Ask the Bear & Badger - The Bear and Badger

kestla

stonechalice wrote:

kestla wrote:

Would like to book 1 or 2 nights in a cottage somewhere with the wife and 4 year old daughter. Near a beach is preferable, but not make or break. Some solitude would be grand, though.

Have had a quick gander, but the volume of websites, locations etc is overwhelming. Who knew so many people had holiday homes to rent?!

Does anyone have any recommendations? I'm based in London, but happy to drive 3 or 4 hours if need be.

Going through various websites is a nightmare for this. I've found the best way is to find a location you're keen on, then open Google maps and search for accommodation nearby. You can find some hidden gems this way.

Pretty much I've given up on it for. That's why I originally popped the question here, because I don't have anywhere specific in mind and narrowing it down is usually easier with solid recommendations. The few places I did scout, seemed sold out or too expensive. May try again later in the year.

JMW

Quite a few people have booked a UK holiday and one abroad for the same time, waiting to see what’s open. If things go the way they seem to be, they’re all going to dump the UK holiday at the last minute, so a fair amount of stuff could become available late on, cheaper.

acemuzzy

Just what the UK tourism industry needs, discounts and people fleeing the country...

kestla

Anyone using a VPN?

I'm currently subscribed to ExpressVPN, but it's due to end this week and I'm open to switching. NordVPN currently has a 2 year deal for £73, which is roughly what Express want for a year. Seems like a no brainer but just wondered whether anyone had an personal experiences with alternatives?

Edit: Just adding, that it would predominantly be used for streaming via IPTV.

Diluted Dante

I use Nord. It's absolutely fine for everything I've used it for, though I've never done an IPTV.

bad_hair_day

Nord here too. Easy to use and fast.

kestla

Cheers chaps, much obliged.

Diluted Dante

kestla

equinox_code

Maths questions. It might be stupid and obvious. I just have a B at GCSE level and can't even remember anything I learnt.

I'm playing this game called Factorio. Perhaps some of you have heard of it. At a very basic level it involves extracting resources, crafting things, then finding ways to automate these processes.

In doing so I have found myself utterly stuck for the last days with problem I do not know how to solve. It may actually be unsolvable. if so, please tell me. So here it is, simplified as best I can:

I have 115 units of a material on a single conveyor belt, which needs to be distributed to 2 areas within my factory.

Area A requires 15 units
Area B requires 100 units

The conveyor belt and items on it can be divided as many times as we like. For example, if we split into two separate conveyor belts, there would in theory be 57.5 on each. We could then divide one of those lanes in half, to give 1 belt containing 57.5, and two more carrying 27.25 each.

Using this process we can also divide into thirds if we need, by splitting the items in half, then half again, leaving 4 belts of equal quantity, and then feeding the 4th belt back into the start when there was just one belt. (I have no idea how easy this is to visualise for people who haven't played the game).

Anyway, using these processes, I need to separate 15 of the items from other 100, with no decimal places left at the end.

The simplest/most unlikely solution appeared to be by dividing the items into 23 equally weighted lanes containing 5 items each, which would then be easy to merge to create the desired quantities. But I think this might actually be impossible due to both 23 and 5 being prime numbers.

If that's not the case, would love to hear how it can be achieved.

acemuzzy

Using this process we can also divide into thirds if we need, by splitting the items in half, then half again, leaving 4 belts of equal quantity, and then feeding the 4th belt back into the start when there was just one belt.

I don't understand this bit. Wouldn't that 4th quarter then be split into either two or four again? How do you get from that to thirds? Or can you just loop it round kinda indefinitely so that that quarter goes into 16ths (one 16th onto each of the first three quarters, and then the final sixteenth around again into 1/64ths, etc. etc.)?

I think the question otherwise becomes "can you make 15/115 as a sum of inverse powers of two", to which I think the answer is no. But maybe with cunning looping it is possible...

acemuzzy

I mean if you do it into 32 lanes (so 5 halves and halves again, and then churn 9 of the 32 back in the start), does that do you?

acemuzzy

I think failing that, you can't do it, cos adding powers of two always leaves powers of two as the denominator.

DrewMerson

equinox_code

acemuzzy wrote:

Using this process we can also divide into thirds if we need, by splitting the items in half, then half again, leaving 4 belts of equal quantity, and then feeding the 4th belt back into the start when there was just one belt.

I don't understand this bit. Wouldn't that 4th quarter then be split into either two or four again? How do you get from that to thirds? Or can you just loop it round kinda indefinitely so that that quarter goes into 16ths (one 16th onto each of the first three quarters, and then the final sixteenth around again into 1/64ths, etc. etc.)? I think the question otherwise becomes "can you make 15/115 as a sum of inverse powers of two", to which I think the answer is no. But maybe with cunning looping it is possible...

