Problem 108

[luke]

i don't know if it's correct to consider distinct two solution (x1,y1) (x2,y2)of the diophantine equation if and only if

x1!=x2 and y1!=y2 or x1!=y2 and y1!=x2

is it correct?
thanks
luke

[daniel.is.fischer]

Identify a solution of [frac]1,x[/frac] + [frac]1,y[/frac] = [frac]1,n[/frac] with the set {x,y}. So order doesn't matter, (x,y) is the same as (y,x).

[quilan]

Well, if they're the same, then it's the following logic:

(x1==x2 and y1==y2) or (x1==y2 and y1==x2)

So, take the inverse of that to test if they're distinct, and any of the following tests should work (DeMorgan's Law I think?):

!((x1==x2 and y1==y2) or (x1==y2 and y1==x2))
!(x1==x2 and y1==y2) and !(x1==y2 and y1==x2)
(x1!=x2 or y1!=y2) and (x1!=y2 or y1!=x2)


(1,2) (2,1) for example would be detected as distinct in your example (which it is not, they're the same set)
-> ((1!=2) and (2!=1)) or ((1!=1) and (2!=2))
-> (T and T) or (F and F)
-> T or F
-> T

[daniel.is.fischer]

Well, if they're the same, then it's the following logic:

(x1==x2 and y1==y2) or (x1==y2 and y1==x2)

So, take the inverse of that, and any of the following tests should work (DeMorgan's Law I think?):

!((x1==x2 and y1==y2) or (x1==y2 and y1==x2))
!(x1==x2 and y1==y2) and !(x1==y2 and y1==x2)
(x1!=x2 or y1!=y2) and (x1!=y2 or y1!=x2)


(1,2) (2,1) for example would be detected as distinct in your example (which it is not, they're the same set)
-> ((1!=2) and (2!=1)) or ((1!=1) and (2!=2))
-> (T and T) or (F and F)
-> T or F
-> T

You mean (x1,y1) == (x2,y2) [equiv] (x1 == x2 /\ y1 == y2) \/ (x1 == y2 /\ y1 == x2) ?
Yes. But in the example you gave, you swapped 'and' and 'or', the test should unfold:
(1,2) != (2,1) [equiv] (x1 != x2 \/ y1 != y2) /\ (x1 != y2 \/ y1 != x2)
[equiv] (1 != 2 \/ 2 != 1) /\ (1 != 1 \/ 2 != 2)
[equiv] (T \/ T) /\ (F \/ F) [equiv] T /\ F = F.

[quilan]

You mean (x1,y1) == (x2,y2) [equiv] (x1 == x2 /\ y1 == y2) \/ (x1 == y2 /\ y1 == x2) ?
Yes. But in the example you gave, you swapped 'and' and 'or', the test should unfold:
(1,2) != (2,1) [equiv] (x1 != x2 \/ y1 != y2) /\ (x1 != y2 \/ y1 != x2)
[equiv] (1 != 2 \/ 2 != 1) /\ (1 != 1 \/ 2 != 2)
[equiv] (T \/ T) /\ (F \/ F) [equiv] T /\ F = F.

Yup. Was showing him an example where the logic he provided would fail to be the correct case (parent logic was: x1!=x2 and y1!=y2 or x1!=y2 and y1!=x2)

[daniel.is.fischer]

Sorry, didn't link it with the first post.

[luke]

Hi,
yesterday I had a lot of thread open, I've made an error negating condition I'm use in my javaCode.
So I agree that (x1==x2 and y1==y2) or (x1==y2 and y1==x2) identify the same solution....

I've posted my question, cause my solution is wrong...so I've a bug in my code (I suppose some stuff related to float2int conversion)

thanks quilan&daniel

luca