| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | True. A graph's score refers to the degree sequence of the graph, which is a list of the degrees of its vertices, usually written in non-increasing order. A degree of a vertex is the number of edges that are incident to the vertex. To determine if there exists a graph with a given degree sequence, we can use the Havel-Hakimi theorem. According to the theorem, a sequence of non-negative integers is graphical (i.e., can be realized as the degree sequence of a simple graph) if and only if the sequence obtained by deleting the first number and subtracting 1 from the next k numbers (where k is the first number) is graphical. Let's apply this theorem to the given sequence (1, 1, 2, 2, 3, 3, 4, 4): 1. Start with (1, 1, 2, 2, 3, 3, 4, 4). 2. Remove the largest number (4) and subtract 1 from the next 4 numbers: (0, 0, 1, 1, 2, 2, 3). 3. Remove the largest number (3) and subtract 1 from the next 3 numbers: (0, 0, 0, 0, 1, 1). 4. Remove the largest number (1) and subtract 1 from the next number: (0, 0, 0, 0, 0). 5. We now have a sequence of all zeros, which indicates that the original sequence is graphical. Thus, there exists a graph with the degree sequence (1, 1, 2, 2, 3, 3, 4, 4). Here is an example of such a graph: ``` 1 | 2 -- 3 | | 4 -- 5 / \ 6 7 ``` In this graph, vertices 6 and 7 have degree 1, vertices 1, 4, and 5 have degree 2, vertex 2 has degree 3, and vertex 3 has degree 4. |
True |
| deepseekr1 | The given problem asks whether there exists a graph with the degree sequence (1, 1, 2, 2, 3, 3, 4, 4). To determine this, we apply the Erdős–Gallai theorem and verify the conditions step-by-step. 1. **Sort the sequence in non-increasing order**: The sequence is already sorted as (4, 4, 3, 3, 2, 2, 1, 1). 2. **Check the sum of degrees**: Sum = \(4 + 4 + 3 + 3 + 2 + 2 + 1 + 1 = 20\), which is even. This satisfies the necessary condition. 3. **Apply the Erdős–Gallai theorem**: For each \(k\) from 1 to 7 (number of vertices - 1), verify if the sum of the first \(k\) degrees \(\leq k(k-1) + \sum_{i=k+1}^n \min(d_i, k)\): - **\(k=1\)**: Sum of first 1 term = 4. RHS = \(0 + \sum_{i=2}^8 \min(d_i, 1) = 7\). \(4 \leq 7\) → True. - **\(k=2\)**: Sum of first 2 terms = 8. RHS = \(2 + \sum_{i=3}^8 \min(d_i, 2) = 12\). \(8 \leq 12\) → True. - **\(k=3\)**: Sum of first 3 terms = 11. RHS = \(6 + \sum_{i=4}^8 \min(d_i, 3) = 15\). \(11 \leq 15\) → True. - **\(k=4\)**: Sum of first 4 terms = 14. RHS = \(12 + \sum_{i=5}^8 \min(d_i, 4) = 18\). \(14 \leq 18\) → True. - **\(k=5\)**: Sum of first 5 terms = 16. RHS = \(20 + \sum_{i=6}^8 \min(d_i, 5) = 24\). \(16 \leq 24\) → True. - **\(k=6\)**: Sum of first 6 terms = 18. RHS = \(30 + \sum_{i=7}^8 \min(d_i, 6) = 32\). \(18 \leq 32\) → True. - **\(k=7\)**: Sum of first 7 terms = 19. RHS = \(42 + \min(d_8, 7) = 43\). \(19 \leq 43\) → True. All conditions are satisfied, confirming the sequence is graphical. Further validation via the Havel–Hakimi algorithm constructs such a graph successfully. **Answer:** \boxed{True} |
True |
| gpto1 | To determine whether there exists a simple graph with the given degree sequence (1, 1, 2, 2, 3, 3, 4, 4), we can use the **Havel-Hakimi algorithm**, which is a process used to determine whether a degree sequence can be graphical (i.e., realized by a simple graph with no loops or multiple edges). **Step 1: Sort the Sequence in Non-Increasing Order** First, sort the degrees in non-increasing order: \[ (4, 4, 3, 3, 2, 2, 1, 1) \] **Step 2: Apply the Havel-Hakimi Algorithm** We will repeatedly apply the following steps: 1. Remove the largest degree \(d\) from the sequence. 2. Subtract 1 from each of the next \(d\) degrees in the sequence. 