rep'nA wrote:JPF wrote:As mentionned before, if A~A' and B~B' it can happen that d(A,B)<>d(A',B') but it doesn't matter.

JPF

Let f be an isomorphism of the grid sending A to A'. So we may write A' = f(A). Is it the case that d(A,B) = d(f(A), f(B))?

Again, in most cases that's not true.

See the examples already given before :

i wrote:Example :d(A,B)= 4

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`A : 500000009020100070008000300040600000000050000000207010003000800060004020900000005 m_b_metcalf`

B : 500000009020100070008000300040002000000050000000706010003000800060004020900000005 StrmCkr

* * * *

using the gsf's canonical forms for these puzzles :d(A',B')= 36

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`A' : 000006000007180000000020005008500001300000900060000040002070008040000060900000300`

B' : 100050000000089200000700000070000003006000040500002900004000060800010500030000007

* ** ** ** ** * *** ** * ** ** *** * * **** * ** ** * *

Of course if h is an automorphism of A, then A'=h(A)=A and

d(A,B)=d(h(A),B) for every B.

Mauricio wrote:I would think that a better definition of distance is this:

D(A,B)=min{d(A,B):A~A', B~B'}, using the d function defined by JPF.

In this new metric it is trivial that D(A,B)=D(A',B') if A~A' and B~B'. We would want this, as properties of sudokus besides aesthetics should be independent of isomorphism.

I think that's a good idea.

It has to be proved that D is a distance on the quotient set S/~

D(A,B)=0 then min{d(A',B'): A'~A, B'~B}=0.

so there are A' and B' such that d(A',B')=0 .

As d is a distance, A'=B' with A'~A and B'~B.

Finally A~B.

I don't have any proof (or counter example) that D(A,B)<=D(A,C)+D(C,B)

Any idea ?

JPF