Answer:
Sure! I'll comment in the comments of this answer on how it was <3
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oh yeah hi im boréd -,-
Explanation:
h r u
Answer:
See explaination
Explanation:
Diverse Subset Problem is NP: When presented with a set of k customers, it can be checked in polynomial time that you wont at any time have two customers in the set have ever bought the same product.
Independent Set is known to be NP-complete.
Independent Set ?P Diverse Subset Problem: Suppose we have a black box for Diverse Subset Problem and want to solve an instance of Independent Set. For our Independent Set Problem, we have a graph G=(V,E) and a number k, and need to find out if G contains an independent set of size (at least) k. We need to reduce the Independent Set Problem to a Diverse Subset Problem. We do this by constructing an array where each v in V is a customer and each e in E is a product, and customer v purchased every product e for which the product edge e touches the customer node v. Then we ask the black box for the Diverse Subset Problem if there is a subset of k customers that is diverse.
The black box for the Diverse Subset Problem will return.
There are about 15,000,006
Answer:
ummmm try the inequality protragathron theorum
Explanation:
ok
