Given an equation
Like 2(p1) + 3(p2) + 7(p3) >= 257
I need to find all possible combinations of p1, p2, p3
such the above statement is true and the resulting sum ( left hand side of the equation ) is minimal where all xn were known.
I tried looking up algorithms for general cases like
(x1)(p1) + (x2)(p2) + (x3)(p4) + ... + (xn)(pn) >= target
And I came across the Knapsack problem and Subset-Sum algorithm solutions, but they weren't exactly like this problem.
I tried before using an algorithm in Python 3.x that has lower-bound values for pn, but it still runs in O( ridiculous ) time complexity.
Obviously all numbers here are natural numbers, otherwise there would be infinite solutions.
I can see two possible approaches, depending on whether the Pi have to be >= 0. The case with Pi >= 0 is more sensible, so I will consider it first.
Treat this as dynamic programming, where you work from left to right along the equation. Looking at the larger equation in your comment, first of all create a list of the contributions from p0: 0, 5, 10, 15... 190384760, and beside them the value of p0 that produces them: 0, 1, 2, ... 190384760/5.
Now use this table to work out the values of 5p0 + 7p1 possible by combining the first two: 0, 5, 7, 10, 12, 14.... and keep the value of p1 needed to produce them.
Working from right to left you will end up with a table of the values up to just over 190384755 that can be created by positive integer combinations of p0..p8. You obviously only care about the largest one >= 190384755. Consider all possible values of the p8 contribution, subtract these from 190384755, and look in the table for p0..p7 to see which of these are possible. This gives you all possible values of p8, and for each of these you can recursively repeat the process to print out all possible values of p7, and so on repeat the recursion to provide all values of p0..p8 that yields the lowest value just over 190384755. This is very similar to the pseudo-polynomial algorithm for subset sum.
If the Pi can be < 0, then the achievable values are all multiples of the gcd of the Pi, which is very likely to be all integers, and there are an infinite number of solutions for this. If this is really what you want, you can start by reading about the http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm.
Maybe the given example is just a toy case.
If not, exhaustive search is quite feasible: the minimal sum is bounded by 259 (combination 0, 0, 37), and there are less than a half million combinations under this bound.
In addition, if you set two variables, say p2 and p3, such that 3(p2) + 7(p3) < 257, it is an easy matter to find the smallest p1 such that 2(p1) + 3(p2) + 7(p3) >= 257. You will just have to try 3200 (p2, p3) combinations or so.
Related
I have a problem and I feel like there should be a well-known algorithm for solving it that's better than just brute force, but I can't think of one, so I'm asking here.
The problem is as follows: given n sorted (from low to high) lists containing m probabilities, choose one index for each list such that the sum of the chosen indexes is less than m. Then, for each list, we flip a coin, where the chance of it landing heads is equal to the probability at the chosen index for that list. Maximize the chance of the coin landing heads at least once.
Are there any algorithms for solving this problem that are better than just brute force?
This problem seems most similar to the knapsack problem, except the value of the items in the knapsack isn't merely a sum of the items in the knapsack. (Written in Python, instead of sum(p for p in chosen_probabilities) it's 1 - math.prod([1 - p for p in chosen_probabilities])) And, there's restrictions on what items you can add given what items are already in the knapsack. For example, if the index = 3 item for a particular list is already in the knapsack, then adding in the item with index = 2 for that same list isn't allowed (since you can only pick one index for each list). So there are certain items that can and can't be added to the knapsack based on what items are already in it.
Linear optimization won't work because the values in the lists don't increase linearly, the final coin probability isn't linear with respect to the chosen probabilities, and our constraint is on the sum of the indexes, rather than the values in the lists themselves. As David has pointed out, linear optimization will work if you use binary variables to pick out the indexes and a logarithm to deal with the non-linearity.
EDIT:
I've found that explaining the motivation behind this problem can be helpful for understanding it. Imagine you have 10 seconds to solve a problem, and three different ways to solve it. You have models of how likely it is that each method will solve the problem, given how many seconds you try that method for, but if you switch methods, you lose all progress on the one you were previously trying. What methods should you try and for how long?
Maximizing 1 - math.prod([1 - p for p in chosen_probabilities]) is equivalent to minimizing math.prod([1 - p for p in chosen_probabilities]), which is equivalent to minimizing the log of this objective, which is a linear function of 0-1 indicator variables, so you could do an integer programming formulation this way.
