suppose I've got groups:
{1=>[1,1,1,1,1,1], 2=>[2,2,2], 3=>[3,3,3,3,3,3], 4=>[4,4,4,4,4,4]}
the keys represent teams, and the values within the arrays represent employees. Imagine I wish to match employees in a semi-random way. I want to make groups of 3's-5's like this:
[1,1,2,3,5], [1,2,3,4], [1,2,3,3], [1,3,4,1,4]
I have the wish to create groups and have a bias for matching team members of opposite teams, but not an absolute bias. Also you must match every member of each team with a group.
How would you solve this?
This is how I've done it:
group_by_team = records.group_by {|x| x.team_id}.values
mixed_groups = group_by_team.each{|x| x.shuffle!}
# take 1 element from each team and mix
# the number of teams is defined as a constant so we don't have to hit the db with a count
for index in (1..(TEAMS-1))
zipped_groups ||= mixed_groups[0]
zipped_groups = zipped_groups.zip(mixed_groups[index])
end
# flatten the arrays to produce one large Array
# remove nil values from ziped steps with compact!
zipped_groups = zipped_groups.flatten!.compact!
lunch_groups = zipped_groups.each_slice(3)
# we can no longer reduce table size, so lets join two small lunch groups
if lunch_groups.any?{|x| x.size<3}
lunch_groups = self.merge_last_two(lunch_groups)
end
But the problems with my implementation are vast. Groups size is fixed at 3. And its no exactly elegant, or efficient.
How would you make semi-random groups happen?
Related
I'm developing an algorithm in Ruby with the following properties:
It works on two objects of type Set, where each element is an Array, where all elements are of type String
Each Array involved has the same number of elements
No two arrays happen to be have the same content (when comparing with ==)
The algorithm involves many operations of moving an array from one Set to the other (or back), storing references to certain Arrays, and testing whether or not that reference is part of the Array
There is no duplication of the Arrays; all Arrays keep their object ID during all the time.
A native implementation would do something like this (to give you the idea); in practice, the arrays here have longer strings and more elements:
# Set up all Arrays involved
master=[
%w(a b c d),
%w(a b c x),
%w(u v w y),
# .... and so on
]
# Create initial sets.
x=Set.new
y=Set.new
# ....
x.add(master[0])
x.add(master[2])
y.add(master[1])
# ....
# Operating on the sets.
i=1
# ...
arr=master[i]
# Move element arr from y to x, if it is in y
if(y.member?(arr)
y.delete(arr)
x.add(arr)
end
# Do something with the sets
x.each { |arr| puts arr.pretty_print }
This would indeed work, simply because the arrays are all different in content. However, testing for membership means that y.member?(arr) tests that we don't have already an object with the same array content like arrin our Set, while it would be sufficient to verify to test that we don't have already an element with the same object_id in our Set, so I'm worried about performance. From my understanding, finding the the object id of an object is cheap, and since it is just a number, maintaining a set of numbers is more performant than maintaining a set of arrays of strings.
Therefore I could try to define my two sets as sets of object_id, and membership test would be faster. However when iterating over a Set, using the object_id to find the array itself is expensive (I would have to search ObjectSpace).
Another possibility would be to not maintain the set of arrays, but the set of indexes into my master array. My code would then be, for example,
x.add(0) # instead of x.add(master[0])
and iterating over a Set would be, i.e.
x.each { |i| puts master[i].pretty_print }
I wonder whether there is a better way - for instance that we can somehow "teach" Set.new to use object identity for maintaining its members, instead of equality.
I think you’re looking for Set#compare_by_identity, which makes the set use the object’s identity (i.e. object ID) of its contents.
x = Set.new
x.compare_by_identity
Say that I have a list (or array) that links Suppliers with the materials they supply. For example, an array of the form
[[Supplier_1, Material_a], [Supplier_2, Material_a], [Supplier_3, Material_a], [Supplier_1, Material_b], [Supplier_2, Material_c], [Supplier_3, Material_b], ...]
