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Originally, this question was way too wide open. I have tried to clarify / narrow it down / restrict the possibilities. Let me know if it is still too wide open.


These are some number sequences. I used the data set $A$ to generate both $B$ & $C$. I then compared those data sets in one way to get $D$ and in another way to get $E$.

Can you find $B$ and $C$?

A: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30
B: ?
C: ?
D: 2,3,0,2,1,2,2,1,1,2,2,1,2,2,2,2,2,2,2,3,2,3,0,2,1,2,2,1,1,3
E: 0,1,3,4,2,3,5,5,3,3,6,4,8,8,7,7,9,8,8,5,6,7,9,10,8,9,11,11,9,5

The methods to generate $B$ & $C$ do not use any mathematical operators.

The methods to generate $D$ & $E$ use the operators $+$ & $-$ and nothing else.

All numbers are whole numbers.


If you were to look at it another way, it might looks like this:

Find $f_B$ & $f_C$ where: $$n=1\to30$$ $$A_n=n$$ $$B_n=f_B(A_n)$$ $$C_n=f_C(A_n)$$ $$D_n=f_D(B_n,C_n)$$ $$E_n=f_E(B_n,C_n)$$

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  • $\begingroup$ There are too many ways to generate two sequences and an algorithm that fits them, we need more information to unequivocally solve the problem $\endgroup$ – leoll2 May 1 '15 at 14:07
  • $\begingroup$ The numbers are integers or naturales? $\endgroup$ – leoll2 May 1 '15 at 14:28
  • $\begingroup$ @leoll2 All numbers are whole numbers. $\endgroup$ – Engineer Toast May 1 '15 at 14:43
  • $\begingroup$ I'm on my way to an answer. Just to demonstrate to Engineer Toast, one of the sequences starts with 3, 3, 5. $\endgroup$ – Glen O May 1 '15 at 15:13
  • $\begingroup$ @GlenO I'm very curious to see that answer. It's not mine but it may prove that the question is still too broad. $\endgroup$ – Engineer Toast May 1 '15 at 15:16
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The relevant sequences aren't actually numbers, but words.

Sequence B is

One, Two, Three, Four, Five...

And Sequence C is

First, Second, Third, Fourth, Fifth...

The operations to get to the final sequence involve differences, but not mathematically. Sequence D is obtained by...

Counting how many more letters there are in sequence C than sequence B. That is, "First"=5 and "One"=3, giving 2.

While Sequence E is obtained by...

Counting how many letters are common between the two words. So "one" and "first" have no letters in common, while "two" and "second" have "o" in common, giving 1. Similarly, "three" and "third" both have "t", "h", and "r", making it three.

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Not sure if this fits as a comment, but I think you need to specify a few more rules for your comparison. Indeed, I can simply create two lists:

$2,8,9,14,21,47,27,101,10,11,20,17,58...$
$5,5,3,3,10,15,5,20,3,3,3,4,7...$

And the two relations would be:

$f(a,b) = a\mod b$
$g(a,b) = \lfloor a/b \rfloor $
Essentially, this treats the first list as our numerator, the second as the denominator, and the two resultant sets are the remainder and quotient, respectively. Literally an infinite combination of lists can be generated this way.

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  • $\begingroup$ It might still be terrible but is it any better now? $\endgroup$ – Engineer Toast May 1 '15 at 14:22

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