please empty your brain below
No answers or heavy hints, thanks.

Solution tomorrow.

diamond geezer | Homepage | 15.01.17 - 12:00 p.m. | #

Done!

Gross | 15.01.17 - 12:37 p.m. | #

Completed on second attempt - more of a fluke than a strategy, I think.

Waterhouse | 15.01.17 - 1:05 p.m. | #

Yup, done it in 10 !

You'll have to do a version using the Tube Map.

Gerry | 15.01.17 - 1:43 p.m. | #

I think I've done it in six. Just by guessing. Is that good that I've done it in six, or does it have to be 10?

Chris | 15.01.17 - 3:44 p.m. | #

10 moves - as a hint I'd say try doing this from the zero outwards.

Giles | Homepage | 15.01.17 - 3:46 p.m. | #

Ah. I changed direction mid-move in one of the moves and having had a re-read of the instructions I can see I messed up.

I knew I couldn't beat DG. Sigh.

Chris | 15.01.17 - 3:47 p.m. | #

If changing direction mid-move were allowed, you could do it in 1 move! So it's not allowed :)

diamond geezer | Homepage | 15.01.17 - 3:57 p.m. | #

The hint about going backwards is very useful.

Malcolm | 15.01.17 - 4:14 p.m. | #

Working backwards - done

Martin Bayes | 15.01.17 - 5:06 p.m. | #

Solved in 5 minutes before looking here.

Too many heavy hints above though. Suggest DG deletes them.

Ian | 15.01.17 - 7:30 p.m. | #

Anyone who solves it backwards, and who didn't think of the idea themselves before reading it here, only scores half points.

diamond geezer | Homepage | 15.01.17 - 8:16 p.m. | #

I like that ruling. It does require one to be honest with oneself though. For instance, I suspect that being convinced (as I am) that I would have thought of it myself if I hadn't read it here, probably does not count.

I can also solve it in 12, 14, 16, 18 etc moves, if that helps.

Malcolm | 15.01.17 - 8:31 p.m. | #

I also worked backwards, but I am struggling to work out why this gives such an advantage on working forward.

I think it must just be the limited options make the target one move shorter (ie 9 moves not 10 moves), but that seems a bit unsatisfactory to me for the level of difference it seems to make.

Messiah | 15.01.17 - 10:46 p.m. | #

Working backwards drastically limits the search space. For instance there are up to 8 ways to leave a square, but there are at most 1 or 2 ways to arrive at any given square, and commonly no way at all. So if you work backwards most paths reach a dead end or a loop leaving only one possible way to continue, and the problem stops being a puzzle and just becomes a "join the dots".

Ian | 15.01.17 - 11:02 p.m. | #

Er, the way I see it, there are at most 3 ways to leave a square, but there are potentially up to 12 ways to arrive at any given square.

diamond geezer | Homepage | 15.01.17 - 11:17 p.m. | #

Sorry, I got confused. I should have said there are 4 ways to leave a square: up, down, left, right.

But there is, for example, no way to arrive at the square 4th across on the top row.

Ian | 15.01.17 - 11:30 p.m. | #

<snip>

timbo | 16.01.17 - 7:51 a.m. | #

I don wonder what view one is hoping to see from the window of a W&C or Victoria line train anyway...

andy c | 16.01.17 - 9:47 a.m. | #

@Andy C

One can see which station one is at. Seeing the tunnel wall moving past may also help avoid motion sickness (so may travelling forwards or rearwards rather than sideways)

The original Victoria Line trains also worked on the Woodford/Hainault shuttle, which is quite scenic.

Picture of 1898 Drain train: [photo]

And its 1940 successor: [photo]

timbo | 16.01.17 - 10:04 a.m. | #

Given a previous post about random announcements at stations and on trains during disruption - the view through the window provide a visual confirmation about which station you are actually at!

I also hate travelling sideways, is it true that the seats on the Bakerloo Line stock are actually deeper than that on the other refurbished stock - or is it a visual illusion?

Still anon | 16.01.17 - 10:04 a.m. | #

12 ways to arrive at a given square?

I make it a theoretical maximum of 4 ways to arrive and 4 ways to leave (up, down, left, right).

Have I missed something?

Messiah | 16.01.17 - 10:06 a.m. | #

How many ways there are to leave a given square is nearly fixed. There are between 2 and 4 (depending on the number in the square and its position). But the number of ways to arrive at a given square is much more variable, depending very much on the numbers in all the other squares. It could theoretically be as much as 12.

By the handshaking theorem, the total number of ways you can leave all the squares is the same as the total number of ways to arrive at all the squares. But because arrival numbers can be bigger, this means that many of them are smaller, including quite a lot of zeros in this diagram. Which is why the search is much easier going backwards, as all the side paths fizzle out very quickly.

Malcolm | 16.01.17 - 11:25 a.m. | #

D'oh, schoolboy error in my logic.

I think that does make sense - that on average there are less ways to arrive at a square than there are to leave it - or that some of the ways to arrive can be quickly discounted.

Messiah | 16.01.17 - 12:35 p.m. | #

Solution (don't look if you don't want to know)










6↓ 3↑ 2↑ 4→ 4↓ 3← 5→ 2↑ 5← 2→

diamond geezer | Homepage | 16.01.17 - 1:00 p.m. | #

Done.

daveid76 | 20.01.17 - 11:18 a.m. | #











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