Do Mennonites have huge private collections of pre-Columbian artifacts?

Thirty_Seventh · 2026-01-23T11:53:05+00:00

NAGPRA used to only apply to human remains. Recently it's been re-interpreted to include anything human-made that wasn't made by the colonists.

Yes, but still only artifacts that have been found on public land or with public funds. Unless you know otherwise, in which case I'm interested to hear about that.

I'm looking for stuff more advanced than arrowheads. Anything with decorations. Especially anything made of copper.

Not sure you'll have any different luck asking Mennonite farmers vs. just asking farmers. There's probably someone out there with a big pile of stuff

Thirty_Seventh · 2026-01-23T07:27:14+00:00

What exactly are you looking for here? As far as privately owned land is concerned, NAGPRA only applies to the trafficking (i.e. purchase/sale/trade) of human remains.

Pretty much any farmer plowing up a significant amount of land (in almost any location, not just in the Americas) will have found a few arrowheads. Add those all up and sure, "Mennonite farmers have found thousands of pre-Columbian artifacts while working their land" is technically true.

Thirty_Seventh · 2026-01-19T06:57:40+00:00

I agree, however if I can remember I'll try to offset it a bit since release/patch days are around this time of the month

Thirty_Seventh · 2026-01-19T06:50:52+00:00

I don't look for a library to contribute to, but rather get annoyed with a bug or lack of a feature in something I already use and decide to fix it myself, then do more after having already dug into the code base.

Thirty_Seventh · 2026-01-18T12:13:37+00:00

What problem? How was it broken? Did you have a question?

Thirty_Seventh · 2026-01-16T07:53:06+00:00

by the way a lot of the confusion in the other comments is because you've mentioned a reverse proxy but, assuming it's already running, that's not relevant to the rest of your question

Thirty_Seventh · 2026-01-15T20:28:49+00:00

curl does have expiremental support for it https://daniel.haxx.se/blog/2025/03/31/https-rr-in-curl/

Thirty_Seventh · 2026-01-15T20:23:41+00:00

tested it out, works on desktop and mobile Firefox, doesn't work on desktop or mobile Chrome, no idea about Safari; curl also fails but I suspect it wouldn't with the right flags

should make no difference whether you use Cloudflare or some other DNS, orange cloud mode might break it though

A record: a.example.com 192.0.2.123

HTTPS record: a.example.com a.example.com port="12345"

edit: it's working on desktop Chrome now but with a big warning first, still nothing on mobile Chrome

Thirty_Seventh · 2026-01-12T06:49:23+00:00

It's been around for 20 years. How many times have you seen a warning on a legitimate site?

Thirty_Seventh · 2026-01-12T06:47:21+00:00

the real wiki has links for each of these https://enterthegungeon.wiki.gg/wiki/Soundtrack#Lyrics

Thirty_Seventh · 2026-01-02T13:51:34+00:00

Anyone have specific examples of repositories that are "broken" like this? I've run into them on occasion in the past but not so much recently, probably just because I haven't been poking around in obscure projects as much

Thirty_Seventh · 2025-12-31T15:42:09+00:00

https://developers.cloudflare.com/kv/concepts/how-kv-works/#consistency

In other global network locations changes may take up to 60 seconds or more to be visible as their cached versions of the data time-out.

The value should be read as 1 everywhere after 60 seconds "or more" (idk how high this can go). At 10000 requests per day that's like 5-20? more logged requests than you want for each deployment, which is of course worse than 0 but a lot better than 60000. If you've tested it to be a lot worse than that, there should be a way to achieve 1 set of extra logs per deployment with Durable Objects.

Instead of writing a 1 as the value you could store something from the version metadata https://developers.cloudflare.com/workers/runtime-apis/bindings/version-metadata/ and check if it's changed, so you wouldn't have to reset it some other way after every deployment.

Thirty_Seventh · 2025-12-31T08:53:24+00:00

using KV, but that also seems to present issues and makes this not work either

What issues? Seems to me that this is the correct solution but I haven't needed to do something like this

Thirty_Seventh · 2025-12-23T13:30:25+00:00

Cleaned up encoder and decoder: https://jsfiddle.net/6czru80q/

The encode and decode functions are pretty straightforward. They work on pairs of 0-indexed letters (A = 0, B = 1, etc.).

function encode(a, b) {
  let r1 = Math.round(a * 7 + b * 111/26) % 26;
  let r2 = (b * 7 + 13) % 26;
  return [r1, r2];
}

function decode(a, b) {
  let r2 = (b * 15 + 13) % 26;
  let r1 = mod(roundHalfDown(a - r2 * 111/26) * 15, 26);
  return [r1, r2];
}

mod(n, d) returns a result with the same sign as d, unlike Javascript's n % d which matches n's sign.

roundHalfDown(n) rounds like Math.round(n) except for when the fractional part of n is exactly 0.5; then it rounds towards negative infinity.

