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Topic: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE! - page 35. (Read 108519 times)

newbie
Activity: 3
Merit: 0
Since everyone's sharing and I'm stuck...

Code:
# start at inside top, order = length + outside_color + inside_color
flames = [('s', 'y', 'b'), ('l', 'y', 'g'), ('l', 'y', 'g'), ('s', 'r', 'g'),
          ('l', 'y', 'b'), ('l', 'r', 'b'), ('s', 'y', 'g'), ('l', 'r', 'b'),
          ('l', 'y', 'g'), ('s', 'y', 'g'), ('l', 'r', 'b'), ('s', 'y', 'g'),
          ('s', 'r', 'b'), ('s', 'r', 'g'), ('l', 'y', 'b'), ('s', 'r', 'b'),
          ('l', 'r', 'b'), ('l', 'r', 'b'), ('s', 'y', 'g'), ('l', 'y', 'g'),
          ('l', 'r', 'g'), ('s', 'y', 'g'), ('l', 'r', 'b'), ('l', 'r', 'b'),
          ('s', 'y', 'b'), ('s', 'y', 'g'), ('l', 'y', 'b'), ('l', 'y', 'b'),
          ('l', 'y', 'g'), ('l', 'y', 'b'), ('s', 'y', 'b'), ('l', 'y', 'b'),
          ('l', 'r', 'b'), ('l', 'y', 'b'), ('l', 'r', 'g'), ('l', 'r', 'b'),
          ('s', 'r', 'b'), ('l', 'y', 'g'), ('l', 'r', 'g'), ('s', 'r', 'b'),
          ('l', 'r', 'g'), ('l', 'y', 'g'), ('s', 'y', 'g'), ('l', 'y', 'g'),
          ('l', 'r', 'b'), ('l', 'r', 'g'), ('l', 'r', 'b'), ('l', 'r', 'b'),
          ('s', 'y', 'g'), ('s', 'y', 'b'), ('l', 'y', 'g'), ('s', 'y', 'g'),
          ('l', 'r', 'g'), ('s', 'y', 'g'), ('s', 'r', 'g'), ('s', 'y', 'g'),
          ('l', 'r', 'b'), ('l', 'r', 'b'), ('l', 'r', 'g'), ('s', 'y', 'g'),
          ('s', 'y', 'b'), ('s', 'y', 'b'), ('l', 'y', 'g'), ('s', 'r', 'b'),
          ('l', 'r', 'b'), ('l', 'r', 'b'), ('s', 'y', 'g'), ('l', 'y', 'g'),
          ('l', 'y', 'g'), ('l', 'r', 'b'), ('l', 'y', 'b'), ('l', 'r', 'g'),
          ('s', 'r', 'g'), ('l', 'r', 'g'), ('l', 'r', 'b'), ('l', 'r', 'g'),
          ('l', 'r', 'b'), ('l', 'r', 'g'), ('s', 'y', 'b'), ('l', 'r', 'g'),
          ('l', 'y', 'g'), ('l', 'r', 'b'), ('l', 'r', 'b'), ('l', 'r', 'b'),
          ('s', 'y', 'b'), ('s', 'y', 'g'), ('l', 'r', 'g'), ('l', 'r', 'b'),
          ('l', 'r', 'g'), ('s', 'r', 'g'), ('s', 'y', 'g'), ('l', 'r', 'g'),
          ('l', 'y', 'g'), ('s', 'y', 'g'), ('l', 'r', 'b'), ('l', 'r', 'b'),
          ('s', 'y', 'b'), ('l', 'y', 'g'), ('l', 'r', 'b'), ('s', 'y', 'g'),
          ('l', 'r', 'g'), ('l', 'y', 'b'), ('s', 'y', 'g'), ('l', 'r', 'g'),
          ('l', 'r', 'g'), ('s', 'y', 'b'), ('l', 'r', 'g'), ('s', 'r', 'b'),
          ('s', 'y', 'b'), ('s', 'y', 'b'), ('l', 'r', 'b'), ('s', 'r', 'b'),
          ('l', 'y', 'b'), ('l', 'y', 'b'), ('s', 'y', 'b'), ('l', 'r', 'b'),
          ('l', 'r', 'g'), ('l', 'r', 'b'), ('l', 'r', 'b'), ('l', 'y', 'g'),
          ('s', 'y', 'g'), ('s', 'y', 'g'), ('l', 'r', 'g'), ('s', 'r', 'b'),
          ('l', 'y', 'b'), ('s', 'y', 'b'), ('s', 'y', 'b'), ('s', 'y', 'b'),
          ('l', 'r', 'b'), ('l', 'y', 'g'), ('l', 'r', 'b'), ('l', 'r', 'g'),
          ('s', 'r', 'g'), ('s', 'y', 'b'), ('l', 'r', 'g'), ('s', 'y', 'b'),
          ('l', 'r', 'b'), ('l', 'r', 'g'), ('s', 'y', 'b'), ('s', 'y', 'b'),
          ('l', 'y', 'g'), ('l', 'r', 'b'), ('l', 'r', 'g'), ('s', 'r', 'g'),
          ('s', 'y', 'b'), ('l', 'r', 'g'), ('l', 'r', 'b'), ('s', 'r', 'g'),
          ('l', 'y', 'b'), ('l', 'y', 'g'), ('s', 'r', 'g'), ('s', 'y', 'g')]

