64 chars with explanations solution in Clear category for Univocalic davasaan by Ylliw
davasaan=d=lambda n:[n>9,(a:=n>>4)and a+d(n-(a<<3)-(a<<1))][a>0]
# recursively with the greatest power of n (let's call it a) so that 10*a is less than n
# a=n>>4 = n//16 is that greatest power, so we know that 10a is less than n
# Therefore new_n=n-10a is less than n and if we know that davasaan(n)=a+davassan(new_n)
# It's now time to make this work, so let's define d=davassan (less char usage in recursion)
# and a is n>>4 (n//16)
# so main recursion element is d(n-(n>>4<<3)-(n>>4<<1))
# Let's now add the corner cases:
# We can do the recursion only if n>=16, so n>>4 and ensure n>=16 only as and does not evaluate 2nd element if 1st is False
# Finally is n is less than 16, so 15 or less, we need to return 1 if n>=10 and 0 if n<10
# n>9 is the shortest condition and +(n>9) ensure it's considered as int and not boolean
# I finally used a two element list (n<=16, n>15) with a condition (n>15) to choose as 'or' is not acceptable.
# Last trick added (Python 3.8.1): assign n>>4 to p (a:n>>4) to reduce number of chars as it's used 4 times.
# Brackets around a:=n>>4 is mandatory as otherwise it's evaluating the 'and' part before assignment.
# Finally and thanks to Phil15 comments, I remove the unnecessary + sign to convert n>9 to int and the parenthesis left around one 'a' occurence
Jan. 22, 2020