Thinking about it:
if the 3 5 and 6 are in the top row of the first block
then they can't be in the top row of the second block
or the top row of the third block.
Taking the second block first:
3,5 & 6 can be positioned so that
all are in the second row
or they can be split so that one of them is in the second row and two in the
third row
or 2 of them in the second row and one of them in the third row
or all are in the third row.
Similarly with the third block (but with more restrictions based on what has
happened in the second block).
So, yes, it is inevitable that at least 2 of them will stay together in each
of the second and third blocks, because there aren't enough rows to spread
them any other way.
Sue
----- Original Message -----
From: "Avital" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, April 18, 2006 3:18 AM
Subject: RE: [lace-chat] Sudoku (again)
Very interesting! I had a quick look at some Sudoku solutions and I think
you
may be right. Thanks!
Best wishes,
Avital
Susan Webster wrote:
By studying solutions (and frankly looking for some trick), I have
discovered an “ugly” little truth that may or may not be well known (but
in the event it is not, I thought I would share it here).
Consider a 3-row block which consists of 27 little squares (within the
bold separators) – it happens that numbers travel together. What I mean
is that if, for example, the first 3 numbers across are 3 5 6, then you
can be sure that at least two of them will move across the 3-row block
together. That is, the 3 and the 5 OR the 3 and the 6 OR the 5 and
the 6 will appear in the same 3-square block in the rest of the section.
They may not be in the same order, or they have a separator between
them, but they will both be in the same little 3-square section across
those 27 squares.
In the puzzle I have just finished, the top 3 rows are as follows:
3 5 6 4 9 7 2 8 1
9 2 1 5 6 8 4 3 7
4 8 7 1 3 2 5 6 9
<snip>
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