🚀 go-pugleaf

hagadorn@amsaa-cleo.brl.mil (Thomas Hagadorn) writes:
>erick@fraser.sfu.ca (Erick Bryce Wong) writes:
>|freitag@elrond.toppoint.de (Claus Schoenleber) writes:
>|>[a  is digit a at position i, i=10 is leftmost]
                                 ^^^^
>|>     [         / 10         \          ]
>|>     |        /  __          \         |
>|>c =  | 11 -  (   \   a  * i   ) mod 11 | mod 11
>|>     |        \  /_   i      /         |
>|>     [         \ i=2        /          ]
>|
>|You can put this in slightly simpler terms if you define i=1 to be the leftmo
                                                           ^^^
>|       / #digits    \  
>|      /  __          \ 
>|c =  (   \   a  * i   ) mod 11
>|      \  /_   i      / 
>|       \ i=1        /  
>|
>   Almost, make that:
>
>            / #digits    \  
>           /  __          \ 
>11 - c =  (   \   a  * i   ) mod 11
>           \  /_   i      / 
>            \ i=1        /  

You apparently didn't read both articles carefully.  If the original formula
is correct, then (using pseudo-TeX notation to cut the size down a bit):
(I'll use the same meanings for a_i and i as the original article.)

11 - \sum_2^10{a_i * i} == - \sum_2^10{a_i * i}
                        == \sum_2^10{a_i * {-i}}
                        == \sum_2^10{a_i * {11-i}}
                        == \sum_1^9{a_{11-i} * i} (mod 11).

Which is exactly what my formula corresponds to, as I define the leftmost
position to be 1 with the rest ascending, not 10 with the rest descending.

-- 
 -- Erick, the perfect square  :-)