All sequence riddles are wrong .why?

[ahmedhassan72]

All sequence riddles are wrong...why?

Thanks for consideration!

if you get a sequence like that 1,2,3,... so you will say 4 how??
the relation between 1 and 2 is that between 2 and 3(it is a first degree function) (y=mx+b)
so you can get two equations to solve a+b=2 2a+b=3 therefore a=1 b=1 so you multiply by one and add one
another sequence 2,-10,110,11870,... so you can try first degree relation as previous and it will not make sense so try the second degree equation f(x)=ax^2+bx+c
4a+2b+c=-10 100a-10b+c=110 2100a+110b+c=11870 solve these equations you get
a=1 b=-2 c=-10 so you square and subtract twice the number and subtract ten to get the next term
then another sequence can't be applied in the second degree so apply for the third degree and so on
so if in any sequence i can get a relation so it is not a riddle as you can't tell me that i am wrong
?I can also make the same thing between the term number and it's value i.e in term 1 value 2, term 2 value -10, term 3 value 110...and get equations and i can get another relation...

so 0,1,1,1,1,5,1,7,2,... can be solved maximum by the seventh degree it is too long but easily solved
so it will have 3 answers my 2 answers and another answer says that the relation is that you take 0/12,1/12,1/6,1/4,1/3,5/12,1/2,7/12,2/3,5/6,... so the riddle is wrong as it has three answers ..!

 

[Pere Callahan]

Great observation !

There is of course no unique answer because there is no question to be answered ..

You are just given some numbers and asked to continue the sequence.. which you could do in whatever way you like .. and be itonly for asthetic reasons, "because it looks nice".

 

[DaveC426913]

All 'fill in the missing number' riddles can have a virtually unlimited number of possible answers.

The instructions are to find the 'best' one - usually the simplest (or more pedantically, the one with the fewest required transformations).

If you decide you have found an answer, and it turns out there is a simpler answer, it's because you weren't able to visualize a whole host of possible answers and pick the best one.

Intelligent-types can see these patterns better than others; consistently picking the simplest one (among many possible ones) is regarded as a sign of high intelligence.

 

[CRGreathouse]

All 'fill in the missing number' riddles can have a virtually unlimited number of possible answers.

I'll do you one better -- there is a literally unlimited number of possible answers, because you can fit a polynomial of n+1 points to any given data (1, k_1), (2, k_2), ..., (n, k_n), (n+1, x) for any x.

 

[DaveC426913]

I'll do you one better -- there is a literally unlimited number of possible answers, because you can fit a polynomial of n+1 points to any given data (1, k_1), (2, k_2), ..., (n, k_n), (n+1, x) for any x.

I'll take you word for it, but I don't really follow.

If you start with a sequence, say

1,1,2,3,5,8...

I'll grant that there are an arbitrarily large number of possible next numbers, but unlimited? Surely the applicability of such solutions must rapidly approach zero.

 

[bassplayer142]

If you have points on a graph. A line or curve of infinite possiblities may be created to fit the points.

 

[HallsofIvy]

My freshman year, a professor gave as an example sequence the numbers 24, 23, 22, 21, 20 and asked for the next number. No, it wasn't 19! It was 12 because those are the numbers of subway stations on his way to work- and his train changed from one line to another at that point.

 

[DaveC426913]

If you have points on a graph. A line or curve of infinite possiblities may be created to fit the points.

While that may be true, it does not mean that every line on that graph represents an answer to the riddle.

 

[dst]

While that may be true, it does not mean that every line on that graph represents an answer to the riddle.

If you have a vaguely defined sequence and you are asked for the next term then every curve is valid. I don't see why not?

 

[ahmedhassan72]

you can't get any curve equation you can get at maximum an equation of nth degree where n is equal the number of given terms-2 i.e you can't get a fourth degree equation from less than 5 points which is at least represented by 6 terms in a sequence

 

[DaveC426913]

If you have a vaguely defined sequence and you are asked for the next term then every curve is valid. I don't see why not?

The unspoken requirement is for the simplest solution.

 

[quadraphonics]

The unspoken requirement is for the simplest solution.

Define "simplest." :]

What's interesting about these problems is that there IS an accepted right answer, even though it depends on aesthetic judgements.

 

[ahmedhassan72]

I think complication is that you think when you don't need to...

 

[CRGreathouse]

you can't get any curve equation you can get at maximum an equation of nth degree where n is equal the number of given terms-2 i.e you can't get a fourth degree equation from less than 5 points which is at least represented by 6 terms in a sequence

Sure you can. Given points (0, 0) and (1, 1) I can say "these points are on y = x^7" as easily as "these points are on y = x".

 

[Crosson]

This is a skeptical paradox that first arose in the 20th century. Suppose I gave the sequence as:

2 , 4, 6, ... , 2 n , ...

And I asked "What is the fourth term of the sequence?" Most of you are probably thinking the answer is unambiguously 8, but what if the student misinterprets the rule 2*n. Should I produce another rule to explain that rule?

 

[DaveC426913]

This is a skeptical paradox that first arose in the 20th century. Suppose I gave the sequence as:

2 , 4, 6, ... , 2 n , ...

And I asked "What is the fourth term of the sequence?" Most of you are probably thinking the answer is unambiguously 8, but what if the student misinterprets the rule 2*n. Should I produce another rule to explain that rule?

I'm confused. What is your point?

8 seems the simplest answer. What alternate answer were you suggesting?

 

[DaveC426913]

Define "simplest." :]

Least unmber of transformations, lowest order of deviation, etc.

