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March 13 9x9 cages 25+, 27+, 27+
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Posted on: Thu Mar 15, 2012 9:59 pm

Posts: 36
Joined: Fri May 13, 2011 10:23 am
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

Posted on: Fri Mar 16, 2012 1:24 am

Posts: 700
Joined: Fri May 13, 2011 6:51 pm
Re: March 13 9x9 cages 25+, 27+, 27+
rickt wrote:
yes I do agree with the contents of the 25+ cage 57193 (top to bottom)

The content is OK but we must define the exact position of all numbers based on a precise analysis. Let’s continue then from the previous graphic: c7 = [1,5] (3, 7 and 9 already present in rows 6 and 7; g7 = [1,5] (3, 6 and 7 not possible, due to the previous note and g1 = 6).

Now let’s go with the cages “27+” (h1) and “27+” (i2). First we observe that the numbers that compose those cages are: 9, 8, 8, 7, 7, 5, 4, 3, 2, 1 (the pending numbers in the three rightmost columns). If we take five numbers among these for one of the cages, the other five numbers, obviously, will go to the other cage (the cages are “paired”).

In the arjen’s post they appeared five combinations for each of those cages but at that moment we had less information than now. I suggest to use, for instance, an auxiliary program like the jomapil’s … (otherwise the calculation must be obtained “manually”). So, let’s obtain the possiblities for the “27+” (i2) that we have named “27+” (lower), by carefully scrolling and supressing the wrong possibilities, considering that for this cage: no 2’s are allowed, no 4’s are allowed, no 6’s are alloved, no more than a 1, no more than one 3, no more than one 5, no more than two 7’s, no more than two 8’s and no more than one 9. In these conditions:

“27+” (lower) = [1,3,7,8,8] >>> “27+” (upper) = [2,4,5,7,9]
“27+” (lower) = [1,3,7,7,9] >>> “27+” (upper) = [2,4,5,8,8]

I have writen both possibilities, in the cages, in two lines, one in balck and the other in red because I am going to inmediately demonstrate that the one in red is forbidden. First observe that the upper cage does not contain a 1 in any case so in row 1, e1 = 1, then e2 = 2, e3 = 7 (green). We need a 2 in row 1, h1 = 2 and a 4 in row 2, h2 = 4 (green).

Let’s suppose that “27+” (upper) = [2,4,5,8,8]: since there are two 8’s in the combination it implies i9 = 8 and, since the “27+” (lower) does not contain now any 8 (it is 13779) the 8 of row 3 should go to g3 (as shown in red) >>> g2 = 5 >>> c3 = 5 (shown in red). And, this is the nuclear point, now we would have two 5’s in c3 and g2 making impossible to place any 5 in row 7 (like with the X-Wing in the sudoku). So we must take as the only valid the first line of the combinations:

“27+” (lower) = [1,3,7,8,8] (and “27+” (upper) = [2,4,5,7,9]).

Now (in violet colour) we place the 8’s in h3 and i2, the 3 of the cage goes to i3. In the cage “11664x”, g6 = 8 and, obviously, h6 = 3 and i7 = 6. Now the 3 of cage “630x” goes to g8 and then the 3 of cage “25+” goes to b9.
Also in the “27+” (upper) i1 = 7, g2-g3 = [5,9]. And b1 = 8.
Finally, and using a new graphic:

In green colour: The [1,7] in g4-h4 produce a 9 in f4 (f3 = 1). The [5,9] in g2-g3 force a 1 in g7 and a 5 in c7 >>> b2 = 5. We fill the other green numbers in the cages “10+” and “1-“.

In brown colour: The 1 of cage “25+” to c8, b8 = 7 (due to, i.e., 9 in i8), a9 = 9, b5 = 9, a5 = 7. Also g2 = 9, g3 = 5; and c3 = 9, c2 = 7.

In blue colour: The 1 in d9 forces i9 = 5 >>> e9 = 6, e8 = 5. Additionally h8 = 6 and h9 =7. Also i9 = 5 >>> i5 = 1, h5 = 5.

And that’s it: g4 = 7, h4 = 1.

Posted on: Fri Mar 16, 2012 6:56 pm

Posts: 49
Joined: Sun Oct 16, 2011 10:58 am
Re: March 13 9x9 cages 25+, 27+, 27+
Thanks for the explanation.
I totally overlooked the 2 sixs in 23+.
After minutes staring at the numbers I gave up, that happens relatively rarely.

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