Re: March 13 9x9 cages 25+, 27+, 27+

picklepep wrote:

arjen wrote:

It might help if you bolded a few more things. The double 6's in the 23+ cage can be bolded

as they can not be in any other position.

You can also look at rows 6 and 7. You can figure out that the 5 and 1 in row 7 can only be

in 2 spots, hence they are fixed that makes row 7 columnn 5 be 8, thus r8c1=8. I think there is

more...

I agree with Pickelepep, to solve the last cages you'll need to find and fix more numbers.

Cages 23+ and 128+ all green:

Since there is no '1' or '2' in 11664x, [b6,c6] = [1,2]

There is no [2,4] in the 25+, 19+, 11664x , 630x cages >>> [a7,b7] = [2,4] >>> a8=8.

From this and since d9=1, the '1' in column a can only be in a2 >>> a2=1.

Because i4=2 >> b3=2 >> c4 =8 >> b6=1 >> c6 =2 >> a7=2 >> b7=4 , the cage 23+ is all green

Going to the top 27+:

Then looking at column a, a1 = 4 because in 23+ and 25+ there are no '4's

Knowing a1=4, the last '4' to be placed must be in h2 >>> h2=4.

Since we know b3=2 and i4=2 there can be no '2' in the lower 27+. In fact because g9=2 and h2=4 >>> h1 =2!

Now we know the top 27+ is either [2,4,5,8,8] or [2,4,5,9,7]

Because f1=5, the 5 of the top 27+ must be in [g2,g3]

Going down to the lower half of the puzzle:

In row 6 the '8' must be in [g6, h6] since the '8'in 128x is already placed in a8

In row 7 the '8' must be in e7 since the '8' in 128x is in row 8 and the '8'of 11664 in row 6!

Because of the 10+, the 1- and the 11664 are sharing the '7', the '3' and the '6', c7 and g7 are [1,5].

We know there is a 5 in [g2,g3] >>> g7=1 >>> c7 = 5

From this follows c8 = 1, since 10080x has no '1' in it an column c needs a '1'

Now we know [c2,c3] = [9,7]!

Fixing the 14+:

The '1' in the first row, must be placed in e1, since the cages 10080x and 27+upper contain no '1'

e1=1 and b3=2 >>> e2=2 >>> e3=7

Leads to fixing [c2,c3] and the [1,9] cage:

e3=7 leads to c2=7 >>> c3=9 >>> f3=1 >>> f4=9!

But then in row 4 all we miss is an '1' and a '7' >>> [g4,h4] = [7,1]

We know g7=1 >>> g4 = 7 >>> h5=1 >> h6=5 >>> i5=1!

This will fix the 630x cage:

[g2,g3] = 5 and h6=5 thus the 5 in 630x must be in i9 >>> i9=5 >>> e9=6 >>> e8=5

Because g1=6 and e9=6 the '6' in 630x must be in h8 >>> h8=6

We know g4=7, h8 and i9 are occupied >>> the '7' in 630x is in h9 >>> h9=7

This leaves g8=3

Go to the 11664x cage:

Since h8=6 and g1=6 >>> i7=6

And since g8=3 >>> g6=8 >>> h6 =3

Go to the upper 27+ cage:

We know column g is only missing a '9' and c3=9 >>> g2=9 >>> g3=5 >>> i1=7

This makes the lower 27+ cage [1,3,7,8,8]...

Now the rest is easy....

I did not include any pictures, I hope you can follow. If there is need for pictures I'll try make some... :-D