I have a set of many (10000+) items, from which have I have to choose exactly 20 items. I can only choose each item once. My items have profits, and costs, as well as several boolean properties (such as colour). I need to output the results in a specific order: in particular I need the first and third items to be blue, and the second and fourth items to be red.
Each item is represented as a tuple:
item = ('item name', cost, profit, is_blue, is_red)
as an example
vase = ['Ming Vase', 1000, 10000, 0, 1]
plate = ['China Plate', 10, 5, 1, 0]
and the total set of items is a list of lists:
items = [item1, item2, ..., itemN].
My profits and costs are also lists:
profits = [x[2] for x in items]
costs = [x[1] for x in items]
For each item chosen, it needs to have a minimum value, and a minimum of 5 items must have the property (is_blue) flag set to 1.
I want to choose the 20 cheapest items with the highest value, such that 5 of them have the is_blue flag set to 1, and the first and third items are blue (etc).
I'm having trouble formulating this using google OR tools.
from ortools.linear_solver import pywraplp
solver = pywraplp.Solver('SolveAssignmentProblemMIP',
pywraplp.Solver.CBC_MIXED_INTEGER_PROGRAMMING)
x = {}
for i in range(MAX_ITEMS):
x[i] = solver.BoolVar('x[%s]' % (i))
#Define the constraints
total_chosen = 20
solver.Add(solver.Sum([x[i] for i in range(MAX_ITEMS)]) == total_chosen)
blues = [x[3] for x in items]
solver.Add(solver.Sum([blues[i] * x[i] for i in .
range(MAX_ITEMS)]) >= 5)
max_cost = 5.0
for i in range(MAX_ITEMS):
solver.Add(x[i] * cost[i] <= max_cost)
solver.Maximize(solver.Sum([profits[i] * x[i] for i in range(total_chosen)]))
sol = solver.Solve()
I can get the set of items I've chosen by:
for i in range(MAX_ITEMS):
if x[i].solution_value() > 0:
print(item[i].item_name)
This works fine - it chooses the set of 20 items which maximise the profits subject to the cost constraint, but I'm stuck on how to extend this to choosing items in way that guarantees that the first is blue etc.
Any help in formulating the constraints and objective would be really helpful. Thanks!
Instead of expressing chosen items with BoolVar, consider making a list of 20 IntVar with domain of 0..MAX_ITEMS. From there it should be fairly easy to do something like this:
solver.Add(chosens[0].IndexOf(all_items)[3] == 1)
solver.Add(chosens[2].IndexOf(all_items)[3] == 1)
chosens[i].IndexOf(all_items) simply means all_items[IndexOfChosen], I.E: whichever item is chosen for the Ith place. If you go with this approach, do not forget to MakeAllDifferent!
Related
I want to create a program where a user can input what the maximum size of the groups is that they want. These groups are formed from a list of names from a submission form. The idea is that there are multiple rounds in which the names are paired in the requested maximum group size and each round does not create previously formed groups. Also, no one should be left out, so no groups of 1 person.
I have two problems: first off: if I have a list of 10 names and I input that I want max size groups of 3 persons, I get 3 groups of 3 persons and 1 of 1, but it should be 3, 3, 2, 2. I used two different functions I found on here, but both have the same problem.
Secondly, I have no idea how to make sure that in a new round there won't be any groups from previous round.
I am pretty new to programming, so any tips are welcome.
This is the first function I have:
members = group_size()
def teams(amount, size):
for i in range(0, len(amount), size):
yield amount[i:i + size]
participants = Row_list_names
random.shuffle(participants)
print("These are your groups:")
print(list(teams(participants, members)))
And this is the second:
members = group_size()
participants = Row_list_names
random.shuffle(participants)
for i in range(len(participants) // members + 1):
print('Group {} consists of:'.format(i+1))
group = participants[i*members:i*members + members]
for participant in group:
print(participant)
group_size() returns an integer number for how many people should be in the group.
For the second problem, shuffling as you do should do the trick nicely.
