Imagine I have person 1, 2, 3 and 4, then I have shirt styles A, B, C, D and I want to distribute the shirt styles to the people such that 25% of them get style A, 25% get style B, 25% get style C and 25% get style D but some of the people refuse to wear certain styles, these people are represented by Fs. How can I randomly match all the people with the styles they are willing to wear to get the approximate distribution?
A B C D
1 T F T T
2 T F F F
3 T T T T
4 T T T F
In this case this is easy and 25% is can be fully achieved, just give each person a different style. However, I intend to take this problem beyond this simple situation, my solution has to be generic. The number or styles, the number of people, and the distribution is all variable. Sometimes, the distribution will be impossible to create 100% accurately, approximate/close/best effor is expected. The selection process should be random and attempt to maintain the distribution.
I'm pretty agnostic to the language here, I'm just seeking the algorithm. Though preferably it would be able to be distributed.
Finding a solution when you are hampered by the Fs is https://en.wikipedia.org/wiki/Assignment_problem. One way to select an arbitrary assignment when there are many would be to set random costs where a style is acceptable to a person and then let it find the assignment with lowest possible cost. However it is not obvious that this will fit any natural definition of random. One (very inefficient) natural definition of random would be to select from all possible assignments at random until you get one that is acceptable to everybody. The distribution you get from this might not be the same as the one you would get by setting up random costs and then solving the resulting assignment problem.
You are using the term 'randomly match' which should be used with caution. The correct interpretation, I believe, is a random selection from the set of all valid solutions, so basically if we could enumerate all valid solution - we could trivially solve the problem.
You are looking for a close-enough solution, so we need to better define what a valid solution is. I suggest to define some threshold (say 1% error at most).
In your example, there are 4 groups (those assigned with shirt style A/B/C/D). Therefore, there are 2^4-1 possible person archetypes (love/hate each of A/B/C/D, with the assumption that anyone loves at least one shirt style). Each archetype population has a given population size, and each archetype can be assigned to some of the 4 groups (1 or more).
The goal is to divide the population of each archetype between the 4 groups, such that say, the size of each group is between L and H.
Lets formalize it.
Problem statement:
Denote A(0001),...,A(1111): the population size of each of the 15 archetypes
Denote G1(0001): the size of A(0001) assigned to G1, etc.
Given
L,H: constants
A(0001),...,A(1111): 15 constants
Our goal is to find all integer solutions for
G1(0001),G1(0011),G1(0101),G1(0111),G1(1001),G1(1011),G1(1101),G1(1111),
G2(0010),G2(0011),G2(0110),G2(0111),G2(1010),G2(1011),G2(1110),G2(1111),
G3(0100),G3(0101),G3(0110),G3(0111),G3(1100),G3(1101),G3(1110),G3(1111),
G4(1000),G4(1001),G4(1010),G4(1011),G4(1100),G4(1101),G4(1110),G4(1111)
subject to:
G1(0001) = A(0001)
G2(0010) = A(0010)
G2(0011) + G1(0011) = A(0011)
G3(0100) = A(0100)
G3(0101) + G1(0101) = A(0101)
G3(0110) + G2(0110) = A(0110)
G3(0111) + G2(0101) + G1(0101) = A(0111)
G4(1000) = A(1000)
G4(1001) + G1(1001) = A(1001)
G4(1010) + G2(1010) = A(1010)
G4(1011) + G2(1011) + G1(1011) = A(1011)
G4(1100) + G3(1100) = A(1100)
G4(1101) + G3(1101) + G1(1101) = A(1101)
G4(1110) + G3(1110) + G2(1110) = A(1110)
G4(1111) + G3(1111) + G2(1111) + G1(1111) = A(1111)
L < G1(0001)+G1(0011)+G1(0101)+G1(0111)+G1(1001)+G1(1011)+G1(1101)+G1(1111) <
H
L < G2(0010)+G2(0011)+G2(0110)+G2(0111)+G2(1010)+G2(1011)+G2(1110)+G2(1111) <
H
L < G3(0100)+G3(0101)+G3(0110)+G3(0111)+G3(1100)+G3(1101)+G3(1110)+G3(1111) <
H
L < G4(1000)+G4(1001)+G4(1010)+G4(1011)+G4(1100)+G4(1101)+G4(1110)+G4(1111) <
H
Now we can use an integer programming solver for the job.