s = ''
for f in flames:
    s += ''.join('0' if f[i] in ('l', 'y', 'g') else '1' for i in range(len(f)))

points = list(s[j:j+6] for j in range(0, len(s), 6))

This gives you an ordering where both points on either side of the halfway point == '011010'. Leaves pointing at individual flames also seems to indicate this interpretation of '0' and '1' is correct. The halfway point also corresponds to a pretty unique pair of flames that overlap with a kind of heart pattern.

There might be a pattern with the even flame lengths (that I can't quite figure out) but it may also just be a 'residue' of the encoding algorithm.

If you skip the first repeating pattern bit then:  5 bits == base32 == number of white squares on the board. I tried a few base32 encodings based on the position of the key with the ribbons (as hex) and the ordering of chessboard numbering schemes and they seem promising but have no way to determine the correct encoding (and brute force isn't an option Grin). I also think that this might not be the correct path since base32 is overkill for hex but not enough for much else but the whole range from '00000' to '11111' is represented (though only 30 unique ones are used).

I don't know is what the 17 leaves is all about, everything else (including the poem; 19 verses) is based upon 19, I tried a few theories but not much came of it.

This is where I'm stuck, I'm pretty sure the decoded flames (as two seperate units) need to be recombined in some way that is indicated in the painting but I just can't see it. I kind of think they are points on an elliptic curve but that's just my current working theory based on watching a random unrelated youtube video where I saw 19 and 38 appearing in an elliptical curve algorithm generated by a proof checker...like I said, totally unrelated. Haven't fully explored this elliptical curve theory but also ran out of time to work on this so...

--edit--

There is definitely some more complex pattern beyond the '100' even length pattern which is why I had the flames grouped in lengths of 4, I can almost see it but not quite there yet.
hero member
Activity: 694
Merit: 500
the popularity of bitcoin have made it to gained global attraction and many people around the world have liked bitcoin also believed that it is the best cryptocurrency you can think about.


your Toilet might be the best place to post this.
hero member
Activity: 694
Merit: 500
to the trolling number: 802529er

this trolling face from the image is for you:

https://imgur.com/a/XClBx
full member
Activity: 140
Merit: 100
Ethereum All The Way Baby
3 years and this is what most of people here see and talk about:
https://imgur.com/a/91nrE



Share what you see then?

I think everyone’s come to the conclusion that the flames are the only logical place for the key to be hidden, while you and white rabbit seem intent on returning attention to the painting.

I think disregarding the entire painting is wrong, I feel that the correct way to decode the flames will be hidden in the bigger picture, but knowing that regardless of what’s inside the panting, the flame border is the most likely place for the private key to of been painted.

Please provide some of your own critiques or information of kindly leave this thread, I’m guessing you’ve got notifications turned on, and that speaks to the kind of person you are, more dedicated to trolling and shitposting then to solving this puzzle.

Maybe try and apply yourself to the task at hand at see where that gets you.

3+ years and your still at the same step everyone else is, or did we all miss you moving the BTC out of the wallet when you solved it with your clearly superior intellect /s

newbie
Activity: 41
Merit: 0
Bi-gram analysis is showing LRPLRP at high frequencies (7) versus about average across the board for the rest.
hero member
Activity: 694
Merit: 500
3 years and this is what most of people here see and talk about:
https://imgur.com/a/91nrE

full member
Activity: 140
Merit: 100
Ethereum All The Way Baby
Seriously...... really ?
People is still trying to solve this riddle?
It is obviously a hoax, just let it go.