For the first problem, the functions are doing what you tell them to: you skip ahead and slice the list in chunks that contain exactly member participants. You do not notice that the last slice is out of bound because python is lenient on that:
>>> l = [0,1,2,3,4]
>>> l[:40]
[0, 1, 2, 3, 4]
The point is that not all groups should be of the same size:
from math import ceil
from math import floor
def split_groups(group_size, part):
# first compute the number of groups given the requested size
group_num = ceil(len(part) / group_size)
print(f"group_num {group_num}")
# compute a fractional length of the groups
group_size_frac = len(part) / group_num
print(f"group_size_frac {group_size_frac}")
total_assigned = 0
for i in range(group_num):
# get the start and end indexes using the fractional length
start = floor(i * group_size_frac)
end = floor((i + 1) * group_size_frac)
group = part[start:end]
print(f"start {start} end {end} -> {group}")
print(f"number of participants in this group {len(group)}")
total_assigned += len(group)
# check that we assigned all of the participants
assert total_assigned == len(part)
I have not tested any edge case, but a quick check by running
for group_size in range(1, 5):
for num_participants in range(10, 50):
part = list(range(num_participants))
split_groups(group_size, part)
shows that every participant was assigned to a group.
Plug in the shuffling you did before and you have random groups.
Cheers!
I am stuck at this question where I am required to update all largest and smallest numbers in the list with the average value as a way to remove the extreme values in the given list.
For example:
def remove_extreme( [ [0,4], [1,4], [-1,2] ] ) would return [ [0,1.5], [1,1.5], [1.5,2] ].
The function will need to use the average value to modify the elements in the lists, which, in this case the smallest and largest values are -1 and 4 respectively. Then, all the largest and smallest values present in the list need to be changed to the average(mean) of these two values, which is (4+-1)/2 = 1.5
Here's my code:
def remove_extreme(datalist):
for numlist in datalist:
for index, number in enumerate(numlist):
largest = max(number)
smallest = min(number)
average = (largest - smallest)/2
if number == largest or smallest:
num_list[index] = average
return datalist
May I know what went wrong? I keep getting 'int' object is not iterable.
What you asked about
To answer your immediate question, the built in functions max and min return for you the maximum and minimum number from an iterable.
https://docs.python.org/3/library/functions.html#max
So it throws a TypeError when you pass it an integer. Run it on a list/iterable instead.
But your code has more issues than just that.
Your if statement, though syntactically correct, is probably not what you want. More than likely you wanted to do this:
if number == largest or number == smallest:
Like Tomerikoo pointed out, you want to put your max and min outside the loop. As an aside, you do not need to return the list as lists are mutable and you are modifying it freely inside the function.
def remove_extreme(datalist):
for numlist in datalist:
largest = max(numlist)
smallest = min(numlist)
average = (largest - smallest)/2
for index, number in enumerate(numlist):
if number == largest or number == smallest:
numlist[index] = average
return datalist
What your problem is actually asking you
Looking at your original question I think you're a little off from the correct answer if your lists need to look like your given answer. The first hint is that your given answer shows only one of the values changed, and it's not always the average of the inner list. Take [0, 4] for instance. 1.5 is not the average of 0 and 4, yet that is what you say it should return. It seems that you are really desiring to change the most extreme number for each inner list based off the average of all the lists. Taking the average of the numbers of all the inner lists yields 1.66, so I'm not sure of this precisely, but I think one of your numbers might be off by 1 (I think so because 10/6 yields 1.66 while 9/6 yields 1.5).
If the above assumptions are all correct you will want to calculate the average (which is usually the sum/number of elements) and then find the most extreme element within each list.