I can't promise that this will be much better than brute force. The problem is that math.log(1 - p) is well approximated by -p when p is close to zero. My intuition is that for nontrivial instances it will be qualitatively similar to using integer programming to solve subset sum, which doesn't go particularly well.
If you're willing to settle for a bicriteria approximation scheme (get an answer such that the sum of the chosen indexes is less than m, that is at least as good as the best answer summing to less than (1 − ε) m) then you can round up the probability to multiples of ε and use dynamic programming to get an algorithm that runs in time polynomial in n, m, 1/ε.
Here is working code for David Eisenstat's solution.
To understand the implementation, I think it helps to go through the math first.
As a reminder, there are n lists, each with m options. (In the motivating example at the bottom of the question, each list represents a method for solving the problem, and you are given m-1 seconds to solve the problem. Each list is such that list[index] gives the chance of solving the problem with that method if the method is run for index seconds.)
We let the lists be stored in a matrix called d (named data in the code), where each row in the matrix is a list. (And thus each column represents an index, or, if following the motivating example, an amount of time.)
The probability of the coin landing heads, given that we chose index j* for list i, is computed as
We would like to maximize this.
(To explain the stats behind this equation, we're computing 1 minus the probability that the coin doesn't land on heads. The probability that the coin doesn't land on heads is the probability that each flip doesn't land on heads. The probability that a single flip doesn't land on heads is just 1 minus the probability that does land on heads. And the probability it does land on heads is the number we've chosen, d[i][j*]. Thus, the total probability that all the flips land on tails is just the product of the probability that each one lands on tails. And then the probability that the coin lands on heads is just 1 minus the probability that all the flips land on tails.)
Which, as David pointed out, is the same as minimizing:
Which is the same as minimizing:
Which is equivalent to:
Then, since this is linear sum, we can turn it into an integer program.
We'll be minimizing:
This lets the computer choose the indexes by allowing it to create an n by m matrix of 1s and 0s called x where the 1s pick out particular indexes. We'll then define rules so that it doesn't pick out invalid sets of indexes.
The first rule is that you have to pick out an index for each list:
The second rule is that you have to respect the constraint that the indexes chosen must sum to m or less:
And that's it! Then we can just tell the computer to minimize that sum according to those rules. It will spit out an x matrix with a single 1 on each row to tell us which index it has picked for the list on that row.
In code (using the motivating example), this is implemented as:
'''
Requirements:
cvxopt==1.2.6
cvxpy==1.1.10
ecos==2.0.7.post1
numpy==1.20.1
osqp==0.6.2.post0
qdldl==0.1.5.post0
scipy==1.6.1
scs==2.1.2
'''
import math
import cvxpy as cp
import numpy as np
# number of methods
n = 3
# if you have 10 seconds, there are 11 options for each method (0 seconds, 1 second, ..., 10 seconds)
m = 11
# method A has 30% chance of working if run for at least 3 seconds
# equivalent to [0, 0, 0, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3]
A_list = [0, 0, 0] + [0.3] * (m - 3)
# method B has 30% chance of working if run for at least 3 seconds
# equivalent to [0, 0, 0, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3, 0.3]
B_list = [0, 0, 0] + [0.3] * (m - 3)
# method C has 40% chance of working if run for 4 seconds, 30% otherwise
# equivalent to [0.3, 0.3, 0.3, 0.3, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4, 0.4]
C_list = [0.3, 0.3, 0.3, 0.3] + [0.4] * (m - 4)
data = [A_list, B_list, C_list]
# do the logarithm
log_data = []
for row in data:
log_row = []
for col in row:
# deal with domain exception
if col == 1:
new_col = float('-inf')
else:
new_col = math.log(1 - col)
log_row.append(new_col)
log_data.append(log_row)
log_data = np.array(log_data)
x = cp.Variable((n, m), boolean=True)
objective = cp.Minimize(cp.sum(cp.multiply(log_data, x)))
