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548 points nsagent | 1 comments | | HN request time: 0.205s | source
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lukev ◴[] No.44567263[source]
So to make sure I understand, this would mean:

1. Programs built against MLX -> Can take advantage of CUDA-enabled chips

but not:

2. CUDA programs -> Can now run on Apple Silicon.

Because the #2 would be a copyright violation (specifically with respect to NVidia's famous moat).

Is this correct?

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saagarjha ◴[] No.44567309[source]
No, it's because doing 2 would be substantially harder.
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lukev ◴[] No.44567356[source]
There's a massive financial incentive (billions) to allow existing CUDA code to run on non-NVidia hardware. Not saying it's easy, but is implementation difficulty really the blocker?
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fooker ◴[] No.44568123[source]
Existing high performance cuda code is almost all first party libraries, written by NVIDIA and uses weird internal flags and inline ptx.

You can get 90% of the way there with a small team of compiler devs. The rest 10% would take hundreds of people working ten years. The cost of this is suspiciously close to the billions in financial incentive you mentioned, funny how efficient markets work.

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lcnielsen ◴[] No.44568168[source]
> funny how efficient markets work.

Can one really speak of efficient markets when there are multiple near molopolies at various steps in the production chain with massive integration, and infinity amounts of state spending in the process?

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bigyabai ◴[] No.44568219[source]
Sure they can. CUDA used to have a competitor, sponsored by Apple. It's name is OpenCL.
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1. dannyw ◴[] No.44569174[source]
And after Apple dropped NVIDIA, they stopped caring about openCL performance on their GPUs.