I'm practicing for my entrance exam. There are some sample tests available for candidates and I want to solve them before appearing for the exam. But facing problem with the following question.
If anyone can help me solve this, would be highly appreciated.
There are 15 main cities in one country. Some of them are connected by roads.
Besides,
a) You can get from each one city to any other one by a single route;
b) There is only one city with 2 outgoing roads. The other cities have either
1 or 3 outgoing roads.
What is the number of cities with one outgoing road?
Thanks in advance.
Because of a) we can say that this is a tree structure. And in trees the number of edges are one less than number of nodes. Therefore there are only 14 roads. If the number of cities with one outgoing road is x, then 1 * x + 2 * 1 + 3 * (14 - x) = 2 * 14 Which gives us x = 8.
Related
I recently began working on r for social network analysis. Everything goes well and up until now, I found answers to my questions here or on google. But not this time!
I am trying to find a way to calculate "vertex reciprocity" (% of reciprocal edges of each actor of the network). On igraph, reciprocity(g) works fine to calculate the reciprocity of the whole network, but it doesn't help me with the score per actor. Does anybody know what I could do?
Thank you!
I am going to assume that you have a simple graph, that is no loops and no multiple links between nodes. In that case, it is fairly easy to compute this. What does it mean for a link to be reciprocated? When there is a link from a to b, there is a link back from b to a. That means that there is a path of length two from a to itself a->b->a. How many such paths are there? If A is the adjacency matrix, then the entries of AA gives the number of paths of length two. We only want the ones from a node to itself, so we want the diagonal of AA. This will only count a->b->a as one path, but you want to count it twice: once for the link a->b and once for b->a. So for each node you can get the number of reciprocated links from 2*diag(A*A). You want to divide by the total number of links to and from a which is just the degree.
Let me show the computation with an example. Since you do not provide any data, I will use the Enron email data that is available in the 'igraphdata' package. It has loops and multiple links which i will remove. It also has a few isolated vertices, which I will also remove. That will leave us with a connected, directed graph with no loops.
library(igraph)
library(igraphdata)
data(enron)
enron = simplify(enron)
## remove two isolated vertices
enron = delete_vertices(enron, c(72,118))
Now the reciprocity computation is easy.
EnronAM = as.matrix(as_adjacency_matrix(enron))
Path2 = diag(EnronAM %*% EnronAM)
degree(enron)
VertRecip = 2*Path2 / degree(enron)
Let's check it by walking through one node in detail. I will use node number 1.
degree(enron,1)
[1] 10
ENDS = ends(enron, E(enron))
E(enron)[which(ENDS[,1] == 1)]
+ 6/3010 edges from b72ec54:
[1] 1-> 10 1-> 21 1-> 49 1-> 91 1->104 1->151
E(enron)[which(ENDS[,2] == 1)]
+ 4/3010 edges from b72ec54:
[1] 10->1 21->1 105->1 151->1
Path2[1]
[1] 3
Node 1 has degree 10; 6 edges out and 4 edges in. Recip shows that there are three paths of length 2 from 1 back to itself.
1->10->1
1->21->1
1->151->1
That makes 6 reciprocated links and 4 unreciprocated links. The vertex reciprocity should be 6/10 = 0.6 which agrees with what we computed above.
VertRecip[1]
[1] 0.6
An undirected graph is given (as an adjacency list or incidence matrix). For multiple queries, check if a path of length exactly x exists between two nodes. Same nodes can be visited more than once.
I know that for single queries it's easy to check for this, simply by raising the incidence matrix to the power x (number of steps) and checking if the value at [first node][second node] is greater that 0. This takes too long, and for bigger matrices it takes too much memory. Even more so for multiple queries.
How can I solve this problem using as little space and time as possible?
Example:
Graph
Queries:
Is it possible to reach 3 from 2 in 1 step? yes
Is it possible to reach 4 from 1 in 1 step? no
Is it possible to reach 5 from 5 in 8 steps? yes
Is it possible to reach 8 from 1 in 10 steps? no
Thank you in advance.
I am struggling optimising this past amazon Interview question involving a DAG.
This is what I tried (The code is long and I would rather explain it)-
Basically since the graph is a DAG and because its a transitive relation a simple traversal for every node should be enough.
So for every node I would by transitivity traverse through all the possibilities to get the end vertices and then compare these end vertices to get
the most noisy person.
In my second step I have actually found one such (maybe the only one) most noisy person for all the vertices of the traversal in step 2. So I memoize all of this in a mapping and mark the vertices of the traversal as visited.
So I am basically maintaining an adjacency list for the graph, A visited/non visited mapping and a mapping for the output (the most noisy person for every vertex).
In this way by the time I get a query I would not have to recompute anything (in case of duplicate queries).
The above code works but since I cannot test is with testcases it may/may not pass the time limit. Is there a faster solution(maybe using DP) to this. I feel I am not exploiting the transitive and anti symmetric condition enough.
Obviously I am not checking the cases where a person is less wealthy than the current person. But for instance if I have pairs like - (1,2)(1,3)(1,4)...etc and maybe (2,6)(2,7)(7,8),etc then if I am given to find a more wealthy person than 1 I have traverse through every neighbor of 1 and then the neighbor of every neighbor also I guess. This is done only once as I store the results.
