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 Killer Skyscrapers (Posted on 2008-02-20)
In each row and column, there are eight skyscrapers, all of different sizes, ranging in height from 1 to 8 units.

The numbers outside the grid show how many buildings can be seen when looking towards the grid. So, for example, the 2 at the beginning of the fifth row means that the following combinations could be found in that row:
51874362
64873125

The small figures in the shaded rectangles show the total of the numbers in that rectangle.

"The shaded rectangles are similar to those found in "Killer Sudoku".
 7 21 12 6 15 15 3 10 21 13 6 7 6 10 11 11 2 5 19 1 8 15 5 4 10 12

 No Solution Yet Submitted by Josie Faulkner No Rating

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 Solution with partial walkthrough | Comment 1 of 6
The first column contains a skyscraper of three squares with a sum total of 21. As each square section of each skyscraper is of a height that is distinct for the column and row, we know the unit heights for the three squares must be 6, 7, and 8. Below it, the three squares with a sum total of 6, must have heights of 1, 2, and 3. this leaves heights 4 and 5. As the two square skyscraper at the bottom has a sum total of 10, we know that it must contain the 4 with a 6 in the second column. The unit square above the bottom skyscraper in the first column would then have a height of 5.

On the right hand side of the grid, where we are given the number 1 as the number of buildings viewable in that row from that side, we know the first building must have a height of 8 to hide all other buildings in the row from that direction. As it is an 8, we know the skyscraper at the top of the row with a sum total of 15 to have heights 8 in column 7 and 7 in column 8. With numbers 7 and 8 now used in the first row, we know that the three square skyscraper in the top left column to have number 6 at the top of the column.

In the second column, where we find the two square skyscraper of sum total height 15, we know it has heights 7 and 8. Below it, the four square skyscraper of sum total height 10, must be comprised of heights 1, 2, 3 and 4. The skyscraper at the top of the second column must then have a height of 5.

We know from the number 2 on the left hand side of the grid before row 5 that the the bottom square of the three square skyscraper with sum total height of 21 must have a height of 8.

With continued deduction and some trial and error we eventually can fill the grid as follows:

6 5 4 3 2 1 8 7
7 8 1 2 3 5 6 4
8 7 6 5 4 3 1 2
1 3 7 8 5 2 4 6
3 2 8 1 6 4 7 5
2 1 3 4 7 6 5 8
5 4 2 6 8 7 3 1
4 6 5 7 1 8 2 3

 Posted by Dej Mar on 2008-02-20 13:11:10

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