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  32019-11-15T00:04:24  <theStack> anyone here from review group 1? :)
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 752019-11-15T17:16:50  <waxwing> In "using a k of k script for every combination" (*script BIP, rationale 6), i note that each k of k is implemented with CHECKSIG, CHECKSIGVERIFY and not CHECKSIGADD; i can see that you don't need the counting mechanism that CHECKSIGADD has (because these are always k of k) but, is it just a choice? could you use *ADD anyway? is it just a bit less economical?
 762019-11-15T17:18:22  <sipa> waxwing: yeah you'd need to push a 0 at the beginning
 772019-11-15T17:19:13  <waxwing> also i didn't get why e.g. 2 of n is only more economical that way if n>= 6; say if n=5, then there are 5 choose 2, or 10 branches of 2 of 2 each, but wouldn't each of those 2 of 2 branches be better than a big script with 5 keys? hmm not sure how to count this.
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 792019-11-15T17:20:32  <waxwing> oh .. well you would need 5 keys in the script the other way.
 802019-11-15T17:20:53  <sipa> waxwing: i did the math :)
 812019-11-15T17:21:27  <waxwing> :)
 822019-11-15T17:24:04  <waxwing> i guess it's interesting for people to observe that k of n policies are feasible now with huge n, as long as k is small. not sure it'll be obvious to people that that design window has opened up.
 832019-11-15T17:24:14  <waxwing> well i mean as if anything here is obvious ...
 842019-11-15T17:25:01  <sipa> personally i think most interesting theshold policies are k-of-n with k>n/2
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 882019-11-15T18:23:38  <instagibbs_> if k is almost n it's also possible
 892019-11-15T18:23:40  <instagibbs_> :)
 902019-11-15T18:30:52  <instagibbs_> point made in the review slack: "Let ej = c[33+32j:65+32j]" can be confusing if you don't know python slicing
 912019-11-15T18:31:03  <instagibbs_> could seem 33 bytes rather than 32
 922019-11-15T18:31:33  <sipa> it's not python slicing!
 932019-11-15T18:31:46  <sipa> the [:] is actually defined in bip-schnorr ;)
 942019-11-15T18:32:01  <instagibbs_> hah!
 952019-11-15T18:32:07  <sipa> it just coincidentally matches python notation
 962019-11-15T18:33:52  <instagibbs_> ok worth a link :) thanks
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1002019-11-15T19:11:07  <waxwing> instagibbs_, yeah the combinatorial symmetry, but: with 99 out of 100 you need 99 sigs, not 1, even though the combs are the same.
1012019-11-15T19:12:40  <waxwing> oh but: musig?
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1042019-11-15T19:56:01  <aj> waxwing: yeah, just 100 leafs each with one 99-of-99 musig
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1072019-11-15T20:53:49  <waxwing> gotcha so by symmetry we're ok at the high end, but only to the extent people are willing and able to do large musigs.
1082019-11-15T20:54:19  <waxwing> well i mean ignoring any 'native' threshold secret sharing shenanigans
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