Related
I am trying to find a solution in which a given resource (eg. budget) will be best distributed to different options which yields different results on the resource provided.
Let's say I have N = 1200 and some functions. (a, b, c, d are some unknown variables)
f1(x) = a * x
f2(x) = b * x^c
f3(x) = a*x + b*x^2 + c*x^3
f4(x) = d^x
f5(x) = log x^d
...
And also, let's say there n number of these functions that yield different results based on its input x, where x = 0 or x >= m, where m is a constant.
Although I am not able to find exact formula for the given functions, I am able to find the output. This means that I can do:
X = f1(N1) + f2(N2) + f3(N3) + ... + fn(Nn) where (N1 + ... Nn) = N as many times as there are ways of distributing N into n numbers, and find a specific case where X is the greatest.
How would I actually go about finding the best distribution of N with the least computation power, using whatever libraries currently available?
If you are happy with allocations constrained to be whole numbers then there is a dynamic programming solution of cost O(Nn) - so you can increase accuracy by scaling if you want, but this will increase cpu time.
For each i=1 to n maintain an array where element j gives the maximum yield using only the first i functions giving them a total allowance of j.
For i=1 this is simply the result of f1().
For i=k+1 consider when working out the result for j consider each possible way of splitting j units between f_{k+1}() and the table that tells you the best return from a distribution among the first k functions - so you can calculate the table for i=k+1 using the table created for k.
At the end you get the best possible return for n functions and N resources. It makes it easier to find out what that best answer is if you maintain of a set of arrays telling the best way to distribute k units among the first i functions, for all possible values of i and k. Then you can look up the best allocation for f100(), subtract off the value this allocated to f100() from N, look up the best allocation for f99() given the resulting resources, and carry on like this until you have worked out the best allocations for all f().
As an example suppose f1(x) = 2x, f2(x) = x^2 and f3(x) = 3 if x>0 and 0 otherwise. Suppose we have 3 units of resource.
The first table is just f1(x) which is 0, 2, 4, 6 for 0,1,2,3 units.
The second table is the best you can do using f1(x) and f2(x) for 0,1,2,3 units and is 0, 2, 4, 9, switching from f1 to f2 at x=2.
The third table is 0, 3, 5, 9. I can get 3 and 5 by using 1 unit for f3() and the rest for the best solution in the second table. 9 is simply the best solution in the second table - there is no better solution using 3 resources that gives any of them to f(3)
So 9 is the best answer here. One way to work out how to get there is to keep the tables around and recalculate that answer. 9 comes from f3(0) + 9 from the second table so all 3 units are available to f2() + f1(). The second table 9 comes from f2(3) so there are no units left for f(1) and we get f1(0) + f2(3) + f3(0).
When you are working the resources to use at stage i=k+1 you have a table form i=k that tells you exactly the result to expect from the resources you have left over after you have decided to use some at stage i=k+1. The best distribution does not become incorrect because that stage i=k you have worked out the result for the best distribution given every possible number of remaining resources.
SO recently, I have been attempting to solve a code challenge and can not find the answer. The issue is not the implementation, but rather what to implement. The prompt can be found here http://pastebin.com/DxQssyKd
the main useful information from the prompt is as follows
"Write a function answer(w, h, s) that takes 3 integers and returns the number of unique, non-equivalent configurations that can be found on a star grid w blocks wide and h blocks tall where each celestial body has s possible states. Equivalency is defined as above: any two star grids with each celestial body in the same state where the actual order of the rows and columns do not matter (and can thus be freely swapped around). Star grid standardization means that the width and height of the grid will always be between 1 and 12, inclusive. And while there are a variety of celestial bodies in each grid, the number of states of those bodies is between 2 and 20, inclusive. The answer can be over 20 digits long, so return it as a decimal string."
