Bonetown 1.0.4 Trainer
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bonetown trainer 1.0.4 trainer is published and supported by bonetown software.
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Bonetown trainer 1.0.4
bonetown trainer 1.0.4 and bonetown trainer 1.1.1 are the most important updates and what you will have to do. The trainer updates, bonetown trainer 1.0.4 and bonetown trainer 1.1.1 patches are available for download.
are you a bonetown trainer fan? and also did you know it will have bonetown trainer 1.0.4 plus 1.1.1 updates on march 31st.Q:
Parametrized relationship between 2 strings
My question is how to:
create a string which includes a given number of strings.
create a relation between the strings so that one string can be substring of another.
For example:
(i) If stringA contains substrings (a1, b2, c3) and stringB contains the string (a1, b1, a2) then stringA is a substring of stringB.
(ii) If stringA contains strings (a1, b2, c3) and stringB contains the string (a1, c2, b3) then stringA is equal to stringB.
Note that (i) and (ii) depend on the order of substrings contained in stringA and stringB.
Any suggestion is greatly appreciated, for example:
def firstSubstring(self, text, substrings):
return (substrings[0])
def strEqual(self, textA, textB):
# How to write the logic for checking substring relation?
# comment if you have not seen this question
A:
using sets
>>> from sets import set
>>> a = set(['a1', 'a2', 'b1', 'b2', 'c3', 'c4'])
>>> b = set(['a1', 'a2', 'b1', 'b3', 'c2 0b46394aab
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