Question Part 1
Question Part 2
Edit(Added question Text)-
Rounaq is graduating this year. And he is going to be rich. Very rich. So rich that he has decided to have
a structured way to measure his richness. Hence he goes around town asking people about their wealth,
and notes down that information.
Rounaq notes down the pair (Xi; Yi) if person Xi has more wealth than person Yi. He also notes down
the degree of quietness, Ki, of each person. Rounaq believes that noisy persons are a nuisance. Hence, for
each of his friends Ai, he wants to determine the most noisy(least quiet) person among those who have
wealth more than Ai.
Note that "has more wealth than"is a transitive and anti-symmetric relation. Hence if a has more wealth
than b, and b has more wealth than c then a has more wealth than c. Moreover, if a has more wealth than
b, then b cannot have more wealth than a.
Your task in this problem is to help Rounaq determine the most noisy person among the people having
more wealth for each of his friends ai, given the information Rounaq has collected from the town.
Input
First line contains T: The number of test cases
Each Test case has the following format:
N
K1 K2 K3 K4 : : : Kn
M
X1 Y1
X2 Y2
. . .
. . .
XM YM
Q
A1
A2
. . .
. . .
AQ
N: The number of people in town
M: Number of pairs for which Rounaq has been able to obtain the wealth
information
Q: Number of Rounaq’s Friends
Ki: Degree of quietness of the person i
Xi; Yi: The pairs Rounaq has noted down (Pair of distinct values)
Ai: Rounaq’s ith friend
For each of Rounaq’s friends print a single integer - the degree of quietness of the most noisy person as required or -1 if there is no wealthier person for that friend.
Perform a topological sort on the pairs X, Y. Then iterate from the most wealthy down the the least wealthy, and store the most noisy person seen so far:
less wealthy -> most wealthy
<- person with lowest K so far <-
Then for each query, binary search the first person with greater wealth than the friend. The value we stored is the most noisy person with greater wealth than the friend.
UPDATE
It seems that we cannot rely on the data allowing for a complete topological sort. In this case, traverse sections of the graph that lead from known greatest to least wealth, storing for each person visited the most noisy person seen so far. The example you provided might look something like:
3 - 5
/ |
1 - 2 |
/ |
4 --
Traversals:
1 <- 3 <- 5
1 <- 2
4 <- 2
4 <- 5
(Input)
2 1
2 4
3 1
5 3
5 4
8 2 16 26 16
(Queries and solution)
3 4 3 5 5
16 2 16 -1 -1
Hello Stack Overflow Community,
I'm attempting to solve this problem:
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1040
The problem is to find the best path based on capacity between edges. I get that this can be solved using Dynamic Programming, I'm confused by the example they provide:
According to the problem description, if someone is trying to get 99 people from city 1 to 7, the route should be 1-2-4-7 which I get since the weight of each edge represents the maximum amount of passengers that can go at once. What I don't get is that the description says that it takes at least 5 trips. Where does the 5 come from? 1-2-4-7 is 3 hops, If I take this trip I calculate 4 trips, since 25 is the most limited hop in the route, I would say you need 99/25 or at least 4 trips. Is this a typo, or am I missing something?
Given the first line of the problem statement:
Mr. G. works as a tourist guide.
It is likely that Mr. G must always be present on the bus, thus the equation for the number of trips is:
x = (ceil(x) + number_of_passengers) / best_route
rather than simply:
x = number_of_passengers / best_route
or, for your numbers:
x = (ceil(x) + 99) / 25
Which can be solved with:
x == 4.16 (trips)
I am not good at probability and I know it's not a coding problem directly. But I wish you would help me with this. While I was solving a computation problem I found this difficulty:
Problem definition:
The Little Elephant from the Zoo of Lviv is going to the Birthday
Party of the Big Hippo tomorrow. Now he wants to prepare a gift for
the Big Hippo. He has N balloons, numbered from 1 to N. The i-th
balloon has the color Ci and it costs Pi dollars. The gift for the Big
Hippo will be any subset (chosen randomly, possibly empty) of the
balloons such that the number of different colors in that subset is at
least M. Help Little Elephant to find the expected cost of the gift.
Input
The first line of the input contains a single integer T - the number
of test cases. T test cases follow. The first line of each test case
contains a pair of integers N and M. The next N lines contain N pairs
of integers Ci and Pi, one pair per line.
Output
In T lines print T real numbers - the answers for the corresponding test cases. Your answer will considered correct if it has at most 10^-6 absolute or relative error.
Example
Input:
2
2 2
1 4
2 7
2 1
1 4
2 7
Output:
11.000000000
7.333333333
So, Here I don't understand why the expected cost of the gift for the second case is 7.333333333, because the expected cost equals Summation[xP(x)] and according to this formula it should be 33/2?
Yes, it is a codechef question. But, I am not asking for the solution or the algorithm( because if I take the algo from other than it would not increase my coding potentiality). I just don't understand their example. And hence, I am not being able to start thinking about the algo.
Please help. Thanks in advance!
There are three possible choices, 1, 2, 1+2, with costs 4, 7 and 11. Each is equally likely, so the expected cost is (4 + 7 + 11) / 3 = 22 / 3 = 7.33333.