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[lillyplop]

Silver #231 Cast Adrift 2

[MartinM]

Im pretty sure that you have to cut out each 3*3 square and arrange them so that all the paths fit. I would be grateful if someone who has this card could transcribe the numbers on each island, as i cannot make them out off the scans.
M

[lillyplop]

Going from left to right, in rows

No numbers in first row
4 16
13 6
21 22
5 18
12 25 19 11
D 9 3
27 7
17 20 23
24
C 8 B 26
15 14 A 10 2
No numbers
1
No numbers

Hope this helps

[lillyplop]

I think we have to cut out the squares, then rearrange them so that all 4 boats paths work. With 16 squares, rearranging them could take while in itself so I am now wondering if, to truly make this worthy of a silver, if we have to figures out where north is too.

I am going to cut them out, throw them onto the table and randomly put them back together. It may be worth numbering each box (3x3) 1-16 so we can tell which box goes where and which permutations have been done.

[sixsidedsquare]

Well for anyone that didn't realise, the first Cast Adrift was a puzzle at the Clapham Commons event.
That involved arranging 9 different squares together, and it was a bitch to do, so I'd say this one will be much harder, having 16 and all.

So good luck to anyone trying it, if anyones up to doing it by hand.

[c1023]

I've just done a few quick calculation to figure out the size of this problem.
If the orientation of all of the tiles stay the same, there are 16! (1e13) possible orders of the tiles.

If a computer evaluated a million posibilities a second, it will still take 242 days to evaluate all of the posibilities. Of course one million tests per second is just a top of the head eastimate. If it can only do 100,000/s then it would be 6.6 years.

There are a few restrictions that would cut the problem down a little, like:
Boat B cannot be in the top row, as it must sail north.

Still. it's not as bad as 13th labour!

[Curlytek]

Forgive me if I misinterpret the question here, but isn't the problem a little less complex than is being made out? Since the boats have to arrive at an island at each step of their journeys, then only certain of the cards can be placed next to any of the boat cards. For example, boat C is in the middle of a card, and goes W first up. Therefore, the card that is placed on the left of the card containing boat C must have an island in the middle right row (below, X can be anything, I represents an island):

XXX
XXI
XXX

Only 3 cards match this description. Similarly, for boat B there are only 5 cards that can be placed above its card. This of course assumes (almost certainly naively) that the individual 3x3 cards are all oriented correctly from the start. Even so, there are still only 6 cards that could accomodate the first move of boat C if initial orientations are not defined.

Other limits, in the first instance, apply for boats A and D, where one orientation is eliminated by the presence of an adjacent island.

So while this is clearly not a dead easy puzzle, I do not think it is as devestatingly impossible as is being implied.

[c1023]

[echidna]

[kookaburrah]

What does 'M' mean in the boat D directions?
A

[doublecross]

I suspect that M means N!

Would it be safe to assume that the squares are arranged 4x4 and that no boats fall off the edge, i.e. they will always hit the next island before they reach the edge of the grid of squares?

[Kvasir]

[echidna]

Given the amount of times that D travels North it would also be logical to assume that it starts from the bottom row.

[Emptynuke]

lol, it looks like a deforme soduku.

[Gibbet]

I got a bit bored of revising today and have knocked up a "solver" of sorts.

It looks pretty and in the short while it has been running has found a solution which matches 30/42 island sailings. So possibly given long enough it might get lucky.

Certainly 1,000,000 possibilities a second is far far away. Currently the un-optimised code runs at a few thousand a minute! However I am sure that object based flash coding, however fun it is to watch my boats actually sailing about is not really the best method to use, and I imagine someone with some real coding knowledge will be able to better this.