Thanks.  All those Euler problems had me fearing something worse!
Mike

Please reply to [email protected].      
Sent from my iPad

> On 20 Jun 2017, at 21:56, Raul Miller <[email protected]> wrote:
> 
> Yes.
> 
> One way of thinking about that would be that the elements have
> orthogonal sides, and slanted "rectangles" would thus include empty
> space on the edges.
> 
> Thanks,
> 
> -- 
> Raul
> 
> On Tue, Jun 20, 2017 at 4:49 PM, 'Mike Day' via Programming
> <[email protected]> wrote:
>> I've only just seen this.
>> I expect it's obvious,  but you ARE looking only for rectangles with 
>> "vertical" and "horizontal" sides, i.e. no rotations other than multiples of 
>> pi/2 involved?
>> Thanks,
>> Mike
>> 
>> Please reply to [email protected].
>> Sent from my iPad
>> 
>>> On 20 Jun 2017, at 16:29, Raul Miller <[email protected]> wrote:
>>> 
>>> Something I stumbled over today.
>>> 
>>> If we have a series of bars of varying height, what's the largest
>>> rectangle that can be drawn over the bars without covering any empty
>>> space.
>>> 
>>> For example:
>>> 
>>>  '*'#"0~2 6 7 4 1 7
>>> **
>>> ******
>>> *******
>>> ****
>>> *
>>> *******
>>> 
>>> I'll post a solution later, and I'll be interested in seeing if it's
>>> basically the only obvious approach or if there's a variety of good
>>> approaches. (I have reason to believe, though, that there's a better
>>> way than what I came up with.)
>>> 
>>> Thanks,
>>> 
>>> --
>>> Raul
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>> 
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