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Latest comment: 7 September by GHXX in topic Improve further
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:@[[User:GHXX|GHXX]], yes to all, except one thing. Why LUTs should output only a single bit? I thought using it to produce a pattern of bits is a common application. [[User:DjSapsan|DjSapsan]] ([[User talk:DjSapsan|talk]]) 20:45, 7 September 2025 (UTC)
:@[[User:GHXX|GHXX]], yes to all, except one thing. Why LUTs should output only a single bit? I thought using it to produce a pattern of bits is a common application. [[User:DjSapsan|DjSapsan]] ([[User talk:DjSapsan|talk]]) 20:45, 7 September 2025 (UTC)
::No, usually it only produces one output bit (<math>Output1</math>), because all those chains get OR ed together, and unless the other output (<math>Output2</math>)) can be computed by <math>Output2 = Output1 \quad OR \quad (other chains)</math>, there is no way of computing that other output, as far as i am aware.
::I suppose you could maybe re-use the intermediate NOT outputs, but that is far from the general case from my experience. [[User:GHXX|GHXX]] ([[User talk:GHXX|talk]]) 20:55, 7 September 2025 (UTC)

Revision as of 20:55, 7 September 2025

Improve further

Hi @GHXX and hi all, I made a change to the article to make it simpler and clearer. However, I suspect it needs some improvements. How it can be improved in terms of phrasing and/or additional information ? DjSapsan (talk) 17:36, 7 September 2025 (UTC)Reply

@DjSapsan,

Imo we should add that the AND-inputs may also be negated.

And that a lut can only output a single bit, in the general case, and not a batch of outputs (instead of 'The output may be a single bit or a pattern of bits.')

The "we can avoid delays by not using buffers" paragraph should imo be replaced with one that states that the not-inputs are generally fed through a fastbuffer first, so that no backpropagation happens.

Lmk what you think, please GHXX (talk) 20:31, 7 September 2025 (UTC)Reply

@GHXX, yes to all, except one thing. Why LUTs should output only a single bit? I thought using it to produce a pattern of bits is a common application. DjSapsan (talk) 20:45, 7 September 2025 (UTC)Reply
No, usually it only produces one output bit (Output1), because all those chains get OR ed together, and unless the other output (Output2)) can be computed by Output2=Output1OR(otherchains), there is no way of computing that other output, as far as i am aware.
I suppose you could maybe re-use the intermediate NOT outputs, but that is far from the general case from my experience. GHXX (talk) 20:55, 7 September 2025 (UTC)Reply