hi Doru,

haa yes I understand now ! Yes it should be the same complexity as we traverse
each line twice =).

Thanks !!

Pierre

2014-10-23 16:04 GMT-03:00 Tudor Girba <tudor@tudorgirba.com>:

Hi Pierre,

You pretty much doing the same thing as I said only in the reverse order. My idea was to do a normal flow layout with top alignment and traverse again each line and shift the nodes to the bottom.

But, your algorithm has about the same complexity, so it should be quite fine.

Cheers,

Doru

On Thu, Oct 23, 2014 at 8:38 PM, Pierre CHANSON <chans.pierre@gmail.com> wrote:

I am sorry Doru but in fact I did not well understood what you meant by "shift the smaller boxes",

But I finally changed the solution, as another problem came with this first solution: as I put the last element at 0:0 position then the positions to the fist element are in negative x and y.

So now I browse the elements in the normal order, I just pre-compute the next line to get the next line biggest element height before creating the line. Not a miracle solution but it works.

thanks !!

pierre

2014-10-23 15:32 GMT-03:00 Alexandre Bergel <alexandre.bergel@me.com>:

Hi Pierre!

I agree with Doru, the layout should remain the same. Maybe we could have a look at your algorithm next week, but it is doable.

Alexandre


> Le 23-10-2014 à 10:26, Pierre CHANSON <chans.pierre@gmail.com> a écrit :

>
> hello !
>
> thank you Doru, the thing is the max width is not based on number of elements but their size in pixel so we can't predict what will be the elements (and so the biggest element height) in the next line so, as a reversed align top, I put the elements from the bottom right corner to the top left. (Just put the elements in the reverse order) The last line is the fist one and less filled. But there is certainly a simple trick to fix this. What about the extra step to move at the end ?
>
> yes I will take a look at the center aligned version, I thought this one was already working but with a closer look, it's not :)
>
> Pierre
>

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