The equivalency is in a way that
00
01
is equivalent to
01
00
and so on.
The problem is, what algorithm(s) should I use? i know this is somewhat related to permutations, combinations, and group theory, but I can not find anything specific.
The key weapon is Burnside's lemma, which equates the number of orbits of the symmetry group G = Sw × Sh acting on the set of configurations X = ([w] × [h] → [s]) (i.e., the answer) to the sum 1/|G| ∑g∈G |Xg|, where Xg = {x | g.x = x} is the set of elements fixed by g.
Given g, it's straightforward to compute |Xg|: use g to construct a graph on vertices [w] × [h] where there is an edge between (i, j) and g(i, j) for all (i, j). Count c, the number of connected components, and return sc. The reasoning is that every vertex in a connected component must have the same state, but vertices in different components are unrelated.
Now, for 12 × 12 grids, there are far too many values of g to do this calculation on. Fortunately, when g and g' are conjugate (i.e., there exists some h such that h.g.h-1 = g') we find that |Xg'| = |{x | g'.x = x}| = |{x | h.g.h-1.x = x}| = |{x | g.h-1.x = h-1.x}| = |{h.y | g.y = y}| = |{y | g.y = y}| = |Xg|. We can thus sum over conjugacy classes and multiply each term by the number of group elements in the class.
The last piece is the conjugacy class structure of G = Sw × Sh. The conjugacy class structure of this direct product is really just the direct product of the conjugacy classes of Sw and Sh. The conjugacy classes of Sn are in one-to-one correspondence with integer partitions of n, enumerable by standard recursive methods. To compute the size of the class, you'll divide n! by the product of the partition terms (because circular permutations of the cycles are equivalent) and also by the product of the number of symmetries between cycles of the same size (product of the factorials of the multiplicities). See https://groupprops.subwiki.org/wiki/Conjugacy_class_size_formula_in_symmetric_group.
Im trying to determine the "difficultly" of a quiz object.
My ultimate goal is to be able to create a "difficulty score" (DS) for any quiz. This would allow me to compare one quiz to another accurately, despite being made up of different questions/answers.
When creating my quiz object, I assign each question a "difficulty index" (DI), which is number on a scale from 1-15.
15 = most difficult
1 = least difficult
Now a strait forward way to measure this "difficulty score" could be to add up each question's "difficulty index" then divide by maximum possible "difficulty index" for the quiz. ( ex. 16/30 = 53.3% Difficulty )
However, I also have multiple "weighting" properties associated to each question. These weights are again one a scale of 1-5.
5 = most impact
1 = least impact
The reason I have (2) instead of the more common (1) is so I can accommodate a scenario as follows...
If presenting the student with a very difficult question (DI=15) and the student answers "incorrect", don't have it hurt their score so much BUT if they get it "correct" have it improve their score greatly. I call these my "positive" (PW) and "negative" (NW) weights.
Quiz Example A:
Question 1: DI = 1 | PW = 3 | NW = 3
Question 2: DI = 1 | PW = 3 | NW = 3
Question 3: DI = 1 | PW = 3 | NW = 3
Question 4: DI = 15 | PW = 5 | NW = 1
Quiz Example B:
Question 1: DI = 1 | PW = 3 | NW = 3
Question 2: DI = 1 | PW = 3 | NW = 3
Question 3: DI = 1 | PW = 3 | NW = 3
Question 4: DI = 15 | PW = 1 | NW = 5
Technically the above two quizzes are very similar BUT Quiz B should be more "difficult" because the hardest question will have the greatest impact on your score if you get it wrong.
My question now becomes how can I accurately determine the "difficulty score" when considering the complex weighting system?
Any help is greatly appreciated!
The challenge of course is to determine the difficulty score for each single question.
