Place the numbers 1-9 in the boxes so that the difference of each pair of numbers joined by a line is at least three. The number 8 has already been placed.

(In reply to re: solution by reasoning by ed bottemiller)

It is 03:25 AM and this is  probably the best  excuse for so many errors in my explanation . <o:p></o:p>

It was G=5, not C and now I write   while viewing both the diagram and my solution.                         <o:p></o:p>

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837  UPPER ROW        ABC<o:p></o:p>

296   MID                    DEF   <o:p></o:p>

514  BOTTOM ROW     GHI<o:p></o:p>

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Start H=1 <o:p></o:p>

 G is between 1 and 8,
so 4 or  5 are candidates, <o:p></o:p>

Choosing  G=4 leaves no choice for D , <o:p></o:p>

so G=5 and D=4<o:p></o:p>

We now have <o:p></o:p>

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8 B C<o:p></o:p>

5 E F<o:p></o:p>

2 1 I<o:p></o:p>

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There are only two digits left with only 2 possible neighbors: 4(neighboring 6 or 9) and 6(3 or 9) .<o:p></o:p>

One of those has to be placed in I,  the only box with 2 links only.   <o:p></o:p>

Placing 6 leads to a dead end forcing C=9 thus leaving no choice for B  . so the choice is I=4 ,C=7 and B =3.

6 and 9 go to E and F or F and E.

 

Hopr there are no errors .

 

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es no choice for c therefore i=4 and c f<o:p></o:p>

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Posted by Ady TZIDON on 2010-08-05 00:17:08