[Guest]

The rules are simple, fill out the 6x6 grid with the numbers 1-6, which will

only occur once each per row or column. The numbers will represent the

'height' of that square. The numbers provided around the border of the

grid will be how many pillars you can see, looking down that column or

across that row. Any shorter pillar behind a larger pillar cannot be seen.

Eg, if a row was 1 4 5 2 6 3, a 4 would be placed on the left, to indicate the

1, 4, 5 and 6 height pillars are visible, and a 2 on the right, to indicate the

3 and the 6 height pillars can be seen.

And the grid is:

1 6 3

--------------

|O O O O O O|

3 |O O O O O O|

5 |O O O O O O|

4 |O O O O O O|

|O O O O O O|

|O O O O O O|

--------------

2 1 3

Not sure why the box is messed up, but I think you can figure out that those 3 lines are supposed to by moved over one space to the left.

Edited October 28, 2010 by savidbor

[Guest]

Hi i tried to solve this. i felt this very interesting also.

But i reached a stage that this particular problem cannot have a solution. Or am i wrong?

[witzar]

There is no way to put number 6 in 3rd column.

[Guest]

Sorry, stupid notepad doesn't paste well here. Let me try to fix this for you:

XX 1 6 XX 3

--------------

X |O O O O O O|

3 |O O O O O O|

5 |O O O O O O|

4 |O O O O O O|

X |O O O O O O|

X |O O O O O O|

--------------

XX 2 1 XX 3

I think I fixed it, but I had to used some tape (the X's). The X's are just spaces that the post made me use for correct alignment--so ignore them. Top: 1 is over the first column, 6 the second, 3 the fourth. Bottom: 2 is under the first, 1 the second, 3 the fourth.

Edited October 28, 2010 by savidbor

[Guest]

Givens in Black. My answer in Blue.

[Guest]

Nice!

[araver]

Nice!

at least 5 solutions so far:

6 1 3 2 4 5

1 2 6 5 3 4

2 3 1 4 5 6

3 4 5 6 1 2

4 5 2 3 6 1

5 6 4 1 2 3

6 1 4 2 3 5

1 2 6 5 4 3

2 3 1 4 5 6

3 4 5 6 1 2

4 5 2 3 6 1

5 6 3 1 2 4

6 1 2 4 3 5

1 2 6 5 4 3

2 3 4 1 5 6

3 4 5 6 2 1

4 5 1 3 6 2

5 6 3 2 1 4

6 1 3 4 2 5

1 2 6 5 3 4

2 3 4 1 5 6

3 4 5 6 1 2

4 5 2 3 6 1

5 6 1 2 4 3

6 1 3 4 2 5

1 2 6 5 4 3

2 3 4 1 5 6

3 4 5 6 1 2

4 5 2 3 6 1

5 6 1 2 3 4

Am I missing something or is it OK that the solution is not unique?

Edited October 28, 2010 by araver

[Guest]

The rules are simple, fill out the 6x6 grid with the numbers 1-6, which will

only occur once each per row or column. The numbers will represent the

'height' of that square. The numbers provided around the border of the

grid will be how many pillars you can see, looking down that column or

across that row. Any shorter pillar behind a larger pillar cannot be seen.

Eg, if a row was 1 4 5 2 6 3, a 4 would be placed on the left, to indicate the

1, 4, 5 and 6 height pillars are visible, and a 2 on the right, to indicate the

3 and the 6 height pillars can be seen.

And the grid is:

1 6 3

--------------

|O O O O O O|

3 |O O O O O O|

5 |O O O O O O|

4 |O O O O O O|

|O O O O O O|

|O O O O O O|

--------------

2 1 3

Not sure why the box is messed up, but I think you can figure out that those 3 lines are supposed to by moved over one space to the left.

I am not sure what you mean by "The grid is 163 or 213." When I wrote this, I saw something different but why the gap between the 3 and 6 on top and the 1 and 3 on the bottom?

Edited October 28, 2010 by BSUtutor

[Guest]

Sorry, stupid notepad doesn't paste well here. Let me try to fix this for you:

XX 1 6 XX 3

--------------

X |O O O O O O|

3 |O O O O O O|

5 |O O O O O O|

4 |O O O O O O|

X |O O O O O O|

X |O O O O O O|

--------------

XX 2 1 XX 3

Let me begin by saying if we identify the columns from left to right as A,B,C,D,E,F and the rows from top to bottom as I,II,III,IV,V,VI then we can use ordered pairs to identify each box in the grid.

1.) First, column B has to be 1-6 from top to bottom.

2.) Row III has to start 2,3,4,5,6, with 1 in (F,III).

3.) (A,II) has to be a 1 because otherwise the number to the left of the grid there would be a 1, not a 3.

4.) (C,III), (D,III), and (E,III) all have to be 4,5,6 respectively with (F,III) being a 1.

5.) (D,IV) must be a 6 in order for the number at the bottom of the column to be a 3, but then the top of that column will show a 4 and not a 3.

Therefore, I say it is unsolvable. The little table shows where I stopped at, after 4.):

.........1.6...3

...(I)---X 1 X X X X

..3(II)--1 2 X X X X

..5(III)-2 3 4 5 6 1

..4(IV)--X 4 X X X X

...(V)---X 5 X X X X

...(VI)--X 6 X X X X

.........2.1...3

I think I fixed it, but I had to used some tape (the X's). The X's are just spaces that the post made me use for correct alignment--so ignore them. Top: 1 is over the first column, 6 the second, 3 the fourth. Bottom: 2 is under the first, 1 the second, 3 the fourth.

Edited October 28, 2010 by BSUtutor

[araver]

@BSUtutor -

2.)Row III has to start 2,3,4,5,6, with 1 in (F,III).

Also, I found that choosing Courier New as a font works with aligning dots and letters

Edited October 28, 2010 by araver

[bonanova]

Hint:

Use code tags to align tables and grids (any font is ok):

(code)

a b c d e

f g h i j

k l m n o

p q r s t

u v w x y

(/code)

But use [] brackets instead of parentheses.

a b c d e

f g h i j

k l m n o

p q r s t

u v w x y

[/code]

[araver]

Hint:

Use code tags to align tables and grids (any font is ok):

Thanks a lot. It's very useful!