I suggest the following model:
Hardness (H): Define a hard question such that chances of answering it correctly are lower. The hardest question is such that (1) the chance of answering it correctly are equal to random choice (because it is inherently very hard), and (2) it has the largest number of possible answers. We'll define such question as (H = 15). On the other end of the scale, we'll define (H = 0) for a question where the chances of answering it correctly are 100% (because it is trivial) (I know - such question will never appear). Now - define the hardness of each question by subjective extrapolation (remember that one can always guess between the given options). For example, if a (H = 15) question has 4 answers, and another question with similar inherent hardness has 2 answers - it would be (H = 7.5). Another example: If you believe that an average student has 62.5% of answering a question correctly - it would also be a (H = 7.5) question (this is because a H = 15 has 25% of correct answer, while H = 0 has 100%. The average is 62.5%)
Effect (E): Now, we'll measure the effect of PW and NW. For questions with 50% chance of answering correctly - the effect is E = 0.5*PW - 0.5*NW. For questions with 25% chance of answering correctly - the effect is E = 0.25*PW - 0.75*NW. For trivial question NW doesn't matter so the effect is E = PW.
Difficulty (DI): The last step is to integrate the hardness and the effect - and call it difficulty. I suggest DI = H - c*E, where c is some positive constant. You may want to normalize again.
Edit: Alternatively, you may try the following formula: DI = H * (1 - c*E), where the effect magnitude is not absolute, but relative to the question's hardness.
Clarification:
The teacher needs to estimate only one parameter about each question: What is the probability that an average student would answer this question correctly. His estimation, e, will be in the range [1/k, 1], where k is the number of answers.
The hardness, H, is a linear function of e such that 1/k is mapped to 15 and 1 is mapped to 0. The function is: H = 15 * k / (k-1) * (1-e)
The effect E depends on e, PW and NW. The formula is E = e*PW - (1-e)*NW
Example based on OP comments:
Question 1:
k = 4, e = 0.25 (hardest). Therefore H = 15
PW = 1, NW = 5, e = 0.25. Therefore E = 0.25*1 - 0.75*5 = -3.5
c = 5. DI = 15 - 5*(-3.5) = 32.5
Question 2:
k = 4, e = 0.95 (very easy). Therefore H = 1
PW = 1, NW = 5, e = 0.95. Therefore E = 0.95*1 - 0.05*5 = 0.7
c = 5. DI = 1 - 5*(0.7) = -2.5
I'd say the core of the problem is that mathematically your example quizzes A and B are identical, except that quiz A awards the student 4 gratuitous bonus points (or, equivalently, quiz B arbitrarily takes 4 points away from them). If the same students take both of them, the score distribution will be the same, except shifted by 4 points. So while the two quizzes may feel different psychologically (because, let's face it, getting extra points feels good, and losing points feels bad, even if you technically did nothing to deserve it), finding an objective way to distinguish them seems tricky.
That said, one reasonable measure of "psychological difficulty" could simply be the average score (per question) that a randomly chosen student would be expected to get from the quiz. Of course, that's not something you can reliably calculate in advance, although you could estimate it from actual quiz results after the fact.
However, if you could somehow relate your (presumably arbitrary) difficulty ratings to the fraction of students likely to answer the question correctly, then you could use that to estimate the expected average score. So, for example, we might simply assume a linear relationship with the question difficulty as the success rate, with difficulty 1 corresponding to a 100% expected success rate, and difficulty 15 corresponding to a 0% expected success rate. Then the expected average score S per question for the quiz could be calculated as:
S = avg(PW × X − NW × (1 − X))
where the average is taken over all questions in the quiz, and where PW and NW are the point weights for a correct and an incorrect answer respectively, DI below is the difficulty rating for the question, and X = (15 − DI) / 14 is the estimated success rate.
