Published February 2001,February 2011.

Many years ago, when I was still knee high to a grasshopper (er,
when I was about 10 probably), I was given a fiendish jigsaw puzzle
called **Impuzzable** . It has only 9 pieces and yet
it's the hardest jigsaw puzzle I've ever had. The reason it's so
difficult is that there are no edges and no pictures. Here's what
it looks like:

My family and I tried to solve this puzzle for many months before eventually giving up. Before reading the rest of this article, you might like to have a go at solving it yourself by printing out the jigsaw puzzle and cutting out the shapes. Hopefully this will give you an idea of how fiendish it is!

A few years ago I found the puzzle lying about my house and resolved to defeat it. Rather than spend inordinate amounts of time trying to solve it by hand, I decided to cheat and write a computer program to solve it for me. The rest of this article describes how the program works. It does require some knowledge of programming, but it's written in the easiest programming language of all, BASIC. If you have an oldish PC, it should have come with a copy of QBASIC (at an MSDOS prompt, type "QBASIC" to see if you have this program). If not, you can download it here .

So, how do you you write a computer program to solve a jigsaw
puzzle? The most obvious method is to try out every possible way of
laying down the 9 pieces in a three by three grid, and checking to
see if the edges fit together. Unfortunately, this method could
take a very long time, because there are 2,380,855,680 ways of
doing this. If you've done any probability or stats at school, you
should be able to work out why this is the number of ways of laying
the pieces down; if not, don't worry about the next bit. There are
9!=9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 ways of choosing which piece to
put in which position. For each piece, there are 4 different ways
to rotate it, so for each way of putting the pieces in the
available positions, there are 4^{9} different choices of
ways to arrange the pieces. So, there are 9! x 4^{9} ways
to put down all the pieces.

So that method is too slow, is there a better method? There are
probably lots of ways of doing it, but I'll describe the way I
chose to do it (a friend of mine has told me that this method is
called *alpha-beta pruning* , which is nice). You lay the
first piece down in the first available position. Then you lay the
second piece down in the next available position (just underneath
it for instance). If it doesn't fit, you rotate it by 90 degrees
and try again. If you find a fit, then you pick up the next
available piece (the third piece) and place this below the second
piece and try different rotations of this piece. If you didn't find
a fit for the second piece, you try the third piece underneath the
first piece, and so on. Obviously this saves checking millions of
ways of laying down all the pieces. In fact, with my program, it
only had to check 56,592 arrangements to find every solution to the
puzzle! So we've gone from 2 billion arrangements to check all the
way down to 60 thousand arrangements to check.

The first thing we need to do is to turn our puzzle into a form the computer can understand. I've decided the number the pieces as follows:

Now, we need to tell the computer the characteristics of each piece. The important thing about each piece in this puzzle is that it has four sides, and each side can be one of 8 different types; clubs, spades, diamonds and hearts pointing inwards and outwards. I've numbered them in the following way:

This is a useful way to number pieces, because if we have two
edges numbered *a* and *b* next to one another, they
will fit together if *a+b=0* (an edge numbered *a*
only fits with the edge numbered *-a* ). We'll use this
later when writing the actual program.

Now we can label each piece using the key above, here is an example for piece number 1:

We'll also give each piece a rotation number, this is a number (0 to 3) which tells us how many times it has been rotated anticlockwise. If the number is 0, it has not been rotated at all, if it's 1 it's rotated 90 degrees anticlockwise, if it's 2 it's been rotated 180 degrees, if it's 3 it's been rotated 90 degrees clockwise, or 270 degrees anticlockwise.

We also number each of the edges of a piece, so that the top edge is edge 1, the right hand edge is edge 2, the bottom edge is edge 3 and the left hand edge is edge 4.

Finally, we number the positions which each piece can be placed in as follows:

Don't worry about why I've numbered them in that counterintuitive way. I got a bit confused when I originally wrote the program, but it doesn't matter, it's a perfectly good way of numbering the positions.

We've now numbered everything we could possibly want to number. I can specify any way of laying down these pieces purely in terms of numbers. For instance, if I said that piece 1 was in position 5 with rotation number 0, piece 8 in position 4 with rotation number 0 and piece 3 in position 7 with rotation number 1, the board would look like this:

In this example, edge 3 of piece 8 doesn't fit together with edge 1 of piece 1, but edge 4 of piece 8 does fit with edge 2 of the rotated piece 3, which is the same as edge 3 of piece 3 before it's rotated.

