1
$\begingroup$

So I have been using a site called Brain Bashers, which is a site filled with logical deduction puzzles, because I like trying to master different logical deduction puzzle types.

I have finished the first two Skyscrapers puzzles currently working on the 5x5 (Easy) Daily Skyscrapers. This is the puzzle:

    3   2   2   3   1
  +---+---+---+---+---+
4 |   |   |   |   |   | 1
  +---+---+---+---+---+
2 |   |   |   |   |   | 2
  +---+---+---+---+---+
2 |   |   |   |   |   | 2
  +---+---+---+---+---+
1 |   |   |   |   |   | 5
  +---+---+---+---+---+
2 |   |   |   |   |   | 3
  +---+---+---+---+---+
    2   3   1   2   4

This is my progress so far on this puzzle:

+-----------------------------------------------------------------+
|Note: A missing number means that the number is already satisfied|
+-----------------------------------------------------------------+
            2   3
  +---+---+---+---+---+
4 | 2 | 3 |   |   | 5 |
  +---+---+---+---+---+
  | 1 | 5 |   |   |   | 2
  +---+---+---+---+---+
2 | 4 | 2 |   |   |   | 2
  +---+---+---+---+---+
  | 5 | 4 | 3 | 2 | 1 |
  +---+---+---+---+---+
2 | 3 | 1 |   |   |   | 3
  +---+---+---+---+---+
            1   2   4

My question: What could I deduce right now from here about this puzzle? I know I could probably solve it as sort of a 5x5 KenKen, but I want to know what I could deduce the solution from how it is actually supposed to solved.

Improve this question
$\endgroup$
2
  • $\begingroup$ Column 3 looks very constrained. $\endgroup$
    – codewarrior0
    Commented Mar 18 at 18:41
  • $\begingroup$ So does column 5 $\endgroup$
    – xnor
    Commented Mar 18 at 18:44

1 Answer 1

Reset to default
4
$\begingroup$

Columns

  • From the bottom of column 3, you can see only a single stack. That must mean the 5 is in the bottom-most row (otherwise you would be able to see it behind whatever other value was there).

  • From the bottom of column 4, you can see two stacks. That means that 1 and 5 are not possibilities for the bottow row, because you would see either >2 or 1 stacks respectively with those values. That leaves 3 and 4. 3 is already in that row, so the only possible number remaining is 4.

  • From the bottom of column 5, you can see 4 stacks. That means at least 3 stacks must be visible behind whatever you put into the bottow row. Thus you obviously need to put the 2 there, so that there are still 3 bigger numbers behind it.

Rows

  • From the right of row 2, you know you will be able to see the 5, and you only need to see one more, so the 4 must go in the rightmost column. (Otherwise you would see 5, 4, and whatever is to the right of 4.) That leaves 3 and 2, which can only be arranged one way so as not to conflict with row 4.

  • From the right of row 5, you can see 3 stacks. You know won't be able to see the 3 or the 1 at the left of the row, because a 5 has to go in there somewhere, and it will block both of them. But you still need to be able to see 3 stacks, so the 5 must go as far to the left as possible, and then the remaining numbers in descending order.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.