We show that every atom a that is added to C by the algorithm is a logical consequence of KB ∪ Percepts. Suppose there is an atom h ∈ C that is not a logical consequence. Let h be the first ever such atom to be added in C. There must be some clause in KB ∪ Percepts, in the form h ← a1 ∧ · · · ∧ am such that a1, . . . , am are all in C. By assumption, KB ∪ Percepts |= ai for all 1 ≤ i ≤ m. Then it must be that KB ∪ Percepts |= h. Contradiction.是什么意思
时间: 2024-04-28 17:24:35 浏览: 18
这段文字描述了一个关于知识库(KB)和感知(Percepts)的算法,在该算法中,每个添加到集合C中的原子a都是KB和Percepts的逻辑结果。如果有一个原子h在C中,但不是它们的逻辑结果,那么假设h是第一个这样的原子被添加到C中,那么一定存在一个KB和Percepts中的子句,形式为h ← a1 ∧ · · · ∧ am,其中a1,...,am都在C中。由于假设KB和Percepts对于1 ≤ i ≤ m中的所有ai都成立,因此KB和Percepts也应该对h成立。但是这与假设相矛盾,即h不是KB和Percepts的逻辑结果。因此,原命题的结论是每个添加到C中的原子a都是KB和Percepts的逻辑结果。
相关问题
Try to write an algorithm to Calculate the WPL of a Huffman Tree.
Here's an algorithm to calculate the weighted path length (WPL) of a Huffman tree:
1. Traverse the Huffman tree in a depth-first manner.
2. Assign a binary code of 0 to the left child and 1 to the right child of each node.
3. For each leaf node, multiply its depth (i.e., the length of the binary code) by its weight (i.e., the frequency of the corresponding character) to get its contribution to the WPL.
4. Sum up the contributions of all leaf nodes to get the total WPL.
Here's the pseudocode for the algorithm:
```
function calculate_wpl(huffman_tree):
wpl = 0
stack.push(huffman_tree.root)
while !stack.empty():
node = stack.pop()
if node.is_leaf():
wpl += node.depth * node.weight
else:
if node.left_child:
node.left_child.depth = node.depth + 1
node.left_child.code = node.code + '0'
stack.push(node.left_child)
if node.right_child:
node.right_child.depth = node.depth + 1
node.right_child.code = node.code + '1'
stack.push(node.right_child)
return wpl
```
Note that this algorithm assumes that the Huffman tree has already been constructed. If you need to construct the Huffman tree from scratch, you can use the Huffman coding algorithm.
Given that 5 is a primitive element modulo 47, solve 10 ≡ 5 a mod 47 by using Shank’s algorithm
Shank's algorithm is an algorithm for solving the discrete logarithm problem. In this case, we want to solve the equation 10 ≡ 5^a (mod 47).
First, we need to find the order of 5 modulo 47. Since 47 is a prime number, we know that any integer a such that 1 ≤ a ≤ 46 is a primitive element modulo 47 if and only if gcd(a,47)=1 and a^(46) ≡ 1 (mod 47). Since 5 is a primitive element modulo 47, we know that the order of 5 modulo 47 is 46.
Next, we need to compute the values of 5^0, 5^1, 5^2, ..., 5^22 mod 47 and store them in a table. We can use the repeated squaring method to compute these values efficiently. For example:
5^0 ≡ 1 (mod 47)
5^1 ≡ 5 (mod 47)
5^2 ≡ 25 (mod 47)
5^3 ≡ 43 (mod 47)
5^4 ≡ 38 (mod 47)
5^5 ≡ 6 (mod 47)
5^6 ≡ 30 (mod 47)
5^7 ≡ 41 (mod 47)
5^8 ≡ 19 (mod 47)
5^9 ≡ 24 (mod 47)
5^10 ≡ 37 (mod 47)
5^11 ≡ 45 (mod 47)
5^12 ≡ 7 (mod 47)
5^13 ≡ 35 (mod 47)
5^14 ≡ 31 (mod 47)
5^15 ≡ 18 (mod 47)
5^16 ≡ 33 (mod 47)
5^17 ≡ 44 (mod 47)
5^18 ≡ 3 (mod 47)
5^19 ≡ 15 (mod 47)
5^20 ≡ 32 (mod 47)
5^21 ≡ 14 (mod 47)
5^22 ≡ 42 (mod 47)
Now, we can use Shank's algorithm to solve the equation 10 ≡ 5^a (mod 47). We can rewrite this equation as 5^a ≡ 10 (mod 47). We want to find a such that 0 ≤ a ≤ 45.
We start by dividing the range [0, 45] into two equal parts: [0, 22] and [23, 45]. We compute 10 × 5^-0 ≡ 10 (mod 47) and 5^0, 5^1, 5^2, ..., 5^22 mod 47 using the table. We find that 10 ≡ 10 (mod 47) and 10 ≢ 1, 5, 25, 43, 38, 6, 30, 41, 19, 24, 37, 45, 7, 35, 31, 18, 33, 44, 3, 15, 32, 14, 42 (mod 47). Therefore, 10 is not in the first half of the table.
Next, we compute 10 × 5^-23 ≡ 39 (mod 47) and 5^23, 5^24, 5^25, ..., 5^45 mod 47 using the table. We find that 39 ≢ 1, 5, 25, 43, 38, 6, 30, 41, 19, 24, 37, 45, 7, 35, 31, 18, 33, 44, 3, 15, 32, 14, 42 (mod 47). Therefore, 10 is not in the second half of the table.
Since we have exhausted all the possibilities, we conclude that there is no solution to the equation 10 ≡ 5^a (mod 47).