Got confused with subnetting - subnet

I opened this question:
Something wrong with my understanding in subnetting
and got an answer.
But, a minute after i thought - for example I have this IP : 155 . 115 . 51 . 68 / 29 so I can know that 13 bits are for the subnets IDs (I can tell it by telling this IP is class B, therefor require 16 bits for the net ID and the rest of the mask (13) used for the subnets.) What I do if i get in the cmd the next IP: 10 . 115 . 51 . 68 / 29 Then, I couldnt really tell you how many bits are for the net and how many for the subnets


CIDR means Classless Inter-Domain Routing, the notion of IP class is no longer in use. (To be honest, it's my first time hearing about it, but I'm young).
From what I could read here and there, the notion of net and subnet are no longer relevant. Here is a class A IP:
1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|0| Net | Subnet | Host |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Nowadays, the net part has disappeared and is included in the subnet. The following would match the subnet mask 255.255.0.0:
1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|0| Subnet | Host |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
The CIDR notation says it all: an IP ending in /25 has a subnet defined by 25 bits and your host is defined by the 7 bits left.
In your case of 10.115.51.68/29, your subnet requires 29 bits: the subnet mask is 255.255.255.248 .

Related

Pairing People in R with Loop

I am trying to create pen-pal pairs in R. The problem is that I can't figure out how to loop it so that once I pair one person that person and their pair are eliminated from the pool and the loop continues until everyone has a pair.
I have already rated the criteria to pair them and found a score for every person for how well they would pair for the other person. I think added every pair score together to get a sense of how good the pair is overall (not perfect, but good enough for these purposes). I have found each person's ideal match then and ordered these matches by most picky person to least picky person (basically person with the lowest best-paired score to highest best-paired score). I also found their 2nd-8th best match (there will probably be about 300 people in the data).
A test of the best-matches is below:
indexed_fake apply.fin_fake..1..max. X1 X2 X3 X4 X5 X6 X7 X8
14 14 151 3 9 8 4 10 12 2 6
4 4 177 9 5 8 7 11 3 10 12
9 9 177 4 11 3 6 10 7 12 5
5 5 179 7 4 11 3 12 10 8 5
10 10 179 12 10 2 9 3 5 6 4
13 13 182 8 1 12 11 10 5 3 2
1 1 185 7 1 3 8 6 13 2 11
7 7 185 1 12 5 7 4 6 9 11
3 3 187 12 3 8 5 9 1 2 10
8 8 190 8 12 13 3 4 11 1 6
2 2 191 6 12 11 10 3 4 5 1
6 6 191 2 11 7 1 6 9 10 8
11 11 193 12 6 9 5 2 8 11 4
12 12 193 11 3 8 7 12 10 2 5
Columns X1-X8 are the 8 best pairs for the people listed in the first columns. With this example every person would ideally get paired with someone in their top 8, ideally maximizing the pair compatibility as another user mentioned. Every person would get one pair.
Any help is appreciated!

This is not a specific answer. But it's easier to write in this space. You have a classic assignment optimization problem. These problems can be solved using packages in R. You have to assign preference weights to your feasible pairings. So for example 14-3 could be assigned 8 points, 14-9; 7 points, 14-8; 6 points...14-6; 1 point. Note that 3-14 would be assigned no points because while 14 likes 3, 3 does not like 14. The preference score for any x-y, y-x pairing could be the weight for the x-y preference plus the weight of the y-x preference.
The optimization model would choose the weighted pairs to maximize the total satisfaction among all of the the pairings.
If you have 300 people I can't think of an alternative algorithm that could be simply implemented.

R Studio: Time Syncing Data Sets

I have a simple problem, and a bit more complicated twist at the end.
I have 2 datasets A & B (Separate when imported into R):
Dataset A is pulled from a DAQ that is sampling at 2000 times a second, while dataset B is pulled from a scope at 500 times a second. I have a test that records data from the DAQ and Scope for 5 seconds.
In R Studio I want to time synchronize this data and, for the sake of learning, how can I do it in both of the following ways?
1) Without duplicating values so filtering doesn't stair step:
A B
1 1 1
2 2 NA
3 3 NA
4 4 NA
5 5 2
6 6 NA
7 7 NA
8 8 NA
9 9 3
10 10 NA
11 11 NA
12 12 NA
2) With duplicating numbers if I don't want NA's in the functions I apply to the frame:
A B
1 1 1
2 2 1
3 3 1
4 4 1
5 5 2
6 6 2
7 7 2
8 8 2
9 9 3
10 10 3
11 11 3
12 12 3
Now here is the twist where it becomes a very unique problem I have. Lets say Dataset A records a bit before & after the 5 second test. Dataset A also has an extra column for "Trigger" which is either a 0 or a 1. 1 is a high that represents recording and basically where Dataset B starts. When it switches back to 0, Dataset B has finished recording.
Is there a way I can strategically do the above time sync in Dataset A? The reason I want to keep the data before & after the "true" recording section, is to make sure a filter or a filtfilt sweep will level out before the data truly starts.
Thanks for any help!