3. If at any point we get a negative degree or the degree sequence cannot be satisfied, the sequence is not graphical. **Iteration 1:** - **Remove \(d = 4\):** \[ \text{Sequence after removal: } (4, 3, 3, 2, 2, 1, 1) \] - **Subtract 1 from the next 4 degrees:** \[ (4-1, 3-1, 3-1, 2-1) = (3, 2, 2, 1) \] - **Updated sequence:** \[ (3, 2, 2, 1, 2, 1, 1) \] - **Re-sort:** \[ (3, 2, 2, 2, 1, 1, 1) \] **Iteration 2:** - **Remove \(d = 3\):** \[ \text{Sequence after removal: } (2, 2, 2, 1, 1, 1) \] - **Subtract 1 from the next 3 degrees:** \[ (2-1, 2-1, 2-1) = (1, 1, 1) \] - **Updated sequence:** \[ (1, 1, 1, 1, 1, 1) \] **Iteration 3:** - **Remove \(d = 1\):** \[ \text{Sequence after removal: } (1, 1, 1, 1, 1) \] - **Subtract 1 from the next 1 degree:** \[ (1-1) = 0 \] - **Updated sequence:** \[ (0, 1, 1, 1, 1) \] - **Re-sort:** \[ (1, 1, 1, 1, 0) \] **Iterations 4-6:** Repeat the process for the remaining degrees: - **Remove \(d = 1\) and subtract from next degree:** \[ (1, 1, 1, 0) \] - **Subtract 1:** \[ (0, 1, 1, 0) \] - **Updated and re-sorted sequence:** \[ (1, 1, 0, 0) \] - Continue until all degrees are reduced to zero. **Conclusion:** Since we successfully reduced all degrees to zero without encountering a negative degree or an impossible situation, the original degree sequence is **graphical**. **Answer: True. There exists a graph with the given degree sequence.** |
True |
| GPT-4o | To determine if there exists a graph with the given degree sequence (1, 1, 2, 2, 3, 3, 4, 4), we can use the Havel-Hakimi algorithm or the Erdős–Gallai theorem. Here, I will use the Havel-Hakimi algorithm to find the solution. The Havel-Hakimi algorithm proceeds as follows: 1. Sort the degree sequence in non-increasing order. 2. Remove the first element \( d \) (the largest degree) from the sequence. 3. Reduce the next \( d \) elements by 1. 4. Repeat the process with the new sequence. Let's apply the algorithm step by step to the sequence (1, 1, 2, 2, 3, 3, 4, 4): 1. Initial sequence: \( (4, 4, 3, 3, 2, 2, 1, 1) \) 2. Remove the first element (4), which leaves \( (4, 3, 3, 2, 2, 1, 1) \). 3. Reduce the next 4 elements by 1: \( (3, 2, 2, 1, 2, 1, 1) \). New sequence: \( (3, 2, 2, 2, 1, 1, 1) \) 1. Sort the sequence: \( (3, 2, 2, 2, 1, 1, 1) \) 2. Remove the first element (3), which leaves \( (2, 2, 2, 1, 1, 1) \). 3. Reduce the next 3 elements by 1: \( (1, 1, 1, 1, 1) \). New sequence: \( (2, 1, 1, 1, 1) \) 1. Sort the sequence: \( (2, 1, 1, 1, 1) \) 2. Remove the first element (2), which leaves \( (1, 1, 1, 1) \). 3. Reduce the next 2 elements by 1: \( (0, 0, 1, 1) \). New sequence: \( (1, 1, 0, 0) \) 1. Sort the sequence: \( (1, 1, 0, 0) \) 2. Remove the first element (1), which leaves \( (1, 0, 0) \). 3. Reduce the next 1 element by 1: \( (0, 0, 0) \). New sequence: \( (0, 0, 0) \) Since we are left with all zeros, the original sequence satisfies the Havel-Hakimi algorithm. Therefore, it is possible to construct a graph with the degree sequence (1, 1, 2, 2, 3, 3, 4, 4). Thus, the statement is **true**. |
True |
| llama3.1-70B-Instruct | True. |
True |
| llama3.1-405B-Instruct | The answer is false. To see why, let's assume that such a graph exists. Since the score of a vertex is the number of edges incident to it, we can start by drawing the vertices with their corresponding scores: Vertex 1: 1 edge Vertex 2: 1 edge Vertex 3: 2 edges Vertex 4: 2 edges Vertex 5: 3 edges Vertex 6: 3 edges Vertex 7: 4 edges Vertex 8: 4 edges Now, let's try to connect the vertices to satisfy the given scores. We can start by connecting Vertex 1 to Vertex 2, since they both have a score of 1. This creates one edge. Next, we need to connect Vertex 3 to two other vertices, since it has a score of 2. We can connect it to Vertex 4 and Vertex 5. This creates two more edges. Now, Vertex 4 has a score of 2, so it needs to be connected to one more vertex. We can connect it to Vertex 6. This creates another edge. Vertex 5 has a score of 3, so it needs to be connected to two more vertices. We can connect it to Vertex 7 and Vertex 8. This creates two more edges. At this point, we have used up all the edges, but Vertex 6 still needs to be connected to two more vertices to satisfy its score of 3. However, there are no more edges available to connect it to other vertices. Therefore, we have reached a contradiction, and our assumption that such a graph exists must be false. Hence, the answer is false. |
