No way! That would be incredible. That would bring this project from secure and private to legit science fiction level anonymous. You could have hundreds or even thousands mixin partners. Correct me if im wrong but wouldn't this be the cypherpunk holy grail? Is it really within our grasp and just need "review"?
Would there be a decrease in mix-in costs as the available mix-in levels increased?
No its just O(log(n)). Every additional mixin partner increases the size of your signature but less than the one before it.
look at O(log(n)) compared to other big o notations on this chart. Particularly look at it in comparison to O(n). Its a big difference. For example: If the limit(log(n)) < some reasonable signature size that can be affordably stored on the blockchain than you can use every single other key ever published on the entire blockchain to produce your ring signature. Infact if this were the case we could set a mixin minimum of like 1000 or something crazy.
So in cost terms: if this is implemented, the costs of a 99 level mix-in would be less than the cost of a 99 mix-in as it currently stands? Correct?
Probably. It's possible that this wouldn't be the case if n=2 in the new scheme were sufficiently more resource intensive than n=2 in the old scheme. But its highly unlikely that the difference between n=2 in the current scheme vs n=2 in the new scheme would be great enough to make mixin 99 in the new scheme cost more than mixin 99 in the old. Its a complicated way of saying that O(log(n)) only talks about the shape of the curve, it doesnt say anything about where that curve is placed on the graph.
*edit* sorry that was needlessly complicated. yes. the answer is yes. i cant imagine that the authors of that paper would have even bothered to produce it if the answer were no.
Thank you for the answer--enjoyed the complication. I was assuming a 99 mix-in would be located in latter parts of the O(log n) line for the sake of simplicity. Probably should have stated that.
Ill need to look over that paper and see if i can understand any of it. Didn't bother yet because its too late tonight. Anyone have link?
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