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General Tech Technology & Software 2 years ago

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manpreet Tuteehub forum best answer Best Answer 2 years ago

The function given is: g=ACFG + ADEG + DHJ + BCFI

and i am asked to perform the reconverge theorem on this function which means taking all possible pairs of LUT that share the same input merge them together and apply chortle-crf to these possibilities and see which one is the best. K is given to be 4.

My most important question is how can a pair be merged together if initillay the LUT is full(has 4 inputs)? Any help or tip is much appreciated, thanks!

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manpreet 2 years ago

I cannot give a detailed answer, but if you are thinking that ACFG fills a LUT and therefore you cannot combine any other term into the LUT, then I think you have to break the pattern of thinking that equates a 4-input AND term with a 4-input LUT.

To get you started : looking at ACFG, you may note that CF is shared with another term, and so is AG. So you could economise by computing CF elsewhere, to share between both terms that use it, and feed CF as one input into this LUT : this frees up an input. Repeating with AG frees another input; the term ACFG is now AG.CF requiring just 2 inputs, so this LUT could (for example) compute AG.CF + B.CF.I.

I am not convinced that there are any solutions that require fewer than the 5 LUTs the obvious solution takes - but on such a small example, that's probably not the point...


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manpreet 2 years ago

I cannot give a detailed answer, but if you are thinking that ACFG fills a LUT and therefore you cannot combine any other term into the LUT, then I think you have to break the pattern of thinking that equates a 4-input AND term with a 4-input LUT.

To get you started : looking at ACFG, you may note that CF is shared with another term, and so is AG. So you could economise by computing CF elsewhere, to share between both terms that use it, and feed CF as one input into this LUT : this frees up an input. Repeating with AG frees another input; the term ACFG is now AG.CF requiring just 2 inputs, so this LUT could (for example) compute AG.CF + B.CF.I.

I am not convinced that there are any solutions that require fewer than the 5 LUTs the obvious solution takes - but on such a small example, that's probably not the point...


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