Talk:Sparkline in unicode: Difference between revisions

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==A Point Is That Which Has No Part==
Due to fence posts it is claimed above that it is impossible to distribute xMax and xMin into uniform buckets. I don't consider a day worth living if the impossible has not been achieved before breakfast. The title above is accepted widely as the first line of The Elements. I would also draw attention to the Platonic view that a point is a monad with position attached because it is often coded thus, and the Aristotelian view that a point is not the thing but produces the thing by motion because this rules out some of the interpretations above. Sticking with The Elements the second line is 'A line is a breadth less length'. I define the line as the length xMax-xMin. I shall use nB as the number of buckets (8 in this tasks case). I want to distribute these buckets equally along the line so I find the width of each bucket using division as dX=(xMax-xMin)/nB. I now need a function which allocates each of the numbers in the test sequence to a bucket. I number the buckets placed along the line from 0 to nB-1 and assign each number x to bucket n=floor((x-xMin)/nB). For the first test xMin(=1)->0/8 (=0). xMax(=8)->56/8 (=7). For the second test xMin(=0.5)->0/3 (=6) and xMax(=6.5)->24/3 (=8). So the impossible to devise a scheme including xMin and xMax is achieved and today is a day worth living (assuming a late breakfast or a different time zone).--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 12:19, 1 March 2019 (UTC)
For those who want to look at the logic more closely the proof of the above comes (better if I could draw a picture) by finding the center (say C or 0) of the line and distributing the buckets either side of C. The above comes by finding the center of which bucket is closest to x from C. If we adjust the scale so that C is 0 then xMin must be negative and xMax must be positive. Some of the confusion above can now be seen to be trying to answer the question is 0 positive or negative. This question has been answered it is neither positive or negative nor either odd or even. What does the suggested algorithm do? C(=4.5)->28/7 (=4) so positive which is usually a good answer.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 12:36, 1 March 2019 (UTC)
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