Talk:Known Issues
Origin Bias
There is no bias towards the origin. This is a naive mistake involving a misunderstanding of the algorithm. Consider a simplified MD5 algorithm that produces a 16 bit (4 digit) hash, ranging from 0000 to FFFF. Split this in half so that each coordinate is two digits, from 00 to FF. To encounter the mistake, one would convert that number to an integer, then prepend a decimal point, so that we get 0.0 0.1 0.2 ... 0.9 0.10 0.11 0.12 ... 0.98 0.99 0.100 0.101 ... 0.254 0.255. This yields a distribution that is biased by 50% towards the bottom quarter of the graticule, and by 4% towards the 1/10th divisions of the graticule. This is not the correct implementation of the algorithm. What actually happens, as is illustrated on the algorithm page, is that the hexadecimal number is, in effect, converted to an integer that is a number of 1/256ths (in our small example, actually 1/18446744073709551616ths in the real algorithm), yielding an even distribution of .0000 .0039 .0078 ... .9883 .9922 .9961. Sparr 07:59, 31 July 2008 (UTC)
- OK, so I lied. There is a very very tiny bias, equal to 1/36893488147419103232th of one degree towards the origin because there is never a result at the 'top' of the graticule. In the example above consider that you can get a point at 0/256ths but not at 256/256ths. This only applies when considering a lone graticule, as the point in question could be at the 'bottom' of the next graticule. Sparr 08:07, 31 July 2008 (UTC)
- Ah, but you CAN get 256/256ths, just as with any other distribution problem: the 33° graticule, at the 256/256ths mark is represented by the 34° graticule, at 0/256ths. It all works out. Honest. :) 64.74.213.98 19:08, 30 September 2009 (UTC)
- Yes. That's what the last bit above was saying :-| Actually, there are of course two places of ambiguous status on the planet - the poles. You could look at them two ways: (1) There is no graticule of latitude +90 or -90, in which case there will never be a hash point exactly on a pole (although, to be honest, what's a few femtometres between friends?), or (2) There IS a latitude 90, in which case there could theoretically be either 210 or 150 hashpoints on each pole, but only in the extraordinarily unlikely event that the latitude fraction is exactly zero. If the latitude fraction were exactly zero for two days running you'd get 360 of them on each pole - but in such a situation, I would probably begin to suspect that the Heart Of Gold were real, and that Zaphod had just driven it through our local sector. --macronencer 20:31, 30 September 2009 (UTC)
- Ah, but you CAN get 256/256ths, just as with any other distribution problem: the 33° graticule, at the 256/256ths mark is represented by the 34° graticule, at 0/256ths. It all works out. Honest. :) 64.74.213.98 19:08, 30 September 2009 (UTC)
- The sample size is still pretty low, but you can check out a visual representation of the distribution here. Btw, feel free to use this URL (altered for your Lat/Long, of course) on your home graticule page. One example is under "notable dates" on the Santa Cruz, California page.
- Nice idea - but currently doesn't work East of W30 :( --macronencer 18:57, 30 September 2009 (UTC)
- And it shows the wrong side of the longitude east of 0°. -- Pne 18:43, 2 June 2010 (UTC)