Okay, that's the easy bit done. Everything is numbered, now we need to actually write the program that does it. The first thing to do is to set up the pieces, in BASIC, here's how we do it:

DECLARE SUB SetPiece (n!, A!, b!, c!, d!) DIM SHARED P(9, 7) AS INTEGER SUB SetPiece (n, A, b, c, d) P(n, 1) = A P(n, 2) = b P(n, 3) = c P(n, 4) = d 'Keep further copies for rotation! P(n, 5) = A P(n, 6) = b P(n, 7) = c END SUB

Whoa there! What's going on here? Let's go over that and see
what we're doing. You can ignore the `DECLARE SUB` bit,
that's just programming details. The `DIM SHARED` line just
sets up an array of integers (whole numbers) in which we're going
to store the information about the pieces. So `P(4,2)` will
be the edge type of the 2nd edge of the 4th piece. We want this to
be -1 (it's a club pointing inwards). The reason we have 7 edges
rather than 4 is for rotations, so that we can work out the edge
type of the j-th edge of the i-th piece with rotation number k
easily, it's the number `P(i,j+k)` . Now, the
`SetPiece` subroutine just sets up the piece, the first
argument (`n`) is the number of the piece we're setting up,
the next 4 arguments (`A,b,c,d`) are the edge types of the 4
edges. We can now set up all the pieces like so (you might like to
check that you've understood everything so far by working out what
the next 9 lines should say yourself):

SetPiece 1, -3, -2, 3, 1 SetPiece 2, 2, 1, -4, -2 SetPiece 3, 4, 2, -3, -2 SetPiece 4, -1, 1, 4, -4 SetPiece 5, -4, 4, 2, -1 SetPiece 6, 1, -1, -2, 2 SetPiece 7, 3, 1, -2, -1 SetPiece 8, -3, -4, 2, 3 SetPiece 9, -4, 3, 3, -2

Next, we set up another array which defines the current positions and rotations of the pieces on the board.

DIM SHARED A(9, 2) AS INTEGER

The number `A(n,1)` is the number of the piece in
position n, the number `A(n,2)` is the rotation number of
this piece. So, for the example board above (included again just
below this paragraph), we have `A(4,1)=8 A(4,2)=0 A(5,1)=1
A(5,2)=0 A(7,1)=3 A(7,2)=1` .

Next, we define a function to check whether or not two pieces fit together or not.

DECLARE FUNCTION CheckIntersection! (pia!, pa!, pib!, pb!) FUNCTION CheckIntersection (pia, pa, pib, pb) CheckIntersection = ABS(SGN(P(A(pia, 1), pa + A(pia, 2)) + P(A(pib, 1), pb + A(pib, 2)))) END FUNCTION

Ugh! That's a very ugly looking expression, what does it do?
First of all, ignore the `ABS` and `SGN` bits.
`ABS` just takes the positive value of its input, so
`ABS(-3)=3 ABS(2)=2`. `SGN` returns the sign of the
input, which is -1 if the input is negative, +1 if the input is
positive and 0 if the input is 0, `SGN(0)=0 SGN(12)=1
SGN(-24)=-1`. So, `ABS(SGN(x))` is 0 if x is 0 or 1 if x
is not 0. Let's just look at this bit of the expression,
`P(A(pia, 1), pa + A(pia, 2))`. This is the edge type of
edge `pa+A(pia,2)` of piece `A(pia,1)`. In other
words, this is just the edge type of edge `pa` of the piece
placed in position `pia` (because `A(n,1)` is the
piece in position n, and `A(n,2)` is the rotation number of
that piece). The next bit, `P(A(pib, 1), pb + A(pib, 2))`,
is just the edge type of edge `pb` of the piece in position
`pib`. These are added together, so if the two edges fit,
the function will return 0, and if they don't fit the function will
return 1.

The next bit is a bit tedious, we need another function which will tell us whether or not everything fits after we've put pieces in positions 1 to u. Here goes:

FUNCTION CheckPoss (u) tot = tot + 1 f = 0 SELECT CASE u CASE 2 f = f + CheckIntersection(1, 3, 2, 1) 'A CASE 3 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B CASE 4 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C CASE 5 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(4, 3, 5, 1) 'F CASE 6 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G CASE 7 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G f = f + CheckIntersection(4, 4, 7, 2) 'H CASE 8 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G f = f + CheckIntersection(4, 4, 7, 2) 'H f = f + CheckIntersection(5, 4, 8, 2) 'I f = f + CheckIntersection(7, 3, 8, 1) 'K CASE 9 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G f = f + CheckIntersection(4, 4, 7, 2) 'H f = f + CheckIntersection(5, 4, 8, 2) 'I f = f + CheckIntersection(6, 4, 9, 2) 'J f = f + CheckIntersection(7, 3, 8, 1) 'K f = f + CheckIntersection(8, 3, 9, 1) 'L END SELECT IF f = 0 THEN CheckPoss = 1 IF u = 9 THEN PrintSolution END IF END IF END FUNCTION