True |
| DeepSeek-Coder-V2 | To determine whether there exists a graph with the score (1, 1, 2, 2, 3, 3, 4, 4), we need to understand what the "score" of a graph refers to. Typically, the score of a graph refers to the degree sequence of the graph, where the degree of a vertex is the number of edges connected to that vertex. Given the score (1, 1, 2, 2, 3, 3, 4, 4), we are looking for a graph with 8 vertices, where the degrees of the vertices are 1, 1, 2, 2, 3, 3, 4, and 4 respectively. To determine if such a graph exists, we can use the Havel-Hakimi algorithm, which is a procedure to determine if a given sequence of integers is graphical (i.e., can represent the degree sequence of some graph). Here is the Havel-Hakimi algorithm applied to the sequence (1, 1, 2, 2, 3, 3, 4, 4): 1. Sort the sequence in non-increasing order: (4, 4, 3, 3, 2, 2, 1, 1). 2. Remove the first number (4) from the sequence. 3. Subtract 1 from the next 4 numbers in the sequence: (4, 3, 3, 2, 2, 1, 1). 4. Sort the resulting sequence in non-increasing order: (4, 3, 3, 2, 2, 1, 1). 5. Repeat the process: - Remove the first number (4) from the sequence. - Subtract 1 from the next 4 numbers in the sequence: (3, 2, 2, 1, 1). - Sort the resulting sequence in non-increasing order: (3, 2, 2, 1, 1). 6. Continue the process: - Remove the first number (3) from the sequence. - Subtract 1 from the next 3 numbers in the sequence: (2, 1, 1). - Sort the resulting sequence in non-increasing order: (2, 1, 1). 7. Continue the process: - Remove the first number (2) from the sequence. - Subtract 1 from the next 2 numbers in the sequence: (1). - Sort the resulting sequence in non-increasing order: (1). 8. Continue the process: - Remove the first number (1) from the sequence. - Subtract 1 from the next 1 number in the sequence: (). - The sequence is now empty. Since the sequence has been reduced to an empty sequence, the original sequence (1, 1, 2, 2, 3, 3, 4, 4) is graphical, and there exists a graph with this degree sequence. Therefore, the answer is true: there exists a graph with the score (1, 1, 2, 2, 3, 3, 4, 4). |
True |
| Qwen2-72B-Instruct | True. A graph's score refers to the degree sequence of the graph, which is a list of the degrees of its vertices, usually written in non-increasing order. A degree of a vertex is the number of edges that are incident to the vertex. To determine if there exists a graph with a given degree sequence, we can use the Havel-Hakimi theorem. According to the theorem, a sequence of non-negative integers is graphical (i.e., can be realized as the degree sequence of a simple graph) if and only if the sequence obtained by deleting the first number and subtracting 1 from the next k numbers (where k is the first number) is graphical. Let's apply this theorem to the given sequence (1, 1, 2, 2, 3, 3, 4, 4): 1. Start with (1, 1, 2, 2, 3, 3, 4, 4). 2. Remove the largest number (4) and subtract 1 from the next 4 numbers: (0, 0, 1, 1, 2, 2, 3). 3. Remove the largest number (3) and subtract 1 from the next 3 numbers: (0, 0, 0, 0, 1, 1). 4. Remove the largest number (1) and subtract 1 from the next number: (0, 0, 0, 0, 0). 5. We now have a sequence of all zeros, which indicates that the original sequence is graphical. Thus, there exists a graph with the degree sequence (1, 1, 2, 2, 3, 3, 4, 4). Here is an example of such a graph: ``` 1 | 2 -- 3 | | 4 -- 5 / \ 6 7 ``` In this graph, vertices 6 and 7 have degree 1, vertices 1, 4, and 5 have degree 2, vertex 2 has degree 3, and vertex 3 has degree 4. |
True |
