Should I convert the frontend generated UUIDv6 to binary(16) for use in SQL Server as the clustered primary key?

Background

As suggested by the front-end developer, I looked into using UUID as the primary key for a bunch of tables in our new system. From learning the pros and cons of random vs. sequential UUIDs, to the use of a non-clustered primary key in combination of a clustered index with a sort-able type, my research pointed me to UUIDv6, and an implementation of it.

It is able to generate UUIDs like the below (that is sequential):

UUIDv1                               UUIDv6
------------------------------------ ------------------------------------
5714f720-1268-11e7-a24b-96d95aa38c32 1e712685-714f-6720-a23a-c90103f70be6
68f820c0-1268-11e7-a24b-96d95aa38c32 1e712686-8f82-60c0-ac07-7d6641ed230d
7ada38f0-1268-11e7-a24b-96d95aa38c32 1e712687-ada3-68f0-93f8-c1ebf8e6fc8c
8cc06fd0-1268-11e7-a24b-96d95aa38c32 1e712688-cc06-6fd0-a828-671acd892c6a
9ea6a6b0-1268-11e7-a24b-96d95aa38c32 1e712689-ea6a-66b0-910c-dbcdb07df7a4

Which I thought SQL Server would gladly sort them for me in the clustered primary key (uniqueidentifier) column.

Little did I know how SQL Server would sort an uniqueidentifier column. Here’s the ascending sort result:

UUIDv6 
------------------------------------ 
1e712688-cc06-6fd0-a828-671acd892c6a 
1e712686-8f82-60c0-ac07-7d6641ed230d 
1e712687-ada3-68f0-93f8-c1ebf8e6fc8c 
1e712685-714f-6720-a23a-c90103f70be6 
1e712689-ea6a-66b0-910c-dbcdb07df7a4

Which is causing fragmentation as if using random UUIDs. This post explains how they were actually sorted.

The real question

Luckily, the system is still in development. Which of these options should I go for next?

  1. reorder the bytes so that the most/least significant bytes are where SQL Server expects them to be
  2. convert the UUIDv6 to binary(16) and use that instead

Problem with option 1

The UUID standard embeds a 4-bit version field inside the ID. UUIDv6 (still non-standard) also follows that rule. The way I will reorder them is going to break this.

Problem with option 2

I’m not sure. Can hardly find anyone talking about it except this, which is going against the idea. Are there other pitfalls that I should be aware of in using the binary(16) type?

Thanks!

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Author: oopoopoop