Talk:Lookup Table: Difference between revisions
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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
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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)
@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)
- @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)
- No, usually it only produces one output bit (), because all those chains get OR ed together, and unless the other output ()) can be computed by , 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)