SPARC register names to binary

So in SPARC V8, the destination register (rd) occupies 5 bits of the instruction (25th-29th). My question is: is there a document with map associating each register name, say %i1, to its respective 5-bit binary, say, 01010? I can't find such thing...

Register numbers 0-7 are %g0-%g7, 8-15 are %o0-%o7, 16-23 are %l0-%l7 and 24-31 are %i0-%i7

http://www.gaisler.com/doc/sparcv8.pdf
Note that Sparc uses register windows.... so you are addressing the register window not the register file itself.
The data formats are defined on page 18 of the Sparc V8 manual. Sparc uses LSB 0 bit notation. The order of the register window locations names are on page 24.
A byte for instance is 7 6 5 4 3 2 1 0 with 0 being least significant.
This in turn means the instructions map as follows.
0 00000 %g0
1 00001 %g1
2 00010 %g2
3 00011 %g3
4 00100 %g4
5 00101 %g5
6 00110 %g6
7 00111 %g7
8 01000 %o0
9 01001 %o1
10 01010 %o2
11 01011 %o3
12 01100 %o4
13 01101 %o5
14 01110 %o6
15 01111 %o7
16 10000 %l0
17 10001 %l1
18 10010 %l2
19 10011 %l3
20 10100 %l4
21 10101 %l5
22 10110 %l6
23 10111 %l7
24 11000 %i0
25 11001 %i1
26 11010 %i2
27 11011 %i3
28 11100 %i4
29 11101 %i5
30 11110 %i6
31 11111 %i7

R code is not creating objects?

