A Quick Puzzle to Test Your Problem Solving

Post by drewcrawford » Thu, Jul 02, 2015, 9:34 PM UTC

scarmig wrote: post #192

nothrabannosir wrote: post #173

My mathematical logic is rusty, but if I recall correctly, Gödel's incompleteness theorem basically states that it is impossible to solve this kind of question. No matter how many tests you run, there will always be an uncertainty. An incredibly stupid example is that the rule could be "yes for strictly increasing, OR if one of the numbers is -18273192783127897981." You'll never know. I understand this is contrived,…

Remotely related: I've been interested for awhile in how the same initial terms of a sequence could possibly be generated by multiple rules. For example, you might have 2,3... And the rest of the sequence might look like either 2,3,4,5,6... or 2,3,5,8,13... or 2,3,5,7,11... or even 2,3,5,10,20... Clearly, on some level those sequences are all much less complicated than one defined as "The first term is 2, the second…

    {0, 1, 3, 8, 19, 43, 94, 201, 423, 880}

    1/10 (5 2^(1 + x) - 5 (1/2 - Sqrt[5]/2)^x + 
    3 Sqrt[5] (1/2 - Sqrt[5]/2)^x - 5 (1/2 + Sqrt[5]/2)^x - 
    3 Sqrt[5] (1/2 + Sqrt[5]/2)^x)

    {-(1/6), 2/15, -(13/140), 23/315, -(83/1386), 305/6006, -(2269/
     51480), 4259/109395, -(16103/461890), 30616/969969}

    (2^(-2 - 3 x)
    x! (Sqrt[\[Pi]] (1 + x)! + 
    3 (-1)^x 2^(
    2 + 3 x) (1/2 (1 + 2 x))! Hypergeometric2F1[1, 3/2 + x, 
    2 + x, -8]))/((1/2 + x)! (1 + x)!)

Post by eli » Thu, Jul 02, 2015, 9:37 PM UTC

perlgeek wrote: post #161

ubernostrum wrote: post #82

Earlier quoted context omitted.

My process was: [3 5 7] [7 5 3] [8 4 2] [5 7 3] [1 2 3] [1 1 1] [0 1 2] At that point I could've done some more to be really certain, but felt confident enough and guessed (correctly).

So you limited yourself to non-negative integers, and had no idea what negative numbers or fractions would do.

Post by mjevans » Thu, Jul 02, 2015, 9:56 PM UTC

SiVal wrote: post #196

tptacek wrote: post #168

Earlier quoted context omitted.

I tried floating point numbers. Also, at 28 decimal places, the test breaks; it's not arbitrary precision. So, technically, the answer isn't simply "any ascending sequence of numbers".

I actually avoided going that far to avoid getting bad data. I was trying to answer the question, "What does the experimenter THINK his rule is?" rather than what will the computer do. Since the computer can't be infinite, it will inevitably fail with overflow, underflow, and such. I was relieved, in fact, when it worked with negatives and floats in a "safe" range. I also tested with 1,1,2 and 1,2,2 to make sure that…

Post by bbcbasic » Thu, Jul 02, 2015, 10:19 PM UTC

perlgeek wrote: post #161

ubernostrum wrote: post #82

Earlier quoted context omitted.

My process was: [3 5 7] [7 5 3] [8 4 2] [5 7 3] [1 2 3] [1 1 1] [0 1 2] At that point I could've done some more to be really certain, but felt confident enough and guessed (correctly).

So you limited yourself to non-negative integers, and had no idea what negative numbers or fractions would do.

Post by bovik » Thu, Jul 02, 2015, 10:27 PM UTC

drewcrawford wrote: post #211

scarmig wrote: post #192

Earlier quoted context omitted.

Remotely related: I've been interested for awhile in how the same initial terms of a sequence could possibly be generated by multiple rules. For example, you might have 2,3... And the rest of the sequence might look like either 2,3,4,5,6... or 2,3,5,8,13... or 2,3,5,7,11... or even 2,3,5,10,20... Clearly, on some level those sequences are all much less complicated than one defined as "The first term is 2, the second…

Fun fact: Mathemetica has a method FindSequenceFunction which does exactly what you describe. In my experience, however, it generally requires at least 4 terms. I'm not totally sure how the math behind it works (maybe it's similar to Eureqa?) but the results speak for themselves and are rather incredible. For example, if I run FindSequenceFunction on this input: {0, 1, 3, 8, 19, 43, 94, 201, 423, 880} Which is the nu…