LOL, this post from July 2015

and i completely agree with him.

+1

Feedo  are you just here so when someone else or a collective solve this you can be present to witness it, or do you get some rise out of trolling and trying to subvert the discussion?

Anyway, I’m really impressed with the progress everyone’s made just in the last 2 pages, coming together and sharing ideas and information was always at the heart of these types of challenges and puzzles.

While they can be solved independantly, together is the way forwards.
 
hero member
Activity: 694
Merit: 500
Seriously...... really ?
People is still trying to solve this riddle?
It is obviously a hoax, just let it go.

LOL, this post from July 2015

and i completely agree with him.

+1
newbie
Activity: 13
Merit: 0
- DNA/RNA has 4 symbols ('atcg' or 'aucg'), is read three units (called bases) at a time, and there are 21 proteins (indicated by one letter abbreviations).
- The three drips on the chess piece could indicate a reading frame, which tells you how to chunk the bases. A different start point will lead to different amino acids, so this is important.
- DNA/RNA has stop codons, which indicate when to stop making a protein. These can either be delimiters between words, or be considered by their pigment based names (amber, ochre, opal).

seriously? don't tell me that coin-artist have a Phd in medicine or biochemistry Grin
your post could be a good lecture, but of course not here.

It's literally done for you here, no brain power or PhD involved.
https://www.dcode.fr/codons-genetic-code

I don't think it's that far-fetched.
hero member
Activity: 694
Merit: 500
- DNA/RNA has 4 symbols ('atcg' or 'aucg'), is read three units (called bases) at a time, and there are 21 proteins (indicated by one letter abbreviations).
- The three drips on the chess piece could indicate a reading frame, which tells you how to chunk the bases. A different start point will lead to different amino acids, so this is important.
- DNA/RNA has stop codons, which indicate when to stop making a protein. These can either be delimiters between words, or be considered by their pigment based names (amber, ochre, opal).

seriously? don't tell me that coin-artist have a Phd in medicine or biochemistry Grin
your post could be a good lecture, but of course not here.

Rob Myers is the mastermind behind the puzzle.

I read a lot of his work and found he mentioned Peter Saville in one of his blogs.

http://yewknee.com/blog/peter-saville-colour-code/

I know, but i don't think he wrote something about DNA/RNA stop codons!
donator
Activity: 1057
Merit: 1021
- DNA/RNA has 4 symbols ('atcg' or 'aucg'), is read three units (called bases) at a time, and there are 21 proteins (indicated by one letter abbreviations).
- The three drips on the chess piece could indicate a reading frame, which tells you how to chunk the bases. A different start point will lead to different amino acids, so this is important.
- DNA/RNA has stop codons, which indicate when to stop making a protein. These can either be delimiters between words, or be considered by their pigment based names (amber, ochre, opal).

seriously? don't tell me that coin-artist have a Phd in medicine or biochemistry Grin
your post could be a good lecture, but of course not here.

Rob Myers is the mastermind behind the puzzle.

I read a lot of his work and found he mentioned Peter Saville in one of his blogs.

http://yewknee.com/blog/peter-saville-colour-code/
hero member
Activity: 694
Merit: 500
- DNA/RNA has 4 symbols ('atcg' or 'aucg'), is read three units (called bases) at a time, and there are 21 proteins (indicated by one letter abbreviations).
- The three drips on the chess piece could indicate a reading frame, which tells you how to chunk the bases. A different start point will lead to different amino acids, so this is important.
- DNA/RNA has stop codons, which indicate when to stop making a protein. These can either be delimiters between words, or be considered by their pigment based names (amber, ochre, opal).

seriously? don't tell me that coin-artist have a Phd in medicine or biochemistry Grin
your post could be a good lecture, but of course not here.
hero member
Activity: 694
Merit: 500
yari shogi

Fixed for you..., blind and stupid.

to the blind double standard moderator who deleted my comment:
why you didn't see the comment above and delete it?
jr. member
Activity: 51
Merit: 1
There is a short colored flame, and it appears as though it was originally a long colored flame. In both the tiff and the jpeg it looks like this flame was corrected. This flame also happens to be an odd numbered flame, which leads me to believe that the lengths of the odd numbered flames are still important.

