Puz 112177 page 1

 

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98.7.....7...6......5..97..5....84...9.6...3...4.2...1.5....8.....3....6....1..2.

;12177; 10.80;10.80;10.50 ;GP;kz0;11;12346;

 

This is an interesting puzzle.

The floors 1236 show a very high potential

 

Here is the reduced PM with in red potential eliminations

 

X X 1236 |X 3+ 123+ |1236+ 16+ 23+

X 123+ 123 |12+ X 123+ |123+ 1+ 23+

1236+ 1236+ X |12+ 3+ X |X 16+ 23+

 

X 1236+ 1236+ |1+ 3+ X |X 6+ 2+

12+ X 12+ |6g X 1+ |2+ 3g 2+

36+ 36+ X |X 2g 3+ |6+ 6+ 1g

 

1236+ X 1236+ |2+ X 26+ |X 1+ 3+

12+ 12+ 12+ |3g X 2+ |1+ 1+ 6g

36+ 36+ 36+ |X 1g 6+ |3+ 2g 3+

 

At that point, to ease next steps, we study all direct effects of a start in the base. Usually, 2 sub scenarios are possible; here 2 in the base requires a specific process

 

16r12c3 not valid conflict in r4c2

 

We now study 12r12c3

 

12r12c3 +1r3c4 +2r3c9 not valid conflict r4c2

 

12r12c3 +2r3c4 +1r3c8

=> in row 7 1r7c1 2r7c6 6r7c3 3r7c9

1r7c1=>1r4c2 6r4c8 6r1c7 3r2c7 2r5c7

In conflict with 2r5c1 last 2 in box 4

12r12c3 +1r3c8 +2r3c9

=> 2r4c2 1r7c1 no possibility for 1 in box 4

 

12r12c3 is not valid

 

We now study 13r12c3 a classical pattern for an exocet

 

13r12c3 +1r4c2 +3r7c1

=>1r3c4 3r3c9 3r4c5 3r6c2 1r7c8 1r8c1

 

13r12c3 +3r4c2 +1r7c1

no possibility for 1 in box 4 not valid

 

we keep in memory

13r12c3 {1r4c2 3r7c1 1r3c4 3r3c9 3r4c5 3r6c2 1r7c8 1r8c1}

 

We now study 23r12c3 due to digit 2, we look in row 3

 

23r12c3 + 2r3c9 + 3r3c5 conflict r4c2

23r12c3 + 2r3c4 + 3r3c5

3r4c2 3r6c6 2r5c1 2r7c6 6r9c6 2r4c9

UR <3>r12c7 3r9c7 3r7c1 6r7c3 1r7c8

HP12r12c7 6r6c6 =>6r4c3 conflict

23r12c3 + 2r3c4 + 3r3c9 3r9c7

3r7c1 2r7c6 6r7c3 1r7c8 6r9c6

We can not have 6r4c8 2r4c9 (no 1 in box 3)

So r4c2 must be 2 or 6

3r4c5

We keep in memory

23r12c3 { 2r3c4 3r3c9 3r9c7 3r7c1 2r7c6 6r7c3 1r7c8 6r9c6 2|6r4c2 3r4c5}

 

We now study 26r12c3 due to digit 2, we look in row 3

Can not be 2r3c9 => conflict r4c2

26r12c3 +6r3c8 +2r3c4 6r6c7 6r4c2 2r4c9 2r1c7 2r5c1

2r7c6 6r7c1 6r9c6 2r8c2

That we keep in memory

26r12c3{6r3c8 2r3c4 6r4c2 6r6c7 2r4c9 2r1c7 2r5c1 2r7c6 6r7c1 6r9c6 2r8c2 }

 

We now study 36r12c3 a classical exocet pattern

Can not be 3r3c5 (conflict r4c2)

 

36r12c3 6r3c8 3r3c9 6r4c2 3r7c1 6r9c1 6r7c6

3r6c2 3r4c5 3r1c6

That we keep in memory

36r12c3 {6r3c8 3r3c9 6r4c2 3r7c1 6r9c1 6r7c6 3r6c2 3r4c5 3r1c6}

 

====================================================================================

 

13r12c3 {3r7c1 1r4c2 1r3c4 3r3c9 3r4c5 3r6c2 1r7c8 1r8c1}

 

23r12c3 {3r7c1 2|6r4c2 2r3c4 3r3c9 3r9c7 2r7c6 6r7c3 1r7c8 6r9c6 3r4c5}

 

26r12c3 {6r7c1 6r3c8 2r3c4 6r4c2 6r6c7 2r4c9 2r1c7 2r5c1 2r7c6 6r9c6 2r8c2}

 

36r12c3 {3r7c1 6r4c2 6r3c8 3r3c9 6r9c1 6r7c6 3r6c2 3r4c5 3r1c6}

 

We see in that summary that

<1>r7c1 (3r7c1 3r7c1 6r7c1 3r7c1)

<3>r4c2 (1r4c2 2|6r4c2 6r4c2 6r4c2)

