Myst without finishing Ahnonay or Ercanna??

Alien · **Posted:** Thu May 16, 2013 7:50 pm

Charura wrote:

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:

It's the sort of mind numbing boring useless thing I tend to do

(I'll work it all out and post it here tomorrow)

angelmyst · **Posted:** Thu May 16, 2013 10:17 pm

WOW....ask a simple question get a super complicated answer...lol
I guess it can be done....thanks. The mathematical possibilities are not the issue for me...but the possibilities of randomly changing the symbols one time and getting the glyph are astronomical....so I am one lucky angel..lol
I do not gamble, so I guess I will never know about the lottery possibilities.

Alien · **Posted:** Thu May 16, 2013 10:28 pm

angelmyst wrote:

WOW....ask a simple question get a super complicated answer...lol
I guess it can be done....thanks. The mathematical possibilities are not the issue for me...but the possibilities of randomly changing the symbols one time and getting the glyph are astronomical....so I am one lucky angel..lol
I do not gamble, so I guess I will never know about the lottery possibilities.

I have the results and they might surprise you.
but it's late for me and I need sleep ... I'll explain it all tomorrow.

chrissifniotis · **Posted:** Fri May 17, 2013 11:00 am

angelmyst wrote:

WOW....ask a simple question get a super complicated answer...lol
I guess it can be done....thanks. The mathematical possibilities are not the issue for me...but the possibilities of randomly changing the symbols one time and getting the glyph are astronomical....so I am one lucky angel..lol
I do not gamble, so I guess I will never know about the lottery possibilities.

I'd say so. With the numbers we have now you achieved an outcome with a probability of less than half of one percent; <0.5%. Nice one.
Maybe you should try your luck a bit with the lotteries. :wink:

Just don't figure out the chances, they're in the millions at least.

DLordofTime wrote:

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

No you don't. In the equation -1 already represents the four rotations of the same combination, because the rotations have been removed via the denominator of the fraction. Placing the negative number into the numerator means that the single combination is taken into account - dividing minus one by four; this results in a decimal number of possibilities. I know this because I've been typing and balancing the equations in this very reply box for the past two hours; one can balance this out by changing -1 to -4 and then replacing the negative number into the numerator, but the last equation - 3 known symbols - produces 0.
The negative number has to detached from the fraction altogether. I didn't get 100% in probability for nothing.

Annabelle · **Posted:** Fri May 17, 2013 11:30 am

chrissifniotis wrote:

The negative number has to detached from the fraction altogether. I didn't get 100% in probability for nothing.

I'll go with his explanations then :lol:

I'll bet on him I won't loose

Alien wrote:

I have the results and they might surprise you.
but it's late for me and I need sleep ... I'll explain it all tomorrow.

I'm not sure it was necessary Alien. But since you did it, let's see

Alien · **Posted:** Fri May 17, 2013 12:04 pm

OK ... So to work out the possible combinations, we have to follow three rules :
1) No symbols are repeated in any of the slots
2) The order must be different even if shifted around 90 or 180 degrees
3) No combination to be repeated

First I tried to work out 3 known symbols, I thought that if there are 7 symbols, and you know three, then that leaves four possible combinations. I realised later, this was wrong and there are in fact more possible combinations.

Then I set to working out 2 known symbols and arranged them in a grid, like so :

[spoiler]

[/spoiler]

As you can see, each square represents the Cleft imager with each slot.
Each symbol has a number (it doesn't matter what symbol has which number).
Each row has the same symbol in slot 1, and each column has the same symbol in slot 2.
There is a gap diagonally as that would be the same symbol in slot 1 & 2 (breaking rule 1)

This focuses ONLY on the two known symbols and ignores the two unknown symbols (I'll get to those in a minute)
As you can see there are 6 combinations per row, and 7 rows, therefore 6 x 7 combinations of 2 known symbols (= 42)

Let's just take the first combination with 1 & 2 in the first two slots (top row, left side)
Here are every possible combination of symbols for that combination of 2 known symbols :

[spoiler]

[/spoiler]

This time the row is for the 9 o'clock slot and the columns for the 6 o'clock slot.
This again follows the three rules set out above to ensure no combinations are repeated.
As you can see here, there are 20 possible combinations just for the two known symbols being 1 & 2 in the 12 o'clock and 3 o'clock positions respectively.

