a program I'm writing requires me to simulate the mutation of genes. If a certain instance (J) of the gene is selected for mutation, it must randomly become one of a predefined subset of alternative genes. These discrete genes are marked by the integers in the range 0 to K-1, where K equals the total number of alleles.
So basically, I need help writing the code to randomly select an integer in the range 0 to K-1 that does not equal J. Any help is greatly appreciated!
ANOTHER EDIT: Thanks for all the help so far, I've come up with the following, which ignores the requirement that i cannot equal J. However, a further problem I've encountered is that the process must be repeated multiple times, and due to the same seed being used each time the value of i is always the same. How should I go about ensuring the seed value doesn't remain the same throughout every generation? Again, any help is greatly appreciated!
program randtest
implicit none
real*4 :: u(5)
integer :: i(5)
integer :: k = 4
integer, dimension (1) :: seed = (/2817/)
call random_seed(put=seed)
call random_number(u)
i = floor((k+1)*u)
print *, i
end program randtest
One can indeed use the method given in the answers to the other question, for the set {0, 1, ..., K-1}. The complication that J (0<=J<=K-1) can be handled by rejecting J when it is sampled.
However, a better approach is to note that there are K-2 valid integers: {0, 1, ..., J-1, J+1, ..., K-1}. So, use the method to generate an integer over {0, 1, ..., K-2} and just shift up anything that would be affected by rejecting J.
call random_number(u)
i = FLOOR((K-1)*u) !this would be how I would define the range
if (i.ge.J) i=i+1
Related
I want to formulate a flexible job shop scheduling problem with MIP instead of CP.
If there is an array indicating number of operations of each job.
num_op = [3, 2, 5]
And Xijk is the decision variable indicating if the j-th operation of job i is processed on machine k or not.
My question is that I don't know how t initiate the 3-D array with different number of operations of each job.
I wrote this dvar boolean x[i in Jobs][j in][k in Machs];, and I don't know to how to complete it.
Please help me. Thanks!!
If I understand your example correctly then your definition of num_op indicates that you have three different jobs, the first job has 3 operations, the second has 2 and the last job has 5 operations. That would mean that the second dimension of Xijk has to change depending on the first dimension. Such a flexible array size is not possible with CPLEX.
Here is an alternative:
Define M to be the maximum number of operations (5 in your case).
Define dvar boolean x[i in Jobs][j in 1..M][k in Machs];
Explicitly fix all variables to 0 that correspond to non-existing operations:
forall (i in Jobs, j in 1..M, k in Machs) if (j > num_op[i]) X[i][j][k] == 0;
The last step is even optional: You can define the variables for non-existing operations but just not use them anywhere in your model (this may give a warning for unused variables, though).
Another option is to create a tuple tuple { int i; int j; int k }, then create a tuple set that contains all the valid combinations of i, j, k and index X by this tuple set.
I was recently doing a project euler problem (namely #31) which was basically finding out how many ways we can sum to 200 using elements of the set {1,2,5,10,20,50,100,200}.
The idea that I used was this: the number of ways to sum to N is equal to
(the number of ways to sum N-k) * (number of ways to sum k), summed over all possible values of k.
I realized that this approach is WRONG, namely due to the fact that it creates several several duplicate counts. I have tried to adjust the formula to avoid duplicates, but to no avail. I am seeking the wisdom of stack overflowers regarding:
whether my recursive approach is concerned with the correct subproblem to solve
If there exists one, what would be an effective way to eliminate duplicates
how should we approach recursive problems such that we are concerned with the correct subproblem? what are some indicators that we've chosen a correct (or incorrect) subproblem?
When trying to avoid duplicate permutations, a straightforward strategy that works in most cases is to only create rising or falling sequences.
In your example, if you pick a value and then recurse with the whole set, you will get duplicate sequences like 50,50,100 and 50,100,50 and 100,50,50. However, if you recurse with the rule that the next value should be equal to or smaller than the currently selected value, out of those three you will only get the sequence 100,50,50.
So an algorithm that counts only unique combinations would be e.g.:
function uniqueCombinations(set, target, previous) {
for all values in set not greater than previous {
if value equals target {
increment count
}
if value is smaller than target {
uniqueCombinations(set, target - value, value)
}
}
}
uniqueCombinations([1,2,5,10,20,50,100,200], 200, 200)
Alternatively, you can create a copy of the set before every recursion, and remove the elements from it that you don't want repeated.