# the current solver doesn't work with equalities, so each equality must be split into two inequalities.
# see https://github.com/cvxgrp/cvxpy/issues/1112
one_choice_per_method_constraint = [cp.sum(x[i]) <= 1 for i in range(n)] + [cp.sum(x[i]) >= 1 for i in range(n)]
# constrain the solution to not use more time than is allowed
# note that the time allowed is (m - 1), not m, because time is 1-indexed and the lists are 0-indexed
js = np.tile(np.array(list(range(m))), (n, 1))
time_constraint = [cp.sum(cp.multiply(js, x)) <= m - 1, cp.sum(cp.multiply(js, x)) >= m - 1]
constraints = one_choice_per_method_constraint + time_constraint
prob = cp.Problem(objective, constraints)
result = prob.solve()
def compute_probability(data, choices):
# compute 1 - ((1 - p1) * (1 - p2) * ...)
return 1 - np.prod(np.add(1, -np.multiply(data, choices)))
print("Choices:")
print(x.value)
'''
Choices:
[[0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]]
'''
print("Chance of success:")
print(compute_probability(data, x.value))
'''
Chance of success:
0.7060000000000001
'''
And there we have it! The computer has correctly determined that running method A for 3 seconds, method B for 3 seconds, and method C for 4 seconds is optimal. (Remember that the x matrix is 0-indexed, while the times are 1-indexed.)
Thank you, David, for the suggestion!
I have a problem statement which says: if you have an array of elements {x1,x2,x3,...x10}, find the combination of elements such that it just sums up above a threshold value (say the threshold value is 100).
So if there exists a combination like x2+x5+x8 = 105, x3+x5+x8=103, and x4+x5 = 101, then the algorithm should output X4, X5.
The knapsack algorithm emits a value that is near but on the lesser side of the threshold (which is 100 here). I want the opposite, that is the smallest sum of selected elements that is greater than 100.
Is there any set of algorithms or any special case of any algorithm which might solve this problem?
I'll start out by noting that you are asking for the smallest value strictly greater than some target. In general "strictly greater than" and "strictly less than" constraints are much harder than "greater than or equal to" or "less than or equal to" constraints. If you have all integer values, then you could simply translate your constraint "the sum exceeds 100" to "the sum is greater than or equal to 101". I'll assume that you've made such a transformation for the rest of the problem.
One approach would be to treat this as an integer optimization problem, in which the binary decision variable y_i for each number is whether or not we include it. Then our goal is to minimize the sum of the numbers, which can be modeled as:
min x_1*y_1 + x_2*y_2 + ... + x_n*y_n
The constraint in this case is that the sum of the numbers is at least 100:
x_1*y_1 + x_2*y_2 + ... + x_n*y_n >= 100
In general this is a hard problem (note that it is at least as hard as the subset sum problem, which is NP-complete). However modern optimization solvers may be efficient enough for your problem instances.
To test the scalability of a free solver for this problem, consider the following implementation with the lpSolve package in R (it returns the selected subset if the problem is feasible and NA otherwise):
library(lpSolve)
min.subset <- function(x, min.sum) {
mod <- lp("min", x, matrix(x, nrow=1), ">=", min.sum, all.bin=TRUE)
if (mod$status == 0) {
which(mod$solution >= 0.999)
} else {
NA
}
}
min.subset(1:10, 43.5)
# [1] 2 3 4 5 6 7 8 9
min.subset(1:10, 88)
# [1] NA
To test the scalability, I'll select n elements randomly from [1, 2, ..., 1000], setting the target sum to be half the sum of the elements. The runtimes were:
With n=100, it ran in 0.01 seconds
With n=1000, it ran in 0.1 seconds
With n=10000, it ran in 8.7 seconds
It appears you can solve this problem for more than 10k elements (with the selected distribution) without too many computational challenges. If your problem is too big for the free solver I've used here, you might consider Gurobi or cplex, two commercial solvers that are free for academic use but otherwise not free.
Suppose X is the sum of all x_i. Then equivalently, you are asking for a minimum subset of your x_i that sum up to at most X - 100 (as the complement of these x_i will be the optimum solution to your problem). So all Knapsack theory can be applied here.
In practice (really large instances), I'd suggest this form of Nemhauser-Ullman generalization which can solve instances with millions of objects.
I am having issues with understanding dynamic programming solutions to various problems, specifically the coin change problem:
"Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter.
For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5."
There is another variation of this problem where the solution is the minimum number of coins to satisfy the amount.
These problems appear very similar, but the solutions are very different.