As I said, this bit is very tedious. If we label the intersection points (i.e. the places where we need to check if two edges fit together) as follows:

Now, if we have three pieces on the board in positions 1,2 and
3, then we only need to check intersection points A and B. If we
have 4 pieces, we need to check A,B and C. If we have 5 pieces, we
need to check A,B,C,D and F. And so on. Where it says
`tot=tot+1` , this is just keeping count of how many
possible solutions we've checked so far. If any intersection points
don't match, then `f` will be bigger than 0, so that's why
we have `IF f=0 THEN CheckPoss = 1` . If we've put 9 pieces
down (u=9) then we've found a solution and we should print it out,
here is where the `PrintSolution` subroutine comes in:

DECLARE SUB PrintSolution () SUB PrintSolution tsol = tsol + 1 c$ = "" FOR z = 1 TO 9 c$ = c$ + CHR$(A(z, 1) + 64) + CHR$(A(z, 2) + 97) NEXT PRINT c$; " "; PRINT tot END SUB

You don't really need to worry about this bit, if you're
interested you can look up the `CHR$` function in your BASIC
manual or help file. What it does is it prints out a sequence of
two letter strings, the first letter is in capitals and tells you
the number of the piece, the second letter is in lower case and
tells you the rotation number of the piece. So, if it printed out
"AaBbCcDdEaFbGcHdIa", then piece in position 1 is "Aa", which is
piece number 1 with rotation number 0, the second piece "Bb" is
piece number 2 with rotation number 2. The 9th piece ("Ia") is
piece number 9 with rotation number 1.

Finally, we come to the interesting bit. Here is the function that does the clever stuff:

DECLARE SUB Recurse (n!) SUB Recurse (n) m = n + 1 FOR k = 1 TO 9 f = 0 FOR z = 1 TO n IF A(z, 1) = k THEN f = 1 NEXT IF f = 0 THEN FOR r = 0 TO 3 A(m, 1) = k A(m, 2) = r i = CheckPoss(m) IF m < 9 AND i = 1 THEN Recurse m NEXT END IF NEXT END SUB

Now how does this work? Suppose that we have already laid down n
pieces and they all fit together. What this subroutine does is to
try all the remaining pieces in the (n+1)th position. The clever
bit is that if it can fit a piece into the (n+1)th position, the
subroutine runs itself to try and fit a piece in the (n+2)th
position, and so on. This is called a recursive function by
programmers, which is why the subroutine is called `Recurse`
. The variable k loops through 1 to 9 (`FOR k = 1 TO 9`),
then the program checks if piece k has already been used or not by
checking pieces 1 to n (`FOR z = 1 TO n, IF A(z,1) = k THEN f =
1, NEXT`). If it hasn't been used (`IF f=0 THEN`) the
program checks through all the 4 possible rotation numbers r
(`FOR r = 0 TO 3`). It places the piece in position k with
rotation number r (`A(m,1) = k, A(m,2) = r`) and checks if
this fits. If it does fit, it tries to place the next piece by
running itself. And that's it! To find the solutions we just need
to put the following line in our program:

Recurse 0

Which just means, run the recurse subroutine given that I've placed no pieces so far. Here is the complete listing for this program then, click here to go to the end of the listing:

