[math-fun] The final Google Turing puzzle: what is 11010110110101101011011...
I've been trying to "solve" the 3-line, 39-symbol Turing machine that you get at the "end" of the Google doodle puzzles. See [1] below for background. After solving the 6 puzzles, click on the little color "dog" icon in the upper right to see the 39-step program I'm investigating. After working through the normal puzzles, the Google doodle starts replaying the solutions of the Turing puzzles, and the "=" box in the upper right changes to something that looks like a little dog, or maybe the Trojan Rabbit from Monty Python and the Holy Grail [4], on a colored background. If you click on the doggie/bunny, the puzzle playback is replaced with a much more complex Turing program. The pattern on the tape grows to the right, here is how the program starts (with "." for an erased cell): at step 1: 1 at step 2: . at step 5: .1 at step 8: .10 at step 11: 110 at step 13: 1.0 at step 17: 1.01 at step 20: 1.010 at step 24: 11010 at step 26: 11.10 at step 31: 11.101 at step 35: 110101 at step 37: 110.01 at step 42: 110.011 at step 45: 110.0110 at step 50: 11010110 at step 52: 1101.110 at step 58: 1101.1101 at step 63: 110101101 at step 65: 11010.101 at step 71: 11010.1011 at step 74: 11010.10110 at step 80: 11010110110 etc... I'm counting head movements and writing or erasing a symbol as "steps". When a 0 or 1 symbol in the middle gets erased, it seems to be always replace it with the same symbol again. Thus it makes sense to ignore these temporary erasures and just look at the whole binary string. It grows only to the right, and it's an irregular pattern. As the pattern grows, the successive values are 110, 11010, 110101, 11010110, 110101101, 11010110110, ... or in decimal: 6, 26, 53, 214, 429, 1718, 6874, 13749, ... which is not presently in OEIS (even just the first 3 terms give no match). By step 49720, it has written: 1101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101011011010110110101101011011010110110101101011011010110101101101011011010110... It looks like any initial substring, no matter how long, occurs arbitrarily many times in the rest of the digits, but the occurrences are irregularly spaced. Taken as a binary fraction (0.1101011011010...) it is equivalent to 0.8392137714451652585671495977783023880500088230714420678280105786051... in decimal. It is irrational; the sequence of approximating fractions is: 0 / 1 1 / 1 5 / 6 21 / 25 26 / 31 47 / 56 167 / 199 214 / 255 1665 / 1984 5209 / 6207 6874 / 8191 438271 / 522240 1321687 / 1574911 1759958 / 2097151 3603955713 / 4294443008 ... It doesn't fit any simple algebraic formula; my RIES program finds nothing [2]. I got nothing with ISC+ either [3] (but I might be using it wrong, I think you need to try different numbers of digits: if 0.83921377144516 finds nothing, try 0.8392137714451, and so on.) Let me know if you have ideas for other, fancier tests I can apply to this sequence, the fraction, etc. - Robert Munafo [1] Background: If you go to Google today (Sat June 23rd) the Google Doodle is a Turing machine. Initially it's running a binary counter program, until you click the green "Play" button to pay the Turing game. It gives you 6 "easy" turing programming puzzles, which you solve by pushing the yellow buttons to set certain program steps, then press Play to see if you got it. If you solve the puzzles and start over (or reload, I'm not sure) it gives you 6 "hard" puzzles. [2] See http://mrob.com/pub/ries/ries.php?target=0.839213771445165 I also did a much longer RIES run on my own machine. [3] http://isc.carma.newcastle.edu.au/index [4] http://www.youtube.com/watch?v=T2PdyxMtiYM Okay it's not that close a resemblance, but I couldn't resist. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
Well, I think I found it in OEIS, but I'll wait a bit in case anyone wants to treat it as a puzzle... Also I made a webpage, which does not yet tell more than my previous email: http://mrob.com/pub/math/seq-google-turing.html On 6/23/12, Robert Munafo <mrob27@gmail.com> wrote:
I've been trying to "solve" the 3-line, 39-symbol Turing machine that you get at the "end" of the Google doodle puzzles. [...]