Number of possible ways to make change: the optimal substructure for this is DP(m,n) = DP(m-1, n) + DP(m, n-Sm) where DP is the number of solutions for all coins up to the mth coin and amount=n.
Minimum amount of coins: the optimal substructure for this is
DP[i] = Min{ DP[i-d1], DP[i-d2],...DP[i-dn] } + 1 where i is the total amount and d1..dn represent each coin denomination.
Why is it that the first one required a 2-D array and the second a 1-D array? Why is the optimal substructure for the number of ways to make change not "DP[i] = DP[i-d1]+DP[i-d2]+...DP[i-dn]" where DP[i] is the number of ways i amount can be obtained by the coins. It sounds logical to me, but it produces an incorrect answer. Why is that second dimension for the coins needed in this problem, but not needed in the minimum amount problem?
LINKS TO PROBLEMS:
http://comproguide.blogspot.com/2013/12/minimum-coin-change-problem.html
http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/
Thanks in advance. Every website I go to only explains how the solution works, not why other solutions do not work.
Lets first talk about the number of ways, DP(m,n) = DP(m-1, n) + DP(m, n-Sm). This in indeed correct because either you can use the mth denomination or you can avoid it. Now you say why don't we write it as DP[i] = DP[i-d1]+DP[i-d2]+...DP[i-dn]. Well this will lead to over counting , lets take an example where n=4 m=2 and S={1,3}. Now according to your solution dp[4]=dp[1]+dp[3]. ( Assuming 1 to be a base case dp[1]=1 ) .Now dp[3]=dp[2]+dp[0]. ( Again dp[0]=1 by base case ). Again applying the same dp[2]=dp[1]=1. Thus in total you get answer as 3 when its supposed to be just 2 ( (1,3) and (1,1,1,1) ). Its so because
your second method treats (1,3) and (3,1) as two different solution.Your second method can be applied to case where order matters, which is also a standard problem.
Now to your second question you say that minimum number of denominations can
be found out by DP[i] = Min{ DP[i-d1], DP[i-d2],...DP[i-dn] } + 1. Well this is correct as in finding minimum denominations, order or no order does not matter. Why this is linear / 1-D DP , well although the DP array is 1-D each state depends on at most m states unlike your first solution where array is 2-D but each state depends on at most 2 states. So in both case run time which is ( number of states * number of states each state depends on ) is the same which is O(nm). So both are correct, just your second solution saves memory. So either you can find it by 1-D array method or by 2-D by using the recurrence
dp(n,m)=min(dp(m-1,n),1+dp(m,n-Sm)). (Just use min in your first recurrence)
Hope I cleared the doubts , do post if still something is unclear.
This is a very good explanation of the coin change problem using Dynamic Programming.
The code is as follows:
public static int change(int amount, int[] coins){
int[] combinations = new int[amount + 1];
combinations[0] = 1;
for(int coin : coins){
for(int i = 1; i < combinations.length; i++){
if(i >= coin){
combinations[i] += combinations[i - coin];
//printAmount(combinations);
}
}
//System.out.println();
}
return combinations[amount];
}
I have a following problem I want to solve efficiently. I am given a set of k-tuples of Boolean values where I know in advance that some fraction of each of the values in each of the k-tuples is true. For example, I might have the following 4-tuples, where each tuple has at least 60% of it's Boolean values set to true:
(1, 0, 1, 0)
(1, 1, 0, 1)
(0, 0, 1, 0)
I am interested in finding sets of indices that have a particular property: if I look at each of the values in the tuples at the indicated indices, at least the given fraction of those tuples have the corresponding bit set. For example, in the above set of 4-tuples, I could consider the set {0}, since if you look at the zeroth element of each of the above tuples, two-thirds of them are 1, and 2/3 ~= 66% > 60%. I could also consider the set {2} for the same reason. However, I could not consider {1}, since at index 1 only one third of the tuples have a 1 and 1/3 is less than 60%. Similarly, I could not use {0, 2} as a set, because it is not true that at least 60% of the tuples have both bits 0 and 2 set.
My goal is to find all sets for which this property holds. Does anyone have a good algorithm for solving this?
Thank you.