I am interested in finding the the list of suppliers that supply at least k materials that a particular supplier say Supplier_1 supplies.
One way that I can think of is to pair all suppliers with Supplier_1 for each material Supplier_1 supplies
[[Supplier_1, Supplier_2, Material_a], [Supplier_1, Supplier_3, Material_a], [Supplier_1, Supplier_3, Material_b]...]
and then count the number of times each pair is present
[[Supplier_1, Supplier_2, 1], [Supplier_1, Supplier_3, 2]...]
The problem is that this approach can be very time consuming since the list provided can be quite long. I was wondering if there is a better way to do this.
You would put the materials of Supplier_1 in a hash set, so that you can verify for any material whether it is supplied by Supplier_1 in constant time.
Once you have that you can iterate the data again, and in a dictionary (hash map) keep a count per supplier which you increment each time the material is in the above mentioned set.
In Python it would look like this:
def getsuppliers(pairs, selected_supplier, k):
materialset = set()
countmap = {} # a dictionary with <key=supplier, value=count> pairs
for supplier, material in pairs:
if supplier == selected_supplier:
materialset.add(material)
countmap[supplier] = 0
# An optional quick exit: if the selected provider does not have k materials,
# there is no use in continuing...
if countmap[selected_supplier] < k:
return [] # no supplier meets the requirement
for supplier, material in pairs:
if material in materialset:
countmap[supplier] = countmap[supplier]+1
result = []
for supplier, count in countmap.items():
if count >= k:
result.append(supplier)
return result
NB: this would include the selected supplier also, provided it has at least k materials.
All operations within each individual loop body, have a constant time complexity, so the overall time complexity is O(n), where n is the size of the input list (pairs).
I'm attempting to solve http://projecteuler.net/problem=1.
I want to create a method which takes in an integer and then creates an array of all the integers preceding it and the integer itself as values within the array.
Below is what I have so far. Code doesn't work.
def make_array(num)
numbers = Array.new num
count = 1
numbers.each do |number|
numbers << number = count
count = count + 1
end
return numbers
end
make_array(10)
(1..num).to_a is all you need to do in Ruby.
1..num will create a Range object with start at 1 and end at whatever value num is. Range objects have to_a method to blow them up into real Arrays by enumerating each element within the range.
For most purposes, you won't actually need the Array - Range will work fine. That includes iteration (which is what I assume you want, given the problem you're working on).
That said, knowing how to create such an Array "by hand" is valuable learning experience, so you might want to keep working on it a bit. Hint: you want to start with an empty array ([]) instead with Array.new num, then iterate something num.times, and add numbers into the Array. If you already start with an Array of size num, and then push num elements into it, you'll end up with twice num elements. If, as is your case, you're adding elements while you're iterating the array, the loop never exits, because for each element you process, you add another one. It's like chasing a metal ball with the repulsing side of a magnet.
To answer the Euler Question:
(1 ... 1000).to_a.select{|x| x%3==0 || x%5==0}.reduce(:+) # => 233168
Sometimes a one-liner is more readable than more detailed code i think.
Assuming you are learning Ruby by examples on ProjectEuler, i'll explain what the line does:
(1 ... 1000).to_a
will create an array with the numbers one to 999. Euler-Question wants numbers below 1000. Using three dots in a Range will create it without the boundary-value itself.
.select{|x| x%3==0 || x%5==0}
chooses only elements which are divideable by 3 or 5, and therefore multiples of 3 or 5. The other values are discarded. The result of this operation is a new Array with only multiples of 3 or 5.
.reduce(:+)
Finally this operation will sum up all the numbers in the array (or reduce it to) a single number: The sum you need for the solution.
What i want to illustrate: many methods you would write by hand everyday are already integrated in ruby, since it is a language from programmers for programmers. be pragmatic ;)
I have two arrays and I want to see the total number of matches, between the arrays individual items that their are.
