v,a))),sets,IF(LEFT(Table1[Column1],24)=bwcs,TEXTAFTER(Table1[Column1],"'"),""),sets_filled,SCAN("",sets,filldown),rptfilter,sets_filled<>"",result,PIVOTBY(band,sets_filled,sets_filled,COUNTA,,,rptfilter)
Theorem 1 There does not exist any algorithm which, given a dimension , a periodic subset of , and a finite subset of , determines in finite time whether there is a translational tiling of by . The caveat is that we have to work with periodic subsets of , rather than all of ; we ...
The compact group is quite large, and in particular is likely to be inseparable; but as with the case of graphons, when one is only studying at most countably many functions , one can cut down the size of this group to be separable (or equivalently, second countable or metrisable) if...
v,a))),sets,IF(LEFT(Table1[Column1],24)=bwcs,TEXTAFTER(Table1[Column1],"'"),""),sets_filled,SCAN("",sets,filldown),rptfilter,sets_filled<>"",result,PIVOTBY(band,sets_filled,sets_filled,COUNTA,,,rptfilter)
(MathOverflow users with enough privileges to see deleted answers will find that there are no less than seventeen deleted attempts at a proof in response to this question!) On the other hand, the one surviving response to the question does point out this paper of Poonen which shows that ...
Also, I noticed that , there seems some issue in the formula as if there's 3 Feature set , then in Col-M it must have 3 entries , but seems there are 4.. Br, Anupam anupambit1797 If you'd like the aggregation in output 2 to be like the one in output 1 then I switch from ...
(even if one works up to equivalence), making it problematic to give this class the structure of a measurable space; furthermore, even once one does so, one needs to take additional care to pin down what it would mean for a random vector lying in a random vector space to depend “...