Of course, we might want to also account for the fact that, even if a student doesn't know the answer to a question, they can still guess. Basically this means that the estimated success rate X should not range from 0 to 1, but from 1/N to 1, where N is the number of options for the question. So, taking that into account, we can adjust the formula for X to be:
X = (1 + (N − 1) × (15 − DI) / 14) / N
One problem with this estimated average score S as a difficulty measure is that it isn't bounded in either direction, and provides no obvious scale to indicate what counts as an "easy" quiz or a "hard" one. The fundamental problem here is that you haven't specified any limits for the question weights, so there's technically nothing to stop someone from making a question with, say, a positive or negative weight of one million points.
That said, if you do impose some reasonable limits on the weights (even if they're only recommendations), then you should be able to also establish reasonable thresholds on S for a quiz to be considered e.g. easy, moderate or hard. And even if you don't, you can still at least use it to rank quizzes relative to each other by difficulty.
Ps. One way to present the expected score in a UI might be to multiply it by the number of questions in the quiz, and display the result as "par" for the quiz. That way, students could roughly judge their own performance against the difficulty of the quiz by seeing whether they scored above or below par.
I edited my question trying to make it as short and precise.
I am developing a prototype of a facial recognition system for my Graduation Project. I use Eigenface and my main source is the document Turk and Pentland. It is available here: http://www.face-rec.org/algorithms/PCA/jcn.pdf.
My doubts focus on step 4 and 5.
I can not correctly interpret the number of thresholds: If two types of thresholds, or only one (Notice that the text speaks of two types but uses the same symbol). And again, my question is whether this (or these) threshold(s) is unique and global for all person or if each person has their own default.
I understand the steps to be calculated until an matrix O() of classes with weights or weighted. So this matrix O() is of dimension M'x P. Since M' equal to the amount of eigenfaces chosen and P the number of people.
What follows and confuses me. He speaks of two distances: the distance of a class against another, and also from a distance of one face to another. I call it D1 and D2 respectively. NOTE: In the training set there are M images in total, with F = M / P the number of images per person.
I understand that threshold(s) should be chosen empirically. But there must be a way to approximate. I was initially designing a matrix of distances D1() of dimension PxP. Where the row vector D(i) has the distances from the vector average class O(i) to each O(j), j = 1..P. Ie a "all vs all."
Until I came here, and what follows depends on whether I should actually choose a single global threshold for all. Or if I should be chosen for each individual value. Also not if they are 2 types: one for distance classes, and one for distance faces.
I have a theory as could proceed but not so supported by the concepts of Turk:
Stage Pre-Test:
Gender two matrices of distances D1 and D2:
In D1 would be stored distances between classes, and in D2 distances between faces. This basis of the matrices W and A respectively.
Then, as indeed in the training set are P people, taking the F vectors columns D1 for each person and estimate a threshold T1 was in range [Min, Max]. Thus I will have a T1(i), i = 1..P
Separately have a T2 based on the range [Min, Max] out of all the matrix D2. This define is a face or not.
Step Test:
Buid a test set of image with a 1 image for each known person
Itest = {Itest(1) ... Itest(P)}
For every image Itest(i) test:
Calculate the space face Atest = Itest - Imean
Calculate the weight vector Otest = UT * Atest
Calculating distances:
dist1(j) = distance(Otest, O (j)), j = 1..P
Af = project(Otest, U)
dist2 = distance(Atest, Af)
Evaluate recognition:
MinDist = Min(dist1)
For each j = 1..P
If dist2 > T2 then "not is face" else:
If MinDist <= T1(j) then "Subject identified as j" else "subject unidentified"
Then I take account of TFA and TFR and repeat the test process with different threshold values until I find the best approach gives to each person.
Already defined thresholds can put the system into operation unknown images. The algorithm is similar to the test.
I know I get out of "script" of the official documentation but at least this reasoning is the most logical place my head. I wondered if I could give guidance.
EDIT:
i No more to say that has not already been said and that may help clarify things.
Could anyone tell me if I'm okay tackled with my "theory"? I'm moving into my project, and if this is not the right way would appreciate some guidance and does not work and you wrong.