Sorry. I could have included a picture to illustrate what I mean.

This splits into 4 parts. if we take any one of those 4 exits at the top and loop it back into the start, we get thirds on the output. So yes we can use as many loops as we like.

DrewMerson

acemuzzy

My answer above is valid, isn't it??

equinox_code

DrewMerson wrote:

For example, if we split into two separate conveyor belts, there would in theory be 57.5 on each.

Maybe the ‘in theory’ is throwing me here, but what does the game actually do when you split 115 items down two belts? 57.5 each, or 58 on one and 57 on the other?

There are no half items. However if, for example, I took a lane with 10 items and split in half twice, there would 'in theory' be 2.5 on each. When I merge any two of these lanes there will always be 5 in total on that lane. There will not occur a situation where the merged lane has 6 or 4 items, as one might expect. I don't know how or why it seems to work this way but it does.

DrewMerson

equinox_code

acemuzzy wrote:

My answer above is valid, isn't it??

Probably. I'm not equipped to answer that. Thanks though

I'm not sure if I'm happy or sad to hear this. I was determined to figure this out myself and have been working on it for the best part of 3 days.

equinox_code

DrewMerson wrote:

Okay, but if the game can’t split units, then splitting an odd number must put different numbers on each belt. No theoretical .5 units anywhere. In which case, follow my guide above.

If I split 115 there would be no way to know which lane had 57 and which lane had 58, which renders the rest of the processes dysfunctional.

DrewMerson

equinox_code

And even if we did know which lane had 57 and which lane had 58, the next time we feed through a further 115 units it would change.

If it helps- although I'm pretty sure it doesn't- we can double or quadruple the starting number. It's the ratio that is important.

equinox_code

I perhaps should have mentioned that it needs to be a process that can be repeated. I overlooked the relevance of this bit of info at the time of posting.

LivDiv

5 is your lowest common denominator so you need 3 lanes to get the 15 and 20 lanes to get the 100.
23 lanes total with 5 units on each.
That obviously isn't a power of 2.

You could find a number that is a power of 2 and divisible by 23 but that will split your units.

If you can increase your units in line with whatever that number is it would be doable.

DrewMerson

acemuzzy

If what I wrote above wasn't clear, maybe a ~diagram...

1
1/2 1/2
1/4 1/4 1/4 1/4
1/8 1/8 ... 1/8 1/18
1/16 1/16 ......... 1/16 1/16
1/32 1/32 .................... 1/32 1/32 -> from this lot, take 3 to the lot you want 15 for, 20 for the lot you want 100 for, and put the other 9 back in the top. That gives a 3:20 ratio out the bottom, which is the key thing, AIUI.

equinox_code

DrewMerson wrote:

And even if we did know which lane had 57 and which lane had 58, the next time we feed through a further 115 units it would change.

Again, have you tested that?

Not your exact example, but I don't think it follows any logic that it would work. Imagine each item being sent left then right when dividing a lane in two. Cycle 1: the 1st and 115th unit will be sent left. Cycle 2: the 1st and 115th unit are sent right.

Thanks though

acemuzzy wrote:

If what I wrote above wasn't clear, maybe a ~diagram... 1 1/2 1/2 1/4 1/4 1/4 1/4 1/8 1/8 ... 1/8 1/18 1/16 1/16 ......... 1/16 1/16 1/32 1/32 .................... 1/32 1/32 -> from this lot, take 3 to the lot you want 15 for, 20 for the lot you want 100 for, and put the other 9 back in the top. That gives a 3:20 ratio out the bottom, which is the key thing, AIUI.

Great, thanks! Sorry when you said your answer was valid I thought you meant the other answer; that it was impossible. the diagram def helped. Cheers.

Will try this out. I suspect it's going to bottleneck my throughput quite a lot but right now I'd welcome a different sort of headache.

edit- thanks to livdiv for the suggestion too

acemuzzy

You'll maybe be able to speed it up by taking some bits from higher layers actually. Eg 20/32 = 1/2 + 1/8 the big way, 9/32 = 1/4 + 1/32 back in the top, 3/32 = 1/16 + 1/32 the little way. I e. Prob no point subdividing if all subdivisions will go there same way.

Not sure if that's useful though, haven't played it myself.

equinox_code

Another thanks. I feel better equipped to figure further problems out myself already.

Do you have this game?

I'd really recommend it to anyone.

Howdy, Stranger!

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