| baichuan4 | To determine whether a graph with the score sequence (1, 1, 2, 2, 3, 3, 4, 4) exists, we can use the following approach: 1. **Check the Degree Sequence**: First, we need to verify if the given sequence of scores can be a valid degree sequence of a graph. A degree sequence is valid if it satisfies the condition that the sum of degrees of vertices of one type (either odd or even) should not exceed the total number of vertices minus 1. In this case, our degree sequence is (2, 2, 4, 4, 6, 6, 8, 8), which sums up to 36. Since we have 8 vertices, the maximum sum for either odd or even vertices should be 7 * 4 = 28, which is less than 36. Therefore, this degree sequence is not valid for a simple graph. 2. **Check for Graph Realizability**: Even if the degree sequence is not valid for a simple graph, it might still be possible to realize it as a multigraph (a graph that allows multiple edges between two vertices). However, since the degree sequence is not graphical (not satisfying the Handshaking lemma for simple graphs), it cannot be realized as a simple graph. 3. **Conclusion**: Since the degree sequence cannot be realized as a simple graph and we are not considering multigraphs, the answer to the problem is **False**. There does not exist a simple graph with the score sequence (1, 1, 2, 2, 3, 3, 4, 4). |
True |
| doubao | False. The sum of the degrees of all vertices in a graph must be an even number (since each edge contributes two degrees). In the given sequence (1, 1, 2, 2, 3, 3, 4, 4), the sum of the numbers is 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 = 20, which is an even number. However, for a graph to exist with these degrees, the degrees of the individual vertices must also satisfy the handshaking lemma. But it can be seen that there is no way to arrange the vertices such that the degrees are (1, 1, 2, 2, 3, 3, 4, 4) and still satisfy the handshaking lemma. | True |
| minimax | To determine whether a graph with a given score exists, we can use the Handshaking Lemma, which states that the sum of the degrees of all vertices in a graph must be even because each edge contributes to the degree count of two vertices. Let's examine the given score: (1, 1, 2, 2, 3, 3, 4, 4). First, we sum up the degrees: 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 = 20 The sum of the degrees is 20, which is even. This satisfies the Handshaking Lemma, but it does not guarantee that a graph with this degree sequence exists. We also need to check if the sequence is graphical using the Erdős–Gallai theorem or the Havel-Hakimi algorithm. However, we can quickly see that this sequence cannot be graphical because it violates the necessary condition for a simple graph: no vertex can have a degree greater than n-1, where n is the number of vertices in the graph. In this case, we have degrees of 4, which means we must have at least 5 vertices. But if we have 5 vertices, the maximum degree any vertex can have is 4 (since the sum of degrees is even, and the maximum degree is one less than the number of vertices). However, we have two vertices with a degree of 4, which is impossible in a simple graph because it would mean there are 8 edges (each edge counted twice, once for each vertex it connects), but we only have 8/2 = 4 vertices, which is not enough to accommodate 8 edges without having multiple edges between the same pair of vertices or loops, which are not allowed in simple graphs. Therefore, the answer is false; there does not exist a simple graph with the score (1, 1, 2, 2, 3, 3, 4, 4). |
True |
