Going a step further... (see below)

----- Original Message ----- 
From: "Brent Meeker" <meeke...@dslextreme.com>
To: <everything-list@googlegroups.com>
Sent: Wednesday, July 22, 2009 12:57 PM
Subject: Re: The seven step series


> 
> m.a. wrote:
>> Hi Brent,
>>                 I really appreciate the help and I hate to impose on 
>> your patience but...(see below)
>>  
>> ----- Original Message -----
>> From: "Brent Meeker" <meeke...@dslextreme.com 
>> <mailto:meeke...@dslextreme.com>>
>> To: <everything-list@googlegroups.com 
>> <mailto:everything-list@googlegroups.com>>
>> Sent: Tuesday, July 21, 2009 5:24 PM
>> Subject: Re: The seven step series
>> 
>>  >
>>  > Take all strings of length 2
>>  > 00             01                   10               11
>>  > Make two copies of each
>>  > 00      00      01      01      10      10      11      11
>>  
>>  > Add a 0 to the first and a 1 to the second
>>  > 000    001      010   011      100   101   110      111
>>  > and you have all strings of length 3.
>> *I can see where adding 0 to the first and 1 to the second gives 000 and 
>> 001 and I think I see how you get 010 but the rest of the permutations 
>> don't seem obvious to me. P-l-e-a-s-e  explain,  Best,*
>> ** 
>>                                                                         
>>                                                                         
 They aren't permutations.  They're just sticking a 0 or 1 on the end.  One 
copy 
> of 01 becomes 010 and the other become 011.

Then I assume the next step would be making two copies of each of those:

000    000       001     001      010      010       011     011     100      
100       101         101         110           110             111          111

...and sticking a 0 or 1 at the end:

0000   0001    0010    0011     0100    0101    0110    0111    1000     1001   
  1010       1011         1100         1101           1110      1111

and this is the binary sequence of length 4.

How do these translate into ordinary numerals? 1,2,3,4...

> 
> Brent
> 
> >
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