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Peaceful superqueens (Posted on 2015-12-31) Difficulty: 4 of 5
A "superqueen" is a chess queen that also moves like a knight.
Place four superqueens on a five-by-five board so that no piece attacks another.

If you solve this, try arranging 10 superqueens on a 10-by-10 board so that no piece attacks another.

Both solutions are unique if rotations and reflections are ignored.

by Hilario Fernandez Long.

No Solution Yet Submitted by Ady TZIDON    
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Solution computer solutions Comment 2 of 2 |
The ordinary Queens problem is typically used in introducing computer students to the power of recursive programming. The Superqueens problem adds an extra wrinkle.

Part 1:

1's mark the locations of the superqueens, and 0's empty squares:

 0 0 0 1 0
 1 0 0 0 0
 0 0 0 0 0
 0 0 0 0 1
 0 1 0 0 0
 
This same tableau was actually shown multiple times--once for each position of the superqueen left out ot the middle column. The mirror image was avoided by not allowing the first-column superqueen to be lower than the middle row.

DefDbl A-Z
Dim crlf$, board As Variant, n, emptycol

Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 n = 5
 
 ReDim board(-n To n * 2, -n To n * 2)
   
 For emptycol = 1 To n
 addOn 1
 Next
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Sub addOn(col)
  DoEvents
  If col = 1 Then mx = -Int(-n / 2) Else mx = n
  For row = 1 To mx
    good = 1
    If col <> emptycol Then
    board(row, col) = 1
    
    
    For c = 1 To col - 1
      If board(row, c) Then good = 0: Exit For
      If board(row - (col - c), c) Then good = 0: Exit For
      If board(row + (col - c), c) Then good = 0: Exit For
    Next
    If board(row - 1, col - 2) Then good = 0
    If board(row + 1, col - 2) Then good = 0
    If board(row - 2, col - 1) Then good = 0
    If board(row + 2, col - 1) Then good = 0
    
    End If
    
    If good Then
      If col = n Then
        For r = 1 To n
        For c = 1 To n
          Text1.Text = Text1.Text & Str(board(r, c))
        Next
        Text1.Text = Text1.Text & crlf
        Next
        Text1.Text = Text1.Text & crlf
      Else
        addOn (col + 1)
      End If
    End If
    
    board(row, col) = 0
  Next
End Sub

Part 2:

With appropriate modification (n=10 and no empty column):

 0 0 0 1 0 0 0 0 0 0
 0 0 0 0 0 0 0 1 0 0
 1 0 0 0 0 0 0 0 0 0
 0 0 0 0 1 0 0 0 0 0
 0 0 0 0 0 0 0 0 1 0
 0 1 0 0 0 0 0 0 0 0
 0 0 0 0 0 1 0 0 0 0
 0 0 0 0 0 0 0 0 0 1
 0 0 1 0 0 0 0 0 0 0
 0 0 0 0 0 0 1 0 0 0


 0 0 1 0 0 0 0 0 0 0
 0 0 0 0 0 1 0 0 0 0
 0 0 0 0 0 0 0 0 1 0
 1 0 0 0 0 0 0 0 0 0
 0 0 0 1 0 0 0 0 0 0
 0 0 0 0 0 0 1 0 0 0
 0 0 0 0 0 0 0 0 0 1
 0 1 0 0 0 0 0 0 0 0
 0 0 0 0 1 0 0 0 0 0
 0 0 0 0 0 0 0 1 0 0

which are reflections of each other.  Reflections were avoided in the first instance, by not having the queen in column 1 be below midlevel. Here, the reflected tableau has the queen in column 1 also above midlevel, so the restriction didn't rule out the reflection.

  Posted by Charlie on 2015-12-31 15:16:31
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