I have written some code for a university assignment. The assignment is based on various concrete samples and their tensile strengths. There are 20 types of concrete mixtures (made from four different accelerators, and five different plasticisers). Our job is to do a statistical analysis on this data frame:
TStrength accelerator plasticiser
1 3.417543 1 1
2 2.887113 1 2
3 3.600988 1 3
4 3.702631 1 4
5 3.686944 1 5
6 3.699785 1 1
7 3.112972 1 2
8 3.918160 1 3
9 3.600538 1 4
10 2.748832 1 5
11 3.404498 1 1
12 3.735437 1 2
13 3.347577 1 3
14 3.101556 1 4
15 3.527621 1 5
16 3.856831 1 1
17 3.492118 1 2
18 3.928343 1 3
19 3.511689 1 4
20 3.371985 1 5
21 3.069794 2 1
22 3.168010 2 2
23 3.316657 2 3
24 3.455162 2 4
25 2.818250 2 5
26 4.054507 2 1
27 3.065984 2 2
28 3.201351 2 3
29 3.417554 2 4
30 3.364320 2 5
31 3.218677 2 1
32 2.647151 2 2
33 3.222705 2 3
34 3.145210 2 4
35 3.636642 2 5
36 3.317620 2 1
37 3.645922 2 2
38 2.556071 2 3
39 3.177663 2 4
40 3.014374 2 5
41 3.838183 3 1
42 4.155951 3 2
43 3.886330 3 3
44 3.723898 3 4
45 4.425442 3 5
46 3.738460 3 1
47 3.217834 3 2
48 3.942241 3 3
49 3.699851 3 4
50 3.797089 3 5
51 3.652456 3 1
52 4.851609 3 2
53 3.359099 3 3
54 4.089559 3 4
55 4.282991 3 5
56 3.803784 3 1
57 3.519551 3 2
58 3.935084 3 3
59 3.890324 3 4
60 4.611936 3 5
61 3.343098 4 1
62 3.713952 4 2
63 3.629883 4 3
64 3.082509 4 4
65 3.346548 4 5
66 3.277845 4 1
67 3.509506 4 2
68 3.490567 4 3
69 3.235009 4 4
70 3.970925 4 5
71 3.504646 4 1
72 3.270798 4 2
73 3.547298 4 3
74 3.278489 4 4
75 3.322743 4 5
76 2.975010 4 1
77 3.384996 4 2
78 3.399486 4 3
79 3.703567 4 4
80 3.214973 4 5
My first step was to attempt to find out the means of the Tstrength values for each of the 20 concrete types (there are four types of each unique concrete sample). I am very new to R, and my code is certainly not beautiful, but this is the code I wrote to find the means:
#Setting the correct directory
setwd("C:/Users/Matthew/Desktop/Work/Engineering")
#Creating the data frame object, Concrete.
#Note that this will only work if the file
#s...-CW.dat is in the current working directory
#Therefore for this code to work, CreateData.r must
#be run on the individual computer with the
#given matriculation number, and the file must be saved
#in the specified directory
Concrete<-read.table(file='s...-CW.dat',header=TRUE)
#Since the samples of concrete are made from 4 different accelerators and
#5 different plasticisers there will be 4*5=20 unique combinations from
#which concrete samples can come from (i.e. 1,1; 1,2; 4,5 etc).
# There are four samples of each combination
#The next section of code is used to find the mean of the four samples,
#for each combination (20 total)
#creating a list with Tstrength from all (1,1) combinations
#Then finding average
combo1 = list(Concrete[1,1],Concrete[6,1],Concrete[11,1],Concrete[16,1])
combo1mean = mean(unlist(combo1))
#Repeating for (1,2)
combo2 = list(Concrete[2,1],Concrete[7,1],Concrete[12,1],Concrete[17,1])
combo2mean = mean(unlist(combo2))
#Repeating for (1,3)
combo3 = list(Concrete[3,1],Concrete[8,1],Concrete[13,1],Concrete[18,1])
combo3mean = mean(unlist(combo3))
#Repeating for (1,4)
combo4 = list(Concrete[4,1],Concrete[9,1],Concrete[14,1],Concrete[19,1])
combo4mean = mean(unlist(combo4))
#Repeating for (1,5)
combo5 = list(Concrete[5,1],Concrete[10,1],Concrete[15,1],Concrete[20,1])
combo5mean = mean(unlist(combo5))
#Repeating for (2,1)
combo6 = list(Concrete[21,1],Concrete[26,1],Concrete[31,1],Concrete[36,1])
combo6mean = mean(unlist(combo6))
#Repeating for (2,2)
combo7 = list(Concrete[22,1],Concrete[27,1],Concrete[32,1],Concrete[37,1])
combo7mean = mean(unlist(combo7))
#Repeating for (2,3)
combo8 = list(Concrete[23,1],Concrete[28,1],Concrete[33,1],Concrete[38,1])
combo8mean = mean(unlist(combo8))
#Repeating for (2,4)
combo9 = list(Concrete[24,1],Concrete[29,1],Concrete[34,1],Concrete[39,1])
combo9mean = mean(unlist(combo9))
#Repeating for (2,5)
combo10 = list(Concrete[25,1],Concrete[30,1],Concrete[35,1],Concrete[40,1])
combo10mean = mean(unlist(combo10))
#Repeating for (3,1)
combo11 = list(Concrete[41,1],Concrete[46,1],Concrete[51,1],Concrete[56,1])
combo11mean = mean(unlist(combo11))
#Repeating for (3,2)
combo12 = list(Concrete[42,1],Concrete[47,1],Concrete[52,1],Concrete[57,1])
combo12mean = mean(unlist(combo12))
#Repeating for (3,3)
combo13 = list(Concrete[43,1],Concrete[48,1],Concrete[53,1],Concrete[58,1])
combo13mean = mean(unlist(combo13))
#Repeating for (3,4)
combo14 = list(Concrete[44,1],Concrete[49,1],Concrete[54,1],Concrete[59,1])
combo14mean = mean(unlist(combo14))
#Repeating for (3,5)
combo15 = list(Concrete[45,1],Concrete[50,1],Concrete[55,1],Concrete[60,1])
combo15mean = mean(unlist(combo15))
#Repeating for (4,1)
combo16 = list(Concrete[61,1],Concrete[66,1],Concrete[71,1],Concrete[76,1])
combo16mean = mean(unlist(combo16))
#Repeating for (4,2)
combo17 = list(Concrete[62,1],Concrete[67,1],Concrete[72,1],Concrete[77,1])
combo17mean = mean(unlist(combo17))
#Repeating for (4,3)
combo18 = list(Concrete[63,1],Concrete[68,1],Concrete[73,1],Concrete[78,1])
combo18mean = mean(unlist(combo18))
#Repeating for (4,4)
combo19 = list(Concrete[64,1],Concrete[69,1],Concrete[74,1],Concrete[79,1])
combo19mean = mean(unlist(combo19))
#Repeating for (4,5)
combo20 = list(Concrete[65,1],Concrete[70,1],Concrete[75,1],Concrete[80,1])
combo20mean = mean(unlist(combo20))
A few notes about the code: "s..." is just my matriculation number. I have triple checked that I have not made a mistake here regarding either the file name or the directory with where it is stored. CreataData.r is just a script provided to us the generates the data used to create 'Concrete' based on our matriculation number (so we're not just blindly copying each other I suppose).
The problem I am having with the code is that whenever it runs, the object Concrete is created, as is combo1mean, combo2mean and combo3mean. However, I just cannot figure out why the rest of the objects aren't being created.
I have had no success using running the script in the Rgui. After running the script, it tells I check that Concrete has initialised, and I check to see if the combo4mean and above have initialised too, but they never do. I thought it maybe had to do with running the wrong file, or that I hadn't saved the data properly, but the script definitely contains all the code, and I created a new file to see if that would work, but unfortunately it didn't. Also, I have read an introduction to R by W.N. Venables, D.M. Smith and the R Core Team, but nothing there has helped me figure this out.
PS I am not doing this as an easy way out of homework. I have genuinely tried to figure out what is going wrong but I cannot seem to find the problem. I also apologise if the question is inaccurate in anyway, or if I have had misunderstandings, I am very new to R and am trying my best to learn it! Cheers in advance.
EDIT: Just in case anyone is curious, I managed to get the exact same code to work on a different computer, starting from an empty workspace. I'm still not very sure why it didn't work on the first computer, but thanks 42 for the code suggestions.