DECLARE SUB PrintSolution () DECLARE FUNCTION CheckIntersection! (pia!, pa!, pib!, pb!) DECLARE FUNCTION CheckPoss! (u!) DECLARE SUB Recurse (n!) DECLARE SUB SetPiece (n!, A!, b!, c!, d!) ' ' Impuzzable Solver ' COMMON SHARED tot, tsol tp = 4 ^ 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 DIM SHARED P(9, 7) AS INTEGER DIM SHARED A(9, 2) AS INTEGER SetPiece 1, -3, -2, 3, 1 SetPiece 2, 2, 1, -4, -2 SetPiece 3, 4, 2, -3, -2 SetPiece 4, -1, 1, 4, -4 SetPiece 5, -4, 4, 2, -1 SetPiece 6, 1, -1, -2, 2 SetPiece 7, 3, 1, -2, -1 SetPiece 8, -3, -4, 2, 3 SetPiece 9, -4, 3, 3, -2 CLS PRINT "IMPUZZABLE SOLVER - by Dan Goodman" PRINT STRING$(80, "-") Recurse 0 PRINT STRING$(80, " ") PRINT LTRIM$(RTRIM$(STR$(tot))); " possibilities checked." PRINT LTRIM$(RTRIM$(STR$((tp - tot) / 1000000))); " million possibilites discarded." PRINT LTRIM$(RTRIM$(STR$(tsol))); " solutions found." FUNCTION CheckIntersection (pia, pa, pib, pb) CheckIntersection = ABS(SGN(P(A(pia, 1), pa + A(pia, 2)) + P(A(pib, 1), pb + A(pib, 2)))) END FUNCTION FUNCTION CheckPoss (u) tot = tot + 1 f = 0 SELECT CASE u CASE 2 f = f + CheckIntersection(1, 3, 2, 1) 'A CASE 3 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B CASE 4 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C CASE 5 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(4, 3, 5, 1) 'F CASE 6 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G CASE 7 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G f = f + CheckIntersection(4, 4, 7, 2) 'H CASE 8 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G f = f + CheckIntersection(4, 4, 7, 2) 'H f = f + CheckIntersection(5, 4, 8, 2) 'I f = f + CheckIntersection(7, 3, 8, 1) 'K CASE 9 f = f + CheckIntersection(1, 3, 2, 1) 'A f = f + CheckIntersection(2, 3, 3, 1) 'B f = f + CheckIntersection(1, 4, 4, 2) 'C f = f + CheckIntersection(2, 4, 5, 2) 'D f = f + CheckIntersection(3, 4, 6, 2) 'E f = f + CheckIntersection(4, 3, 5, 1) 'F f = f + CheckIntersection(5, 3, 6, 1) 'G f = f + CheckIntersection(4, 4, 7, 2) 'H f = f + CheckIntersection(5, 4, 8, 2) 'I f = f + CheckIntersection(6, 4, 9, 2) 'J f = f + CheckIntersection(7, 3, 8, 1) 'K f = f + CheckIntersection(8, 3, 9, 1) 'L END SELECT IF f = 0 THEN CheckPoss = 1 IF u = 9 THEN PrintSolution END IF END IF END FUNCTION SUB PrintSolution tsol = tsol + 1 c$ = "" FOR z = 1 TO 9 c$ = c$ + CHR$(A(z, 1) + 64) + CHR$(A(z, 2) + 97) NEXT PRINT c$; " "; PRINT tot END SUB SUB Recurse (n) m = n + 1 FOR k = 1 TO 9 f = 0 FOR z = 1 TO n IF A(z, 1) = k THEN f = 1 NEXT IF f = 0 THEN FOR r = 0 TO 3 A(m, 1) = k A(m, 2) = r i = CheckPoss(m) IF m < 9 AND i = 1 THEN Recurse m NEXT END IF NEXT END SUB SUB SetPiece (n, A, b, c, d) P(n, 1) = A P(n, 2) = b P(n, 3) = c P(n, 4) = d 'Keep further copies for rotation! P(n, 5) = A P(n, 6) = b P(n, 7) = c END SUB

And now we come to the solution, when we run it, the program dutifully chugs and whirs and outputs the following message:

IMPUZZABLE SOLVER - by Dan Goodman -------------------------------------------------------------------------------- AaHaGcFaCbEcIcDcBb 1103 BdDaIaEaCdFcGaHcAc 12229 GdEdBcHbCcDdAbFbId 44352 IbFdAdDbCaHdBaEbGb 53782 56592 possibilities checked. 95126.758128 million possibilites discarded. 4 solutions found.

But what does this mean? In what sense have I now solved the puzzle? How much more work do we need to do? Well, not much, we just need to decode the somewhat cryptic "AaHaGcFaCbEcIcDcBb" into an actual solution. This string just means (see above for an explanation of why) that in position 1 we have piece 1 with rotation number 0, in position 2 we have piece 8 with rotation number 0, in position 3 we have piece 7 with rotation number 2, and so on. After a bit of work, we get the solution:

The other solutions are just what you get when you rotate the whole board around.

I hope you enjoyed that little excursion into the world of programming and jigsaw puzzles. The details can be a bit boring at times (especially checking all the intersection points), but hopefully it's been fun anyway. If you've understood all of the above then you should have got the general idea of how to solve lots of similar puzzles. If you didn't understand some of it, you can post a message on the "Ask NRICH" board.