Consider a set of length, each associated with a probability. i.e.
X={a1=(100,1/4),a2=(500,1/4),a3=(200,1/2)}
Obviously, the sum of all the probabilities = 1.
Arrange the lengths together on a line one after the other from a starting point.
For example: {a2,a1,a3} in that order from start to finish.
Define the cost of an element a_i as its the total length from the starting line to the end of this element multiplied by its probability.
So from the previous arrangement:
cost(a2) = (500)*(1/4)
cost(a1) = (500+100)*(1/4)
cost(a3) = (500+100+200)*(1/2)
Define the total cost as the sum of all costs. e.g. cost(X) = cost(a2) + cost(a1) + cost(a3). Give an algorithm that finds an arrangement that minimizes cost(X)
My thoughts:
This looks like an greedy algorithm, since the last element in the arrangement always has the same sum multiplied by its probability, but I'm can't think of an heuristics that accomplishes this. It goes without saying that sorting by probability or length will not work.
I've decided to learn deeper the concept of backtracking and I have following task:
Given N investors, M cities, N by M matrix P of investors preferences (P[i, j] = 1 when i-th investor would like the pool to be built in the j-th city; P[i, j] = 0 then he's neutral and when P[i, j] = -1 he's sceptical) and acceptance level L (if for a given choice of places, sum of investors preferences is greater or equal to L then we consider him as convinced). Find maxmimal number of investors that can be convinced and cities in which pools should be built.
I have tried using backtracking but I wonder if it is possible to optimize it more. For now, on each recursion level I keep track of how many people can possibly be convinced. If this number is less or equal to my current maximum then I return (there will be no better answer).
I'm not sure if this is what you're looking for, but with a little trick, you can express the problem as an integer linear program (ILP). Then you can use an integer linear programming solver (for example, GLPK) to find an optimal solution.
Let s[i] be 0-1 integer variables (i ranging over investors), and c[j] 0-1 integer variables ranging over cities and K be a large number (L + the number of investors will do).
Then, your problem is to minimize sum(s[i]) such that for each i, sum(P[i, j]*c[j]) + s[i] * K >= L. The value of sum(s[i]) in the optimal solution is the number of dissatisfied investors, and c[j] indicates whether to build a pool in city j.
This formulation of the problem is in a standard form for ILPs, so you're good to go.
I've found similar questions but this is a bit more complicated.
I have a large number n(I actually have more, but it doesn't matter now), (>40 digits), and I want to find a*b*c=n triplets. n's prime factorisation is done. It has no large prime divisors, but many of small prime divisors. The sum of all prime divisors (included multiple divisors) is greater than 50.
I'd like to find a*b*c=n triplets, where a<=b<=c. I don't want all the triplets, because there are too much of them. I'm searching for special ones.
For example:
the triplet(s) where c-a is minimal,
the triplet(s) where c/a minimal,
the one where a,b and c has the maximal common divisor,
these conditions combined.
This can be a little easier to solve if we know that n=k!(factorial). Solving could lead to a general method.
Computing all these triplets with brute force is not an option because of the size of n, so i need a good algorithm or some special tools to help me implement a solution for this.
Sorry for my bad English,
Thanks for the answers!
You can achieve it with a simple, O(|D|^2) algorithms, where D is an ordered list of all the numbers dividing n, which you already have.
Note that you only have to find a,b, because c=n/(a*b), so the problems boils down to finding all the pairs (a,b) in D so that a<b and n/(a*b) ∈ D.
Pseudocode:
result = empty_list
for (int i=0; i<D.size-1, i++) { // O(|D|)
for (j=i+1; j<D.size, j++) { // O(|D|)
a, b = D[i], D[j]
c = n/(a*b)
if (D.contains(c) && c>b) { // O(1)
result.append( (a,b,c) )
}
}
} // O(|D|)*O(|D|)=O(|D|^2)
I might have the solution, but I have no time to implement it today. I write it down, so maybe somebody will agree with me or will spot the weak point of my algorithm.
So let's see the first or second case, where c/a or c-a should be minimal.