Adding code that should bypass issues related to reading a text file. This shouls succeed on any R installation:
Concrete <- read.table(text="TStrength accelerator plasticiser
1 3.417543 1 1
2 2.887113 1 2
3 3.600988 1 3
4 3.702631 1 4
5 3.686944 1 5
6 3.699785 1 1
7 3.112972 1 2
8 3.918160 1 3
9 3.600538 1 4
10 2.748832 1 5
11 3.404498 1 1
12 3.735437 1 2
13 3.347577 1 3
14 3.101556 1 4
15 3.527621 1 5
16 3.856831 1 1
17 3.492118 1 2
18 3.928343 1 3
19 3.511689 1 4
20 3.371985 1 5
21 3.069794 2 1
22 3.168010 2 2
23 3.316657 2 3
24 3.455162 2 4
25 2.818250 2 5
26 4.054507 2 1
27 3.065984 2 2
28 3.201351 2 3
29 3.417554 2 4
30 3.364320 2 5
31 3.218677 2 1
32 2.647151 2 2
33 3.222705 2 3
34 3.145210 2 4
35 3.636642 2 5
36 3.317620 2 1
37 3.645922 2 2
38 2.556071 2 3
39 3.177663 2 4
40 3.014374 2 5
41 3.838183 3 1
42 4.155951 3 2
43 3.886330 3 3
44 3.723898 3 4
45 4.425442 3 5
46 3.738460 3 1
47 3.217834 3 2
48 3.942241 3 3
49 3.699851 3 4
50 3.797089 3 5
51 3.652456 3 1
52 4.851609 3 2
53 3.359099 3 3
54 4.089559 3 4
55 4.282991 3 5
56 3.803784 3 1
57 3.519551 3 2
58 3.935084 3 3
59 3.890324 3 4
60 4.611936 3 5
61 3.343098 4 1
62 3.713952 4 2
63 3.629883 4 3
64 3.082509 4 4
65 3.346548 4 5
66 3.277845 4 1
67 3.509506 4 2
68 3.490567 4 3
69 3.235009 4 4
70 3.970925 4 5
71 3.504646 4 1
72 3.270798 4 2
73 3.547298 4 3
74 3.278489 4 4
75 3.322743 4 5
76 2.975010 4 1
77 3.384996 4 2
78 3.399486 4 3
79 3.703567 4 4
80 3.214973 4 5", header=TRUE)
This probably does what you are attempting with about 1/10th (or less) code (and more importantly no errors):
> means.by.type <- with( Concrete, tapply(TStrength,
list( acc=accelerator, plas=plasticiser),
FUN=mean))
> means.by.type
plas
acc 1 2 3 4 5
1 3.594664 3.306910 3.698767 3.479103 3.333845
2 3.415150 3.131767 3.074196 3.298897 3.208397
3 3.758221 3.936236 3.780689 3.850908 4.279364
4 3.275150 3.469813 3.516808 3.324893 3.463797
Importantly, you forgot to offer str or dput on Concrete, so cannot really tell whether you problem is data-prep or coding.

Finding Bridges in undirected Graph?

A bridge in a graph means if we remove it the graph will be disconnected !
so i want to know if there is way to find all bridges in a graph :
here is an example :
input
12 15
1 2
1 3
2 4
2 5
3 5
4 6
6 7
6 10
6 11
7 8
8 9
8 10
9 10
10 11
11 12
Output :
2 4
4 6
11 12
PLEASE DO NOT GIVE ME THE SOLUTION JUST A HINT !
Thanks

If you have the visiting number vn[v] and low number low[v] for each vertex v in graph G, then you can find if an edge is bridge of not (while unwinding the dfs recursive calls) using the following condition
if (low[w] > vn[v]) then (v,w) is a bridge

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