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Get them up (Posted on 2014-10-01) Difficulty: 2 of 5
10 playing cards are placed face-down on a table, forming a circle.
Your are allowed to turn over simultaneously 4 cards that either:
a.)all 4 are consecutive,
or
b.) 2 are on the left and 2 on the right of any specific card chosen by you (i.e. out of 5 consecutive cards only the middle one remains unturned) .

By executing a final number of the above (a. or b.) procedures can you reach an "all face-up" state?

If yes, explain how.
Otherwise, prove why not.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Unsatisfying computer proof | Comment 1 of 5
DefDbl A-Z
Dim crlf$, used(1023), state, h(260), sth(260), mask(20), maxlvl

Private Sub Form_Load()
 ChDir "C:Program Files (x86)DevStudioVBprojects looble"
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 Text1.Text = Text1.Text & base$(1023, 2) & crlf
 
 DoEvents
 
 For i = 0 To 9
   v = 15 * Int(2 ^ i + 0.5)
   If v > 1023 Then
    xs = v 1024
    v = v Mod 1024 + xs
   End If
   mask(i + 1) = v
 Next
 For i = 0 To 9
   v = 27 * Int(2 ^ i + 0.5)
   If v > 1023 Then
    xs = v 1024
    v = v Mod 1024 + xs
   End If
   mask(i + 11) = v
 Next

 seq = 1
 used(0) = 1
 addOn seq
 
 For i = 0 To 1023
  If used(i) Then
    Text1.Text = Text1.Text & "  " & base$(i, 2) & "   "
    uct = uct + 1
  End If
  DoEvents
 Next
 
 
 Text1.Text = Text1.Text & crlf & maxlvl & Str(uct) & " done"
End Sub

Sub addOn(wh)
  For i = 1 To 20
    state = state Xor mask(i)
    If used(state) = 0 Then
      used(state) = 1
      h(wh) = mask(i)
      sth(wh) = state
      If state = 2047 Then
        For j = 1 To wh
          Text1.Text = Text1.Text & base$(h(j), 2) & "   " & base$(sth(j), 2) & crlf
        Next j
        Text1.Text = Text1.Text & crlf
        DoEvents
        Exit Sub
      Else
       If wh > maxlvl Then maxlvl = wh
       If wh < 260 Then
        addOn wh + 1
       End If
      End If
    End If
    state = state Xor mask(i)
  Next
End Sub

Function base$(n, b)
  v$ = ""
  n2 = n
  Do
    d = n2 Mod b
    n2 = n2 b
    v$ = LTrim(Str(d)) + v$
  Loop Until Len(v$) > 9 And n2 = 0
  base$ = v$
End Function

finds only sets where the number of upturned cards is even, and not all of those either: only 256 of the 2^10 possible configurations are achievable:

0000000000     0000000101     0000001010     0000001111     0000010001     0000010100     0000011011     0000011110     0000100010     0000100111     0000101000     0000101101     0000110011     0000110110     0000111001     0000111100     0001000001     0001000100     0001001011     0001001110     0001010000     0001010101     0001011010     0001011111     0001100011     0001100110     0001101001     0001101100     0001110010     0001110111     0001111000     0001111101     0010000010     0010000111     0010001000     0010001101     0010010011     0010010110     0010011001     0010011100     0010100000     0010100101     0010101010     0010101111     0010110001     0010110100     0010111011     0010111110     0011000011     0011000110     0011001001     0011001100     0011010010     0011010111     0011011000     0011011101     0011100001     0011100100     0011101011     0011101110     0011110000     0011110101     0011111010     0011111111     0100000001     0100000100     0100001011     0100001110     0100010000     0100010101     0100011010     0100011111     0100100011     0100100110     0100101001     0100101100     0100110010     0100110111     0100111000     0100111101     0101000000     0101000101     0101001010     0101001111     0101010001     0101010100     0101011011     0101011110     0101100010     0101100111     0101101000     0101101101     0101110011     0101110110     0101111001     0101111100     0110000011     0110000110     0110001001     0110001100     0110010010     0110010111     0110011000     0110011101     0110100001     0110100100     0110101011     0110101110     0110110000     0110110101     0110111010     0110111111     0111000010     0111000111     0111001000     0111001101     0111010011     0111010110     0111011001     0111011100     0111100000     0111100101     0111101010     0111101111     0111110001     0111110100     0111111011     0111111110     1000000010     1000000111     1000001000     1000001101     1000010011     1000010110     1000011001     1000011100     1000100000     1000100101     1000101010     1000101111     1000110001     1000110100     1000111011     1000111110     1001000011     1001000110     1001001001     1001001100     1001010010     1001010111     1001011000     1001011101     1001100001     1001100100     1001101011     1001101110     1001110000     1001110101     1001111010     1001111111     1010000000     1010000101     1010001010     1010001111     1010010001     1010010100     1010011011     1010011110     1010100010     1010100111     1010101000     1010101101     1010110011     1010110110     1010111001     1010111100     1011000001     1011000100     1011001011     1011001110     1011010000     1011010101     1011011010     1011011111     1011100011     1011100110     1011101001     1011101100     1011110010     1011110111     1011111000     1011111101     1100000011     1100000110     1100001001     1100001100     1100010010     1100010111     1100011000     1100011101     1100100001     1100100100     1100101011     1100101110     1100110000     1100110101     1100111010     1100111111     1101000010     1101000111     1101001000     1101001101     1101010011     1101010110     1101011001     1101011100     1101100000     1101100101     1101101010     1101101111     1101110001     1101110100     1101111011     1101111110     1110000001     1110000100     1110001011     1110001110     1110010000     1110010101     1110011010     1110011111     1110100011     1110100110     1110101001     1110101100     1110110010     1110110111     1110111000     1110111101     1111000000     1111000101     1111001010     1111001111     1111010001     1111010100     1111011011     1111011110     1111100010     1111100111     1111101000     1111101101     1111110011     1111110110     1111111001     1111111100   

The program would have gone 2047 levels deep, but always found a repetition after 255 levels of flips.

Edited on October 2, 2014, 10:53 am
  Posted by Charlie on 2014-10-01 11:21:11

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