Hello swaatacba,
The eliminations you mentioned are correct, but the
pattern you mentioned does not justify them.
- Code: Select all
.----------------------------------------------.
| 4 1 25 | 9 3 8 | 6 25 7 |
| 8 25 3 | 6 57 47 | 9 1 24 |
| 79 69 67 | 2 1 45 | 8 3 45 |
|------------------+------------+--------------|
| 1 45 457 | 57 2 6 | 3 8 9 |
| 6 3 9 | 8 4 15 | 2 7 15 |
| 257 8 257 | 3 9 17 | 14 45 6 |
|------------------+------------+--------------|
| 259 2459 8 | 57 6 3 | 147 249 12 |
| 29 7 1 | 4 8 29 | 5 6 3 |
| 3 24569 2456 | 1 57 29 | 47 249 8 |
'----------------------------------------------'
Before mentioning directly the skyscraper, I take the opportunity
to show how to make eliminations with colouring.
I have coloured the 5s using two pairs of opposite colors (',")
and (!,!!). For instance, choosing the color (') for (5)r1c3, we
can color (5")r1c8 because one of these two 5s must be true (strong link).
We have two clusters (',") and (!,!!), using the two pairs of opposite
colors, as shown below.
In each pair of opposite colors only one color must be true.
- Code: Select all
.-------------------------------.
| . . '5 | . . . | . "5 . |
| . "5 . | . '5 . | . . . |
| . . . | . . "5 | . . '5 |
|-----------+---------+---------|
| . -5 -5 |"5 . . | . . . |
| . . . | . . '5 | . . "5 |
| !5 . 5 | . . . | . '5 . |
|-----------+---------+---------|
|!!5 5 . |'5 . . | . . . |
| . . . | . . . | . . . |
| . 5 -5 | . "5 . | . . . |
'-------------------------------'
Now, (5)r4c3 sees ('5)r1c3 and ("5)r4c4.
Since one of the opposite colors must be true, one of these
colored 5s must be true, so (5)r4c3 can be eliminated (marked
with a minus sign at the grid). Something similar happens to (5)r9c3,
that sees ('5) r1c3 and ("5)r9c5.
A key observation is that since (!!5)r7c1 and ('5)r7c4 are in
the same row, both colors cannot be true, so one of the colors
(!,") must be true.
This is useful, since (5)r4c2 and (5)r4c3 both see (!5)r6c1 and ("5) r4c4.
This implies that (5)r4c2,(5)r4c3 can both be eliminated. One of
these eliminations was already justified, but I mention this
second way to get both elimination because the pattern abcd below
is the skyscraper.
- Code: Select all
.-------------------------------.
| . . 5 | . . . | . 5 . |
| . 5 . | . 5 . | . . . |
| . . . | . . 5 | . . 5 |
|-----------+---------+---------|
| . -5 -5 |d5 . . | . . . |
| . . . | . . 5 | . . 5 |
| a5 . 5 | . . . | . 5 . |
|-----------+---------+---------|
| b5 5 . |c5 . . | . . . |
| . . . | . . . | . . . |
| . 5 5 | . 5 . | . . . |
'-------------------------------'
Here abcd is not a notation for colors. I am using labels a,b,c,d just to show the cells that are part of the
skyscraper pattern: two walls "ab" and "cd", floor "bc" and tilted roof "da".
The logic to justify the skyscraper elimination is simple:
If (5)r6c1 (marked a) is true, it eliminates (5)r4c23.
If (5)r6c1 is false, then (5)r7c1 (marked b) must be true
(column 1 has only two 5s), so (5)r7c4 must be false, but this implies
that (5)r4c4 must be true (column 4 has only two 5s), which eliminates
the same (5)r4c23. This reasoning show that one of (5)r6c1,(5)r4c4 must
be true. Any cell that sees these two cells and has a 5 in it, must have
a false 5 that can be eliminate. So,(5)r4c23 can eliminated.
When you get used to skyscrapers, you will see these eliminations instantly.
Did you noticed something neat and powerful in that colouring argument
shown in the second grid? ('5)r6c8 sees (!5)r6c1 and (5")r5c9, so it must
be false. But that implies that the color (') is false. So, all digits marked (') are
false and all digits marked (") are true!
Regards,
JCO
Edit: improved text at the beginning.