Kids Cube/3_Recess.iso/fiddle/methods.txt

Text File | 1992-01-12 | 9KB | 189 lines

                      F I D D L E  by Doug Beeferman
                 Systematic ways to solve selected levels


NOTATION:

               @  - standard unit square
               X  - center of rotation unit square

 --> When I say "2-edge", "3-edge", etc., I am referring to a piece's edge
     that has length 2, 3, etc.

 --> The First (and only!) Fundamental Rule of Fiddle is as follows:  When
     a piece is rotated, its center of rotation does not move.


LEVEL E, "Jagged edge":

The first ten levels of Fiddle rely heavily on visualization, and the
remainder rely heavily on logic.  But the two aren't unrelated.  Look at piece
(1).  Our goal is to construct a solid square- no holes are to be left
unfilled.  A region of (1) encloses a small "bay"- that is, it surrounds a
region of unfilled squares on three sides.  Clearly this "bay" must be filled
for a solid square to be constructed, regardless of what we know of the other
pieces.  It must be filled by another piece, or possibly by more than one.
Look around at the other pieces and determine which piece can singlehandedly
fill (1)'s bay.  Only (2) can.

Now, on the side of (1) opposite of where you've placed (2) there's an empty
region of 2 unit squares length.  There are two possibilities:  (a) this side
of (1) is an EDGE of the 11x11 square, or (b) this side of (1) is not an edge,
and there is at least one more piece in this direction.  Inspection will tell
you that (b) is true, because if this side were an edge then the only piece
that could fill the 2-square void (piece (6)) would cause the edge to have
length 13, which exceeds the desired 11.  Piece (4) is the only remaining
piece that fits nicely next to this side of (1).  After (4) is placed, the
correct positions of (6) and (5) are visually apparent.  Then the final piece,
(3), can be placed.

--------------------

LEVEL M, "One good turn":

1=  @@@X    3= @         ...
       @       @  @@@
   @@@ @       @  @ @
     @@@       @@X@ @
                    @
                   @@


The area that piece (1) encloses must be filled, and the only piece that can
fill it is (3).  A corollary to the First Fundamental Rule is that the proper
position for a piece that is to be rotated into another can be found by
considering how far apart the centers of rotation are to be. For our rotation
of (3) into (1), we want the end result to look like:

        111X <-- C(1)
      222221               If the center of (3) can be placed 3 squares to the
      211121               left and 4 squares below the center of (1), it can
      222111               be rotated to fill the hole as shown.  Manipulate
        X <-- C(3)         (3) so that its center is in this position and
        2                  rotate as necessary.
     2222

After (1) and (3) are joined, the remaining pieces can quickly be put into
place to complete the square.

--------------------

LEVEL T, "Spacemaker":

1=  @   2=  @    3= @@@   4= @ @  ...
   @@      @X@        X      @X@
   @X

The outer piece (6) has dimensions of the completed square, so the solution
entails filling it exactly.  Piece (5) must enter the square somehow.  Its
center of rotation must be within the square in the solution, so by the First
Fundamental Rule it must be within the square before it is rotated into place.
It can only enter the square by way of (6)'s 2-square void at the upper left.

After (6) is rotated into the square it will fill the two upper, undivided
quadrants.  The divided quadrants below have to be filled, therefore, with
pieces 1 through 4.  Two solid 3x3 boxes need to made; (2) joins with (4) to
form such a box, and (3) joins with (1) to form another.

To move (2) from the upper-right quadrant of (6) to its destination in the
lower left, rotate (6) clockwise so that (2) can first be moved to the lower
right quadrant.  Follow a similar procedure to join (2) with (4), and again to
join (3) with (1), rotating pieces 1 through 4 as necessary to fit inside the
3x3 quadrants.  Soon you'll be able to rotate (5) into place.

--------------------

LEVEL W, "Holey Piece":

This level is one of three in the game in which the final positions of the
pieces are clear from the beginning.  Pieces (1) and (2) both have holes to
fill.  Since the holes are completely within the pieces (they are "lakes"
rather than "bays"), filling them can only be accomplished by rotating the
"mainland" pieces to surround the "island" pieces.  Which do we fill first,
(1) or (2)?  These pieces have to ultimately attach to each other (it's clear
how) and we know that a piece cannot slide if it has an embedded "island";
therefore, when whatever we choose to be the second piece is filled it must be
attached to the other piece.  Filling requires rotation, and we should notice
that (2) cannot be rotated at all when they're attached;  (1) can.  We should
therefore fill (2) first.  (I won't describe that here.)

