If you have a problem or need to report a bug please email : support@dsprobotics.com
There are 3 sections to this support area:
DOWNLOADS: access to product manuals, support files and drivers
HELP & INFORMATION: tutorials and example files for learning or finding pre-made modules for your projects
USER FORUMS: meet with other users and exchange ideas, you can also get help and assistance here
NEW REGISTRATIONS - please contact us if you wish to register on the forum
Users are reminded of the forum rules they sign up to which prohibits any activity that violates any laws including posting material covered by copyright
A series of numbers: Help needed
15 posts
• Page 1 of 2 • 1, 2
A series of numbers: Help needed
I hope somebody will be able to help with this. There's a series of numbers. The series will have a minimum of 2 layers, but can go as high as imaginable. I evaluated the progression of the first five evolutions.
For 2 layers: 0, 1
For 3 layers: 0, 2, 3
For 4 layers: 0, 4, 6, 7
For 5 layers: 0, 16, 24, 28, 30
For 6 layers: 0, 32, 48, 56, 60, 62
I know it is somehow a reversed base 2 exponential. The differences from layer to layer can be expressed as 2^x. If I only write down those differences as x, it looks like this:
0
1, 0
2, 1, 0
4, 3, 2, 1
5, 4, 3, 2, 1
Damn, it was so close! But it still isn't a linear progressing series. Look at the break between 4 and 5 layers. For 4 layers it is 2, 1, 0 and not 3, 2, 1. But then again they both progress linear towards their other neighbours...
Is there anybody out there who could help me getting to the formula behind this? I need it to calculate any of the numbers for any amount of layers (Like 32 for the second layer if there are six in total, but 2 for the second layer if there are only three in total). Probably I don't see the tree for the woods.
For 2 layers: 0, 1
For 3 layers: 0, 2, 3
For 4 layers: 0, 4, 6, 7
For 5 layers: 0, 16, 24, 28, 30
For 6 layers: 0, 32, 48, 56, 60, 62
I know it is somehow a reversed base 2 exponential. The differences from layer to layer can be expressed as 2^x. If I only write down those differences as x, it looks like this:
0
1, 0
2, 1, 0
4, 3, 2, 1
5, 4, 3, 2, 1
Damn, it was so close! But it still isn't a linear progressing series. Look at the break between 4 and 5 layers. For 4 layers it is 2, 1, 0 and not 3, 2, 1. But then again they both progress linear towards their other neighbours...
Is there anybody out there who could help me getting to the formula behind this? I need it to calculate any of the numbers for any amount of layers (Like 32 for the second layer if there are six in total, but 2 for the second layer if there are only three in total). Probably I don't see the tree for the woods.
"There lies the dog buried" (German saying translated literally)
- tulamide
- Posts: 2714
- Joined: Sat Jun 21, 2014 2:48 pm
- Location: Germany
Re: A series of numbers: Help needed
Are you sure that layer 5 is not 0, 8, 4, 2, 1, and layer 6: 0, 16, 24, 28, 30, 31?
- Youlean
- Posts: 176
- Joined: Mon Jun 09, 2014 2:49 pm
Re: A series of numbers: Help needed
what u doing?
- djbrynte
- Posts: 613
- Joined: Mon Jun 22, 2009 10:51 am
Re: A series of numbers: Help needed
Youlean wrote:Are you sure that layer 5 is not 0, 8, 4, 2, 1, and layer 6: 0, 16, 24, 28, 30, 31?
Yes, unfortunately It is exactly as I wrote it down. That's why I have an issue.
@djbrynte could you please stop pushing your own posts every few minutes to be shown on top, and give others a chance to be noticed, too? Thanks.
"There lies the dog buried" (German saying translated literally)
- tulamide
- Posts: 2714
- Joined: Sat Jun 21, 2014 2:48 pm
- Location: Germany
Re: A series of numbers: Help needed
I asked what r u trying to do now? Is it something unison related?
- djbrynte
- Posts: 613
- Joined: Mon Jun 22, 2009 10:51 am
Re: A series of numbers: Help needed
Hm, can you write down more layers? It is hard to assume math behind it if the has change occured only once. If there is another change it will be much easier to find solution...
- Youlean
- Posts: 176
- Joined: Mon Jun 09, 2014 2:49 pm
Re: A series of numbers: Help needed
If you write it down in binary representation, the structure becomes evident, including the irregularity in the transition from layer 4 to layer 5 (a factor 2):
- Code: Select all
layer 2: 0 1
layer 3: 0 10 11
layer 4: 0 100 110 111
layer 5: 0 10000 11000 11100 11110
layer 6: 0 100000 110000 111000 111100 111110
-
martinvicanek - Posts: 1328
- Joined: Sat Jun 22, 2013 8:28 pm
Re: A series of numbers: Help needed
I know, Martin. That's what's making me crazy. But the numbers are correct!
For 7 layers: 0, 64, 96, 112, 120, 124, 126
For 8 layers: 0, 128, 192, 224, 240, 248, 252, 254
And here's a rough sketch of what I'm trying to do:
Green is just an example. I need it for every first circle per row. Now, the only constant in this is that the last row will always have the circles directly aligning. For that graphic x is 4 for row 1, 6 for row 3 and 7 for row 4. I've listed x for trees up to 8 layers now, and there is this break between 4 and 5
For 7 layers: 0, 64, 96, 112, 120, 124, 126
For 8 layers: 0, 128, 192, 224, 240, 248, 252, 254
And here's a rough sketch of what I'm trying to do:
Green is just an example. I need it for every first circle per row. Now, the only constant in this is that the last row will always have the circles directly aligning. For that graphic x is 4 for row 1, 6 for row 3 and 7 for row 4. I've listed x for trees up to 8 layers now, and there is this break between 4 and 5
"There lies the dog buried" (German saying translated literally)
- tulamide
- Posts: 2714
- Joined: Sat Jun 21, 2014 2:48 pm
- Location: Germany
Re: A series of numbers: Help needed
Are you sure that you're not mixing up r and d (= 2r)?
-
martinvicanek - Posts: 1328
- Joined: Sat Jun 22, 2013 8:28 pm
Re: A series of numbers: Help needed
if i understand your sketch correctly
row 5 should read: 0,8,12,14,15
row 6 should read: 0,16,24,28,30,31
which would be easy to solve...?
row 5 should read: 0,8,12,14,15
row 6 should read: 0,16,24,28,30,31
which would be easy to solve...?
- stw
- Posts: 111
- Joined: Tue Jul 13, 2010 11:09 am
15 posts
• Page 1 of 2 • 1, 2
Who is online
Users browsing this forum: No registered users and 94 guests