Proof that check_errors() is sufficient.

[originally from svn r8813]
This commit is contained in:
Simon Tatham
2010-01-05 23:40:42 +00:00
parent dd4c8ceb1f
commit 2500531423

View File

@ -1047,6 +1047,34 @@ static int check_errors(game_state *state, long *errors)
digit *grid = state->grid; digit *grid = state->grid;
int i, j, k, x, y, errs = FALSE; int i, j, k, x, y, errs = FALSE;
/*
* To verify that we have a valid group table, it suffices to
* test latin-square-hood and associativity only. All the other
* group axioms follow from those two.
*
* Proof:
*
* Associativity is given; closure is obvious from latin-
* square-hood. We need to show that an identity exists and that
* every element has an inverse.
*
* Identity: take any element a. There will be some element e
* such that ea=a (in a latin square, every element occurs in
* every row and column, so a must occur somewhere in the a
* column, say on row e). For any other element b, there must
* exist x such that ax=b (same argument from latin-square-hood
* again), and then associativity gives us eb = e(ax) = (ea)x =
* ax = b. Hence eb=b for all b, i.e. e is a left-identity. A
* similar argument tells us that there must be some f which is
* a right-identity, and then we show they are the same element
* by observing that ef must simultaneously equal e and equal f.
*
* Inverses: given any a, by the latin-square argument again,
* there must exist p and q such that pa=e and aq=e (i.e. left-
* and right-inverses). We can show these are equal by
* associativity: p = pe = p(aq) = (pa)q = eq = q. []
*/
if (errors) if (errors)
for (i = 0; i < a; i++) for (i = 0; i < a; i++)
errors[i] = 0; errors[i] = 0;