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What comes next...

Started by Will, Thu, 26 Jan 2017 20:55

Robert Staudinger

I don't Play Skat either and don't know the rules, but a query with the search machine helped.
Robert

martin

QuoteFind the continuous pattern in this sequence and continue this pattern.
But I keep thinking that first the tester would have to prove that there exists "one and only one" pattern at all.
And I have a feeling that cannot be proved, certainly not within the finite (and usually short) length of the given sequence.

Cheers,
Martin

Will

Yes, it is indeed the bidding sequence for Skat. And players who speak German will often omit the tens digit so that the bidding goes faster. "Zwei" is less to say than "zewiundzwanzig."

I did not expect that anyone would come up with a mathematical rule, since there isn't one. The sequence starts with 18 and is successive multiples of 9, 10, 11, and 12, with four outliers thrown in: 23, 35, 46, and 59, representing null, null hand, null ouvert, and null ouvert hand.

But I did think that perhaps a German-speaking Skat player would recognize it. :-)

I learned how to play Skat this past year and I really fell in love with it. The problem is that nobody here knows how to play, so I'm stuck using a pretty good app or else finding online games to join. I've tried to teach my friends, but their eyes glaze over after the 25th rule and the 25th exception... There is always Trump, except when there isn't. Jacks are always highest, except when they aren't. Tens are higher than kings, except when kings are higher than tens. Getting 61 points wins you the hand, but you only get 18 points to add to your score. And so on.

I sure love it though. It's very fun.
Will /Chicago /USA

Hardy Heinlin

Quote from: martin on Sun, 29 Jan 2017 19:27
QuoteFind the continuous pattern in this sequence and continue this pattern.
But I keep thinking that first the tester would have to prove that there exists "one and only one" pattern at all.
And I have a feeling that cannot be proved, certainly not within the finite (and usually short) length of the given sequence.

Cheers,
Martin

By "continuous" I mean a continuously increasing or decreasing or static value, not a toggling value like 2, 3, 2, 3 ... or 1, 2, 3, 1, 2, 3 ... If that is defined, I think it's a fair task. If one just says "write down the next number" there isn't any definition, and one could write down any random number.

If you look at the examples on the previous page, and define the task like I did, I see just one solution:

8, 0, 2, 3, 4, 7, 0, 3, 5, 6 ... <-- Starting sample

8, 0, 2, 3, 4, 7, 0, 3, 5, 6, 6, 0, 4, 7, 8, 5, 0, 5, 9, 10 ...

To see the whole picture a bit better, split the sequence in 5-packs, and look at the vertical pattern in each column:

8, 0, 2, 3, 4
7, 0, 3, 5, 6
6, 0, 4, 7, 8
5, 0, 5, 9,10


Or the other example:

3, 7, 13, 21, 31 ...

Check the intervals, they go 4, 6, 8, 10 ... So it continues with intervals of 12, 14, 16 etc.:

3, 7, 13, 21, 31, 43, 57, 73, 91, 111 ...

What other pattern would work without breaking the original continuity? I can't see any alternative ...


Cheers,

|-|ardy

Will

8, 0, 2, 3, 4, 7, 0, 3, 5, 6 ... <-- Starting sample

Hardy's guess: 8, 0, 2, 3, 4, 7, 0, 3, 5, 6, 6, 0, 4, 7, 8, 5, 0, 5, 9, 10 ...

Actual sequence: 8, 0, 2, 3, 4, 7, 0, 3, 5, 6, 0, 4, 5, 6, 8, 0, 4, 5, 9, 0...


The sequence is successive multiples of 9, 10, 11, and 12, starting with 18, with four semi-random values thrown in (23, 35, 46, and 59, chosen because they are just short of a multiple of 12), and then with the tens digit removed.

This is one that needs to be solved with familiarity of the sequence, not through a mathematical rule.

Kind of like this:

7, 8, 5, 5, 3, 4, 4, 6, ...?

Those are the number of letters in the English words for the months of the year. The next in sequence would be 9 (September), 7 (October), 8 (November), etc...
Will /Chicago /USA

Hardy Heinlin

Continue this sentence:

"Hello echo!"

Zapp

Quote from: Hardy Heinlin on Sun, 29 Jan 2017 20:59
.........

Or the other example:

3, 7, 13, 21, 31 ...

Check the intervals, they go 4, 6, 8, 10 ... So it continues with intervals of 12, 14, 16 etc.:

3, 7, 13, 21, 31, 43, 57, 73, 91, 111 ...

What other pattern would work without breaking the original continuity? I can't see any alternative ...
.....................

Well as I said, in this case the key info is that it is a polynomial sequence. As you correctly spotted, the first differences in this sequence are 4, 6, 8, 10, and the second differences are 2,2,2, hence the third differences are 0,0. It follows that there exists one (and only one) second degree polynomial through the given points.

Obviously, as a mathematician, once I have proved that a solution exists and that it is unique, I lose interest and forget to give the actual solution ..... :-))))

But wait .... what about third degree polynomials? Well, we can rule them out by specyfying that we are looking for the simplest solution.

Bye

Andrea

martin

#27
Quote from: HardyOr the other example:                                                                             
3, 7, 13, 21, 31 ...                                                                                           
Check the intervals, they go 4, 6, 8, 10 ... So it continues with intervals of 12, 14, 16 etc.:               
3, 7, 13, 21, 31, 43, 57, 73, 91, 111 ...                                                                     
What other pattern would work without breaking the original continuity? I can't see any alternative ...