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
On Fri, Jun 22, 2012 at 9:40 PM, Robert Munafo <mrob27@gmail.com> wrote:
By step 49720, it has written:
1101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101011011010110110101101011011010110110101101011011010110101101101011011010110...
That has period 90. It repeats this binary string: 11010110110101101011011010110110101101011011010110101101101011011010110101101101011011010 It looks remarkably like this sequence that grows like Fibonacci's rabbits: 0 1 10 101 10110 10110101 1011010110110 101101011011010110101 ... -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
On 6/23/12, Mike Stay <metaweta@gmail.com> wrote:
That has period 90. It repeats this binary string: 11010110110101101011011010110110101101011011010110101101101011011010110101101101011011010
Not quite. Your string (which has 89 characters, making more sense given your conclusion) occurs twice, then we get 1101011011010110101101101011011010110101101101011011010 which has 55 characters, then another of the 89-character one. If we set A to be your 89-character string, and B = the 55-character string, then the output of the mystery Turing machine starts: A A B A A B A B A A B A A B A B A A B ... which bears some resemblance to the original string, but it's not just simply replacing 1/0 with A/B. On 6/23/12, Mike Stay <metaweta@gmail.com> wrote:
On Fri, Jun 22, 2012 at 9:40 PM, Robert Munafo <mrob27@gmail.com> wrote:
By step 49720, it has written: 1101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101011011010110110101101011011010110110101101011011010110101101101011011010110...
It looks remarkably like this sequence that grows like Fibonacci's rabbits: 0 1 10 101 10110 10110101 1011010110110 101101011011010110101 ...
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
Two observations (i) we can replace 110 with 1 and 10 with 0 in the original sequence and get a reduced sequence and it seems we can reduce the resulting sequence in a similar manner. so it forms a kind of self similar sequence... (ii) 55 and 89 are Fibonacci numbers Christoph ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] On Behalf Of Robert Munafo [mrob27@gmail.com] Sent: Saturday, June 23, 2012 8:31 PM To: math-fun Cc: Mike Stay Subject: Re: [math-fun] The final Google Turing puzzle: what is 11010110110101101011011... On 6/23/12, Mike Stay <metaweta@gmail.com> wrote:
That has period 90. It repeats this binary string: 11010110110101101011011010110110101101011011010110101101101011011010110101101101011011010
Not quite. Your string (which has 89 characters, making more sense given your conclusion) occurs twice, then we get 1101011011010110101101101011011010110101101101011011010 which has 55 characters, then another of the 89-character one. If we set A to be your 89-character string, and B = the 55-character string, then the output of the mystery Turing machine starts: A A B A A B A B A A B A A B A B A A B ... which bears some resemblance to the original string, but it's not just simply replacing 1/0 with A/B. On 6/23/12, Mike Stay <metaweta@gmail.com> wrote:
On Fri, Jun 22, 2012 at 9:40 PM, Robert Munafo <mrob27@gmail.com> wrote:
By step 49720, it has written: 1101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101101011010110110101101011011010110110101101011011010110101101101011011010110101101101011011010110101101101011010110110101101101011010110110101101011011010110110101101011011010110110101101011011010110101101101011011010110...
It looks remarkably like this sequence that grows like Fibonacci's rabbits: 0 1 10 101 10110 10110101 1011010110110 101101011011010110101 ...
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I believe this works: 10 Begin with the empty string 20 Apply L-system rules '1' -> '10', '0' -> '1'. 30 Prepend '1' to the string 40 GOTO 20 Here are the first few iterations: 1 110 110101 11010110110 1101011011010110101 Sincerely, Adam P. Goucher
That's a great complement to the simple description of the machine (adding symbols one at a time), and I'll add it to my page ( mrob.com/pub/math/seq-google-turing.html ) On 6/24/12, Adam P. Goucher <apgoucher@gmx.com> wrote:
I believe this works:
10 Begin with the empty string 20 Apply L-system rules '1' -> '10', '0' -> '1'. 30 Prepend '1' to the string 40 GOTO 20
Here are the first few iterations:
1 110 110101 11010110110 1101011011010110101
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
participants (4)
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Adam P. Goucher -
Mike Stay -
Pacher Christoph -
Robert Munafo