 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  King on the run (Posted on 2016-06-01) A chess king starts at a position A in the top row of a standard chessboard. The number of paths of length 7 to a position B in the bottom row is a perfect square, but not a perfect cube.
The number of paths of length 7 from A to a position C in the bottom row is a perfect cube, but not a perfect square.
The number of paths of length 7 to a position D is both a perfect square and a perfect cube.

How many chess king paths of length 5 are there from B to C?

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) computer assisted solution | Comment 1 of 5
With 7 rows to go in 7 moves, each move must include a vertical component, and the only options are to go diagonally left, right or go straight down at each turn.

The table below lists, for the 8 possible starting positions that A might be, the number of ways of getting to the positions from left to right on the bottom row. If the number is a cube it's marked with a c, if a square it's marked with an s.

Only if the king started in a corner square do the conditions of the puzzle hold and points B and C are separated by 4 spaces (i.e., 3 intervening squares so you can count over 4 squares to get from one to the other).

`column     ways of getting to the differentfor          positions on bottom rowA1         127 196s 189 133 70 27c 7 1sc2         196s 316 329 259 160 77 28 73         189 329 386 356 266 161 77 27c4         133 259 356 393 357 266 160 705         70 160 266 357 393 356 259 1336         27c 77 161 266 356 386 329 1897         7 28 77 160 259 329 316 196s8         1sc 7 27c 70 133 189 196s 127`

One of B and C is next to a corner square and the other is 2 away from the other bottom corner (one intervening square).

Starting at the bottom of the second column, here are the numbers of paths of length 5 to each other place on the board from point B:

`   A   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0  30  46  44  30  15   5   1   0  90 140 140 100  55  20   5   0 230 354 356 250 140  50  14   0 340 520 540 380 225  80  25   0 395 595 625 430 260  90  30   0 255 380 409 280 175  60  21   0      B               C        D`

Specifically, there are 60 ways of getting from point B to point C.

Of course the mirror image case would have the same total.

In the program below I obviously wrote the second part after seeing the results of the first part:

DefDbl A-Z
Dim crlf\$, x, ways(), grid(8, 8), y

Form1.Visible = True

Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)

For a = 1 To 8
ReDim ways(8)
x = a
moveKing 1
Text1.Text = Text1.Text & a & "        "
For i = 1 To 8
Text1.Text = Text1.Text & Str(ways(i))
sr = Int(Sqr(ways(i)) + 0.5)
If sr * sr = ways(i) Then Text1.Text = Text1.Text & "s"
cr = Int(ways(i) ^ (1 / 3) + 0.5)
If cr * cr * cr = ways(i) Then Text1.Text = Text1.Text & "c"
Next
Text1.Text = Text1.Text & crlf
Next a

x = 2: y = 8
moveKingAgain 1
For y = 1 To 8
For x = 1 To 8
Text1.Text = Text1.Text & mform(grid(x, y), "###0")
Next
Text1.Text = Text1.Text & crlf
Next

Text1.Text = Text1.Text & crlf & " done"

End Sub

Sub moveKing(wh)
For newx = x - 1 To x + 1
oldx = x
x = newx

If newx > 0 And newx <= 8 Then
If wh = 7 Then
ways(x) = ways(x) + 1
Else
moveKing wh + 1
End If
End If

x = oldx
Next
End Sub

Sub moveKingAgain(wh)
For dx = -1 To 1
For dy = -1 To 1
If dx <> 0 Or dy <> 0 Then
oldx = x: oldy = y
x = x + dx: y = y + dy

If x > 0 And x <= 8 Then
If y > 0 And y <= 8 Then
If wh = 5 Then
grid(x, y) = grid(x, y) + 1
Else
moveKingAgain wh + 1
End If
End If
End If

x = oldx: y = oldy
End If
Next
Next
End Sub

Function mform\$(x, t\$)
a\$ = Format\$(x, t\$)
If Len(a\$) < Len(t\$) Then a\$ = Space\$(Len(t\$) - Len(a\$)) & a\$
mform\$ = a\$
End Function

 Posted by Charlie on 2016-06-01 15:37:46 Please log in:

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