Then you are the linguo. :) Nice to meet you.
Yes, that part is extremely clear, if there is 20 unknown position left, we can 
enumerate all. Actually in 8 minutes, maybe we can handle 22-23 unknown 
position. :) The complicated part to handle bottom partition. I asked because 
maybe there can be another solution maybe via logarithm vs.
Thanks for your help.

21 Nisan 2018 Cumartesi 05:27:22 UTC+3 tarihinde Luke Pebody yazdı:
> As the only person who solved the large on that question, the analysis nails 
> exactly what I did. The idea is that the first half or second half of a 
> square is enough to determine it. There are 40 missing digits and at most 20 
> of them are in the first half or second half. 2^20 is a million* and so you 
> can run all possibilities and see if they make a square. There are some 
> slight issues with having 00 at the end.
> 
> 
> Interestingly we had 8 minutes to submit and my code gave me about 5 seconds 
> to spare. This is what you get for using python and not using pypy
> 
> 
> On Thu, 19 Apr 2018, 3:37 am Serhat Giydiren, <[email protected]> wrote:
> Hi all.
> 
> 
> 
> I am reviewing past problems. Is there anyone who has another analysis / 
> approach for the following problem.
> 
> 
> 
> https://code.google.com/codejam/contest/1158485/dashboard#s=p3
> 
> 
> 
> I could not get the following analysis.
> 
> 
> 
> https://code.google.com/codejam/contest/1158485/dashboard#s=a&a=3
> 
> 
> 
> Thanks.
> 
> 
> 
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