1: In first step I split the prime factors of n to 3 group with a greedy algorithm.
I will have an initial a,b and c and they will be not very far from each other. The prime factors will be stored in 3 arrays: a_pf,b_pf,c_pf.
2: In next step I compute all the possible factors for a,b and c, I store them in different arrays, then I order these arrays. These will be a_all,b_all and c_all.
3: I compute q=max(a,b,c)/min(a,b,c). (now we can say that a is the smallest, c is the greatest number)
4: I search a_all and c_all for numbers on this condiition: c_all[i]/a_all[j] < q. When I find it, I change the prime factors of these values in a_pf and c_pf. With this method, the largest and the smallest member of the triplet will come closer to each other.
I repeat step 2-3-4, until I can. I think this will end after finite number of steps.
Since the triplet's members are smaller than the original n, I hope this solution will give me the correct triplet at most in a few minutes.
I hope someone can help me answer the following question. Thanks!
Here is a pseudo code of Permute-By-Sorting algorithm:
Permute-By-Sorting (A)
n = A.length
let P[1..n] be a new array
for i = 1 to n
P[i] = Random (1,n^3)
sort A, using P as sort keys
In the above algorithm, the array P represents the priorities of the elements in array A. Line 4 chooses a random number between 1 and n^3.
The question is what is the probability that all priorities in P are unique? and how do I get the probability?
To reconcile the answers already given: for choice i = 0, ..., n - 1, given that no duplicates have been chosen yet, there are n^3 - i non-duplicate choices of n^3 total for the ith value. Thus the probability is the product for i = 0, ..., n - 1 of (1 - i/n^3).
sdcwc is using a union bound to lowerbound this probability by 1 - O(1/n). This estimate turns out to be basically right. The proof sketch is that (1 - i/n^3) is exp(-i/n^3 + O(i^2/n^6)), so the product is exp(-O(n^2)/n^3 + O(n^-3)), which is greater than or equal to 1 - O(n^2)/n^3 + O(n^-3) = 1 - O(1/n). I'm sure the fine folks on math.SE would be happy to do this derivation "properly" for you.
Others have given you the probability calculation, but I think you may be asking the wrong question.
I assume the reason you're asking about the probability of the priorities being unique, and the reason for choosing n^3 in the first place, is because you're hoping they will be unique, and choosing a large range relative to n seems to be a reasonable way of achieving uniqueness.
It is much easier to ensure that the values are unique. Simply populate the array of priorities with the numbers 1 .. n and then shuffle them with the Fisher-Yates algorithm (aka algorithm P from The Art of Computer Programming, volume 2, Seminumerical Algorithms, by Donald Knuth).
The sort would then be carried out with known unique priority values.
(There are also other ways of going about getting a random permutation. It is possible to generate the nth lexicographic permutation of a sequence using factoradic numbers (or, the factorial number system), and so generate the permutation for a randomly chosen value in [1 .. n!].)
You are choosing n numbers from 1...n^3 and asking what is the probability that they are all unique.
There are (n^3) P n = (n^3)!/(n^3-n)! ways to choose the n numbers uniquely, and (n^3)^n ways to choose the n-numbers total.
So the probability of the numbers being unique is just the first equation divided by the second, which gives
n3!
--------------
(n3-n)! n3n
Let Aij be the event: i-th and j-th elements collide. Obviously P(Aij)=1/n3.
There is at most n2 pairs, therefore probability of at least one collision is at most 1/n.
If you are interested in exact thing, see BlueRaja's answer, but in randomized algorithms it is usually enough to give this type of bound.
So the sort part is irrelevant
Assuming the "Random" is real random, the probability is just
n^3!
----------------
(n^3-n)!n^(3n)
We've got some nonnegative numbers. We want to find the pair with maximum gcd. actually this maximum is more important than the pair!
For example if we have:
2 4 5 15
gcd(2,4)=2
gcd(2,5)=1
gcd(2,15)=1
gcd(4,5)=1
gcd(4,15)=1
gcd(5,15)=5
The answer is 5.
