[bonanova]

What's with Alex? Jamie was watching his friend brood

over a square of paper, writing some numbers, erasing

them, trying other numbers and erasing again ...

He had just racked for a game of 8-ball with Ian, Davie

explained; and he noticed the 8 ball was out of position.

He also noted something about the placement of the other

balls in the rack. While he worked out their positions, and

before he could fix the rack, Ian broke. Now he's trying

to reconstruct the layout.

And what was so special? Jamie wondered.

Looking at the rack as Ian saw it, Davie continued,

with the triangle pointing toward him, each ball except

for the five in the back row was the positive difference

of the two balls diagonally behind it.

Huh? Jamie had downed a few, and his mind was not

the clearest.

Look - suppose there were only three balls. Then it could

have been the 1-ball in front of the 2- and 3-balls. Or the

2-ball in front of the 1 and 3. Get it? But with all 15 balls

it gets more complicated.

I still don't get it, there must be a bunch ways that could

happen with that many balls. Why is it taking him so long?

He thought so, too, and there are, for smaller triangles of

3, 6 and 10 balls. But for 15, he's convinced now that there

is only one [not counting the reflected] solution.

With that, Alex broke his silence, passed them both a square

of paper and offered to buy drinks if either could find the

fifteen-ball "layout with a difference" before he did.

Davie and Jamie began to scribble ...

And now here's your paper.

The clock's ticking! Enjoy

[superprismatic]

first row: 5 ; second row: 4,9 ; third row: 7,11,2 ; fourth row: 8,1,12,10 ; fifth row: 6,14,15,3,13

[Guest]

first row: 5 ; second row: 4,9 ; third row: 7,11,2 ; fourth row: 8,1,12,10 ; fifth row: 6,14,15,3,13

Looks correct but how did you go about solving it?

[bonanova]

Alex slides a cold one down the bar to superp!

[bonanova]

Looks correct but how did you go about solving it?

Yes, the process is often the most satisfying part of solving a puzzle.

Think of where the odd numbered balls must lie.

And you might start out solving the 6-ball and 10-ball racks.

[superprismatic]

Looks correct but how did you go about solving it?

Here's how I solved it: Notice that if you get the 5th row, you can construct the rest of it upwards from 4th row to the 1st. Notice also that 15 MUST be in the bottom row and probably 14 and 13 are there as well. I assumed, then, that 13 and 14 are on the bottom. Now, there are 66 choices for the other two 5th row numbers. I started with 6 for one of the numbers (in the middle of the possible 1 thru 12) -- it turned out to be a very lucky start. The other number I guessed starting at 12 and working my way down. Now there are 60 permutations of the bottom row to look at (we don't try the reverse of something we already tried). I laboriousely went through them and tried to reconstruct the rack from each one. Nearly all of them gave me two of the same number very quickly. And sometimes a particular string can't happen (like 13,14,15 which would make two 1s in the fourth row immediately). Checking cases is actually quite fast once you get the hang of it. It took a little more than an hour to find the correct answer. If I hadn't made the lucky start of 6 in the bottom, I probably would have quit by now.

[Guest]

Here's how I solved it: Notice that if you get the 5th row, you can construct the rest of it upwards from 4th row to the 1st. Notice also that 15 MUST be in the bottom row and probably 14 and 13 are there as well. I assumed, then, that 13 and 14 are on the bottom. Now, there are 66 choices for the other two 5th row numbers. I started with 6 for one of the numbers (in the middle of the possible 1 thru 12) -- it turned out to be a very lucky start. The other number I guessed starting at 12 and working my way down. Now there are 60 permutations of the bottom row to look at (we don't try the reverse of something we already tried). I laboriousely went through them and tried to reconstruct the rack from each one. Nearly all of them gave me two of the same number very quickly. And sometimes a particular string can't happen (like 13,14,15 which would make two 1s in the fourth row immediately). Checking cases is actually quite fast once you get the hang of it. It took a little more than an hour to find the correct answer. If I hadn't made the lucky start of 6 in the bottom, I probably would have quit by now.

I would approach the problem from the other end. This would allow for less guesswork. You can still keep the assumptions that 14, 15 and probably 13 are at the end to help speed up the process.

Knowing that the top ball is 5 there are only 10 possible combinations in the second row (1,6..10,15). You can easily rotate through every one of those possible sets.

