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- Thread starter Veracity
- Start date

If we assume that the first letter designates difficulty, the remaining six letters allow 26**6 = 308,915,776 different planets at that difficulty. That is 28.2 bits of information.

If I wanted anomalous plants + anomalous animals at difficulty G, how many of those 308 million planets have both? It's not a common combination; the Wiki knows of none, but my database contains 6 examples, out of 2766 known planets. (152 of which are unvisited by me, yet, and hence unverified). That is .21%, so I'd expect to find more than 670,000 such planets at G difficulty. How many of those have no other life? How can I generate (a set of) coordinates to select such a planet?

Reverse Engineering is a daunting prospect, so I am continuing to collect data until I feel like trying some strategies.

I note that taltimir has, apparently, discovered a correlation between which hostile alien trophy is generated by hostile aliences with particular images (sgalienbX.gif):

spear: 1, 2

blowgun: 3,4

loincloth: 5,6

totem: 7,8

necklace: 9,10

So, that is two planet attributes which draw from the same part of the bitspace. There are only a handful of planets that have been publicly shared that contain trophies. Considering that we know planets for all 5 trophies, there is no competitive advantage to NOT sharing such coordinates, but I suspect that most people are not obsessing on Spacegate like I am. The fact that built in support for it into VMF probably accounts for some of those.

Reverse engineering will consist of saying "76 of the 2767 validated planets (and 5 of the 152 unvalidated planets) contain anomalous animals. What do all of those coordinates have in common?" Not sure how to do that, yet. If I generate a code from 0 to 26**6-1 (or 26**7-1), perhaps they are all divisible by XXXX? Or perhaps they all have bits X,Y,Z set?

Code:

```
[color=green]> ashq print( 26 ** 7 )[/color]
8031810176
```