<1>1r3c8 (1r3c4 1r7c8 6r3c8 6r3c8)

<2>2r3c9 (3r3c9 2r3c4 2r3c4 3r3c9)

 

Giving the new reduced PM

 

X X 1236 |X 3+ 123+ |1236+ 16+ 23+

X 123+ 123 |12+ X 123+ |123+ 1+ 23+

1236+ 1236+ X |12+ 3+ X |X 6+ 3+

 

X 126 1236+ |1+ 3+ X |X 6+ 2+

12+ X 12+ |6g X 1+ |2+ 3g 2+

36+ 36+ X |X 2g 3+ |6+ 6+ 1g

 

36 X 1236+ |2+ X 26+ |X 1+ 3+

12+ 12+ 12+ |3g X 2+ |1+ 1+ 6g

36+ 36+ 36+ |X 1g 6+ |3+ 2g 3+

 

We show now that cell r2c2 must have an extra digit

 

===> <1>r2c2

. 6 in base Forces 1r3c4 6r3c8 conflict in r4c2

. 23 in base Forces 2r3c4 no 1 in row 3

 

 

===> <>2r2c2 forces 2r3c4

. 6 in base 36r12c3

=> 2r7c3 1r7c8 2r4c9 2r1c7 1r2c7

No 1 in box2

. 13 in base no more 2 in r3

==> <3>r2c2 26r12c3

1r3c2 1r8c1 1r7c8 1r2c7 1r1c6

3r6c6 3r4c3 3r7c9 no room for 3 in box 3

 

r2c2=4

And <3>r3c5 (now located in r3c12 for 26r12c3)

 

Here the new reduced PM 1236 after all that and the summary again

 

X X 1236 |X 3+ 123+ |1236+ 16+ 23+

X + 123 |12+ X 123+ |123+ 1+ 23+

1236 1236 X |12+ x X |X 6+ 3+

 

X 126 1236+ |1+ 3+ X |X 6+ 2+

12+ X 12+ |6g X 1+ |2+ 3g 2+

36+ 36+ X |X 2g 3+ |6+ 6+ 1g

 

36 X 1236+ |2+ X 26+ |X 1+ 3+

12+ 12+ 12+ |3g X 2+ |1+ 1+ 6g

36+ 36+ 36+ |X 1g 6+ |3+ 2g 3+

13r12c3 {3r7c1 1r4c2 1r3c4 3r3c9 3r4c5 3r6c2 1r7c8 1r8c1}

 

23r12c3 {3r7c1 2|6r4c2 2r3c4 3r3c9 3r9c7 2r7c6 6r7c3 1r7c8 6r9c6 3r4c5}

 

26r12c3 {6r7c1 6r3c8 2r3c4 6r4c2 6r6c7 2r4c9 2r1c7 2r5c1 2r7c6 6r9c6 2r8c2}

 

36r12c3 {3r7c1 6r4c2 6r3c8 3r3c9 6r9c1 6r7c6 3r6c2 3r4c5 3r1c6}

 

We now show <3>r1c5

Obvious for 13r12c3 23r12c3 36r12c3

26r12c3 (+3r1c5)

3r4c3 3r7c9 3r2c7 1r8c7 1r7c3

No more 1 in box 4

 

r4c5=3

 

We now show <1>r7c3

obvious for 13r12c3 23r12c3

26r12c3 (+3r4c5 +1r7c3) no more 1 in box 4

36r12c3 +1r7c3 and 12r3c12

2r7c4 2r2c6 1r2c4 1r4c3 conflict

 

r7c8=1

 

We now show <3>r79c9

obvious if 3r3c9

26r12c3 (with now r7c8=1) => 1r2c7 3r9c7

 

r9c7=3

 

X X 1236 |X x 123+ |126+ 6+ 23+

X + 123 |12+ X 123+ |12+ x 23+

1236 1236 X |12+ x X |X 6+ 3+

 

X 126 126+ |1+ 3 X |X 6+ 2+

12+ X 12+ |6g X 1+ |2+ 3g 2+

36+ 36+ X |X 2g x |6+ 6+ 1g

 

36 X 236+ |2+ X 26+ |X 1 x

12+ 12+ 12+ |3g X 2+ |x x 6g

6+ 6+ 6+ |X 1g 6+ |3 2g x

 

We have covered most of the potential eliminations

We can still show easily now 36r12c3 not valid

 

36r12c3 {3r7c1 6r4c2 6r3c8 3r3c9 6r9c1 6r7c6 3r6c2 3r4c5 3r1c6}

 

HP 12r2c46 1r1c7 2r1c9 2r5c7 2r4c3 2r7c4

conflict column 4 12r2c4 1r4c4 2r7c4

Leaving 12r2c6

 

no more extra digit in r3c4

no extra digit in r1c6 HP in r12c6 in each scenario

 

here we skip to the full PM where we find

 

r5c6=4 r4c5=1 r6c4=9 r3c4=2 r7c6=2 r9c6=6

 

the rest will not require more than an XWing