Filling (1) subsequently is a matter of recognizing that all three holes are
the same.  Placing one of the islands correctly in the middle (surrounded on
three sides by (2)) will not restrict (1)'s freedom to rotate.  Do this before
bringing (1) over.  Then face (1)'s back (its 15-edge) to the center.  Fitting
the other two islands should then be easy.

--------------------

LEVEL Y, "Revenge of Jigsaw":   (Warning!  Read this at your own risk!)

1=   @  2=  @@  3= @@   4= @@@   5= @ @   6=   @    7=  @   8=  @     9=  @@
   @@@     @@@     @@@     @@@@     @@@      @@@@@     @@@      @@@      @@@@
   @@@      @@     @@@     @@        @@@      @@       @@      @@@@       @ @
   @ @             @                          @@@@     @@@      @@@@


This is essentially a small jigsaw puzzle.  If there were a thousand pieces
instead of nine, how would you go about solving it?  With a square end result
in mind, you'd probably find the corner pieces and edge pieces and isolate them
from the "center" pieces (i.e., the pieces which are to be completely
surrounded by others.)  Among these nine pieces there is a well-defined number
of corner pieces: four.  Let's find them first.  It can be shown (write me if
you want to know precisely how) that only pieces (2), (3), (4), and (7) can be
corners. All of the remaining pieces might be edge pieces-- one or some of them
might instead be in the center.

The position (upper left, upper right, etc.) we choose for the first corner is
arbitrary-- the corners can be rotated to assume any of the four positions.
Piece (3) is peculiar among the four corners in that it has an edge length that
exceeds all edge lengths on all other corners-- 4.  Its other edge length is 2.
The completed square is to be 9 by 9; both sides upon which (3) is incident
must total 9 in length.  There are edge pieces available with lengths 2, 3,
and 4, and there are corner pieces available (other than (3)) with lengths 2
and 3.  For the edge of (3) with length 2, it is clear that it must be next to
an edge of length 4, because anything less would not allow for a total side
with length as high as 9.  (2+3+3<9, 2+2+3<9, 2+2+2<9).  Both pieces (1) and
(8) have edges of length 4 that can be placed next to the 2-edge of (3), but
only (8) in turn allows a corner with edge length 3 ((4) or (7)) to complete
the edge.  So (8) goes next to the 2-edge corner (3).  The corner that goes
next to (8) to complete the edge must be (7), because putting (4) there leaves
a 1x2 void that cannot be filled by any other piece.

Now let's return to the _other_ side of the corner (3)-- that is, the 4-edge
side.  Adjacent to this must be a 3-edge-- the only 2-edge piece that fits
(9) does not allow for a corner piece (there are two remaining at this point,
and we know that they have to be (2) and (4)) to complete the 9-side.  The
only piece that works is (5).  The corner (2) then completes the side.

The placement of the final corner (4) is now obvious.  Pieces (1) and (6) fit
interchangably between the corners (2,4) and (7,4).  But, as it turns out,
piece 9 is the sole center piece, and it can only reside there if (1) is
placed between (7) and (4), and (6) between (2) and (4).  Now that the exact
destinations of the pieces are known, the matter of getting them there is not
that hard.  Rotation might be necessary on whatever you choose to be the last
piece placement.  Only one of the pieces can rotate nicely into its final
position.  Be careful!

If you start with (3) in the lower left, the completed arrangement will look
roughly like:

                        7  1  4
                        8  9  6
                        3  5  2

As the startup configuration might have led you to believe, the solution is a
3 by 3 arrangement of the pieces.  How are you supposed to figure this all out
in 8 minutes?  You're not.  I would have felt bad had I left it out, though.

-------------------

LEVEL Z, "Nightmare":

Just kidding!  This level's worth too much for me to reveal the answer.  Clue:
start with the "what must go where" strategy.  It's actually not nearly as
mightmarish as Y.  Write me if you want the systematic solution.


12-Jan-1992