If I give you th sequence 1,1,2,3,5,8,...


Give me at two valid subsequent numbers, and the logic by which you arrived at them.
We are looking for simple patterns here.

Two possible answers:
13
386,141

I could come up with some logic that shows both those numbers are valid, but one would be somewhat more complex than the other, and I would be hard-pressed to call it a pattern.

 

[ahmedhassan72]

i will give 2 points and give me a 4th degree equation ! in that case you have to draw and make assumed points so that you added to the given terms of the sequence and that is not acceptable i think.
as 2,4,6,... has no answer but 8 but if it is 2,4,6,9,... it also has no answer but 15.375!

 

[CRGreathouse]

i will give 2 points and give me a 4th degree equation ! in that case you have to draw and make assumed points so that you added to the given terms of the sequence and that is not acceptable i think.

Why?

 

[qspeechc]

If you have n points in the seqeunce, you can fit it to an mth degree polynomial, where m>n. Of course, you will have m-n free variables to play with, but that doesn't matter, whatever values you pick for those left over variables, they are all valid, as the polynomial will still go through the n points, and so yes, you will have infinitely many solutions, er, whatever you want to call them.

Hmm, on second thoughts, that sounds a bit dodgy. As long as you make sure that by picking the values of the variables, you don't add a point into the sequence, ie. given:
a0, a1, a2, a3, a4,...
and you fit, say, a seventh degree polynomial to it, you don't choose the values of the variables such that the polynomial will give:
a0, a1, p, a2, a3, a4,...
for instance.

 

[DaveC426913]

If you have n points in the seqeunce, you can fit it to an mth degree polynomial, where m>n.

Right. So the simplest solution would be the one where m is the smallest.

 

[qspeechc]

But then the sequence might not be generated by a polynomial at all. What then is the simplest?

 

[Dragonfall]

You can define "simplest" by the output of the smallest Turing machine which prints the sequence in succession and which contains the predetermined points.

More generally, given a language capable of expressing sequences and sufficient arithmetic, the "simplest" extension to a finite sequence would be the sequence defined by a formula with the least number of characters.

Let's not try to bring Komolgorov complexity into this.

 

[qspeechc]

If you have n points in the seqeunce, you can fit it to an mth degree polynomial, where m>n. Of course, you will have m-n free variables to play with,

Sorry, m-n-1 free (undetermined) constants.

 

[DaveC426913]

You can define "simplest" by the output of the smallest Turing machine which prints the sequence in succession and which contains the predetermined points.

More generally, given a language capable of expressing sequences and sufficient arithmetic, the "simplest" extension to a finite sequence would be the sequence defined by a formula with the least number of characters.

Let's not try to bring Komolgorov complexity into this.

Yeah. That's kind of what I was trying to say.

 

[Dragonfall]

Unfortunately I think that the problem of finding the simplest sequence in that sense is "uncomputable". If I claim that a Turing machine M outputs the required sequence and that M is the smallest machine to do so, then I would be able to prove that every TM of size less than that of M does NOT output the required sequence. Specifically, I'd need to prove that every such TM either HALTS, or not. This may be uncomputable, though it is not exactly the halting problem.

 

[DaveC426913]

Unfortunately I think that the problem of finding the simplest sequence in that sense is "uncomputable". If I claim that a Turing machine M outputs the required sequence and that M is the smallest machine to do so, then I would be able to prove that every TM of size less than that of M does NOT output the required sequence. Specifically, I'd need to prove that every such TM either HALTS, or not. This may be uncomputable, though it is not exactly the halting problem.

And what's astonishing is, not simply that humans can find the solution, but that they find it so intuitively.

 

[qspeechc]

I don't think that's amazing. It would be stupifying if we intuitively picked out extremely difficult solutions. Obviously we're going to see the relatively simple ones first.

 

[Dragonfall]

And what's astonishing is, not simply that humans can find the solution, but that they find it so intuitively.

Ah, but we don't.

 

[quadraphonics]

Ah, but we don't.

That true. We DO tend to pick our solutions that other people will agree are 'right' (and that's something), but I seriously doubt it could be shown that said solutions satisfy any of the specific, formal definitions of simplicity offered up here (Kolmogorov complexity, say). At least outside of the very "easy," "obvious" sequence riddles...

 

[DavidWhitbeck]

You don't even have to do fit to a polynomial fit, sequences do not have to have rules, so the next term in the sequence can be whatever you want it to be.

 

[CRGreathouse]

You don't even have to do fit to a polynomial fit, sequences do not have to have rules, so the next term in the sequence can be whatever you want it to be.

Fair point. So the next member of "1, 2, 4, 8, 16, 32" could be "dragon".

 

[Dragonfall]

Here's an idea for a sequence riddle: given a countable sequence of natural numbers, what ordinals come next?

Say N=(1, 2, 3, 4, ...), I could say what comes next are all ordinals. OR I could say that what comes next are all SINGULAR ordinals, etc.

 

[mubashirmansoor]

Thats good!

Very interesting thread...

I have had some research (not a serious one) on the polynomial sequences when I was in high school. I know this is not exactly what you are looking for but this is what you might find interesting.

I found a single function definition which would do all the work you have been doing in post#1,

I have attached the function to the post as a jpg.

n represents the degree of the polynomial
delta "x" can be anything
"C" represents combinations.

______

comments will be highly appreciated

 

[DavidWhitbeck]

Fair point. So the next member of "1, 2, 4, 8, 16, 32" could be "dragon".

No because sequences are by definition subsets of N.