum size of the ball of radius 1 in centered at is:
This calculation does not require a Python code execution because it is a straightforward combinatorial problem.
[/THOUGHT] \boxed{19900}