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
Related
guys:
I have two matrix as following:
d <- cbind(c(1,2,3,4),c(1,1,1,1),c(1,2,4,8))
v <- cbind(c(2,2,2,2),c(3,3,3,3))
But I want to get a matrix consisted of divj as following:
d1v1 d1v2 d2v1 d2v2 d3v1 d3v2
2 3 2 3 2 3
4 6 2 3 4 6
6 9 2 3 8 12
8 12 2 3 16 24
This is an example of my question,I wonder if you can tell me how to write codes to solve this question.Many thanks.
matrix(apply(v,2,function(x){x*d}),4,6)
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.
I am trying to create an index for a data frame. Each team playing has its own row, but I would like to add a column to use as an index so that the first two teams have the index 'Game 0', the next two teams have the index 'Game 1' until the length of half the list. In python the code would look as follows:
for i in range(0,int(len(teams)/2)):
gamenumber.append('Game '+str(i))
gamenumber.append('Game '+str(i))
I am unfamiliar with R so any help would be appreciated!
This will give you a list of paired index numbers:
> teams=1:100
> data.frame("Games"=sort(c(1:(length(teams)/2), 1:(length(teams)/2))))
Games
1 1
2 1
3 2
4 2
5 3
6 3
7 4
8 4
9 5
10 5
11 6
12 6
13 7
14 7
15 8
16 8
17 9
18 9
19 10
20 10 #etc.
Assuming teams is a data.frame with an even number of rows:
rep(1:(nrow(teams)/2), each=2)
This question already has answers here:
Filtering a data frame by values in a column [duplicate]
(3 answers)
Closed 3 years ago.
I have the following data with the ID of subjects.
V1
1 2
2 2
3 2
4 2
5 2
6 2
7 2
8 2
9 2
10 2
11 2
12 2
13 2
14 2
15 2
16 4
17 4
18 4
19 4
20 4
21 4
22 4
23 4
24 4
I want to subset all the rows of the data where V1 == 4. This way I can see which observations relate to subject 4.
For example, the correct output would be
16 4
17 4
18 4
19 4
20 4
21 4
22 4
23 4
24 4
However, the output I'm given after subsetting does not give me the correct rows . It simply gives me.
V1
1 4
2 4
3 4
4 4
5 4
6 4
7 4
8 4
I'm unable to tell which observations relate to subject 4, as observations 1:8 are for subject 2.
I've tried the usual methods, such as
condition<- df == 4
df[condition]
How can I subset the data so I'm given back a dataset that shows the correct row numbers for subject 4.
You can also use the subset function:
subset(df,df$V1==4)
I've managed to find a solution since posting.
newdf <- subset(df, V1 == 4).
However i'm still very interested in other solutions to this problems, so please post if you're aware of another method.
I would like to import the data into R as intervals, then I would like to count all the numbers falling within these intervals and draw a histogram from this counts.
Example:
start end freq
1 8 3
5 10 2
7 11 5
.
.
.
Result:
number freq
1 3
2 3
3 3
4 3
5 5
6 5
7 10
8 10
9 7
10 7
11 5
Some suggestions?
Thank you very much!
Assuming your data is in df, you can create a data set that has each number in the range repeated by freq. Once you have that it's trivial to use the summarizing functions in R. This is a little roundabout, but a lot easier than explicitly computing the sum of the overlaps (though that isn't that hard either).
dat <- unlist(apply(df, 1, function(x) rep(x[[1]]:x[[2]], x[[3]])))
hist(dat, breaks=0:max(df$end))
You can also do table(dat)
dat
1 2 3 4 5 6 7 8 9 10 11
3 3 3 3 5 5 10 10 7 7 5