Myst without finishing Ahnonay or Ercanna??

angelmyst · **Posted:** Tue May 14, 2013 1:06 pm

I did not think this was possible.
[spoiler]https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn2/969197_169345699896087_623781247_n.jpg[/spoiler]
[spoiler]I had only gone into Ercanna to get the book on my shelf, I had done zero...zip..nadda in the age. ..When I went through fissure and my..angelbuds...fissure party I took a wild guess at imager and got glyph to Myst..has anyone ever done this?[/spoiler]

Calumon · **Joined:** Wed Aug 22, 2007 9:38 pm **Posts:** 1669

If you're randomly entering the code into the imager, chances are you'll stumble upon the Pellet Cave Code by accident. I've read quite a few stories of people incorrectly entering their Pillar Cave Code and getting their Pellet Cave Code instead.

Charura · **Joined:** Wed Oct 14, 2009 4:14 pm **Posts:** 1762

Calumon wrote:

If you're randomly entering the code into the imager, chances are you'll stumble upon the Pellet Cave Code by accident.

If you got it the first time you tried, I would suggest you buy some lottery tickets! :shock:

Ehren · **Posted:** Tue May 14, 2013 8:33 pm

The pellet cave code and the pillar cave code are both randomly generated before even leaving Relto for the first time. So you could technically get to K'veer and/or rainy Cleft after only going through Relto's Cleft totem.

Also, in a way, there aren't that many imager codes, no symbol is ever used twice in any one code and ANY of the three codes (including the intro one which is always the same for everyone) can be rotated any direction and still work, as long as the symbols are correct relative to each other!

Mac_Fife · **Posted:** Tue May 14, 2013 9:19 pm

Yeah, once you know how it works it's pretty easy to "brute force" your way to K'veer, I've done that with a few avvies. And sometimes a nicely timed jump and screen grab in the upper pellet cave can give you a couple of symbols as a starter

.

Annabelle · **Posted:** Tue May 14, 2013 10:05 pm

There are 7 symbols.

7 x 6 x 5 x 4 = 840 possibilities. You were really lucky to find it on the first try :shock:

.

It's easier to turn all your graphic settings to the max, do as Mac_Fife instructed, this way you only have to find 6 x 5 x 4 = 120 or even better 5 x 4 = 20

Ehren · **Posted:** Tue May 14, 2013 10:59 pm

Annabelle wrote:

There are 7 symbols.

7 x 6 x 5 x 4 = 840 possibilities. You were really lucky to find it on the first try :shock:

.

There are 209 possibilities.

((7 x 6 x 5 x 4) \ 4) - 1 = 209

Each code is treated the same as 3 other codes (the same code but rotated differently) so that makes it like guessing 4 codes at once each try. It's like if the code were 1234 but 2341, 3412, and 4123 worked the same. Also need to remove 1 code which is always used for the intro speech, since it in theory should be programmed not to land on the same code or the 3 related to it... (would have to subtract 4 if it was done before the dividing.)

Annabelle · **Posted:** Tue May 14, 2013 11:45 pm

Maths silliness :roll:

Yeah I just made it with oranges and apples and realistic objects.

I went magically from 20 possibilities down to 10.

1234
1235
1236
1237
1245
1246
1247
1256
1257
1267

I typed "Combinatoire" in Wikipédia and it was like mandarin to me. How can one understand those symbols? :lol:

I decided to just go for a dummy (--> me) proof.

Given the fact I know symbol 1 and symbol 2:

I should be able to have 5 x 4 = 20 possibilities. But then entering that on Excel, putting all the 4 on same column, 5..., 6... and looking at my 20 rows...of 7 columns, I saw that 10 of them where identical to another 10.

-----

I guess I have once made a mixed up in my ordinance ending up repeating over and over and over bad tries to end up over 100 tries with 1 symbol. :roll:

So ok let me try to be smart tonight :lol:

:

with no symbol: (7 x 6 x 5 x 4)/4 - 1 = 209 (you gave that one)
with 1 known symbol: (6 x 5 x 4)/3 - 1 = 39 <--- am I right?
with 2 known symbols: (5 x 4)/2 - 1 = 9 (that one, Excel gave me it :wink:

)
with 3 known symbols: ... harder to calculate than to find out :lol:

angelmyst · **Posted:** Wed May 15, 2013 3:39 am

Lol Math gives me a headache...
I was amazed when I changed two symbols and Yeesha appeared with her you have done well speach and the glyph appeared.....just did not think it possible!!

chrissifniotis · **Posted:** Wed May 15, 2013 9:54 am

Annabelle wrote:

...

