Given an NxM array of positive integers, how would one go about selecting integers so that the maximum sum of values is achieved where there is a maximum of x selections in each row and y selections in each column. This is an abstraction of a problem I am trying to face in making NCAA swimming lineups. Each swimmer has a time in every event that can be converted to an integer using the USA Swimming Power Points Calculator the higher the better. Once you convert those times, I want to assign no more than 3 swimmers per event, and no more than 3 races per swimmer such that the total sum of power scores is maximized. I think this is similar to the Weapon-targeting assignment problem but that problem allows a weapon type to attack the same target more than once (in my case allowing a single swimmer to race the same event twice) and that does not work for my use case. Does anybody know what this variation on the wta problem is called, and if so do you know of any solutions or resources I could look to?
Here is a mathematical model:
Data
Let a[i,j] be the data matrix
and
x: max number of selected cells in each row
y: max number of selected cells in each column
(Note: this is a bit unusual: we normally reserve the names x and y for variables. These conventions can help with readability).
Variables
δ[i,j] ∈ {0,1} are binary variables indicating if cell (i,j) is selected.
Optimization Model
max sum((i,j), a[i,j]*δ[i,j])
sum(j,δ[i,j]) ≤ x ∀i
sum(i,δ[i,j]) ≤ y ∀j
δ[i,j] ∈ {0,1}
This can be fed into any MIP solver.
Related
I want to create a dynamic array that returns me X values based on given probabilities. For instance:
Imagine this is a gift box and you can open the box N times. What I want is to have N random results. For example, I want to get randomly 5 of these two rarities but based on their chances.
I have this following formula for now:
=index(A2:A3,randarray(5,1,1,rows(A2:A3),1). And this is the output I get:
The problem here is that I have a dynamic array with the 5 results BUT NOT BASED ON THE PROBABILITIES.
How can I add probabilities to the array?
Here is how you could generate a random outcome with defined probabilities for the entries (Google Sheets solution, not sure about Excel):
=ARRAYFORMULA(
VLOOKUP(
RANDARRAY(H1, 1),
{
{0; OFFSET(C2:C,,, COUNTA(C2:C) - 1)},
OFFSET(A2:A,,, COUNTA(C2:C))
},
2
)
)
This whole subject of random selection was treated very thoroughly in Donald Knuth's series of books, The Art of Computer Programming, vol 2, "Semi-Numerical Algorithms". In that book he presents an algorithm for selecting exactly X out of N items in a list using pseudo-random numbers. What you may not have considered is that after you have chosen your first item the probability array has changed to (X-1)/(N-1) if your first outcome was "Normal" or X/(N-1) if your first outcome was "Rare". This means you'll want to keep track of some running totals based on your prior outcomes to ensure your probabilities are dynamically updated with each pick. You can do this with formulas, but I'm not certain how the back-reference will perform inside an array formula. Microsoft's dynamic array documentation indicates that such internal array references are considered "circular" and are prohibited.
In any case, trying to extend this to 3+ outcomes is very problematic. In order to implement that algorithm with 3 choices (X + Y + Z = N picks) you would need to break this up into one random number for an X or not X choice and then a second random number for a Y or not Y choice. This becomes a recursive algorithm, beyond Excel's ability to cope in formulas.
Introduction
I have written code to give me a set of numbers in '36 by q' format ( 1<= q <= 36), subject to following conditions:
Each row must use numbers from 1 to 36.
No number must repeat itself in a column.
Method
The first row is generated randomly. Each number in the coming row is checked for the above conditions. If a number fails to satisfy one of the given conditions, it doesn't get picked again fot that specific place in that specific row. If it runs out of acceptable values, it starts over again.
Problem
Unlike for low q values (say 15 which takes less than a second to compute), the main objective is q=36. It has been more than 24hrs since it started to run for q=36 on my PC.
Questions
Can I predict the time required by it using the data I have from lower q values? How?
Is there any better algorithm to perform this in less time?
How can I calculate the average number of cycles it requires? (using combinatorics or otherwise).
Can I predict the time required by it using the data I have from lower q values? How?
Usually, you should be able to determine the running time of your algorithm in terms of input. Refer to big O notation.
If I understood your question correctly, you shouldn't spend hours computing a 36x36 matrix satisfying your conditions. Most probably you are stuck in the infinite loop or something. It would be more clear of you could share code snippet.
Is there any better algorithm to perform this in less time?
Well, I tried to do what you described and it works in O(q) (assuming that number of rows is constant).
import random
def rotate(arr):
return arr[-1:] + arr[:-1]
y = set([i for i in range(1, 37)])
n = 36
q = 36
res = []
i = 0
while i < n:
x = []
for j in range(q):
if y:
el = random.choice(list(y))
y.remove(el)
x.append(el)
res.append(x)
for j in range(q-1):
x = rotate(x)
res.append(x)
i += 1
i += 1
Basically, I choose random numbers from the set of {1..36} for the i+q th row, then rotate the row q times and assigned these rotated rows to the next q rows.
This guarantees both conditions you have mentioned.
How can I calculate the average number of cycles it requires?( Using combinatorics or otherwise).
I you cannot calculate the computation time in terms of input (code is too complex), then fitting to curve seems to be right.
Or you could create an ML model with iterations as data and time for each iteration as label and perform linear regression. But that seems to be overkill in your example.
Graph q vs time
Fit a curve,
Extrapolate to q = 36.
You might want to also graph q vs log(time) as that may give an easier fitted curve.
