Maximum number of non-attacking pairs of queens in 8-Queens problem is given by 8 × 7/2 = 28. Can someone explain how it is 8x7/2?
A non attacking pair is when two queens don't attack each other.For max condition no queen attacks any other queen, so number of non attacking pairs
1st queen would have = 7
2nd queen would have = 6
(exclude the pair with the 1st queen as its already counted in step 1)
Similarly, 3rd queen would have = 5
Thus, total number of non attacking pairs for 8 queens would be = 7+6+5+4+3+2+1+0 =28
Here is another thought process:
We have 8 queens and you want to know every possible attacking pair on the board so we have 8 choose 2
or
8!/((8-2)!*2!) = 28
Related
Consider I have this dataframe
Climb no. No. of Climbers Total people reaching Top A is in climb
1 7 2 Yes
2 5 3 No
3 10 1 Yes
How can I find using R the probability one particular person A will reach the top of the mountain for each climb?
Thanks for any direction
Unless I am missing something, the probability that any one person will reach the top is simply the number who reached the top divided by the total number of climbers (assuming that each person is equally likely to reach the top).
dat<-data.frame(Climb=c(1,2,3),Climbers=c(7,5,10),Top=c(2,3,1))
dat$Top/dat$Climbers
[1] 0.2857143 0.6000000 0.1000000
I have below one problem regarding permutation and combination.
I know one solution which I am providing here. But I have another approach to the same problem but it is not giving me same answer as previous one. Can someone tell where am I making mistake here.
Problem: From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. In how many ways can it be done?
First Answer:
We can select 5 men ...(option 1)
Number of ways to do this = 7C5
We can select 4 men and 1 woman ...(option 2)
Number of ways to do this = 7C4 × 6C1
We can select 3 men and 2 women ...(option 3)
Number of ways to do this = 7C3 × 6C2
Total number of ways = 7C5 + (7C4 × 6C1) + (7C3 × 6C2)
= 756.
Below is my new approach, where I am making mistake but not able to understand it.
atleast 3 men should be there. So ways to choose 3 men out of 7 = 7C3
= 35.
Now 2 person has to be selected from remaining 4 men and 6 women. The no of ways it can be done = 10C2 = 45.
Therefore, total no of way = 35*45 = 1575.
Can someone tell me what I am missing in second approach.
Your approach will count some ways more than
Suppose from the 7 men you choose
M1,M2,M3
and from the remaining 10 person you choose a men M4 and remaining women W1,W2,W3...W6
Now suppose you choose M1,M2,M4 men from the 7 men
and from remaining 10 you choose M3,W1,W2...W6
Now both of this represent the same set and should be counted only once but you are counting them as 2 different ways.Thats why your answer is greater than the expected answer
I'm hoping someone may be able to help with a problem I have - trying to solve using R.
Individuals can submit requests for items. The minimum number of requests per person is one. There is a recommended maximum of five, but people can submit more in exceptional circumstances. Each item can only be allocated one individual.
Each item has a 'desirability'/quality score ranging from 10 (high quality) down to 0 (low quality). The idea is to allocate items, in line with requests, such that as many high quality items as possible are allocated. It is less important that individuals have an equitable spread of requests met.
Everyone has to have at least one request met. Next priority is to look at whether we can get anyone who is over the recommended limit within it by allocating requests to others. After that the priority is to look at where the item would rank in each individual's request list based on quality score, and allocate to the person where it would rank highest (eg, if it would be first in someone's list and third in another's, give it to the former).
Effectively I'd need a sorting algorithm of some kind that:
Identifies where an item has been requested more than once
Check all the requests of everyone making said request
If that request is the only one a person has made, give it to them
(if this scenario applies to more than one person, it should be
flagged in some way)
If all requestees have made more than one request, check to see if
any have made more than five requests - if they have it can be taken
off them.
If all are within the recommended limit, see where the request would
rank (based on quality score) and give to the person in whose list it
would rank highest.
The process needs to check that the above step isn't happening to people so many times that it leaves them without any requests...so it
effectively has to check one item at a time.
Does anyone have any ideas about how to approach this? I can think of all kinds of why I could arrange the data to make it easy to identify and see where this needs to happen, but not to automate the process itself. Thanks in advance for any help.
The data (at least the bits needed for this process) looks like the below:
Item ID Person ID Item Score
1 AAG 9
1 AAK 8
2 AAAX 8
2 AN 8
2 AAAK 8
3 Z 8
3 K 8
4 AAC 7
4 AR 5
5 W 10
5 V 9
6 AAAM 7
6 AAAL 7
7 AAAAN 5
7 AAAAO 5
8 AB 9
8 D 9
9 AAAAK 6
9 AAAAC 6
10 A 3
10 AY 3
I need to find the sector with the lowest frequency in my data frame. Using min gives the minimum number of occurrences, but I would like to obtain the corresponding sector name with the lowest number of occurrences...So in this case, I would like it to print "consumer staples". I keep getting the frequency and not the actual sector name. Is there a way to do this?
Thank you.
sector_count <- count(portfolio, "Sector")
sector_count
Sector freq
1 Consumer Discretionary 5
2 Consumer Staples 1
3 Health Care 2
4 Industrials 3
5 Information Technology 4
min(sector_count$freq)
[1] 1
You want
sector_count$Sector[which.min(sector_count$freq)]
The which.min(sector_count$freq) function selects the index or row where the minimum value is found. The sector_count$Sector vector is then subset to the corresponding value.
CompanyName <- c('Kraft', 'Kraft Foods', 'Kfraft', 'nestle', 'nestle usa', 'GM', 'general motors', 'the dow chemical company', 'Dow')
I want to get either:
CompanyName2
Kraft
Kraft
Kraft
nestle
nestle
general motors
general motors
Dow
Dow
But would be absolutely fine with:
CompanyName2
1
1
1
2
2
3
3
I see algorithms for getting the distance between two words, so if I had just one weird name I would compare it to all other names and pick the one with the lowest distance. But I have thousands of names and want to group them all into groups.
I do not know anything about elastic search, but would one of the functions in the elastic package or some other function help me out here?
I'm sorry there's no programming here. I know. But this is way out of my area of normal expertise.
Solution: use string distance
You're on the right track. Here is some R code to get you started:
install.packages("stringdist") # install this package
library("stringdist")
CompanyName <- c('Kraft', 'Kraft Foods', 'Kfraft', 'nestle', 'nestle usa', 'GM', 'general motors', 'the dow chemical company', 'Dow')
CompanyName = tolower(CompanyName) # otherwise case matters too much
# Calculate a string distance matrix; LCS is just one option
?"stringdist-metrics" # see others
sdm = stringdistmatrix(CompanyName, CompanyName, useNames=T, method="lcs")
Let's take a look. These are the calculated distances between strings, using Longest Common Subsequence metric (try others, e.g. cosine, Levenshtein). They all measure, in essence, how many characters the strings have in common. Their pros and cons are beyond this Q&A. You might look into something that gives a higher similarity value to two strings that contain the exact same substring (like dow)
sdm[1:5,1:5]
kraft kraft foods kfraft nestle nestle usa
kraft 0 6 1 9 13
kraft foods 6 0 7 15 15
kfraft 1 7 0 10 14
nestle 9 15 10 0 4
nestle usa 13 15 14 4 0
Some visualization
# Hierarchical clustering
sdm_dist = as.dist(sdm) # convert to a dist object (you essentially already have distances calculated)
plot(hclust(sdm_dist))
If you want to group then explicitly into k groups, use k-medoids.
library("cluster")
clusplot(pam(sdm_dist, 5), color=TRUE, shade=F, labels=2, lines=0)