Let's try*...                                                                                                 
3, 7, 13, 21, 31, 35, 41, 1085, 1093, 1097, 1109, 1199021, 1199069, 1199073, ...         
                                                                                                               
Note monotonous increase as stipulated.                                                                       
                                                                                                               
  ;)                                                                                                           
                                                                                                               
* ...and see if I can get away with it. Pity there are mathematicians around.  :D     

Hardy Heinlin

There is an interval sequence of 10 then 4 in the transition to your continuation. This kind of break doesn't exist in the original sequence. It makes the original pattern inconsistent.

3, 7, 13, 21, 31, 35, ... (intervals of 4, 6, 8, 10, 4)

It would continue the original idea if the original sequence looked like this:

-7, 3, 7, 13, 21, 31, ...

But this information (-7) is not given.


-|-|-

martin

Quote from: HardyThere is an interval sequence of 10 then 4 in the transition to your continuation. This kind of break doesn't exist in the original sequence. It makes the original pattern inconsistent. [...]  (intervals of 4, 6, 8, 10, 4)

Not sure if I understand: The above is correct, but "intervals of 4, 6, 8, 10, 4, ..." would in any case only explain the next two numbers (35, 41) in my continued series, but not the following (1085, 1093, ...)

So the rule "intervals of 4, 6, 8, 10, 4, ..." cannot be a candidate anyway. Or?

Cheers,
Martin

(And no, this time no random numbers are involved. :) )

Hardy Heinlin

It cannot be a candidate, in my opinion. The rest doesn't fit either, in my opinion. There are jumps that the original sequence doesn't have.

Imagine a graph of the original sequence, then continue this graph so that nobody can see where your work starts. Don't add a new pattern. The beginning of a new pattern would reveal where your work starts. You must avoid that.

Zapp

It all depends on the class of functions you can accept as a solution.

For instance, if you consider the class of polynomials up to degree 2, you have just one solution; if you allow for higher degree polynomials, you can have infinite solutions; just for fun, here are some of them:

1. + 1. x + 1. x^2
1. + 1. x + 1. x^2 + 3.69977*10^-16 x^3
1. + 1. x + 1. x^2 - 1.06943*10^-15 x^3 + 8.21197*10^-17 x^4
0.9221 + 1.17787 x + 0.853937 x^2 + 0.0551793 x^3 - 0.00973753 x^4 +
0.000649169 x^5

(coefficients have been rounded but can be calculated with arbitrary precision)

Interestingly, if you fit the restriction of a Fourier series in a specified subdomain, you can have repeating patterns as well. Of course, you'll find sine and cosine (or complex exponential) functions in your answer, which someone would call playing foul, but it all depends on what you can allow.

Andrea

martin

Fourier series, too. Nice!  :D

(Philosophical question: "coefficients have been rounded but can be calculated with arbitrary precision"
Is that the same as having a series of integers?)

What I actually had in mind requires some formatting, therefore was put on this web page. Let's see if it "survives"...

Cheers,
Martin

Hardy Heinlin

Where do you see your four rules in the original sequence? :-)

In scientific words: Your rules are not based on empiric observations, and thus are pure speculation :-)

If Sherlock Holmes would see a shoe, a hat, and a shirt, would he think they belong to a rhino, a bird, and a bee? :-) Where's the link?


Cheers,

|-|ardy

martin

Hmm, have I overlooked something?

Without pretty formatting:

The original sequence is  3, 7, 13, 21, 31, ...

n1 =  3 

Rule #1: n2 =  7 = n1 + 4 
Rule #2: n3 = 13 = 2nd prime after n2
Rule #3: n4 = 21 = n1 * n2
Rule #4: n5 = 31 = 3rd prime after n4


I crafted the rules to fit the original sequence, but left it as an exercise for the reader to check.  :P
Mistake somewhere?

Quote from: HardyIf Sherlock Holmes would see a shoe, a hat, and a shirt, would he think they belong to a rhino, a bird, and a bee?
Of course not. He would think of Irene Adler.

Cheers,
Martin

Hardy Heinlin

Got it!

I think you're right :-)

The solution is not beautiful, but it confirms my very first comment in this thread ... :-)


Cheers,

|-|ardy

Zapp

Quote from: martin on Tue, 31 Jan 2017 16:20
Hmm, have I overlooked something?

Without pretty formatting:

The original sequence is  3, 7, 13, 21, 31, ...

n1 =  3 

Rule #1: n2 =  7 = n1 + 4 
Rule #2: n3 = 13 = 2nd prime after n2
Rule #3: n4 = 21 = n1 * n2
Rule #4: n5 = 31 = 3rd prime after n4


I crafted the rules to fit the original sequence, but left it as an exercise for the reader to check.  :P
Mistake somewhere?

Cool! Of course, it hardly qualifies as a polynomial sequence .... Nice nonetheless.

And I think we all agree on the main point: either you specify what the student is expected to do, and in that case it becomes an exercise of some sort, which ultimately involves solving a system of equations, or you don't, and in that case you should expect the unexpected (and be ready to accept it as a legitimate answer!)

Re:, being able to determine the coefficient to arbitrary precision, it's standard mumbo-jumbo meaning that the algorithm used to obtain the coefficients has asymptotical stability: given any neighborhood of the actual solution, you can get to (and stay in) it in a finite number of steps. This comes typically from being able to show that the problem is actually (at least locally) convex, so that any minimizing algo like gradient, conjugated gradient, Newton-Raphson or whatever will actually "bring you there", and, as a further refinement, that you can estimate the local curvature of the function you are minimizing, thus being able to change the step adaptively to attain the desired precision.

At least in theory; when things get busy I usually sit down, light a cigar and let my engineers write the code and do the  numbers :-)

Andrea