Your function should look a bit more like this:
def remove_extreme(datalist):
# sum len of each list to get total number of elements
num_elements = sum([len(numlist) for numlist in datalist])
# sum the sum of each list to get total
sum_elements = sum([sum(numlist) for numlist in datalist])
# calculate average
average = sum_elements/num_elements
# find the most extreme element in each list and perform substitution
for numlist in datalist:
smallest = min(numlist)
largest = max(numlist)
large_diff = abs(largest - average)
small_diff = abs(average - smallest)
num_to_change = largest if large_diff > small_diff else smallest
for index, number in enumerate(numlist):
if number == num_to_change: # Just look at the most extreme number
numlist[index] = average
return datalist # list will be modified, but returning it doesn't hurt either
Running this function after changing your -1 to -2:
my_list = [
[0,4],
[1,4],
[-2,2]
]
print("Before: ", my_list)
remove_extreme(my_list)
print("After: ", my_list)
Output:
$ python remove_extreme.py
Before: [[0, 4], [1, 4], [-2, 2]]
After: [[0, 1.5], [1, 1.5], [1.5, 2]]
After further clarification
After clarifying what the question was really asking you, the answer is even simpler:
def remove_extreme(datalist):
flattened = [i for numlist in datalist for i in numlist] # flatten list for convenience
largest = max(flattened)
smallest = min(flattened)
average = (largest + smallest)/2
# find the most extreme element in each list and perform substitution
for numlist in datalist:
for index, number in enumerate(numlist):
if number == smallest or number == largest:
numlist[index] = average
return datalist # list will be modified, but returning it doesn't hurt either
Personally I feel like this makes less sense, but that seems to be what you're asking for.
Also, when writing a question, it's helpful to include the stack trace or point to the specific line where the issue is occurring. Just a helpful tip!
You are trying to get max and min of the element not of the list
>>> list = ( [ [0,4], [1,4], [-1,2] ] )
>>> max(list)
Output
[1, 4]
>>> min(list)
Output
[-1, 2]
I would like to constrain locations to be served by the same vehicle.
I used capacity-constraints for achieving this. Say we have l = [[1,2], [3,4]] which means that location 1, 2 must be served by the same vehicle and 3, 4 as well. So 1, 2 ends up on route_1 and 3, 4 on route_2
My code for achieving this is:
for idx, route_constraint in enumerate(l):
vehicle_capacities = [0] * NUM_VEHICLES
vehicle_capacities[idx] = len(route_constraint)
route_dimension_name = 'Same_Route_' + str(idx)
def callback(from_index):
from_node = manager.IndexToNode(from_index)
return 1 if from_node in route_constraint else 0
same_routes_callback_index = routing.RegisterUnaryTransitCallback(callback)
routing.AddDimensionWithVehicleCapacity(
same_routes_callback_index,
0, # null capacity slack
vehicle_capacities, # vehicle maximum capacities
True, # start cumul to zero
route_dimension_name)
The idea is that 1,2 have a capacity demand of each 1 unit (all others have zero). As only vehicle 1 has a capacity of 2 it is the only one able to serve 1,2.
This seems to work fine if len(l) == 1. If greater the solver is not able to find a solution if though I put into l pairs of locations which were on the same route without the above code (hence without the above capacity constraints.
Is there a more elegant way to model my requirement?
Why does the solver fail to find a solution?
I have also considered the possibility of dropping visits (at a high cost) to give the solver the possibility to start from a solution which drops visits such that it will find his way fro this point to a solution without any drops. I had no luck.
Thanks in advance.
Each stop has a vehicle var whose values determine what vehicle is allowed to visit the stop. If you want to have stops 1 and 2 serviced by vehicle 0 use a member constraint on the vehicle var of each stop and set it to [0]. Since you might have other constraints that make stops optional add the value -1 to the list. It is a special value that indicates that the stop is not serviced by a vehicle.
In Python:
n2x = index_manager.NodeToIndex
cpsolver = routing_model.solver()
for stop in [1,2]:
vehicle_var = routing_model.VehicleVar(n2x(stop))
values = [-1, 0]
cpsolver.Add(cpsolver.MemberCt(vehicle_var, values))
I have a task: given a value N. I should generate a list of length L > 1 such that the sum of the squares of its elements is equal to N.