So if every possible combination of 2 known symbols has 20 combinations of 2 unknown symbols,
our maths is now (6 x 7) x 20 = 840

But what about 3 known symbols or 4 known symbols I hear you ask?
I realised that 1 known symbol has the same combinations as no known symbols. Then I realised what I had done is work out the total number of combinations possible regardless of how many symbols

so ...

No known symbols = 840 combinations
1 known symbol = the same 840 combinations
2 known symbols = the same 840 combinations

This was when I suspected that even 3 known symbols or 4 known symbols would still be the same 840 combinations.

The mistake we have fallen into in this thread is to over-complicate this.
If nothing else, I would like to think my grid system is a better way of working out how many combinations than trying to get a formula right.

EDIT: I made a correction ... see my next post in this thread

chrissifniotis · **Posted:** Fri May 17, 2013 1:24 pm

I appreciate the work you've put in to this, but there is a serious flaw in your working out.
Firstly, I know that there are a total of 840 possible combinations in every equation, the numerator of the equation expresses this. The reason the numerator changes as each variable is defined in the equations is exactly that - each variable is defined, thus replacing the highest number in the combination sequence with 1. Your method begins by expressing the first 2 variables as variables, but then their definition changes midway through the working out process as defined numerals. You didn't begin with 2 defined variables, and thus at the start all 4 were undefined. In effect, you inadvertently changed the appearance of the numerator; your example expresses it as (7 x 6)(5 x 4) - 20 combinations in each of the 42 2-known symbol examples. The result is the same as discovering 120 combinations in each of the 7 1-known symbol examples - 7(6 x 5 x 4) - or 4 combinations in each of the 210 3-known symbol examples - (7 x 6 x 5)4, all three sets equal to 840.
Secondly, while your work highlights the fact that there is a constant number of combinations in the equations, which is not at all in dispute, the prior equations express the probability of achieving a certain sequence of symbols while following the rules of the given causality; the given causality is expressed in the remainder of the equation - [(combination equation)/4] - 1.

If I sound offensive or in any other way unpleasant I'm very sorry and I certainly don't mean to be, that's the last thing I want to achieve here, but what you've just done is solve the numerator of the current equation.

Alien · **Posted:** Fri May 17, 2013 1:41 pm

chrissifniotis wrote:

If I sound offensive or in any other way unpleasant I'm very sorry and I certainly don't mean to be, that's the last thing I want to achieve here, but what you've just done is solve the numerator of the current equation.

Not at all ... this is a forum, and forums debate things.

Anyway, I was wrong about something, so let me correct that ...
no known symbols = 840 combinations
1 known symbol = 120 combinations
2 known symbols* = 20 combinations
3 known symbols* = 4 combinations
4 known symbols* = 1 combination

*= this assumes you know the symbols positions relative to each other

I know that how I have worked it out is correct (I know what I mean, explaining it is another matter)
If you all want to continue debating this with fancy formulae after I answered the original question (how many combinations with a given number of known symbols), then you all go ahead, but for me, problem solved.

DLordofTime · **Posted:** Fri May 17, 2013 3:36 pm

Something told me that there was a serious flaw in your logic, Alien, when you said that there were 840 combinations for 4 known symbols. After all, if you know all four, there can only be one combination!
Still, good work on figuring it out.

@Chrissifniotis: Ah, my bad.

Ehren · **Posted:** Fri May 17, 2013 7:50 pm

Alien wrote:

chrissifniotis wrote:

If I sound offensive or in any other way unpleasant I'm very sorry and I certainly don't mean to be, that's the last thing I want to achieve here, but what you've just done is solve the numerator of the current equation.

Not at all ... this is a forum, and forums debate things.

Anyway, I was wrong about something, so let me correct that ...
no known symbols = 840 combinations
1 known symbol = 120 combinations
2 known symbols* = 20 combinations
3 known symbols* = 4 combinations
4 known symbols* = 1 combination

*= this assumes you know the symbols positions relative to each other

I know that how I have worked it out is correct (I know what I mean, explaining it is another matter)
If you all want to continue debating this with fancy formulae after I answered the original question (how many combinations with a given number of known symbols), then you all go ahead, but for me, problem solved.

How did you get up to 840? With no know symbols it is 210 or 209 if you count the original code not being used again.