I'm playing a game that has a weapon-forging component, where you combine two weapons to get a new one. The sheer number of weapon combinations (see "6.1. Blade Combination Tables" at http://www.gamefaqs.com/ps/914326-vagrant-story/faqs/8485) makes it difficult to figure out what you can ultimately create out of your current weapons through repeated forging, so I tried writing a program that would do this for me. I give it a list of weapons that I currently have, such as:
francisca
tabarzin
kris
and it gives me the list of all weapons that I can forge:
ball mace
chamkaq
dirk
francisca
large crescent
throwing knife
The problem is that I'm using a brute-force algorithm that scales extremely poorly; it takes about 15 seconds to calculate all possible weapons for seven starting weapons, and a few minutes to calculate for eight starting weapons. I'd like it to be able to calculate up to 64 weapons (the maximum that you can hold at once), but I don't think I'd live long enough to see the results.
function find_possible_weapons(source_weapons)
{
for (i in source_weapons)
{
for (j in source_weapons)
{
if (i != j)
{
result_weapon = combine_weapons(source_weapons[i], source_weapons[j]);
new_weapons = array();
new_weapons.add(result_weapon);
for (k in source_weapons)
{
if (k != i && k != j)
new_weapons.add(source_weapons[k]);
}
find_possible_weapons(new_weapons);
}
}
}
}
In English: I attempt every combination of two weapons from my list of source weapons. For each of those combinations, I create a new list of all weapons that I'd have following that combination (that is, the newly-combined weapon plus all of the source weapons except the two that I combined), and then I repeat these steps for the new list.
Is there a better way to do this?
Note that combining weapons in the reverse order can change the result (Rapier + Firangi = Short Sword, but Firangi + Rapier = Spatha), so I can't skip those reversals in the j loop.
Edit: Here's a breakdown of the test example that I gave above, to show what the algorithm is doing. A line in brackets shows the result of a combination, and the following line is the new list of weapons that's created as a result:
francisca,tabarzin,kris
[francisca + tabarzin = chamkaq]
chamkaq,kris
[chamkaq + kris = large crescent]
large crescent
[kris + chamkaq = large crescent]
large crescent
[francisca + kris = dirk]
dirk,tabarzin
[dirk + tabarzin = francisca]
francisca
[tabarzin + dirk = francisca]
francisca
[tabarzin + francisca = chamkaq]
chamkaq,kris
[chamkaq + kris = large crescent]
large crescent
[kris + chamkaq = large crescent]
large crescent
[tabarzin + kris = throwing knife]
throwing knife,francisca
[throwing knife + francisca = ball mace]
ball mace
[francisca + throwing knife = ball mace]
ball mace
[kris + francisca = dirk]
dirk,tabarzin
[dirk + tabarzin = francisca]
francisca
[tabarzin + dirk = francisca]
francisca
[kris + tabarzin = throwing knife]
throwing knife,francisca
[throwing knife + francisca = ball mace]
ball mace
[francisca + throwing knife = ball mace]
ball mace
Also, note that duplicate items in a list of weapons are significant and can't be removed. For example, if I add a second kris to my list of starting weapons so that I have the following list:
francisca
tabarzin
kris
kris
then I'm able to forge the following items:
ball mace
battle axe
battle knife
chamkaq
dirk
francisca
kris
kudi
large crescent
scramasax
throwing knife
The addition of a duplicate kris allowed me to forge four new items that I couldn't before. It also increased the total number of forge tests to 252 for a four-item list, up from 27 for the three-item list.
Edit: I'm getting the feeling that solving this would require more math and computer science knowledge than I have, so I'm going to give up on it. It seemed like a simple enough problem at first, but then, so does the Travelling Salesman. I'm accepting David Eisenstat's answer since the suggestion of remembering and skipping duplicate item lists made such a huge difference in execution time and seems like it would be applicable to a lot of similar problems.
Start by memoizing the brute force solution, i.e., sort source_weapons, make it hashable (e.g., convert to a string by joining with commas), and look it up in a map of input/output pairs. If it isn't there, do the computation as normal and add the result to the map. This often results in big wins for little effort.
