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RE: mstack and south carries

From: Daniel Marques <dani@wiris.com>
Date: Tue, 17 Mar 2015 10:48:46 +0100
Message-ID: <bc50695024111bf9522438ceb08ff467@mail.gmail.com>
To: Neil Soiffer <NeilS@dessci.com>, David Carlisle <davidc@nag.co.uk>
Cc: www-math@w3.org, juan@wiris.com, Esteve Badia <esteve@wiris.com>
Thanks for the responses but we are not done!



Assuming that both rows off carries are at south.

<mstack><mscarries><mn location="s">1</mn></mscarries><mscarries

location="s"><mn>2</mn></mscarries><mn>3</mn><mn>4</mn></mstack>



Then,

.

.

3

2

1

4



Where the dots ‘.’ are used only to indicate some amount of vertical space.



Is that correct?



In addition,  south carries move the rows the necessary amount of vertical
space in order to do not overlap (the row with the ‘4’ has been moved two
rows below). Do you agree?



Dani





*From:* neil.soiffer@gmail.com [mailto:neil.soiffer@gmail.com] *On Behalf
Of *Neil Soiffer
*Sent:* lunes, 16 de marzo de 2015 22:14
*To:* David Carlisle
*Cc:* www-math@w3.org
*Subject:* Re: mstack and south carries



I concur with David's reasoning.

    Neil



On Mon, Mar 16, 2015 at 9:12 AM, David Carlisle <davidc@nag.co.uk> wrote:

On 16/03/2015 16:02, Daniel Marques wrote:

Hi all,

The MathML specification allows specifying carries at south positions.
In this case, when other rows of carries exists it is not clear how the
formula must be rendered.

For example, in the following formula

<mstack><mscarries><mn>1</mn></mscarries><mscarries
location="s"><mn>2</mn></mscarries><mn>3</mn><msline/><msrow/></mstack>

What’s is the expected rendering?

1

3

2

Or maybe

1

3

2

?

Dani



personal response but the spec says

> the first row of carries annotates the second (following) row as if
the second row had location="n".

so the position of the 1 isn't affected by the fact that the 2 has
location=s, it is positioned as if the 2 had location=n so just above
where that would have been.

But it goes on to say

> This means that the second row, even if it does not draw, visually
uses some (undefined by this specification) amount of space when displayed.

so the exact amount of space below the 1 is implementation defined.

David


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