From kscott@silicon.csci.csusb.edu Tue Oct 3 18:23 PDT 1995
From: "Katherine A. Scott" <kscott@orion.csci.csusb.edu>
To: "Dr. Richard Botting" <dick@blaze.csci.csusb.edu>
Subject: Re: New program
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Mod5 operator+(Mod5 a, Mod5 b){ return Mod5( (a.v + b.v)%5 ); }
Mod5(int i){ v = i % 5; }
Yes, it does compile and actually works for what I wanted it to.
(Amazing enough, eh?)
While you were talking about operator overloading, I started thinking
about all of the wonderful applications this would have on abstract
algebra. I started writing a program based on the very simple group
< Zsub5, addition mod 5 >. If I enter two integers (a and b), the
computer
will use the group operations to find the group element a*b. I had
planned to write this using operator overload, but found that it was more
efficient not to.
I plan to expand the program later for larger, more difficult groups
(for example, the set of all rotations and reflections on a figure....)
It
would be nice to have a program that would find all of the elements of a
particular group. Or a program that would be able to add the members of
two groups (finally using operation overload!)
++
I have written a brief summary of Group theory and what the program
does. If you want to, you can skip this stuff.
As a reminder, a group is a non-empty set G and a binary operation
(usually multiplication or addition) that satisfies three properties.
1) If a and b are elements of G, then a*b (where * is the group
operation) is also in G. (It is easy to see that division and
subtraction would not work in most cases.)
2) There is an identity e in G such that for all a in G, a*e = e*a = a.
3) For all a in G, there exists some element b such that a*b = b*a = e.
(Not that each element can have a different inverse.)
In this case, my non-empty set G is Zsub5, or {0,1,2,3,4}. (The first
five positive integers) The binary operation is addition mod five.
(You can take a few elements and see that the properties of Group are
satisfied.)
So, if I enter a = 7 and b = 5, I get ( 7+5 mod 5 ) = 12 mod 5 = 2
so 7*5 = 2. It might be interesting to note that 7 is equivalent to 2
mod 5, and 5 = 0 mod 5. 2*0 = 2.
Yours,
Katherine