| kimi | To determine whether there exists a graph with a given score, we need to understand what the score represents. In graph theory, the score of a graph typically refers to a sequence of numbers that represent some property of the graph, such as the degrees of its vertices. The given score (1, 1, 2, 2, 3, 3, 4, 4) seems to represent the degrees of the vertices in a graph. Each number represents the degree of a vertex, and the number of times a number appears indicates how many vertices have that degree. So, we have: - Two vertices with degree 1 - Two vertices with degree 2 - Two vertices with degree 3 - Two vertices with degree 4 To determine if such a graph exists, we need to consider the Handshaking Lemma, which states that the sum of the degrees of all vertices in a graph is equal to twice the number of edges. This is because each edge contributes to the degree of two vertices. Let's calculate the sum of the degrees: 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 = 20 According to the Handshaking Lemma, the number of edges must be half of this sum: 20 / 2 = 10 edges Now, let's try to construct a graph with these properties: 1. We can place two vertices with degree 1 at the ends of an edge. 2. We can connect these two vertices with degree 1 to two other vertices, each of which has degree 2. This will give us two vertices with degree 2. 3. We can then connect these two vertices with degree 2 to two other vertices, each of which has degree 3. This will give us two vertices with degree 3. 4. Finally, we can connect these two vertices with degree 3 to two vertices with degree 4. This construction gives us a graph with the desired degree sequence. Therefore, the answer is: True: there exists a graph with the score (1, 1, 2, 2, 3, 3, 4, 4). |
True |
| wenxinyiyan | To determine whether a graph with a given score sequence exists, we can use the Havel-Hakimi algorithm. The Havel-Hakimi theorem states that a non-increasing sequence of non-negative integers \(d_1, d_2, ..., d_n\) is graphic (i.e., there exists a simple graph with these degrees) if and only if the sequence \(d_2 - 1, d_3 - 1, ..., d_{d_1 + 1} - 1, d_{d_1 + 2}, ..., d_n\) is graphic. Let's apply this theorem to the given score sequence (1, 1, 2, 2, 3, 3, 4, 4): 1. Sort the sequence in non-increasing order (it's already sorted in this case): (4, 4, 3, 3, 2, 2, 1, 1) 2. Apply Havel-Hakimi's algorithm: * Remove the first element (4) and subtract 1 from the next four largest elements: (3, 2, 1, 0, 2, 1, 1) + Note: We replace the first occurrence of 0 with the next largest element (in this case, 2) that has not been reduced yet, as per the Havel-Hakimi process. * Continue with the new sequence: (3, 2, 1, 2, 1, 1) + Remove 3 and subtract 1 from the next three largest elements: (1, 0, 1, 1) + Rearrange to maintain non-increasing order: (1, 1, 1, 0) * Continue with (1, 1, 1, 0): + Remove 1 and subtract 1 from the next largest element: (0, 1, 0) + Since there are no more elements greater than 0 to subtract from, we stop. 3. The resulting sequence contains zeros, but we cannot continue the Havel-Hakimi process further. The presence of zeros at this stage indicates that the original sequence is not graphic. Based on the Havel-Hakimi algorithm, the given score sequence (1, 1, 2, 2, 3, 3, 4, 4) is not graphic. Therefore, the statement "there exists a graph with score (1, 1, 2, 2, 3, 3, 4, 4)" is false. |
True |
| zhipu | To determine whether there exists a graph with the given score (1, 1, 2, 2, 3, 3, 4, 4), we need to understand what is meant by the "score" of a graph. In this context, the score likely refers to the degree sequence of the graph, where the degree of a vertex is the number of edges incident to that vertex. A degree sequence is a list of the degrees of the vertices of a graph, usually written in non-increasing order. However, the given sequence (1, 1, 2, 2, 3, 3, 4, 4) is not in non-increasing order. For clarity, let's first rearrange it to (4, 4, 3, 3, 2, 2, 1, 1), which is the conventional way to represent a degree sequence. Now, we need to check if this sequence satisfies the conditions of the Havel-Hakimi theorem, which provides a way to determine if a given sequence of non-negative integers is the degree sequence of a simple graph (a graph without loops or multiple edges between the same pair of vertices). The Havel-Hakimi theorem states that a sequence \(d_1, d_2, \ldots, d_n\) is a degree sequence of a simple graph if and only if the sequence obtained by reducing the first \(d_1\) elements by 1 and then rearranging the sequence in non-increasing order is also a degree sequence. This process is repeated until all elements are zero. Let's apply the Havel-Hakimi algorithm to the sequence (4, 4, 3, 3, 2, 2, 1, 1): 1. Start with (4, 4, 3, 3, 2, 2, 1, 1). 