Now for the next row we already have 3 balls placed and are assuming that 3 more will end up in the last row (13-15). That only leaves you 9 balls to rotate through.

Continue that process down the rest of the rows using trial and error.

I found that it’s easier to work through the problem with pieces of paper or actual pool balls (it helps keep track of what numbers have/haven't been used. I am going to try it out later to see how long it takes. I am interested in finding out if there are other possible solutions...

[Guest]

first row: 5 ; second row: 4,9 ; third row: 7,11,2 ; fourth row: 8,1,12,10 ; fifth row: 6,14,15,3,13

But doesn't the op say that each ball except 5 is the positive diff between the two balls diagonally across! So if you have 5 followed by 4 & 9, the 5 ball is then also the positive diff of the balls diagonally across it.

Or was it just me who misunderstood the statement

Looking at the rack as Ian saw it, Davie continued, with the triangle pointing toward him, each ball except for the five in the back row was the positive difference of the two balls diagonally behind it

[bonanova]

But doesn't the op say that each ball except 5 is the positive diff between the two balls diagonally across! So if you have 5 followed by 4 & 9, the 5 ball is then also the positive diff of the balls diagonally across it.

Or was it just me who misunderstood the statement

Looking at the rack as Ian saw it, Davie continued, with the triangle pointing toward him, each ball except for the five [balls] in the back row was the positive difference of the two balls diagonally behind it

Apologies for the ambiguous wording...

- bn

[Guest]

Aha! So that is why I was thinking there is no solution! I even made equations using modulus and made a real mess of the problem... thankfully, before toooo long, I decided it wasn't going to work for me.

[plainglazed]

which I think is still incomplete and the level of brute force needed should be less:

There are only four pairs of possibilities for the second row:

{6,1}, {7,2}, {8,3}, and {9,4} (highest possible number in this row is 9 --> 15-1-2-3)

You can also quickly winnow down the third row possibilities (highest possible number in this row is 12 --> 15-1-2):

{6,1} --> (4,10,11), (3,9,10), (4,10,9), (3,9,8), (2,8,9)

{7,2} --> (3,10,12)(4,11,9), (1,8,10), (3,10,8)

{8,3} --> (4,12,9), (1,9,12), (2,10,7), (1,9,3)

{9,4} --> (3,12,8), (2,11,7), (1,10,6)

We can eliminate the ones I've struck through as well:

- There cannot be a 12 in {6,1} or {7,2} because you need 1 and 2 to get 12 in row three (15-1-2=12)

- (1,9,12) to get 12 in row three you have to have use the 1 above (15-1-2)

- (4,10,11) requires a 1 above to get the 10,11 next to one another in the third row

We also logically know that 15 and at least one of 13, and 14 must be on the top row. If 14 is not, than 12 as well as 13 must be.

This is where I put in the brute force. Attacked from both top and bottom starting with two hypotheses:

- 1)decided the most likely scenerio had both 13 and 14 in the top row for what seemed like it would facilitate a more likely/even distribution below

- 2)decided to start with the possible sets int the third row with the higher numbers as I figured they would be the easiest to eliminate.

This did quickly eliminate (3,12,8) and the second I tried (2,11,7) found the answer.

[plainglazed]

Ooops, just checked back and realized I poorly transcribed my notes (edits in bold):

Here's my reasoning which I think is still incomplete and the level of brute force needed should be less:

There are only four pairs of possibilities for the second row:

{6,1}, {7,2}, {8,3}, and {9,4} (highest possible number in this row is 9 --> 15-1-2-3)

You can also quickly winnow down the third row possibilities (highest possible number in this row is 12 --> 15-1-2):

{6,1} --> (4,10,11), (3,9,10), (4,10,9), (3,9,8), (2,8,9), (2,8,7)

{7,2} --> (3,10,12), (4,11,9), (1,8,10), (3,10,8)

{8,3} --> (4,12,9), (1,9,12), (2,10,7), (1,9,6)

{9,4} --> (3,12,8), (2,11,7), (1,10,6)

We can eliminate the ones I've struck through as well:

- There cannot be a 12 in {6,1} or {7,2} because you need 1 and 2 to get 12 in row three (15-1-2=12)

- (1,9,12) to get 12 in row three you have to have use the 1 above (15-1-2)

- (4,10,11) requires a 1 above to get the 10,11 next to one another in the third row

We also logically know that 15 and at least one of 13, and 14 must be on the top row. If 14 is not, than 12 as well as 13 must be.