I wrote a code:
deltas = np.zeros(L)
deltas[0] = (np.random.uniform(-N, N))
i = 1
while i < L and np.sum(np.array(deltas)**2) < N**2:
deltas[i] = (np.random.uniform(-np.sqrt(N**2 - np.sum(np.array(deltas)**2)),\
np.sqrt(N**2 - np.sum(np.array(deltas)**2))))
i += 1
But this approach takes long time, if I generate such list many times. (I think because of loop).
Note, that I don't want my list to consist of just one unique value. The distribution of values does not have to be uniform - I took uniform just for example.
Could you suggest any faster approach? May be there is special function in any lib?
If you didn't mind a few repeating 1s, you could do something like this:
def square_list(integer):
components = []
total = 0
remaining = integer
while total != integer:
component = int(remaining ** 0.5)
remaining -= component ** 2
components.append(component)
total = sum([x ** 2 for x in components])
return components
This code works by finding the taking the largest square, and then decreasing to the next largest square. It continues until the largest square is 1, which could at worse result in 3 1s in a list.
If you are looking for a more random distribution, it might make sense to randomly transform remaining as a separate variable before subtracting from it.
IE:
value = transformation(remaining)
component = int(value ** 0.5)
which should give you more "random" values.
We're doing the classic problem of determining the number of ways that we can make change that amounts to Z given a set of coins.
For example, Amount=5 and Coins={1, 2, 3}. One way we can make 5 is {2, 3}.
The naive recursive solution has a time complexity of factorial time.
f(n) = n * f(n-1) = n!
My professor argued that it actually has a time complexity of O(2^n), because we only choose to use a coin or not. That intuitively makes sense. However how come my recurence doesn't work out to be O(2^n)?
EDIT:
My recurrence is as follows:
f(5, {1, 2, 3})
/ \ .....
f(4, {2, 3}) f(3, {1, 3}) .....
Notice how the branching factor decreases by 1 at every step.
Formally.
T(n) = n*F(n-1) = n!
The recurrence doesn't work out to what you expect it to work out to because it doesn't reflect the number of operations made by the algorithm.
If the algorithm decides for each coin whether to output it or not, then you can model its time complexity with the recurrence T(n) = 2*T(n-1) + O(1) with T(1)=O(1); the intuition is that for each coin you have two options---output the coin or not; this obviously solves to T(n)=O(2^n).
I too was trying to analyze the time complexity for the brute force which performs depth first search:
def countCombinations(coins, n, amount, k=0):
if amount == 0:
return 1
res = 0
for i in range(k, n):
if coins[k] <= amount:
remaining_amount = amount - coins[i] # considering this coin, try for remaining sum
# in next round include this coin too
res += countCombinations(coins, n, remaining_amount, i)
return res
but we can see that the coins which are used in one round is used again in the next round, so at least for 1st coin we have n items at each stage which is equivalent to permutation with repetition n^r for n items available to arrange into r positions at each stage.
ex: [1, 1, 1, 1]; sum = 4
This will generate a recursive tree where for first path we literally have solutions at each diverged subpath until we have the sum=0. so the time complexity is O(sum^n) ie for each stage in the path towards sum we have n different subpaths.
Note however there is another algorithm which uses take/not-take approach and at most there is 2 branch at a node in recursion tree. Hence the time complexity for this algorithm is O(2^(n*m))
ex: say coins = [1, 1] sum = 2 there are 11 nodes/points to visit in the recursion tree for 6 paths(leaves) then complexity is at most 2^(2*2) => 2^4 => 16 (Hence 11 nodes visiting for a max of 16 possibility is correct but little loose on upper bound).
def get_count(coins, n, sum):
if(n == 0): # no coins left, to try a combination that matches the sum
return 0
if(sum == 0): # no more sum left to match, means that we have completely co-incided with our trial
return 1 # (return success)
# don't-include the last coin in the sum calc so, leave it and try rest
excluded = get_count(coins, n-1, sum)
included = 0
if(coins[n-1] <= sum):
# include the last coin in the sum calc, so reduce by its quantity in the sum
# we assume here that n is constant ie, it is supplied in unlimited(we can choose same coin again and again),
included = get_count(coins, n, sum-coins[n-1])
return included+excluded