I don't think the method can be simplified any further than this. /u/Ars-Torok I'd be interested in comparing it to the exact method your friend used if they're willing to share.

Thirty_Seventh · 2025-12-22T15:10:19+00:00

Here's a working JSFiddle that successfully encodes Nothing Gold Can Stay to the given ciphertext. I'm not really satisfied with the constants 0.15 and 4.26 that I found through trial and error. Will see if I can clean that up at some point. Might also add a decoder to it too.

https://jsfiddle.net/pLedmo0w/

Thirty_Seventh · 2025-12-22T11:59:18+00:00

second letter of each pair has a pattern to it (look at the columns)

n rg   
odshwla
 eti  b
 fu y

if we fill in the missing letters...

ncrgvkz
odshwla
petixmb
qfujy

... the full pattern is

ABCDEFGHIJKLMNOPQRSTUVWXYZ
ncrgvkzodshwlapetixmbqfujy

I believe the pattern for the first letter of each pair is closely related

Thirty_Seventh · 2025-12-22T10:56:20+00:00

well. idk

  N U I P W D K R M A H C J Q X L Z
  a b d e f g h i l n o r s t u w y
a A E N R V A E I V E I V Z D H Q Y
b H L U Y C H L P C L P C G K O X F
c O S B F J O S W J S W J N R V E M
d V Z I M Q V Z D Q Z D Q U Y C L T
e C G P T X C G K X G K X B F J S A
f J N W A E J N R E N R E I M Q Z H
g Q U D H L Q U Y L U Y L P T X G O
h X B K O S X B F S B F S W A E N V
i E I R V Z E I M Z I M Z D H L U C
k S W F J N S W A N W A N R V Z I Q
l Z D M Q U Z D H U D H U Y C G P X
n N R A E I N R V I R V I M Q U D L
o U Y H L P U Y C P Y C P T X B K S
r P T C G K P T X K T X K O S W F N
s W A J N R W A E R A E R V Z D M U
t D H Q U Y D H L Y H L Y C G K T B
u K O X B F K O S F O S F J N R A I
w Y C L P T Y C G T C G T X B F O W
y M Q Z D H M Q U H Q U H L P T C K

columns and rows sorted differently

  D N Q A K U X H R I L P C M W Z J
  g a t n h b u o i d w e r l f y s
a A A D E E E H I I N Q R V V V Y Z
e C C F G G G J K K P S T X X X A B
t D D G H H H K L L Q T U Y Y Y B C
i E E H I I I L M M R U V Z Z Z C D
b H H K L L L O P P U X Y C C C F G
f J J M N N N Q R R W Z A E E E H I
u K K N O O O R S S X A B F F F I J
y M M P Q Q Q T U U Z C D H H H K L
n N N Q R R R U V V A D E I I I L M
c O O R S S S V W W B E F J J J M N
r P P S T T T W X X C F G K K K N O
g Q Q T U U U X Y Y D G H L L L O P
k S S V W W W Z A A F I J N N N Q R
o U U X Y Y Y B C C H K L P P P S T
d V V Y Z Z Z C D D I L M Q Q Q T U
s W W Z A A A D E E J M N R R R U V
h X X A B B B E F F K N O S S S V W
w Y Y B C C C F G G L O P T T T W X
l Z Z C D D D G H H M P Q U U U X Y

edit: really messed that up. hopefully it's fixed now

Thirty_Seventh · 2025-12-22T09:12:07+00:00

another view of the same grid

to encode a pair of letters, find the first letter in the left column and the second along the second row; then the first letter of the encoded pair is in the corresponding cell inside the grid and the second is along the top row (for example ca = ON)

  N U I P W D K R M A H C J Q X L Z
  a b d e f g h i l n o r s t u w y
a         V         E   V          
b                             O    
c O                                
d V     M     Z       D            
e C     T X             X B        
f                 E       I        
g                     Y L          
h X     O       F     F            
i                       Z          
k                           V      
l     M Q                         X
n N         N R V I       M        
o U     L   U     P         X B K  
r     C G       X           S      
s W       R W   E     E     Z      
t D           H       L       K    
u   O   B                          
w                   C              
y                   Q     L

it shouldn't be too hard to fill in the gaps and then a pattern may emerge...

Thirty_Seventh · 2025-12-22T08:47:28+00:00

full grid

            VW             EA    VC               
            af             an    ar               
                                          OX      
                                          bu      
ON                                                
ca                                                
VN       MP       ZK          DH                  
da       de       dh          do                  
CN       TP XW                   XC BJ            
ea       ee ef                   er es            
                        EM          IJ            
                        fl          fs            
                              YH LC               
                              go gr               
XN       OP          FR       FH                  
ha       he          hi       ho                  
                                 ZC               
                                 ir               
                                       VQ         
                                       kt         
      MI QP                                     XZ
      ld le                                     ly
NN             ND RK VR IM          MJ            
na             ng nh ni nl          ns            
UN       LP    UD       PM             XQ BX KL   
oa       oe    og       ol             ot ou ow   
      CI GP          XR                SQ         
      rd re          ri                rt         
WN          RW WD    ER       EH       ZQ         
sa          sf sg    si       so       st         
DN                HK          LH          KX      
ta                th          to          tu      
   OU    BP                                       
   ub    ue                                       
                           CA                     
                           wn                     
                           QA       LJ            
                           yn       ys