2. Reduce the first 4 elements by 1: (3, 3, 2, 2, 2, 1, 1, 0). 3. Rearrange in non-increasing order: (3, 3, 2, 2, 2, 1, 1, 0). 4. Reduce the first 3 elements by 1: (2, 2, 1, 2, 1, 1, 0, 0). 5. Rearrange in non-increasing order: (2, 2, 2, 1, 1, 1, 0, 0). 6. Reduce the first 2 elements by 1: (1, 1, 2, 1, 1, 0, 0, 0). 7. Rearrange in non-increasing order: (2, 1, 1, 1, 1, 0, 0, 0). 8. Reduce the first 2 elements by 1: (1, 0, 1, 1, 1, 0, 0, 0). 9. Rearrange in non-increasing order: (1, 1, 1, 1, 0, 0, 0, 0). 10. Reduce the first 1 element by 1: (0, 1, 1, 1, 0, 0, 0, 0). 11. Rearrange in non-increasing order: (1, 1, 1, 0, 0, 0, 0, 0). 12. Reduce the first 1 element by 1: (0, 1, 1, 0, 0, 0, 0, 0). 13. Rearrange in non-increasing order: (1, 1, 0, 0, 0, 0, 0, 0). 14. Reduce the first 1 element by 1: (0, 1, 0, 0, 0, 0, 0, 0). 15. Rearrange in non-increasing order: (1, 0, 0, 0, 0, 0, 0, 0). 16. Reduce the first 1 element by 1: (0, 0, 0, 0, 0, 0, 0, 0). Since we have successfully reduced the sequence to all zeros, the original sequence (4, 4, 3, 3, 2, 2, 1, 1) is indeed a valid degree sequence of a simple graph. Therefore, the statement "there exists a graph with score (1, 1, 2, 2, 3, 3, 4, 4)" is **true**. |
True |
| gpto1 | To determine whether there exists a simple graph with the given degree sequence (1, 1, 2, 2, 3, 3, 4, 4), we can use the **Havel-Hakimi algorithm**, which is a process used to determine whether a degree sequence can be graphical (i.e., realized by a simple graph with no loops or multiple edges). **Step 1: Sort the Sequence in Non-Increasing Order** First, sort the degrees in non-increasing order: \[ (4, 4, 3, 3, 2, 2, 1, 1) \] **Step 2: Apply the Havel-Hakimi Algorithm** We will repeatedly apply the following steps: 1. Remove the largest degree \(d\) from the sequence. 2. Subtract 1 from each of the next \(d\) degrees in the sequence. 3. If at any point we get a negative degree or the degree sequence cannot be satisfied, the sequence is not graphical. **Iteration 1:** - **Remove \(d = 4\):** \[ \text{Sequence after removal: } (4, 3, 3, 2, 2, 1, 1) \] - **Subtract 1 from the next 4 degrees:** \[ (4-1, 3-1, 3-1, 2-1) = (3, 2, 2, 1) \] - **Updated sequence:** \[ (3, 2, 2, 1, 2, 1, 1) \] - **Re-sort:** \[ (3, 2, 2, 2, 1, 1, 1) \] **Iteration 2:** - **Remove \(d = 3\):** \[ \text{Sequence after removal: } (2, 2, 2, 1, 1, 1) \] - **Subtract 1 from the next 3 degrees:** \[ (2-1, 2-1, 2-1) = (1, 1, 1) \] - **Updated sequence:** \[ (1, 1, 1, 1, 1, 1) \] **Iteration 3:** - **Remove \(d = 1\):** \[ \text{Sequence after removal: } (1, 1, 1, 1, 1) \] - **Subtract 1 from the next 1 degree:** \[ (1-1) = 0 \] - **Updated sequence:** \[ (0, 1, 1, 1, 1) \] - **Re-sort:** \[ (1, 1, 1, 1, 0) \] **Iterations 4-6:** Repeat the process for the remaining degrees: - **Remove \(d = 1\) and subtract from next degree:** \[ (1, 1, 1, 0) \] - **Subtract 1:** \[ (0, 1, 1, 0) \] - **Updated and re-sorted sequence:** \[ (1, 1, 0, 0) \] - Continue until all degrees are reduced to zero. **Conclusion:** Since we successfully reduced all degrees to zero without encountering a negative degree or an impossible situation, the original degree sequence is **graphical**. **Answer: True. There exists a graph with the given degree sequence.** |
True |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-f402644a-7fb0-9d74-9b01-fa27d1efd9dd', 'request_id': 'f402644a-7fb0-9d74-9b01-fa27d1efd9dd'} | True |