You can use the Euclidean Algorithm to find the GCD of two numbers.
while (b != 0)
{
int m = a % b;
a = b;
b = m;
}
return a;
If you want an alternative to the obvious algorithm, then assuming your numbers are in a bounded range, and you have plenty of memory, you can beat O(N^2) time, N being the number of values:
Create an array of a small integer type, indexes 1 to the max input. O(1)
For each value, increment the count of every element of the index which is a factor of the number (make sure you don't wraparound). O(N).
Starting at the end of the array, scan back until you find a value >= 2. O(1)
That tells you the max gcd, but doesn't tell you which pair produced it. For your example input, the computed array looks like this:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
4 2 1 1 2 0 0 0 0 0 0 0 0 0 1
I don't know whether this is actually any faster for the inputs you have to handle. The constant factors involved are large: the bound on your values and the time to factorise a value within that bound.
You don't have to factorise each value - you could use memoisation and/or a pregenerated list of primes. Which gives me the idea that if you are memoising the factorisation, you don't need the array:
Create an empty set of int, and a best-so-far value 1.
For each input integer:
if it's less than or equal to best-so-far, continue.
check whether it's in the set. If so, best-so-far = max(best-so-far, this-value), continue. If not:
add it to the set
repeat for all of its factors (larger than best-so-far).
Add/lookup in a set could be O(log N), although it depends what data structure you use. Each value has O(f(k)) factors, where k is the max value and I can't remember what the function f is...
The reason that you're finished with a value as soon as you encounter it in the set is that you've found a number which is a common factor of two input values. If you keep factorising, you'll only find smaller such numbers, which are not interesting.
I'm not quite sure what the best way is to repeat for the larger factors. I think in practice you might have to strike a balance: you don't want to do them quite in decreasing order because it's awkward to generate ordered factors, but you also don't want to actually find all the factors.
Even in the realms of O(N^2), you might be able to beat the use of the Euclidean algorithm:
Fully factorise each number, storing it as a sequence of exponents of primes (so for example 2 is {1}, 4 is {2}, 5 is {0, 0, 1}, 15 is {0, 1, 1}). Then you can calculate gcd(a,b) by taking the min value at each index and multiplying them back out. No idea whether this is faster than Euclid on average, but it might be. Obviously it uses a load more memory.
The optimisations I can think of is
1) start with the two biggest numbers since they are likely to have most prime factors and thus likely to have the most shared prime factors (and thus the highest GCD).
2) When calculating the GCDs of other pairs you can stop your Euclidean algorithm loop if you get below your current greatest GCD.
Off the top of my head I can't think of a way that you can work out the greatest GCD of a pair without trying to work out each pair individually (and optimise a bit as above).
Disclaimer: I've never looked at this problem before and the above is off the top of my head. There may be better ways and I may be wrong. I'm happy to discuss my thoughts in more length if anybody wants. :)
There is no O(n log n) solution to this problem in general. In fact, the worst case is O(n^2) in the number of items in the list. Consider the following set of numbers:
2^20 3^13 5^9 7^2*11^4 7^4*11^3
Only the GCD of the last two is greater than 1, but the only way to know that from looking at the GCDs is to try out every pair and notice that one of them is greater than 1.
So you're stuck with the boring brute-force try-every-pair approach, perhaps with a couple of clever optimizations to avoid doing needless work when you've already found a large GCD (while making sure that you don't miss anything).
With some constraints, e.g the numbers in the array are within a given range, say 1-1e7, it is doable in O(NlogN) / O(MAX * logMAX), where MAX is the maximum possible value in A.
