Saturday, 4 February 2017 and the Guardian reports how the
EU leadership criticise the US president for his ‘lack of respect’. Earlier
this week Jeremy Corbyn had placed a ‘3-line whip’ on Labour MPs to support the
2nd reading of the European Union (Notification of Withdrawal) Bill.
Quite a number defied the whip but not enough. Typically only Ken Clarke from
the Tory side voted against.
It is sad that so many of the UK’s MPs have forgotten that
they were voted into Parliament to act in accordance with the views they
presented at the General Election 2015. Of course they are entitled to change their
mind, but they must be convinced of the rationale for changing it. I wonder whether
they are following what they believe to be the voter arithmetic rather than being
persuaded by the argument. If that is the case, they have failed fundamentally in
their duty as MPs.
But enough of all that. Hard Sudoku is what this blog is about.
I tried the Guardian Sudoku puzzle (No. 3663) and recorded my progress. This
record uses the methodology developed in MS Excel about 10 years ago. The
methodology is quite weird and long winded but, in this context, it is useful
for explaining how and why I got stuck. I describe below the argument I
employed for getting unstuck.
I had got as far as here using conventional arguments and
methods. Assuming no mistakes in recording, this would represent a valid
solution thus far.
The record shows that I had reviewed all the unsolved
positions in the puzzle without making any further progress. This also covered
the examination of individual Sudoku Nos. and their relative positions in each
of the 3 dimensions of the puzzle. There were only four numbers left (1, 2, 6
and 8) and the reasons for not making progress bears some examination.
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Top 3 Rows
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Middle 3 Rows
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Bottom 3 Rows
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Left 3 Columns
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Middle 3 Columns
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Right 3 Columns
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Id 1
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No positions fixed
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All positions fixed
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Only 1 position fixed
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2 positions fixed?
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All positions fixed
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Only 1 position fixed
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Id 2
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No positions fixed
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Only 1 position fixed
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All positions fixed
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2 positions fixed?
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Only 1 position fixed
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All positions fixed
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Id 6
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No positions fixed
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Only 1 position fixed
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Only 1 position fixed
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Only 1 position fixed
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Only 1 position fixed
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No positions fixed
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Id 8
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Only 1 position fixed
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No positions fixed
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2 positions fixed?
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2 positions fixed?
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No positions fixed
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Only 1 position fixed
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Both Id’s 1 & 2 have a combination groups of three such
that one group is solved completely and at least one other group where two positions
are known, but not the third. Unfortunately, the information in the opposing
dimension does not help us limit the position of the 3rd ID. For ID
6, there is altogether too little information to make further progress.
ID 8 has 2 groups of three where the location of two of the
elements is known. At first glance, the unsolved positions in the first column
at rows 4 and 5 look the most promising. After all the sub-grid at the bottom
left hand corner is solved completely and the sub-grid immediately above it has
only these two elements unsolved. Nevertheless, no amount of examination has
revealed a logically provable determination as to which cell should contain
which ID.
I’m left with considering the position of ID 8 in the bottom
3 rows of the puzzle. At first glance this looks considerably less promising
than the column 1 vacancies discussed above, especially since there are 3
unsolved ID positions for row 9 (see the values highlighted in green below) rather
than the just two for column 1. However, in my solution methodology, I
highlight cells where the possible solutions is limited to 1 of 2 IDs (see the
cells circled in red).
The information available to us is therefore is
- · The location of ID 8 is known to be row 8 in for the bottom left sub-group and in row 7 for the bottom right sub-group, thus ID 8 must appear in one of the 3 cells in row 9 of the bottom middle sub-group. The right most cell of this sub-group row 9 has already been solved.
- · There are 3 unallocated cells in row 9 and 3 unallocated IDs.
- · But column 4 of the puzzle has two cells (named 2_43 and 8_48 in my nomenclature) whose ID values are limited 1 or 6. Logically, if these two values must appear in the two named cells, then they cannot appear anywhere else.
I conclude therefore the cell named 8_49 in row 9 must contain
the ID value 8, since that is the ID value left for this column.
This is something of a tortuous argument but it seems to
work. Any thoughts?
