考虑三个决策问题P1，P2和P3。已知P1是可确定的，P2是不可确定的。以下哪一项是TRUE?
(A)如果P1可还原为P3，则可以确定P3
(B)如果P3可还原为P2，则P3不确定
(C)如果P2可还原为P3，则P3不确定
(D)如果P3可还原为P2的补码，则P3是可确定的答案： (C)
解释：
背景：在计算复杂性理论中，决策问题只有两个可能的输出“是”或“否”。
如果存在有效的方法或算法，该方法或算法返回正确的是/否答案，则认为决策问题是可以决定的。
如果不存在总是导致正确的是/否解决方案的单个算法，则决策问题被认为是无法确定的。
在还原性条件：A≤p B表示A是决策问题是还原到B在多项式时间p。这只是意味着A的实例可以转换为B的实例，按照B的解决方案，我们可以得到问题A的解决方案。
因此，在这里我们可以得出一些结论：
1. If A ≤p B and B is decidable then A is also decidable.
This is because if there exists a specific algorithm for solving B and we can
also reduce A to B then we can have a solution of A as well. Hence A is decidable.
However the reverse is not true i.e. if A ≤p B and A is decidable
then B is also decidable because A can have an algorithm existing for its correct
solution but might be the case that B does not.
2. If A ≤p B and A is undecidable then B is also undecidable.
This is because if A is undecidable even when it can be reduced to B that simply
reflects even B cannot provide an algorithm by which we can solve B and hence A.
So decision problem B is also undecidable.
然而，反之则不然这里也即，如果A≤p B和B是不可判定那么A也是不可判定的，因为有可能存在很的算法，可以提供一个解决方案A.
根据上述结论，我们可以说选项1、2和4为假，选项3为真。
Option 1: P1 ≤p P3 and given P1 is decidable gives no conclusion for P3.
Option 2: P3 ≤p P2 and given P2 is undecidable gives no conclusion for P3.
Option 3: P2 ≤p P3 and given P2 is undecidable gives conclusion for P3 to be
undecidable.
Option 4: P3 ≤p P2’s complement and given P2 is undecidable therefore P2’s
complement is also undecidable gives no conclusion for P3.
该解释由Yashika Arora提供。
请访问以下文章以了解更多信息：
不可确定性和可归约性
维基百科：Reduction_(复杂度)这个问题的测验