The rising/falling sequence method also works with iterations. Let's say you want to find all unique combinations of three letters. This algorithm will print results like a,c,e, but not a,e,c or e,a,c:
for letter1 is 'a' to 'x' {
for letter2 is first letter after letter1 to 'y' {
for letter3 is first letter after letter2 to 'z' {
print [letter1,letter2,letter3]
}
}
}
m69 gives a nice strategy that often works, but I think it's worthwhile to better understand why it works. When trying to count items (of any kind), the general principle is:
Think of a rule that classifies any given item into exactly one of several non-overlapping categories. That is, come up with a list of concrete categories A, B, ..., Z that will make the following sentence true: An item is either in category A, or in category B, or ..., or in category Z.
Once you have done this, you can safely count the number of items in each category and add these counts together, comfortable in the knowledge that (a) any item that is counted in one category is not counted again in any other category, and (b) any item that you want to count is in some category (i.e., none are missed).
How could we form categories for your specific problem here? One way to do it is to notice that every item (i.e., every multiset of coin values that sums to the desired total N) either contains the 50-coin exactly zero times, or it contains it exactly once, or it contains it exactly twice, or ..., or it contains it exactly RoundDown(N / 50) times. These categories don't overlap: if a solution uses exactly 5 50-coins, it pretty clearly can't also use exactly 7 50-coins, for example. Also, every solution is clearly in some category (notice that we include a category for the case in which no 50-coins are used). So if we had a way to count, for any given k, the number of solutions that use coins from the set {1,2,5,10,20,50,100,200} to produce a sum of N and use exactly k 50-coins, then we could sum over all k from 0 to N/50 and get an accurate count.
How to do this efficiently? This is where the recursion comes in. The number of solutions that use coins from the set {1,2,5,10,20,50,100,200} to produce a sum of N and use exactly k 50-coins is equal to the number of solutions that sum to N-50k and do not use any 50-coins, i.e. use coins only from the set {1,2,5,10,20,100,200}. This of course works for any particular coin denomination that we could have chosen, so these subproblems have the same shape as the original problem: we can solve each one by simply choosing another coin arbitrarily (e.g. the 10-coin), forming a new set of categories based on this new coin, counting the number of items in each category and summing them up. The subproblems become smaller until we reach some simple base case that we process directly (e.g. no allowed coins left: then there is 1 item if N=0, and 0 items otherwise).
I started with the 50-coin (instead of, say, the largest or the smallest coin) to emphasise that the particular choice used to form the set of non-overlapping categories doesn't matter for the correctness of the algorithm. But in practice, passing explicit representations of sets of coins around is unnecessarily expensive. Since we don't actually care about the particular sequence of coins to use for forming categories, we're free to choose a more efficient representation. Here (and in many problems), it's convenient to represent the set of allowed coins implicitly as simply a single integer, maxCoin, which we interpret to mean that the first maxCoin coins in the original ordered list of coins are the allowed ones. This limits the possible sets we can represent, but here that's OK: If we always choose the last allowed coin to form categories on, we can communicate the new, more-restricted "set" of allowed coins to subproblems very succinctly by simply passing the argument maxCoin-1 to it. This is the essence of m69's answer.
There's some good guidance here. Another way to think about this is as a dynamic program. For this, we must pose the problem as a simple decision among options that leaves us with a smaller version of the same problem. It boils out to a certain kind of recursive expression.
Put the coin values c0, c1, ... c_(n-1) in any order you like. Then define W(i,v) as the number of ways you can make change for value v using coins ci, c_(i+1), ... c_(n-1). The answer we want is W(0,200). All that's left is to define W:
W(i,v) = sum_[k = 0..floor(200/ci)] W(i+1, v-ci*k)
In words: the number of ways we can make change with coins ci onward is to sum up all the ways we can make change after a decision to use some feasible number k of coins ci, removing that much value from the problem.
Of course we need base cases for the recursion. This happens when i=n-1: the last coin value. At this point there's a way to make change if and only if the value we need is an exact multiple of c_(n-1).
W(n-1,v) = 1 if v % c_(n-1) == 0 and 0 otherwise.
We generally don't want to implement this as a simple recursive function. The same argument values occur repeatedly, which leads to an exponential (in n and v) amount of wasted computation. There are simple ways to avoid this. Tabular evaluation and memoization are two.