This is where I put in the brute force. Attacked from both top and bottom starting with two hypotheses:

- 1)decided the most likely scenerio had both 13 and 14 in the top row for what seemed like it would facilitate a more likely/even distribution below and also define 12 as being in the fourth row.

- 2)decided to start with the possible sets in the third row with the higher numbers as I figured they would be the easiest to eliminate.

This did quickly eliminate (4,11,9) and the second try using (2,11,7) found the answer.

[plainglazed]

Going off of the OP's HINT:

- Seems there are only two possible combinations of odd (O) and even (E) numbers--

O O

O

E

E

........

E E

O

O

E

.

E

E

O

E

............

E

O

O

O

..

E O O

.....

and

.....

O E E

....

O E

..................

O E

......

O

.....................

O

Those above in red are the same in both so am thinking must always be as indicated for those balls. How that improves on solving I'm still pondering. Anybody else?

Edited July 23, 2009 by plainglazed

[plainglazed]

So the two (corrected) possible combinations below are for when row two is {7,2}, {9,4}, {1,6} or {3,8}.

O O

O

E

E

........

E E

O

E

O

.

E

E

O

E

............

E

O

O

O

..

E O O

.....

and

.....

O E E

...

O E

...................

O E

....

O

.......................

O

With the possible combinations for {row two} and (row three):

{1,6} - (10,9,3) (9,10,4) (8,9,3) (9,8,2) (7,8,2)

{7,2} - (4,11,9) (1,8,10) (3,10,8)

{3,8} - (9,12,4) (7,10,2) (6,9,1)

{9,4} - (1,10,6) (3,12,8) (2,11,7)

Knowing that the forth row is either E E O E if the leftmost digit in line 3 is even or E O O O if the leftmost digit in line three is odd makes the brute force from here not so brutal.

1,6

10,9,3 (fourth line must be E E O E using 2,4,7,8,11,12,13,14,15)

12,2,11,8

cant fill fifth row

1,6

9,10,4 (fourth line must be E O O O using 1,2,3,7,11,12,13,14,15)

cant fill fourth row

1,6

8,9,3 (fourth line must be E E O E using 2,4,7,10,11,12,13,14,15)

12,4,13,10

cant fill fifth row

1,6

9,8,2 (fourth line must be E O O O using 3,4,7,10,11,12,13,14,15)

cant fill fourth row

1,6

7,8,2 (fourth line must be E O O O using 3,4,9,10,11,12,13,14,15)

10,3,11,9

cant fill fifth row

7,2

4,11,9 (fourth line must be E E O E using 1,3,6,8,10,12,13,14,15)

8,12,1,10

cant fill fifth row

7,2

1,8,10 (fourth line must be E O O O using 3,4,6,9,11,12,13,14,15)

12,11,3,13

cant fill fifth row

7,2

3,10,8 (fourth line must be E O O O using 1,4,6,9,11,12,13,14,15)

cant fill fourth row

3,8

9,12,4 (fourth line must be E O O O using 1,2,6,7,10,11,13,14,15)

cant fill fourth row

3,8

7,10,2 (fourth line must be E O O O using 1,4,6,9,11,12,13,14,15)

cant fill fourth row

3,8

6,9,1 (fourth line must be E E O E using 2,4,7,10,11,12,13,14,15)

10,4,13,14 (but cant have both 13 and 14 in this row)

cant fill fifth row

4,9

1,10,6 (fourth line must be E O O O using 2,3,7,8,11,12,13,14,15)

2,3,13,7

cant fill fifth row

4,9

3,12,8 (fourth line must be E O O O using 1,2,6,7,10,11,13,14,15)

cant fill fourth row

4,9

2,11,7 (fourth line must be E E O E using 1,3,6,8,10,12,13,14,15)

10,12,1,8 or 14,12,1,8 (but cant have both 14 and 1 in this row)

13,3,15,14,6

Cant seem to come up with anything else that further simplifies. Uncle...

Edited July 24, 2009 by plainglazed