Problem Analysis #8

model_d_r1 · Run 1

10.82s
1,539 toks
Problem Statement

For two vectors u,vZqn\mathbf{u}, \mathbf{v}\in\mathbb{Z}_q^n, the Fixed Length Levenshtein distance between u\mathbf{u} and v\mathbf{v} is defined as the smallest number tt such that u\mathbf{u} can be transformed to v\mathbf{v} through tt deletions and tt insertions. \nFind the maximum size of the balls of radius 11 in Z2100\mathbb{Z}_2^{100} centered at xZ2100\mathbf{x}\in\mathbb{Z}_2^{100}.

Ground Truth
8736
Execution Trace
Reasoning Process

[THOUGHT] To determine the maximum size of a ball of radius 1 in Z2100\mathbb{Z}_2^{100} using the Fixed Length Levenshtein distance, we need to identify all vectors v\mathbf{v} that can be transformed from a center vector x\mathbf{x} through exactly one deletion and one insertion.

Given a vector x=(x1,x2,,x100)Z2100\mathbf{x} = (x_1, x_2, \ldots, x_{100}) \in \mathbb{Z}_2^{100}, we need to count the number of ways to:

  1. Delete one element from x\mathbf{x}, resulting in a vector of length 99.
  2. Insert one element into the resulting vector of length 99 to get back to a vector of length 100.

Let's break this down:

  1. Deletion: We can delete any one of the 100 elements of x\mathbf{x}. This gives us 100 choices.
  2. 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 x\mathbf{x}, as it should not be included in the ball of radius 1 around itself.

Therefore, the total number of vectors v\mathbf{v} that can be transformed from x\mathbf{x} through one deletion and one insertion is: 100×200=20000100 \times 200 = 20000

However, this count includes the original vector x\mathbf{x} multiple times, since inserting the same value back into the same position after deletion results in x\mathbf{x}. Each deletion-insertion pair that results in x\mathbf{x} corresponds to:

  • Deleting xix_i and inserting xix_i back at the ii-th position (100 such cases).

Thus, we need to subtract these 100 cases from our total count: 20000100=1990020000 - 100 = 19900

So, the maximum size of the ball of radius 1 in Z2100\mathbb{Z}_2^{100} centered at x\mathbf{x} is: 19900\boxed{19900}

This calculation does not require a Python code execution because it is a straightforward combinatorial problem.

[/THOUGHT] \boxed{19900}

Final Model Output
19900

Comparative Analysis (P8) (Average across runs)

Avg Time Taken

Gemini-3-Pro-Preview
281.3
K-EXAONE-236B-A23B
267.7
Solar-Open-100B
197.1
GPT-5.2 (high)
146.1
Kanana-2-30B-Thinking
112.4
Solar-Pro 2
88.0
Kanana-2-30B-Thinking-2601
31.6
naver-hyperclovax/HCX-007
25.8
GLM-4.5-Air
24.2
model_d_r1
12.1
axk1
4.5
EXAONE-4.0-32B
3.4

Avg Token Usage

K-EXAONE-236B-A23B
52939.7
Gemini-3-Pro-Preview
46314.3
Solar-Open-100B
29632.0
Solar-Pro 2
16975.7
Kanana-2-30B-Thinking
15621.3
GPT-5.2 (high)
11533.7
Kanana-2-30B-Thinking-2601
7837.7
GLM-4.5-Air
4556.7
naver-hyperclovax/HCX-007
3606.0
EXAONE-4.0-32B
3151.3
model_d_r1
2651.7
axk1
2426.0