As you've wrote, that can be assumed that architecture is x86_64 and you are looking for implementation performance, cause asymptotic complexity (as it is not going to go under linear - by definition of problem ;) ), I propose following algorithm (C++ like pseudocode):
/* N=16 -> int16; N=8 -> int8 etc. Select N according to input sizes. (maybe N=24 ;) ) */
count_occurences_intN(vector<intN> t, vector<long> &result_counters){
intN counters[2^N]={};
//first, count bit combinations
for_each(v in t)
++counters[v];
//second, count bit occurrences, using aggregated data
for(column=0; column<N; ++column){
mask = 1 << column;
long *result_counter_ptr = &(result_counters[column]);
for(v=0; v<2^16; ++v)
if( v & mask )
++(*result_counter_ptr);
}
}
Than, split your input k-bit vectors into N-bit vectors, and apply above function.
Depending on input size you might improve performance you choosing N=8, N=16, N=24 or applying naive approach.
As you've wrote, you can not assume anything on client side, just implement N={8,16,24} and naive and select one from four implementations depending on size of input.
Make a k-vector of integers, describing how many passes there were for each index. Loop through your set, for each element incrementing the k-vector of passes.
Then figure out the cardinality of your set (either in a separate loop, or in the above one). Then loop through your vector of counts, and emit a pass/fail vector based on your criteria.
Say S = 5 and N = 3 the solutions would look like - <0,0,5> <0,1,4> <0,2,3> <0,3,2> <5,0,0> <2,3,0> <3,2,0> <1,2,2> etc etc.
In the general case, N nested loops can be used to solve the problem. Run N nested loop, inside them check if the loop variables add upto S.
If we do not know N ahead of time, we can use a recursive solution. In each level, run a loop starting from 0 to N, and then call the function itself again. When we reach a depth of N, see if the numbers obtained add up to S.
Any other dynamic programming solution?
Try this recursive function:
f(s, n) = 1 if s = 0
= 0 if s != 0 and n = 0
= sum f(s - i, n - 1) over i in [0, s] otherwise
To use dynamic programming you can cache the value of f after evaluating it, and check if the value already exists in the cache before evaluating it.
There is a closed form formula : binomial(s + n - 1, s) or binomial(s+n-1,n-1)
Those numbers are the simplex numbers.
If you want to compute them, use the log gamma function or arbitrary precision arithmetic.
See https://math.stackexchange.com/questions/2455/geometric-proof-of-the-formula-for-simplex-numbers
I have my own formula for this. We, together with my friend Gio made an investigative report concerning this. The formula that we got is [2 raised to (n-1) - 1], where n is the number we are looking for how many addends it has.
Let's try.
If n is 1: its addends are o. There's no two or more numbers that we can add to get a sum of 1 (excluding 0). Let's try a higher number.
Let's try 4. 4 has addends: 1+1+1+1, 1+2+1, 1+1+2, 2+1+1, 1+3, 2+2, 3+1. Its total is 7.
Let's check with the formula. 2 raised to (4-1) - 1 = 2 raised to (3) - 1 = 8-1 =7.
Let's try 15. 2 raised to (15-1) - 1 = 2 raised to (14) - 1 = 16384 - 1 = 16383. Therefore, there are 16383 ways to add numbers that will equal to 15.
(Note: Addends are positive numbers only.)
(You can try other numbers, to check whether our formula is correct or not.)
This can be calculated in O(s+n) (or O(1) if you don't mind an approximation) in the following way:
Imagine we have a string with n-1 X's in it and s o's. So for your example of s=5, n=3, one example string would be
oXooXoo
Notice that the X's divide the o's into three distinct groupings: one of length 1, length 2, and length 2. This corresponds to your solution of <1,2,2>. Every possible string gives us a different solution, by counting the number of o's in a row (a 0 is possible: for example, XoooooX would correspond to <0,5,0>). So by counting the number of possible strings of this form, we get the answer to your question.
There are s+(n-1) positions to choose for s o's, so the answer is Choose(s+n-1, s).
There is a fixed formula to find the answer. If you want to find the number of ways to get N as the sum of R elements. The answer is always:
(N+R-1)!/((R-1)!*(N)!)
or in other words:
(N+R-1) C (R-1)
This actually looks a lot like a Towers of Hanoi problem, without the constraint of stacking disks only on larger disks. You have S disks that can be in any combination on N towers. So that's what got me thinking about it.
What I suspect is that there is a formula we can deduce that doesn't require the recursive programming. I'll need a bit more time though.