Other then that, I spent a fair amount of time today trying to apply Bacon's cypher to various portions of the flames in all sorts of different ways and didn't find anything.

In case anyone wants here is the, somewhat lazily written, code I used today. Those who think a bacon cypher was used may find it useful, it has several options that you should be able to figure out how to use.., but if not feel free to ask.

Code:
#include 
#include
#include

char* shift(char* string);
int getlines(FILE* stream, char** line, ssize_t* nread, int* ntot, size_t len);
char* normalize(char* string, int nflag);
void divideString(char* string);
void debacon(char* string, int btype);
void flip(char* string);
int verbose = 0;


int main(int argc, char* argv[]){
FILE* scram;
int nflag = 0;
int btype = 0;
int bitflip = 0;
char* line[16];
int i = 0;
int j = 0;
int k = 0;
size_t len = 0;
ssize_t nread[16];
int ntot = 0;
char buffer[1024];

if (argc < 3 || argc > 6){
printf("\nERROR: Must specify input file and bacon alphabet\n");
printf("\nUSAGE: ./debacon path_to_input_file bacon_alphabet_A_or_B normalize_type_0_1_2_3 flip_bits_t_or_f\n");
printf("\nUSAGE: Specify -v after all options for verbose mode\n");
printf("\nEXAMPLE: ./debacon file a 2 f -v\n");
exit(EXIT_FAILURE);
}

if(argv[2][0] == 'b' || argv[2][0] == 'B'){
btype = 1;
}

if(argc > 3){
if(atoi(argv[3]) < 4){
nflag = atoi(argv[3]);
}
        if(argv[3][0] == 't' || argv[3][0] == 'T'){
bitflip = 1;
}
if(argv[3][1] == 'v' || argv[3][1] == 'V'){
printf("\nTurning on Verbose Mode...\n");
verbose = 1;
}
}

if(argc > 4){
if(argv[4][0] == 't' || argv[4][0] == 'T'){
bitflip = 1;
}
if(argv[4][1] == 'v' || argv[4][1] == 'V'){
printf("\nTurning on Verbose Mode...\n");
verbose = 1;
}
}

if(argc > 5){
if(argv[5][1] == 'v' || argv[5][1] == 'V'){
printf("\nTurning on Verbose Mode...\n");
verbose = 1;
}
}

scram = fopen(argv[1], "r");

i = getlines(scram, line, nread, &ntot, len);

if(verbose == 1){
for(j = 0; j < i; j++){
printf("Retrieved line of length %zu:\n", nread[j]);
printf("line %i is: %s\n", j, line[j]);
}
printf("Total Characters Read: %i\n\n", ntot);
j = 0;
}

for(j = 0; j < i; j++){
if(verbose == 1){
printf("Normalizeing line %i\n", j);
}
if(bitflip == 1){
flip(line[j]);
}
line[j] = normalize(line[j], nflag);
if(verbose == 1){
printf("Line %i is now: %s\n", j, line[j]);
}
}

j = 1;
strcpy(buffer, line[0]);
for(j = 1; j < i; j++){
strcat(buffer, line[j]);
}

if(verbose == 1){
printf("\nCombined String: %s\n", buffer);
}
for(k = 0; k <= strlen(buffer); k++){
divideString(buffer);
if(verbose == 1){
printf("\nString split into 5's for decoding:\n");
printf("%s\n", buffer);

printf("\nDebaconing:\n");
}
debacon(buffer, btype);

shift(buffer);
}

for(; i > 0; i--){
free(line[i-1]);
}
fclose(scram);
exit(EXIT_SUCCESS);

}

void flip(char *string){
char *i;
for(i=string; *i; i++){
if(*i == '1'){
*i = '0';
} else if(*i == '0'){
*i = '1';
}
}
}

void debacon(char *string, int btype){
char buffer[6];
int i = 0;

while(string[i] != '\0'){

buffer[0] = string[i];
i++;
buffer[1] = string[i];
i++;
buffer[2] = string[i];
i++;
buffer[3] = string[i];
i++;
buffer[4] = string[i];
buffer[5] = '\0';
i+=2;