Alternatively, you could do a backward search. Given a multiset of weapons, form predecessors by replacing one of the weapon with two weapons that forge it, in all possible ways. Starting with the singleton list consisting of the singleton multiset consisting of the goal weapon, repeatedly expand the list by predecessors of list elements and then cull multisets that are supersets of others. Stop when you reach a fixed point.
If linear programming is an option, then there are systematic ways to prune search trees. In particular, let's make the problem easier by (i) allowing an infinite supply of "catalysts" (maybe not needed here?) (ii) allowing "fractional" forging, e.g., if X + Y => Z, then 0.5 X + 0.5 Y => 0.5 Z. Then there's an LP formulation as follows. For all i + j => k (i and j forge k), the variable x_{ijk} is the number of times this forge is performed.
minimize sum_{i, j => k} x_{ijk} (to prevent wasteful cycles)
for all i: sum_{j, k: j + k => i} x_{jki}
- sum_{j, k: j + i => k} x_{jik}
- sum_{j, k: i + j => k} x_{ijk} >= q_i,
for all i + j => k: x_{ijk} >= 0,
where q_i is 1 if i is the goal item, else minus the number of i initially available. There are efficient solvers for this easy version. Since the reactions are always 2 => 1, you can always recover a feasible forging schedule for an integer solution. Accordingly, I would recommend integer programming for this problem. The paragraph below may still be of interest.
I know shipping an LP solver may be inconvenient, so here's an insight that will let you do without. This LP is feasible if and only if its dual is bounded. Intuitively, the dual problem is to assign a "value" to each item such that, however you forge, the total value of your inventory does not increase. If the goal item is valued at more than the available inventory, then you can't forge it. You can use any method that you can think of to assign these values.
I think you are unlikely to get a good general answer to this question because
if there was an efficient algorithm to solve your problem, then it would also be able to solve NP-complete problems.
For example, consider the problem of finding the maximum number of independent rows in a binary matrix.
This is a known NP-complete problem (e.g. by showing equivalence to the maximum independent set problem).
We can reduce this problem to your question in the following manner:
We can start holding one weapon for each column in the binary matrix, and then we imagine each row describes an alternative way of making a new weapon (say a battle axe).
We construct the weapon translation table such that to make the battle axe using method i, we need all weapons j such that M[i,j] is equal to 1 (this may involve inventing some additional weapons).
Then we construct a series of super weapons which can be made by combining different numbers of our battle axes.
For example, the mega ultimate battle axe may require 4 battle axes to be combined.
If we are able to work out the best weapon that can be constructed from your starting weapons, then we have solved the problem of finding the maximum number of independent rows in the original binary matrix.
It's not a huge saving, however looking at the source document, there are times when combining weapons produces the same weapon as one that was combined. I assume that you won't want to do this as you'll end up with less weapons.
So if you added a check for if the result_weapon was the same type as one of the inputs, and didn't go ahead and recursively call find_possible_weapons(new_weapons), you'd trim the search down a little.
The other thing I could think of, is you are not keeping a track of work done, so if the return from find_possible_weapons(new_weapons) returns the same weapon that you already have got by combining other weapons, you might well be performing the same search branch multiple times.
e.g. if you have a, b, c, d, e, f, g, and if a + b = x, and c + d = x, then you algorithm will be performing two lots of comparing x against e, f, and g. So if you keep a track of what you've already computed, you'll be onto a winner...
Basically, you have to trim the search tree. There are loads of different techniques to do this: it's called search. If you want more advice, I'd recommend going to the computer science stack exchange.
If you are still struggling, then you could always start weighting items/resulting items, and only focus on doing the calculation on 'high gain' objects...
You might want to start by creating a Weapon[][] matrix, to show the results of forging each pair. You could map the name of the weapon to the index of the matrix axis, and lookup of the results of a weapon combination would occur in constant time.