## Solution to Vidar's Monster #4

Everything about Sudoku that doesn't fit in one of the other sections

### Solution to Vidar's Monster #4

Code: Select all
1.. ..6 ...
.6. 9.. ..8
8.. ..4 36.

..8 ... 4..
..6 .43 9.5
.4. 5.. ...

.2. ... .7.
4.1 .7. ...
..3 .1. 2..

I hope that I've used forcing chain notation correctly.

1) 7=[r2c1]=7=[r6c3]=7=[r4c9]=7=[r2c6]=7=[r2c1] so r2c1<>7

2) 7=[r1c2]=7=[r6c3]=7=[r4c9]=7=[r5c4]=7=[r2c6]=7=[r1c7]=7=[r1c2] so r1c2<>7

3) 1=[r4c2]-7-[r5c2](-7-[r9c1])-2-[r5c1]-9-[r6c3](-5-[r7c3])=9=[r7c1]=6=block 7 so r4c2<>1

4) r5c2=1

5) 7-[r4c2]-7-[r5c1]-2-[r6c3]-9-[r6c1]-5-[r4c1]-5-[r2c1] so r4c2<>7

6) 6-[r9c1]=7=[r9c2]=8=[r8c2]-[naked pair r7c13]=[hidden pair r89c6]=2=[r8c4]-2-[r7c6]=8=[r9c8]=8=[r6c7]=8=[r5c4](=8=[r1c5])(-2-[r5c8](-[naked pair r46c8])=2=[r1c9])-8-[r9c4](-9-[r9c1]-5-[r9c6]-9-[r8c6])=4=[r7c9]=1=[r3c9]=1=[r4c4](-1-[r3c4]-7-[r1c4]-3)-1-[r4c8]=3=[r6c1]=3=block 1 so r9c1<>6

7) r7c1=6

8) locked candidates in block 8 column 4 so r4c4<>6

9) 5-[r2c6]=5=[r7c5](-5-[r3c5]-2-[r2c5]-3-[r1c5]-8-[r1c4]-7-[r3c4]-1)-5-[r7c3]-9-[r7c6]-8-[r9c6]-9-[r8c6]-[naked pair r46c6]-1,7-[r4c4]-2-[r5c4]-8-[r5c8]-2-[r5c1]=7=[r9c2]-7-[r9c1]-5-[r8c2]=8=[r9c8]=8=[r6c7](-8-[r7c7]-1)=6=[r8c7]-6-[r8c4]-3-[r8c9]-9-[r3c9]-7-[r2c7] so r2c6<>5

10) locked candidates in block 2 column 5 so r7c5<>5

11) 2-[r8c4](-[naked pair r34c4]-7-[r5c4]-8-[r5c8]-2-[r5c1]=7=[r9c2]=8=[r8c2])-[naked triple r789c6]-8,9-[r7c5]-3-[r7c4](-4-[r9c4]-6)=3=[r1c4](-3-[r1c5]-8-[naked pair r23c5])=3=[r4c2]=5=[r4c1]=5=[r7c3]-5-[r9c1]-9-[r6c1]-2-[r6c3]=9=[r4c5]-9-[r6c5](=6=[r5c9])=2=[r4c6]-2-[r4c8]-1-[r4c4]=7=[r2c6]=1=[r2c7]-1-[r7c7](-8-[r7c6]-9-[r8c6]-5)-8-[r6c7]-7-[r6c9]-3-[r8c9]-9-[r8c8] so r8c4<>2

12) r8c6=2

13) 3-[r7c5](-[naked pair r13c5]-2,5-[r1c5]-8)(-[naked triple r789c4]-8-[r5c4]=8=[r5c8])=3=[r1c4]=3=[r2c1]=3=[r4c2]=5=[r4c1]=[locked 9s in block 4 row 6]-9-[r6c5]=6=[r4c9]=6=[r8c7]-6-[r8c4]=8=[r9c2]=7=[r9c1]=7=[r6c3]=7=block 6 so r7c5<>3

14) locked candidates in block 8 column 4 so r1c4<>3

15) naked triple r7c356 so r7c479<>5,8,9 and naked triple in block 8 r7c56 and r9c6 so r789c4<>589

16) r7c7=1

17) naked pair r12c7 so r68c7 and r123c9<>5,7

18) 8-[r5c8]-8-[r6c7](-6-[r6c5]=6=[r4c5])-6-[r8c7]=8=[r9c2]=7=[r9c1]=7=[r6c3]-7-[r5c1]-2-[r5c4](=8=[r1c4])=7=[r2c6](=1=[r3c4]=1=[r2c8])-7-[r2c7]-5-[r1c2]=3=[r4c2]-3-[r4c8]-2-[r4c4] so r5c8<>8

19) All singles to the end. I solved it on a piece of paper which I have lost. All I have left are the steps to the solution (which, of course, is better than having the solution and losing the steps to it). So I don't have the solution.
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Once upon a time I was a teenager who was active on here 2007-2011
999_Springs

Posts: 367
Joined: 27 January 2007
Location: In the toilet, flushing down springs, one by one.

999_Springs,

I'm not sure if the notation is correct. For instance,

999_Springs wrote:1) 7=[r2c1]=7=[r6c3]=7=[r4c9]=7=[r2c6]=7=[r2c1] so r2c1<>7

I see that if r2c1 = 7 then r6c3 = 7, so it isn't right to use the '=' notation in this case. In fact, a priori, I don't see a weak or strong link between them on 7, only the one implication I noted. Your deduction seems to be more along the line of a contradiction net than a forcing chain (at least as I understand the terms).
re'born

Posts: 551
Joined: 31 May 2007

1) 7=[r2c1]=7=[r123c3]=[r6c3]...

Is that better?

Anyway is there a sort of fish that eliminates 7 from r1c2 and r2c1? I do not understand any fish except for ordinary fish and turbot fish and finned X-wing, even though I eat a lot of fish.
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Once upon a time I was a teenager who was active on here 2007-2011
999_Springs

Posts: 367
Joined: 27 January 2007
Location: In the toilet, flushing down springs, one by one.

999_Springs wrote:1) 7=[r2c1]=7=[r123c3]=[r6c3]...

Is that better?

Anyway is there a sort of fish that eliminates 7 from r1c2 and r2c1? I do not understand any fish except for ordinary fish and turbot fish and finned X-wing, even though I eat a lot of fish.

The notation still isn't correct. However, you can write [r2c1]-7-[r123c3]=7=[r6c3]...

(the idea is that the first two groups can both be false for 7 so they are not strongly linked on 7. However, if either is true for 7, the other will be false for 7. In other words, you should only write [A]=x=[B] if A<>x => B=x and B<>x => A=x. If A=x => B<>x and B=x =>A<>x, then you should write [A]-x-[B]).

As to whether there is a fish, I'm not sure. However, Nishio rules out 7 from those two cells and where there is Nishio there is often a fish.
re'born

Posts: 551
Joined: 31 May 2007