if(btype == 0){
if(strcmp(buffer, "00000") == 0){
printf("A");
} else if(strcmp(buffer, "00001") == 0){
printf("B");
} else if(strcmp(buffer, "00010") == 0){
printf("C");
} else if(strcmp(buffer, "00011") == 0){
printf("D");
} else if(strcmp(buffer, "00100") == 0){
printf("E");
} else if(strcmp(buffer, "00101") == 0){
printf("F");
} else if(strcmp(buffer, "00110") == 0){
printf("G");
} else if(strcmp(buffer, "00111") == 0){
printf("H");
} else if(strcmp(buffer, "01000") == 0){
printf("J");
} else if(strcmp(buffer, "01001") == 0){
printf("K");
} else if(strcmp(buffer, "01010") == 0){
printf("L");
} else if(strcmp(buffer, "01011") == 0){
printf("M");
} else if(strcmp(buffer, "01100") == 0){
printf("N");
} else if(strcmp(buffer, "01101") == 0){
printf("O");
} else if(strcmp(buffer, "01110") == 0){
printf("P");
} else if(strcmp(buffer, "01111") == 0){
printf("Q");
} else if(strcmp(buffer, "10000") == 0){
printf("R");
} else if(strcmp(buffer, "10001") == 0){
printf("S");
} else if(strcmp(buffer, "10010") == 0){
printf("T");
} else if(strcmp(buffer, "10011") == 0){
printf("V");
} else if(strcmp(buffer, "10100") == 0){
printf("W");
} else if(strcmp(buffer, "10101") == 0){
printf("X");
} else if(strcmp(buffer, "10110") == 0){
printf("Y");
} else if(strcmp(buffer, "10111") == 0){
printf("Z");
} else{
printf(" ");
}
}
if(btype == 1){
if(strcmp(buffer, "00000") == 0){
printf("A");
} else if(strcmp(buffer, "00001") == 0){
printf("B");
} else if(strcmp(buffer, "00010") == 0){
printf("C");
} else if(strcmp(buffer, "00011") == 0){
printf("D");
} else if(strcmp(buffer, "00100") == 0){
printf("E");
} else if(strcmp(buffer, "00101") == 0){
printf("F");
} else if(strcmp(buffer, "00110") == 0){
printf("G");
} else if(strcmp(buffer, "00111") == 0){
printf("H");
} else if(strcmp(buffer, "01000") == 0){
printf("I");
} else if(strcmp(buffer, "01001") == 0){
printf("J");
} else if(strcmp(buffer, "01010") == 0){
printf("K");
} else if(strcmp(buffer, "01011") == 0){
printf("L");
} else if(strcmp(buffer, "01100") == 0){
printf("M");
} else if(strcmp(buffer, "01101") == 0){
printf("N");
} else if(strcmp(buffer, "01110") == 0){
printf("O");
} else if(strcmp(buffer, "01111") == 0){
printf("P");
} else if(strcmp(buffer, "10000") == 0){
printf("Q");
} else if(strcmp(buffer, "10001") == 0){
printf("R");
} else if(strcmp(buffer, "10010") == 0){
printf("S");
} else if(strcmp(buffer, "10011") == 0){
printf("T");
} else if(strcmp(buffer, "10100") == 0){
printf("U");
} else if(strcmp(buffer, "10101") == 0){
printf("V");
} else if(strcmp(buffer, "10110") == 0){
printf("W");
} else if(strcmp(buffer, "10111") == 0){
printf("X");
} else if(strcmp(buffer, "11000") == 0){
printf("Y");
} else if(strcmp(buffer, "11001") == 0){
printf("Z");
} else{
printf(" ");
}
}
}
printf("\n");
}

void divideString(char *string){
int size = strlen(string);
int i = 1;
int j = 1;
int psize;
char buffer[1024];

strcpy(buffer,string);
string[0] = buffer[0];

for(i = 1; j < size; i++){
if(j%5 == 0){
string[i] = ' ';
i++;
}
string[i] = buffer[j];
j++;
}
string[i + 1] = '\0';
}