Inspired from the sieve algorithm, and came across it in a Hackerrank Challenge -- there it is done for two arrays. Check their editorial.
find min(A) and max(A) - O(N)
create a binary mask, to mark which elements of A appear in the given range, for O(1) lookup; O(N) to build; O(MAX_RANGE) storage.
for every number a in the range (min(A), max(A)):
for aa = a; aa < max(A); aa += a:
if aa in A, increment a counter for aa, and compare it to current max_gcd, if counter >= 2 (i.e, you have two numbers divisible by aa);
store top two candidates for each GCD candidate.
could also ignore elements which are less than current max_gcd;
Previous answer:
Still O(N^2) -- sort the array; should eliminate some of the unnecessary comparisons;
max_gcd = 1
# assuming you want pairs of distinct elements.
sort(a) # assume in place
for ii = n - 1: -1 : 0 do
if a[ii] <= max_gcd
break
for jj = ii - 1 : -1 :0 do
if a[jj] <= max_gcd
break
current_gcd = GCD(a[ii], a[jj])
if current_gcd > max_gcd:
max_gcd = current_gcd
This should save some unnecessary computation.
There is a solution that would take O(n):
Let our numbers be a_i. First, calculate m=a_0*a_1*a_2*.... For each number a_i, calculate gcd(m/a_i, a_i). The number you are looking for is the maximum of these values.
I haven't proved that this is always true, but in your example, it works:
m=2*4*5*15=600,
max(gcd(m/2,2), gcd(m/4,4), gcd(m/5,5), gcd(m/15,15))=max(2, 2, 5, 5)=5
NOTE: This is not correct. If the number a_i has a factor p_j repeated twice, and if two other numbers also contain this factor, p_j, then you get the incorrect result p_j^2 insted of p_j. For example, for the set 3, 5, 15, 25, you get 25 as the answer instead of 5.
However, you can still use this to quickly filter out numbers. For example, in the above case, once you determine the 25, you can first do the exhaustive search for a_3=25 with gcd(a_3, a_i) to find the real maximum, 5, then filter out gcd(m/a_i, a_i), i!=3 which are less than or equal to 5 (in the example above, this filters out all others).
Added for clarification and justification:
To see why this should work, note that gcd(a_i, a_j) divides gcd(m/a_i, a_i) for all j!=i.
Let's call gcd(m/a_i, a_i) as g_i, and max(gcd(a_i, a_j),j=1..n, j!=i) as r_i. What I say above is g_i=x_i*r_i, and x_i is an integer. It is obvious that r_i <= g_i, so in n gcd operations, we get an upper bound for r_i for all i.
The above claim is not very obvious. Let's examine it a bit deeper to see why it is true: the gcd of a_i and a_j is the product of all prime factors that appear in both a_i and a_j (by definition). Now, multiply a_j with another number, b. The gcd of a_i and b*a_j is either equal to gcd(a_i, a_j), or is a multiple of it, because b*a_j contains all prime factors of a_j, and some more prime factors contributed by b, which may also be included in the factorization of a_i. In fact, gcd(a_i, b*a_j)=gcd(a_i/gcd(a_i, a_j), b)*gcd(a_i, a_j), I think. But I can't see a way to make use of this. :)
Anyhow, in our construction, m/a_i is simply a shortcut to calculate the product of all a_j, where j=1..1, j!=i. As a result, gcd(m/a_i, a_i) contains all gcd(a_i, a_j) as a factor. So, obviously, the maximum of these individual gcd results will divide g_i.
Now, the largest g_i is of particular interest to us: it is either the maximum gcd itself (if x_i is 1), or a good candidate for being one. To do that, we do another n-1 gcd operations, and calculate r_i explicitly. Then, we drop all g_j less than or equal to r_i as candidates. If we don't have any other candidate left, we are done. If not, we pick up the next largest g_k, and calculate r_k. If r_k <= r_i, we drop g_k, and repeat with another g_k'. If r_k > r_i, we filter out remaining g_j <= r_k, and repeat.
I think it is possible to construct a number set that will make this algorithm run in O(n^2) (if we fail to filter out anything), but on random number sets, I think it will quickly get rid of large chunks of candidates.
pseudocode
function getGcdMax(array[])
arrayUB=upperbound(array)
if (arrayUB<1)
error
pointerA=0
pointerB=1
gcdMax=0
do
gcdMax=MAX(gcdMax,gcd(array[pointera],array[pointerb]))
pointerB++
if (pointerB>arrayUB)
pointerA++
pointerB=pointerA+1
until (pointerB>arrayUB)
return gcdMax