I don't know about all this 1, 2, 3 know symbol stuff, that gets complicated, but I do know that without knowing any symbol there certainly aren't more than 210 combinations. Remembering that rotations are counted as the same code.

I guess if you are just talking about different symbol arrangements rather than different distinct possible solutions, then you have 840.

QUICKEDIT
Looking at your post again you may in fact have been talking about different ways the symbols can be arranged without bothering with the "if it would be counted the same as other arrangements" part...

korovev · **Posted:** Fri May 17, 2013 11:34 pm

To summarize and generalize:

Code:

( (n - r)! / (n - k)! ) / (k - r) - t        n >= k > r

with n: available symbols, k: symbols in the combination, r: known symbols, t: known combinations

In the Cleft, n=7, k=4, r=..., t=1.
This is a fancy formula and now the problem is solved :wink:

Annabelle · **Posted:** Fri May 17, 2013 11:49 pm

korovev wrote:

To summarize and generalize:

Code:

( (n - r)! / (n - k)! ) / (k - r) - t        n >= k > r

with n: available symbols, k: symbols in the combination, r: known symbols, t: known combinations

In the Cleft, n=7, k=4, r=..., t=1.
This is a fancy formula and now the problem is solved :wink:

Problem solved with this? :lol:

I think I will have a migraine soon

chrissifniotis · **Posted:** Sat May 18, 2013 1:03 am

korovev wrote:

To summarize and generalize:

Code:

( (n - r)! / (n - k)! ) / (k - r) - t        n >= k > r

with n: available symbols, k: symbols in the combination, r: known symbols, t: known combinations

In the Cleft, n=7, k=4, r=..., t=1.
This is a fancy formula and now the problem is solved :wink:

Hm, forgot about exclamation points. (says the probability egotist! :oops:

) It'll clean up the numerator nicely, apart from switching the generalization to n > k >= r since k is the symbols in combination and r is the known symbols in combination; both being definitely less than n and potentially equal to each other, I'm okay with that formula.

dtierce · **Posted:** Sat May 18, 2013 6:11 am

FYI...
For those unfamiliar with combinatorial mathematical notation (which I tried to avoid), the exclamation ("!") is called a "factorial" operator. It represents the product of a number with every lesser number down to 1. For example, the number 4! represents 4x3x2x1 which results in a value of 24. This can be described as the total possibilities for the first choice combined with the total possibilities for second, etc, until all others have been chosen and only 1 choice remains.

The "factorial" operator nicely produces the number of combinations of selections where each selection is unique. It is often modified for subsets by dividing by the factorial of the portion NOT used. For example, every possible combination of 7 symbols is 7! Every possible combination of 4 out of 7 symbols is 7! / (7-4)! or otherwise 7!/3!. Spelling it out results in 7x6x5x4 because dividing by 3! cancels out the 3x2x1 portion of the 7! That results in 840 total possible combinations for the four spots on the imager.

As I said, this is EVERY combination. Since the combination rotated four ways really counts as one combination, the total count of possibilities should be divided by 4. That results in 210 effective combinations.

Knowledge of one of the symbols changes the first choice from 7 to 1 and also eliminates that symbol from subsequent choices. Therefore, knowing one symbol changes the "7x6x5x4" into "1x6x5x4" without changing any of the rest of the logic or math. It just reduces the number of possibilities for first choice from 7 to 1. Knowing the second symbol likewise reduces the next number in the sequence from 6 to 1, and so on.

IF you did not know the relative positions of these two known symbols, you would have to multiply the overall equation by the possible relative positions for them. Two symbols, accounting for rotation, introduces a multiplier of 3. The second known symbol can have 3 positions on the imager relative to the first. This consideration is irrelevant, however, since we are talking about knowing the position as well as the value for the second symbol when two symbols are known.

David Tierce

korovev · **Posted:** Sat May 18, 2013 8:40 am

chrissifniotis wrote:

apart from switching the generalization to n > k >= r

I see a little problem if k = r :lol:

Also, n >= k is not really necessary (it's there only to avoid the negative in the factorial): the Golden Dome in Riven had (ignoring the 6th one) 5 colors (n) but 625 positions (k).

dtierce wrote:

For those unfamiliar with combinatorial mathematical notation (which I tried to avoid), the exclamation ("!") is called a "factorial" operator

Yep, sorry if I took that for granted :oops:

But it's such a handy operator

Myst without finishing Ahnonay or Ercanna??

Who is online