char* normalize(char* string, int nflag){
char *i;
switch(nflag){
case 0:
for(i=string; *i; i++){
if(*i == 'y'){
*i = '1';
} else if(*i == 'o'){
*i = '0';
} else if(*i == 'g'){
*i = '1';
} else if(*i == 'p'){
*i = '0';
} else if(*i == '\n'){
*i = '\0';
}
}
return string;
case 1:
for(i=string; *i; i++){
if(*i == 'y'){
*i = '0';
} else if(*i == 'o'){
*i == '1';
} else if(*i == 'g'){
*i = '1';
} else if(*i == 'p'){
*i = '0';
} else if(*i == '\n'){
*i == '\0';
}
}
return string;
case 2:
for(i=string; *i; i++){
if(*i == 'y'){
*i = '1';
} else if(*i == 'o'){
*i = '0';
} else if(*i == 'g'){
*i == '0';
} else if(*i == 'p'){
*i == '1';
} else if(*i == '\n'){
*i = '\0';
}
}
return string;
case 3:
for(i=string; *i; i++){
if(*i == 'y'){
*i = '0';
}else if(*i == 'o'){
*i = '1';
} else if(*i == 'g'){
*i = '0';
} else if(*i == 'p'){
*i = '1';
} else if(*i == '\n'){
*i = '\0';
}
}
return string;
default:
printf("\nERROR: Could not normalize stirng\n");
return string;
}
}

int getlines(FILE* stream, char** line, ssize_t* nread, int* ntot, size_t len){
int i = 0;
for(i = 0; i < 16; i++){
nread[i] = getline(&line[i], &len, stream);
if(nread[i] == -1){
break;
}
*ntot += nread[i];
}

return i;
}

char* shift(char* string){
char *nospace = string;
char *temp = string;
char buffer[1024];
buffer;
char t;
int i = 0;
int size;

if(verbose == 1){
printf("\nRemoving Spaces\n");
}
while(*temp != 0){
*nospace = *temp++;
if(*nospace != ' '){
nospace++;
}
}
*nospace = 0;
if(verbose == 1){
printf("\n%s\n", string);
printf("\nShifting String\n");
}
strcpy(buffer, string);
if(verbose == 1){
printf("Buffer: \n%s\n\n", buffer);
}
t = string[0];
size = strlen(string);
for(i = 0; i < (size-1); i++){
string[i] = buffer[i+1];
}
string[i] = t;
string[i+1] = '\0';
if(verbose == 1){
printf("%s\n", string);
}
}


You can feed the program a text file with up to 16 lines, like below. Note though, I didn't dynamically allocate memory, so you are limited to using a character buffer of size 1024 (Which honestly should be more then enough for anything from this puzzle). Also, I didn't do anything fancy as far as combining the lines go, it just concatenates them together, and shifts the bits by 1 each iteration.

Quote
001110
010001011010
110111
010011001110100010010
oyoooyooyoo
oooogppggppgppgp
pgpgppgg

And it'll output something like this

Code:
HELLOTHISISATEST
OIWW GOREREBGJFG
 RNN M CJCICMSKM
ZC  UZ ESEQEZEUZ
SFVXJTUJEJAJSJJT
ELLOTHISISATESTH
IWW GOREREBGJFGO
RNN M CJCICMSKM
C  UZ ESEQEZEUZZ
FVXJTUJEJAJSJJTS
LLOTHISISATESTHE
WW GOREREBGJFGOI
NN M CJCICMSKM R
  UZ ESEQEZEUZZC
VXJTUJEJAJSJJTSF
LOTHISISATESTHEL
W GOREREBGJFGOIW
N M CJCICMSKM RN
 UZ ESEQEZEUZZC
XJTUJEJAJSJJTSFV
OTHISISATESTHELL
 GOREREBGJFGOIWW
 M CJCICMSKM RNN
UZ ESEQEZEUZZC 
JTUJEJAJSJJTSFVX
THISISATESTHELLO
GOREREBGJFGOIWW
M CJCICMSKM RNN
Z ESEQEZEUZZC  U
TUJEJAJSJJTSFVXJ
HISISATESTHELLOT
OREREBGJFGOIWW G
 CJCICMSKM RNN M
 ESEQEZEUZZC  UZ
UJEJAJSJJTSFVXJT
ISISATESTHELLOTH
REREBGJFGOIWW GO
CJCICMSKM RNN M
ESEQEZEUZZC  UZ
JEJAJSJJTSFVXJTU
SISATESTHELLOTHI
EREBGJFGOIWW GOR
JCICMSKM RNN M C
SEQEZEUZZC  UZ E
EJAJSJJTSFVXJTUJ
ISATESTHELLOTHIS
REBGJFGOIWW GORE
CICMSKM RNN M CJ
EQEZEUZZC  UZ ES
JAJSJJTSFVXJTUJE
SATESTHELLOTHISI
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[the reocurring patterns are as follow:

Actually, on second look, I don't think this is statistically significant.  Out of a sample of 25, you have 16 unique values which is not surprising.  You're bound to get repeats even if you look at random data.  If the sample size was much larger and you still got few values, that would mean something more I think.  But that doesn't mean the thin/thick data can't also be part of the solution.

well, don't forget you are taking it out of 6 bit pool, which has 64 possible values (if I am not mistaken that late at night). If they were taken randomly from a pool of a milion values, would you still expect to have 9 repeats? If so, I should ask you to mark your numbers on my next lotto coupon Cheesy Anyway, I am going to pursue that path for a while and you will do what you choose, but I will just not do it tonight anymore Smiley
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[the reocurring patterns are as follow:

Actually, on second look, I don't think this is statistically significant.  Out of a sample of 25, you have 16 unique values which is not surprising.  You're bound to get repeats even if you look at random data.  If the sample size was much larger and you still got few values, that would mean something more I think.  But that doesn't mean the thin/thick data can't also be part of the solution.
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Merit: 39
It is extremely interesting. Maybe even a breakthrough post, for me. If you divide it in groups of 6 (exactly how alphabetcanary did for lengths earlier) you'll notice that the same patterns reoccure all over again:

Wow, actually that would be a game changer if true!  I will try to verify.

the reocurring patterns are as follow:
Code:

blob bit string hex representation count
010011 13 3
000011 03 2
000111 07 2
010111 17 4
010111 17 4
000011 03 2
100011 23 1
010111 17 4
000100 04 1
001111 0F 2
110001 31 1
100110 26 1
001000 08 1
110100 34 1
000110 06 2
010011 13 3
010011 13 3
000110 06 2
001010 0A 1
010111 17 4
000111 07 2
111101 3D 1
001110 0E 1
110110 36 1
001111 0F 2
00 0 -


not very random
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EDIT: still, they can be a mapping into more random data (a board has been mentioned)

Yes, that's what I'm suspecting too
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OK, I'll share what I've had for a bit in the interest of getting to solve this. I just hope fellow puzzle enthusiast strangers on the internet can acknowledge the contributions of others.
I'm pretty positive it's a DNA/RNA encoding to amino acids (protein building blocks) in some fashion.

- The game of life/life/love topic is relevant.
- The WR portal with the QR code from an earlier stage featured a colored blob. It definitely looks like a volume filled protein structure. Not sure which one.
- DNA/RNA has 4 symbols ('atcg' or 'aucg'), is read three units (called bases) at a time, and there are 21 proteins (indicated by one letter abbreviations).
- The three drips on the chess piece could indicate a reading frame, which tells you how to chunk the bases. A different start point will lead to different amino acids, so this is important.
- DNA/RNA has stop codons, which indicate when to stop making a protein. These can either be delimiters between words, or be considered by their pigment based names (amber, ochre, opal).

Consider only the 4 color combinations of the flames (RB RG YB YG) and treat them as bases (atcg).
Consider the permutations of the bases, reading order of the strips, and reading direction.
Consider reading frames (just in case).

Just looking at the inner colored flames, many combinations will produce keys within the range of a minikey (~29 or 30, with/without omitting stops).
Combining the inner and outer colored flames, all combinations are within range of a privkey, as has been noted (50, without omitting stops).

The question is if this is a key:
What start, order, and direction (and maybe frame, if it changes per strip), and if capitalization is also encoded somehow.

Stuff to rule out:
If DNA/RNA, there is an amino acid that indicates when to start, but I've tried those keys.
I've read according to the spirals, and combos of CCW & CW, as well as many other combos.
Some combinations and starts do indeed encode BIP38 beginnings, but I haven't gotten anywhere fruitful.
Using the ribbons as capitalization doesn't work.
Inner triplets are not consistently one height, if flame height encodes capitalization.
I don't see a pattern in the outer flames that would indicate how to capitalize the inner flames.

If it's not a key:
Translating it could also produce a phrase which points to the next step, and there are legible words (and maybe phrases) in the most likely permutation and frame, but not all of them are. There is one word which relates directly to the painting when unscrambled. Substitution, rearrangement as an anagram, caesar, and vigenère don't go anywhere as a whole that I can see.
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Large scale, green crypto mining ICO
You think very predictably
That was what it was about creating these puzzles

said Yoda
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