Thanks to the analysis of other people, see e.g.
http://mrob.com/pub/math/seq-google-turing.html, it turns out that
fibonacci is involved:
Using these for generating the bits:
;bits=.([:;(-.`(,-.)@.])&.>)&.>^:(<12) <1
1 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 1 0 1 0 1 1 0 1 1 0 ...
+/&> bits
1 1 2 3 5 8 13 21 34 55 89 144
On 23-06-12 17:48, Devon McCormick wrote:
> The link says it's spelling out "Google" but I believe that is
> incorrect - it looks like it's just counting up in binary.
>
> On Sat, Jun 23, 2012 at 10:07 AM, Donna Y <[email protected]> wrote:
>> Thanks for the link. I did not watch long enough. Yes it's the code
>> breaker and not an implementation of his first proof. It simply spells out
>> Google.
>>
>> As the video asks, imagine if Alan Turing could work on one of today's
>> computers to expand on any of the many problems he was pursuing!
>>
>> Donna
>> [email protected]
>>
>>
>> On 2012-06-23, at 7:11 AM, Brian Schott <[email protected]> wrote:
>>
>>> Thanks.
>>> Can the working model be saved somehow?
>>> Btw, I found an interesting explanation at
>>> http://news.cnet.com/8301-17852_3-57459126-71/googles-impossibly-clever-alan-turing-doodle/
>>> That link suggests another purpose of the doodle.
>>>
>>> On Sat, Jun 23, 2012 at 1:26 AM, Donna Y <[email protected]> wrote:
>>>> Its Alan Turing's birthday and that is the proof of the halting problem -
>>>> see Church and Turing.
>>>>
>>>> Donna
>>>> [email protected]
>>>>
>>>>
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>
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Met vriendelijke groet,
@@i = Arie Groeneveld
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