For example arrays with:
1 -- House, Dog, Cat, Car
2 -- Cat, Book, Box, Car
Would return 2.
Any ideas? Thanks!
EDIT/
Basically I have two forms (for two different types of users) that uses nested attributes to store the number of skills they have. I can print out the skills via
current_user.skills.each do |skill| skill.name
other_user.skills.each do |skill| skill.name
When I print out the array, I get: #<Skill:0x1037e4948>#<Skill:0x1037e2800>#<Skill:0x1037e21e8>#<Skill:0x1037e1090>#<Skill:0x1037e0848>
So, yes, I want to compare the two users skills and return the number that match. Thanks for your help.
This works:
a = %w{house dog cat car}
b = %w{cat book box car}
(a & b).size
Documentation: http://www.ruby-doc.org/core/classes/Array.html#M000274
To convert classes to an array using the name, try something like:
class X
def name
"name"
end
end
a = [X.new]
b = [X.new]
(a.map{|x| x.name} & b.map{|x| x.name}).size
In your example, a is current_user.skills and b is other_users.skills. x is simply a reference to the current index of the array as the map action loops through the array. The action is documented in the link I provided.
I work in a consulting organization and am most of the time at customer locations. Because of that I rarely meet my colleagues. To get to know each other better we are going to arrange a dinner party. There will be many small tables so people can have a chat. In order to talk to as many different people as possible during the party, everybody has to switch tables at some interval, say every hour.
How do I write a program that creates the table switching schedule? Just to give you some numbers; in this case there will be around 40 people and there can be at most 8 people at each table. But, the algorithm needs to be generic of course
heres an idea
first work from the perspective of the first person .. lets call him X
X has to meet all the other people in the room, so we should divide the remaining people into n groups ( where n = #_of_people/capacity_per_table ) and make him sit with one of these groups per iteration
Now that X has been taken care of, we will consider the next person Y
WLOG Y be a person X had to sit with in the first iteration itself.. so we already know Y's table group for that time-frame.. we should then divide the remaining people into groups such that each group sits with Y for every consecutive iteration.. and for each iteration X's group and Y's group have no person in common
.. I guess, if you keep doing something like this, you will get an optimal solution (if one exists)
Alternatively you could crowd source the problem by giving each person a card where they could write down the names of all the people they got dine with.. and at the end of event, present some kind of prize to the person with the most names in their card
This sounds like an application for genetic algorithm:
Select a random permutation of the 40 guests - this is one seating arrangement
Repeat the random permutation N time (n is how many times you are to switch seats in the night)
Combine the permutations together - this is the chromosome for one organism
Repeat for how ever many organisms you want to breed in one generation
The fitness score is the number of people each person got to see in one night (or alternatively - the inverse of the number of people they did not see)
Breed, mutate and introduce new organisms using the normal method and repeat until you get a satisfactory answer
You can add in any other factors you like into the fitness, such as male/female ratio and so on without greatly changing the underlying method.
Why not imitate real world?
class Person {
void doPeriodically() {
do {
newTable = random (numberOfTables);
} while (tableBusy(newTable))
switchTable (newTable)
}
}
Oh, and note that there is a similar algorithm for finding a mating partner and it's rumored to be effective for those 99% of people who don't spend all of their free time answering programming questions...
Perfect Table Plan
You might want to have a look at combinatorial design theory.
Intuitively I don't think you can do better than a perfect shuffle, but it's beyond my pre-coffee cognition to prove it.
This one was very funny! :D
I tried different method but the logic suggested by adi92 (card + prize) is the one that works better than any other I tried.
