tag:blogger.com,1999:blog-264226589944705290.post6196594763000117984..comments2023-11-05T03:45:25.001-08:00Comments on God Plays Dice: One hundred thousand! And the Collatz conjectureMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-264226589944705290.post-89703812740944698082008-02-11T13:28:00.000-08:002008-02-11T13:28:00.000-08:00anonymous,the generalizations I was thinking of we...anonymous,<BR/><BR/>the generalizations I was thinking of were just to say that one always gets a cycle. The interesting thing to me is that the sequence never "diverges upward", i. e. to infinity.Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-66861427410385518902008-02-11T12:14:00.000-08:002008-02-11T12:14:00.000-08:00Some simple generalizations of this are actually k...Some simple generalizations of this are actually known to be false: if you consider the "5x+1" conjecture, where you replace odd n with (5n+1)/2 and even n with n/2, then you get the cycle 13, 33, 83, 208, 104, 52, 26, 13.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-68833807465747094922008-02-09T03:01:00.000-08:002008-02-09T03:01:00.000-08:00Thanks John. I figured I had the formula wrong whe...Thanks John. I figured I had the formula wrong when I couldn't get the 3n-1 thingy to work for "7" while driving.<BR/><BR/>I saw that the records were generated by computer, but I couldn't easily find an example computer algorithm. They must be using some sort of theory to speed up the search.<BR/><BR/>The reason I was curious was because I spent many years designing FPGAs which are relatively inexpensive (i.e < $100) and can typically be designed to do this sort of problem about a thousand times faster than a general purpose computer. A typical FPGA raw speed would be 100,000 adders each doing a billion 1-bit arithmetic operations per second. You lose some for input and output operations, etc., but they are fast.CarlBrannenhttps://www.blogger.com/profile/17180079098492232258noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-72479778441737650012008-02-08T19:19:00.000-08:002008-02-08T19:19:00.000-08:00chezmax,of course you're right.Let this be a lesso...chezmax,<BR/><BR/>of course you're right.<BR/><BR/>Let this be a lesson: don't write blog posts that involve notation before drinking coffee. (This started out as just "ooh! look! I have 100,000 hits" while I waited for my coffee to brew this morning; I didn't intend to make a serious post and then somehow it happened.)Michael Lugohttps://www.blogger.com/profile/15671307315028242949noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-76399757710741979022008-02-08T19:16:00.000-08:002008-02-08T19:16:00.000-08:00You wrote:Let an+1 = an/2 if n is even, and an+1 =...You wrote:<BR/><I>Let an+1 = an/2 if n is even, and an+1 = 3an+1 if n is odd.</I>, but looking at your examples, I think you mean: Let an+1 = an/2 if <B>an</B> is even, and an+1 = 3an+1 if <B>an</B> is off....Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-14814002155631397692008-02-08T19:00:00.000-08:002008-02-08T19:00:00.000-08:00Carl, that's not quite it. What you've described ...Carl, that's not quite it. What you've described is the sequence you get if you use 3n-1 instead of 3n+1. When you first shift, you're dropping the LSB, which subtracts one. Then you keep dividing by 2 as you shift.John Armstronghttps://www.blogger.com/profile/15177732626660057584noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-83147839197228493732008-02-08T17:07:00.000-08:002008-02-08T17:07:00.000-08:00It's a pretty simple algorithm in binary. What's g...It's a pretty simple algorithm in binary. What's going on is that you begin with an odd number (which therefore has a 1 in its least significant bit, or LSB), and a stage in the algorithm consists of multiplying the number by 3 (which will leave a 1 in the LSB), and then shifting down by enough bits to bring the next 1 to the LSB.<BR/><BR/>So one might speculate about similar algorithms in other bases. For example, use base 3 and multiply the numbers by 4, then shift down to the next non zero LS3.CarlBrannenhttps://www.blogger.com/profile/17180079098492232258noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-43059091340092386552008-02-08T12:54:00.000-08:002008-02-08T12:54:00.000-08:00Congrats, Isabel! You posts are consistently inte...Congrats, Isabel! You posts are consistently interesting and entertaining, so I'm not surprised. =)<BR/><BR/>By the way, <A HREF="http://godplaysdice.blogspot.com/2007/10/enumerating-multiset-partitions.html" REL="nofollow">I've </A> just been accepted to the PhD program in CS at UPenn, and I'll be there for an <A HREF="http://www.cis.upenn.edu/grad/open-08.shtml" REL="nofollow">open house</A> on 2/29 and 3/1! Think you might have a few minutes to drop by and say hi?Anonymousnoreply@blogger.com