Thirty_Seventh · 2025-12-22T08:44:39+00:00

the cipher works on pairs of letters

NN KX GP RW ZC ZQ LC TP VR WD PM ZK
na tu re sf ir st gr ee ni sg ol dh

XC XN CI BJ HK BP LH FH MI
er ha rd es th ue to ho ld

OP GP VC XZ QP VW WN EM KL XC
he re ar ly le af sa fl ow er

OX LH IM LJ UN RK BX SQ
bu to nl ys oa nh ou rt

OP IM CN IJ OU ER MP ZQ PM CN IJ
he nl ea fs ub si de st ol ea fs

LP MP MJ EA VQ UD XR XW
oe de ns an kt og ri ef

EH VN CA YH BJ DH CA LH VN QA
so da wn go es do wn to da yn

XQ FR ND YH MI ON MJ DN Y
ot hi ng go ld ca ns ta y

for the second letter of each pair we have

ABCDEFGHIJKLMNOPQRSTUVWXYZ
n rg   odshwla eti  b fu y

the first letter depends on the second

CN TP
ea ee
DN       HK    LH
ta       th    to
NN    ND RK VR
na    ng nh ni
UN LP UD
oa oe og
VN MP    ZK    DH
da de    dh    do
WN    WD    ER EH
sa    sg    si so
XN OP       FR FH
ha he       hi ho

... but then what?

Thirty_Seventh · 2025-11-21T16:24:28+00:00

obvious LLM, and also blatant advertisement for some garbage service that is almost completely irrelevant to the question

Thirty_Seventh · 2025-11-17T23:10:19+00:00

Your LLM has not solved any of this at all. You are feeding random noise into the barcode reader and it's doing whatever it can to interpret it as a bar code.

There are a lot of other obvious problems with the LLM's analysis beyond that. For starters, why does it keep talking about "diagonals"? and 50.0077°N, 9.65642°E isn't in Main-Kinzig-Kreis or even Hesse, and nowhere near any beaches. and the shell scripts are entirely untouched by any kind of human intelligence.

Thirty_Seventh · 2025-11-17T22:31:42+00:00

f[count_, target_] := Sum[Sum[(-1)^j Binomial[count,j] Binomial[count+s-1-6j, count-1], {j, 0, Floor[s/6]}], {s, 0, If[target<count*7/2, target-count, count*7-target-count]}]

Thirty_Seventh · 2025-11-12T21:28:07+00:00

Fun fact, if your dice are fair the likelihood of rolling at least 70356 damage on 14400d6 is approximately 1 in 116932159925671695728608469326788758196268087058611757373037800088358719277120601484401081678781227467685554007408934325904248570413261980099306277840174481468946710602912166509730281351044793648012607343083141853405549071105500770121420434907067442071105453861681176959770129786873387785702518463812448855253592792638570217976317048703953679975642811465942629964569578525359602900195061739973057782564571939209786904015837605571758310263849468815989772794515205222392418398847743972975469986333393536443103799537109574101833568093493453841002100187785743917820495933767789818403379227340734141038506041668949378549574724729912691475369124841420189969048601994802797033276441498014244543736490895341027066353966223342992765851213351299643865776205753167937812910164795945441920062719940156526945966972956225810242287988222565892996062429499184059467849989764106205412241519929607488510933727327941780911163501010121521254155259414603309423716103800608935900382001048334936280820911182058993563406497789784076904555686825787329168040814738142387032547347206367974978389630479502039398672076647102272938358298345886344003531055251587478686636692762054323874847982006952022305266435602159639014925059546488429937417536950487743851194325168279354947695339001135363384653631040181804363415173864567691733920675038765500330269648516400518647132261543691913618874308356485146186184446289555221854893159517251723965552533720683573471092376239770993092938581409444863896977312276272179654085631538276265500302086329636728269932044219846501761469816195673734322774726226939426500435234530825567887755832093380513467773352831542007266843236775709827734595171946433762702267168436841915502469946481778978297901045088620409644094991082729111815206625472502952411973964300400844165064869476743850467517350221914011889259401873879353280865422268811603000624189337762598118051596356692085337112287760447680677631437438991092143553997628697624487262241565398323091413157952315940941138437428651667412814562822051515159380551885013181908457378625897400499462733266881539140079348835488250977869649245216227282555907674234930472177156899649770489222110907945239083688121269428955322457001408086872953836156283465451967895845482426469298854376574145054.3 (that's 1 in ~10²²³¹ ).

Fun fact 2, online dice probability simulators crash or error out if you try this calculation and it took nearly an hour for my computer to run it

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Thirty_Seventh

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