> The picture is even worse in fact: the version in chapter 12 cannot
be proved correct!
> The correctness depends on the fact that the propagator A=:B*C
guarantees the
> existence of a solution when A is determined.
I'm not sure I understand what you are stating here? From the code
standpoint, I see you've added B and C to the distribute function.
> Here is a constraint version that solves the same problem as the
relational version in the book:
In my confusion, I was solving the wrong problem from my end as well -
just generating palindromes is what my revised solution was doing.
Whereas chapter 9 states the problem as generating "all four-digit
palindromes that are products of two-digit numbers" - meaning the 4
digit number must be >= 1000 and the two-digit numbers >= 10. The
revised solution you provide matches the result of the code in chapter
9, though I'm not sure if the authors intended to have repeats?
Just for the sake of completeness, I was trying to prove to myself where
the discrepancies arose. In the original code of chapter 12 (translated
to 4 digits), I was getting 46 solutions. This comes from having 36
unique solutions >= 1000, and an additional 10 that are < 1000.
proc {Palindrome_12_4 ?Sol}
sol(A) = Sol
B C X Y
in
A::0#9999 B::0#99 C::0#99
A=:B*C
X::0#9 Y::0#9
A=:X*1000+Y*100+Y*10+X
{FD.distribute ff [X Y]}
end
% 46 unique solutions (includes numbers less than 1000)
[0000 0110 0220 0330 0440 0550 0660 0770 0880 0990
1001 1221 1551 1771 1881 2002 2112 2332 2442 2552
2772 2992 3003 3663 3773 4004 4224 4554 4664 4774
4884 5005 5115 5225 5335 5445 5775 6006 6336 6776
7007 7227 8008 8118 8448 9009]
Filtering out the results < 1000, we would get the 36 unique solutions
>= 1000. (Note: limiting the domain to A::1000#9999 or using the
constraint A>:1000 could be used in lieu of limiting the range of
X::1#9, but then that'd be less efficient. But since the example is
given twice in chapter 12, the A:>1000 is a better approximation to the
original code in chapter 9 for the first version.)
proc {Palindrome_12_4 ?Sol}
sol(A) = Sol
B C X Y
in
A::0#9999 B::0#99 C::0#99
A=:B*C
X::1#9 Y::0#9
A=:X*1000+Y*100+Y*10+X
{FD.distribute ff [X Y]}
end
% 36 unique solutions (excludes numbers less than 1000)
[1001 1221 1551 1771 1881 2002 2112 2332 2442 2552
2772 2992 3003 3663 3773 4004 4224 4554 4664 4774
4884 5005 5115 5225 5335 5445 5775 6006 6336 6776
7007 7227 8008 8118 8448 9009]
In the revised code you supplied, I get 118 solutions which matches the
results of chapter 9 (36 unique solutions & 82 duplicates). Not sure I
understand the distribution of repitition in the numbers? (It must
favor 2772 since it gets repeated 10 times).
proc {Palindrome_12_4 ?Sol}
sol(A) = Sol
B C X Y
in
A::0#9999 B::0#99 C::0#99
A=:B*C
X::1#9 Y::0#9
A=:X*1000+Y*100+Y*10+X
{FD.distribute ff [X Y B C]}
end
% 118 solutions - includes 82 duplicates (note: I've sorted the list)
[1001 1001 1001 1001 1221 1221 1551 1551 1771 1771
1881 1881 1881 1881 2002 2002 2002 2002 2112 2112
2112 2112 2112 2112 2112 2112 2112 2112 2332 2332
2442 2442 2442 2442 2552 2552 2552 2552 2772 2772
2772 2772 2772 2772 2772 2772 2772 2772 2992 2992
2992 2992 3003 3003 3003 3003 3663 3663 3773 3773
4004 4004 4004 4004 4224 4224 4224 4224 4224 4224
4554 4554 4554 4554 4664 4664 4774 4774 4884 4884
5005 5005 5005 5005 5115 5115 5225 5225 5335 5335
5445 5445 5775 5775 6006 6006 6006 6006 6336 6336
6336 6336 6336 6336 6776 6776 7007 7007 7227 7227
8008 8008 8118 8118 8448 8448 9009 9009]
Thanks,
Chris Rathman
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