It works like this:
a guy arrives and examines all the tables
for each table with free seats he counts how many people he has to meet yet, then choose the one with more unknown people
if two tables have an equal number of unknown people then the guy will choose the one with more free seats, so that there is more probability to meet more new people
at each turn the order of the people taking seats is random (this avoid possible infinite loops), this is a "demo" of the working algorithm in python:
import random
class Person(object):
def __init__(self, name):
self.name = name
self.known_people = dict()
def meets(self, a_guy, propagation = True):
"self meets a_guy, and a_guy meets self"
if a_guy not in self.known_people:
self.known_people[a_guy] = 1
else:
self.known_people[a_guy] += 1
if propagation: a_guy.meets(self, False)
def points(self, table):
"Calculates how many new guys self will meet at table"
return len([p for p in table if p not in self.known_people])
def chooses(self, tables, n_seats):
"Calculate what is the best table to sit at, and return it"
points = 0
free_seats = 0
ret = random.choice([t for t in tables if len(t)<n_seats])
for table in tables:
tmp_p = self.points(table)
tmp_s = n_seats - len(table)
if tmp_s == 0: continue
if tmp_p > points or (tmp_p == points and tmp_s > free_seats):
ret = table
points = tmp_p
free_seats = tmp_s
return ret
def __str__(self):
return self.name
def __repr__(self):
return self.name
def Switcher(n_seats, people):
"""calculate how many tables and what switches you need
assuming each table has n_seats seats"""
n_people = len(people)
n_tables = n_people/n_seats
switches = []
while not all(len(g.known_people) == n_people-1 for g in people):
tables = [[] for t in xrange(n_tables)]
random.shuffle(people) # need to change "starter"
for the_guy in people:
table = the_guy.chooses(tables, n_seats)
tables.remove(table)
for guy in table:
the_guy.meets(guy)
table += [the_guy]
tables += [table]
switches += [tables]
return switches
lst_people = [Person('Hallis'),
Person('adi92'),
Person('ilya n.'),
Person('m_oLogin'),
Person('Andrea'),
Person('1800 INFORMATION'),
Person('starblue'),
Person('regularfry')]
s = Switcher(4, lst_people)
print "You need %d tables and %d turns" % (len(s[0]), len(s))
turn = 1
for tables in s:
print 'Turn #%d' % turn
turn += 1
tbl = 1
for table in tables:
print ' Table #%d - '%tbl, table
tbl += 1
print '\n'
This will output something like:
You need 2 tables and 3 turns
Turn #1
Table #1 - [1800 INFORMATION, Hallis, m_oLogin, Andrea]
Table #2 - [adi92, starblue, ilya n., regularfry]
Turn #2
Table #1 - [regularfry, starblue, Hallis, m_oLogin]
Table #2 - [adi92, 1800 INFORMATION, Andrea, ilya n.]
Turn #3
Table #1 - [m_oLogin, Hallis, adi92, ilya n.]
Table #2 - [Andrea, regularfry, starblue, 1800 INFORMATION]
Because of the random it won't always come with the minimum number of switch, especially with larger sets of people. You should then run it a couple of times and get the result with less turns (so you do not stress all the people at the party :P ), and it is an easy thing to code :P
PS:
Yes, you can save the prize money :P
You can also take look at stable matching problem. The solution to this problem involves using max-flow algorithm. http://en.wikipedia.org/wiki/Stable_marriage_problem
I wouldn't bother with genetic algorithms. Instead, I would do the following, which is a slight refinement on repeated perfect shuffles.
While (there are two people who haven't met):
Consider the graph where each node is a guest and edge (A, B) exists if A and B have NOT sat at the same table. Find all the connected components of this graph. If there are any connected components of size < tablesize, schedule those connected components at tables. Note that even this is actually an instance of a hard problem known as Bin packing, but first fit decreasing will probably be fine, which can be accomplished by sorting the connected components in order of biggest to smallest, and then putting them each of them in turn at the first table where they fit.
Perform a random permutation of the remaining elements. (In other words, seat the remaining people randomly, which at first will be everyone.)
Increment counter indicating number of rounds.
Repeat the above for a while until the number of rounds seems to converge.