I’ve used Excel in the past but the calculations including the Poisson-Distribution took a while, that’s why I switched to SQL. Soon I’ve recognized that SQL might not be a proper solution to deal with statistical issues. Finally I’ve decided to switch to Matlab but I’m not used to it at all, my problem Is the following:
I’ve imported a .csv-table and have two columns with values, let’s say A and B (110 x 1 double)
These values both are the input values for my Poisson-calculations. Since I wanna calculate for at least the first 20 events, I’ve created a variable z=1:20.
When I now calculated let’s say
New = Poisspdf(z,A),
it says something like non-scalar arguments must match in size.
Z only has 20 records but A and l both have 110 records. So I’ve expanded Z= 1:110 and transposed it:
Znew = Z.
When I now try to execute the actual calculation:
Results = Poisspdf(Znew,A).*Poisspdf(Znew,B)
I always get only a 100x1 Vector but what I want is a matrix that is 20x20 for each record of A and B (based on my actual choice of z=1:20, I only changed to z=1:110 because Matlab told that they need to match in size).
So in this 20x20 Matrix there should always be in each cell the result of a slightly different calculation (Poisspdf(Znew,A).*Poisspdf(Znew,B)).
For example in the first cell (1,1) I want to have the result of
Poisspdf(0,value of A).*Poisspdf(0,value of B),
in cell(1,2): Poisspdf(0,value of A).*Poisspdf(1,value of B),
in cell(2,1): Poisspdf(1,value of A).*Poisspdf(0,value of B),
and so on...assuming that it’s in the Format cell(row, column)
Finally I want to sum up certain parts of each 20x20 matrix and show the result of the summed up parts in new columns.
Is there anybody able to help? Many thanks!
EDIT:
Poisson Matrix in Excel
In Excel there is Poisson-function: POISSON(x, μ, FALSE) = probability density function value f(x) at the value x for the Poisson distribution with mean μ.
In e.g. cell AD313 in the table above there is the following calculation:
=POISSON(0;first value of A;FALSE)*POISSON(0;first value of B;FALSE)
, in cell AD314
=POISSON(1;first value of A;FALSE)*POISSON(0;first value of B;FALSE)
, in cell AE313
=POISSON(0;first value of A;FALSE)*POISSON(1;first value of B;FALSE)
, and so on.
I am not sure if I completely understand your question. I wrote this code that might help you:
clear; clc
% These are the lambdas parameters for the Poisson distribution
lambdaA = 100;
lambdaB = 200;
% Generating Poisson data here
A = poissrnd(lambdaA,110,1);
B = poissrnd(lambdaB,110,1);
% Get the first 20 samples
zA = A(1:20);
zB = B(1:20);
% Perform the calculation
results = repmat(poisspdf(zA,lambdaA),1,20) .* repmat(poisspdf(zB,lambdaB)',20,1);
% Sum
sumFinal = sum(results,2);
Let me know if this is what you were trying to do.
Excel
Need to find nearest float in a table, for each integer 0..99
https://www.excel-easy.com/examples/closest-match.html explains a great technique for finding the CLOSEST number from an array to a constant cell.
I need to perform this for many values (specifically, find nearest to a vertical list of integers 0..99 from within a list of floats).
Array formulas don't allow the compare-to value (integers) to change as we move down the list of integers, it treats it like a constant location.
I tried Tables, referring to the integers (works) but the formula from the above web site requires an Array operation (F2, control shift Enter), which are not permitted in Tables. Correction: You can enter the formula, control-enter the array function for one cell, copy the formulas, then insert table. Don't change the search cell reference!
Update:
I can still use array operations, but I manually have to copy the desired function into each 100 target cells. No biggie.
Fixed typo in formula. See end of question for details about "perfection".
Example code:
AI4=some integer
AJ4=MATCH(MIN(ABS(Table[float_column]-AI4)), ABS(Table[float_column]-AI4), 0)
repeat for subsequent integers in AI5...AI103
Example data:
0.1 <= matches 0
0.5
0.95 <= matches 1
1.51 <= matches 2
2.89
Consider the case where target=5, and 4.5, 5.5 exist in the list. One gives -0.5 and the other +0.5. Searching for ABS(-.5) will give the first one. Either one is decent, unless your data is non-monotonic.
This still needs a better solution.
Thanks in advance!
I had another problem, which pushed to a better solution.
Specifically, since the Y values for the X that I am interested in can be at varying distances in X, I will interpolate X between the X point before and after. Ie search for less than or equal, also greater than or equal, interpolate the desired X, then interpolate the Y values.
I could go a step further and interpolate N - 1 to N + 1, which will give cleaner results for noisy data.
Would it be reasonable to systematically try all possible placements in a word search?
Grids commonly have dimensions of 15*15 (15 cells wide, 15 cells tall) and contain about 15 words to be placed, each of which can be placed in 8 possible directions. So in general it seems like you can calculate all possible placements by the following:
width*height*8_directions_to_place_word*number of words
So for such a grid it seems like we only need to try 15*15*8*15 = 27,000 which doesn't seem that bad at all. I am expecting some huge number so either the grid size and number of words is really small or there is something fishy with my math.
Formally speaking, assuming that x is number of rows and y is number of columns you should sum all the probabilities of every possible direction for every possible word.
Inputs are: x, y, l (average length of a word), n (total words)
so you have
horizontally a word can start from 0 to x-l and going right or from l to x going left for each row: 2x(x-l)
same approach is used for vertical words: they can go from 0 to y-l going down or from l to y going up. So it's 2y(y-l)
for diagonal words you shoul consider all possible start positions x*y and subtract l^2 since a rect of the field can't be used. As before you multiply by 4 since you have got 4 possible directions: 4*(x*y - l^2).
Then you multiply the whole result for the number of words included:
total = n*(2*x*(x-l)+2*y*(y-l)+4*(x*y-l^2)