使用位操作确定多热编码的有效性

Question

假设我有N个项目和一个二进制数，表示结果中包含这些项目：

N = 4

# items 1 and 3 will be included in the result
vector = 0b0101

# item 2 will be included in the result
vector = 0b0010

我还提供了一个列表冲突，它指示哪些项目不能同时包含在结果中：

conflicts = [
  0b0110, # any result that contains items 1 AND 2 is invalid
  0b0111, # any result that contains AT LEAST 2 items from {1, 2, 3} is invalid
]

鉴于此冲突列表，我们可以确定较早vector s 的有效性：

# invalid as it triggers conflict 1: [0, 1, 1, 1]
vector = 0b0101

# valid as it triggers no conflicts
vector = 0b0010

在这种情况下，如何使用位操作来确定一个向量或大量向量与冲突列表的有效性？

此处提供的解决方案已经帮助我们完成了大部分工作，但我不确定如何使其适应 integer 用例（以避免 numpy arrays 和 numba 完全）。

Answer 1

N = 4

# items 1 and 3 will be included in the result
vector = 0b0101

# item 2 will be included in the result
vector = 0b0010

conflicts = [
  0b0110, # any result that contains items 1 AND 2 is invalid
  0b0111, # any result that contains AT LEAST 2 items from {1, 2, 3} is invalid
]

def find_conflict(vector, conflicts):
    found_conflict = False
    for v in conflicts:
        result = vector & v # do a logical AND operation
        if result != 0: # there are common elements
            number_of_bits_set = bin(result).count("1") # count number of common elements
            if number_of_bits_set >= 2: # check common limit for detection of invalid vectors
                found_conflict = True
                print(f"..Conflict between {bin(vector)} and {bin(v)}: {bin(result)}")
    if found_conflict:
        print(f"Conflict found for {bin(vector)}.")
    else:
        print(f"No conflict found for {bin(vector)}.")

# invalid as it triggers conflict 1: [0, 1, 1, 1]
vector = 0b0101
find_conflict(vector, conflicts)

# valid as it triggers no conflicts
vector = 0b0010
find_conflict(vector, conflicts)

$ python3 pythontest.py
..Conflict between 0b101 and 0b111: 0b101
Conflict found for 0b101.
No conflict found for 0b10.
$