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09:24 <nfo> Hi again, I've just read articles mentioned in 
http://nosql.mypopescu.com/post/524128219/learn-about-conflict-resolution-and-vector-clocks . 
But I'm still struggling with this conflict resolution scenario: 

09:25 <nfo> is it possible to concurrently increment the integer value of an object ? 

09:25 <bingeldac> no 

09:25 <bingeldac> not in the way you're thinking I think 

09:26 <seancribbs> nfo: no, better to write new objects and count them with mapreduce 

09:26 <nfo> bingeldac: ok :) I tried implementing conflict resolution , but I figured 
out that I could not know who incremented the value and by how many clients 

09:26 dysinger joined 

09:28 <pharkmillups> nfo: sidenote - you're best to read the actual posts is you haven't already 

09:28 <pharkmillups> http://blog.basho.com/2010/01/29/why-vector-clocks-are-easy/ 

09:28 <pharkmillups> http://blog.basho.com/2010/04/05/why-vector-clocks-are-hard/ 

09:28 <nfo> pharkmillups: yep, read them. "Automatic" resolution could also work with an 
array of values, not a counter. 

09:30 <nfo> seancribbs: the thing with creating multiple objects, is that I have to find them afterward 

09:31 <nfo> s/with/about 

09:33 <seancribbs> nfo: true 

09:33 <nfo> ok, so with arrays, here is a scenario. Each client adds an element to the array. 
My concern is that I just need to see if the size of the array is equal to a value: 

09:34 <nfo> client "c" updates the array => [a, b, c] 

09:34 <nfo> client "d" updates the array simultaneously => [a, b, d] 

09:35 <nfo> client x gets the object and sees a conflict, it can just fix it by merging 
the arrays, right ? => [a, b, c, d] 

09:37 <nfo> does it seem to be a correct solution ? 

09:42 <seancribbs> set-union is pretty common way to do that, yes 

09:43 <seancribbs> a, b, c, d could be uuids 

09:43 <seancribbs> then for most cases, conflicts are easy to resolve 
09:44 <seancribbs> as long as you're never removing items, heh 
<nfo> seancribbs: ok for the set-union, it will work perfectly for my use case. Thanks!