So ok let me try to be smart tonight :lol:

:

with no symbol: (7 x 6 x 5 x 4)/4 - 1 = 209 (you gave that one)
with 1 known symbol: (6 x 5 x 4)/3 - 1 = 39 <--- am I right?
with 2 known symbols: (5 x 4)/2 - 1 = 9 (that one, Excel gave me it :wink:

)
with 3 known symbols: ... harder to calculate than to find out :lol:

Essentially yes.

The math should be the same; where the equation for no symbols is [(7 x 6 x 5 x 4)/4] - 1, the equation for 3 known symbols should be (4/1) - 1 = 4 - 1 = 3. Should be.

EDIT: Parentheses!

Alien · **Posted:** Wed May 15, 2013 3:48 pm

Annabelle wrote:

...

So ok let me try to be smart tonight :lol:

:

with no symbol: (7 x 6 x 5 x 4)/4 - 1 = 209 (you gave that one)
with 1 known symbol: (6 x 5 x 4)/3 - 1 = 39 <--- am I right?
with 2 known symbols: (5 x 4)/2 - 1 = 9 (that one, Excel gave me it :wink:

)
with 3 known symbols: ... harder to calculate than to find out :lol:

chrissifniotis wrote:

Essentially yes.

The math should be the same; where the equation for no symbols is [(7 x 6 x 5 x 4)/4] - 1, the equation for 3 known symbols should be (4/1) - 1 = 4 - 1 = 3. Should be.

EDIT: Parentheses!

Forgive me if I have this totally wrong ...
3 known symbols = 1 x number of possible symbols (n) - 3 = number of possible combinations
2 known symbols = (n x n) - 2 = number of possible combinations
1 known symbol = (n x n x n) - 1 = number of possible cominations

dtierce · **Posted:** Thu May 16, 2013 6:49 am

Alien wrote:

Forgive me if I have this totally wrong ...
3 known symbols = 1 x number of possible symbols (n) - 3 = number of possible combinations
2 known symbols = (n x n) - 2 = number of possible combinations
1 known symbol = (n x n x n) - 1 = number of possible cominations

Close, but not quite right. Let's look at the case of 1 known symbol.
You start with the one known symbol. Since the symbols don't repeat, you next choose from the n-1 remaining symbols. The third spot is chosen from n-2 symbols because two have already been chosen. Lastly, you chose from the n-3 remaining symbols. That gives you 1x(n-1)x(n-2)x(n-3) possibilities for the 4 spots on the imager with one known symbol. If more symbols are known, you substitute a 1 for the next parenthesis in the equation. If no symbols are known, change the first "1" to an "n" to represent the possible choices for the first symbol.

Since any given pattern of four symbols can be rotated up to 4 steps without changing the effect on the imager, you can divide the whole set of possibilities by 4 resulting in the equation (1x(n-1)x(n-2)x(n-3))/4.

You can also eliminate one pattern for each pattern already known. For example, you know that the unknown pattern will NOT be the same as the pattern for the initial Yeesha speech which is found inscribed on the cleft wall.

The final equation for the number of possible patterns with one known symbol is ( ( 1 x (n-1) x (n-2) x (n-3) ) / 4 ) - 1. I think someone already pointed out that n=7 meaning there are seven possible symbols coded into the machine for each spot.
If you know the pattern from the Bahro pillar cave, you can subtract one more.

I hope this makes the math a little clearer?? :roll:

DLordofTime · **Posted:** Thu May 16, 2013 3:21 pm

You need to subtract one before dividing by four, because the Yeesha speech code can also be rotated...

Alien · **Posted:** Thu May 16, 2013 5:33 pm

Rather than trying to work it out with formulae,
maybe it would be better if someone simply went in, listed all the combinations, and posted here ...

"If you know no symbols, there are ... possible combinations
If you know one symbol, there are ... possible combinations ..." etc.

Charura · **Joined:** Wed Oct 14, 2009 4:14 pm **Posts:** 1762

Alien wrote:

Rather than trying to work it out with formulae,
maybe it would be better if someone simply went in, listed all the combinations, and posted here ...

"If you know no symbols, there are ... possible combinations
If you know one symbol, there are ... possible combinations ..." etc.

I nominate...hmmm...Let's see.... ALIEN !!! :twisted:

Myst without finishing Ahnonay or Ercanna??

Who is online