Another point is that it is more efficient to have the values in descending order. By taking big chunks of value early, the total number of recursive evaluations is minimized. Additionally, since c_(n-1) is now 1, the base case is just W(n-1)=1. Now it becomes fairly obvious that we can add a second base case as an optimization: W(n-2,v) = floor(v/c_(n-2)). That's how many times the for loop will sum W(n-1,1) = 1!
But this is gilding a lilly. The problem is so small that exponential behavior doesn't signify. Here is a little implementation to show that order really doesn't matter:
#include <stdio.h>
#define n 8
int cv[][n] = {
{200,100,50,20,10,5,2,1},
{1,2,5,10,20,50,100,200},
{1,10,100,2,20,200,5,50},
};
int *c;
int w(int i, int v) {
if (i == n - 1) return v % c[n - 1] == 0;
int sum = 0;
for (int k = 0; k <= v / c[i]; ++k)
sum += w(i + 1, v - c[i] * k);
return sum;
}
int main(int argc, char *argv[]) {
unsigned p;
if (argc != 2 || sscanf(argv[1], "%d", &p) != 1 || p > 2) p = 0;
c = cv[p];
printf("Ways(%u) = %d\n", p, w(0, 200));
return 0;
}
Drumroll, please...
$ ./foo 0
Ways(0) = 73682
$ ./foo 1
Ways(1) = 73682
$ ./foo 2
Ways(2) = 73682
I am kind of new in the fortran proramming.
Can anyone please help me out with the solution.
i am having a problem of generating integer random number
in the range [0,5] in fortran random number using
random_seed and rand
To support the answer by Alexander Vogt, I'll generalize.
The intrinsic random_number(u) returns a real number u (or an array of such) from the uniform distribution over the interval [0,1). [That is, it includes 0 but not 1.]
To have a discrete uniform distribution on the integers {n, n+1, ..., m-1, m} carve the continuous distribution up into m+1-n equal sized chunks, mapping each chunk to an integer. One way could be:
call random_number(u)
j = n + FLOOR((m+1-n)*u) ! We want to choose one from m-n+1 integers
As you can see, for the initial question for {0, 1, 2, 3, 4, 5} this reduces to
call random_number(u)
j = FLOOR(6*u) ! n=0 and m=5
and for the other case in your comment {-1, 0, 1}
call random_number(u)
j = -1 + FLOOR(3*u) ! n=-1 and m=1
Of course, other transformations will be required for sets of non-contiguous integers, and one should pay attention to numerical issues.
What about:
program rand_test
use,intrinsic :: ISO_Fortran_env
real(REAL32) :: r(5)
integer :: i(5)
! call init_random_seed() would go here
call random_number(r)
! Uniform distribution requires floor: Thanks to #francescalus
i = floor( r*6._REAL32 )
print *, i
end program
How can I generate a random number that is in the range (1,n) but not in a certain list (i,j)?
Example: range is (1,500), list is [1,3,4,45,199,212,344].
Note: The list may not be sorted
Rejection Sampling
One method is rejection sampling:
Generate a number x in the range (1, 500)
Is x in your list of disallowed values? (Can use a hash-set for this check.)
If yes, return to step 1
If no, x is your random value, done
This will work fine if your set of allowed values is significantly larger than your set of disallowed values:if there are G possible good values and B possible bad values, then the expected number of times you'll have to sample x from the G + B values until you get a good value is (G + B) / G (the expectation of the associated geometric distribution). (You can sense check this. As G goes to infinity, the expectation goes to 1. As B goes to infinity, the expectation goes to infinity.)
Sampling a List
Another method is to make a list L of all of your allowed values, then sample L[rand(L.count)].
The technique I usually use when the list is length 1 is to generate a random
integer r in [1,n-1], and if r is greater or equal to that single illegal
value then increment r.
This can be generalised for a list of length k for small k but requires
sorting that list (you can't do your compare-and-increment in random order). If the list is moderately long, then after the sort you can start with a bsearch, and add the number of values skipped to r, and then recurse into the remainder of the list.
For a list of length k, containing no value greater or equal to n-k, you
can do a more direct substitution: generate random r in [1,n-k], and
then iterate through the list testing if r is equal to list[i]. If it is
then set r to n-k+i (this assumes list is zero-based) and quit.
That second approach fails if some of the list elements are in [n-k,n].
I could try to invest something clever at this point, but what I have so far
seems sufficient for uniform distributions with values of k much less than
n...
Create two lists -- one of illegal values below n-k, and the other the rest (this can be done in place).
Generate random r in [1,n-k]
Apply the direct substitution approach for the first list (if r is list[i] then set r to n-k+i and go to step 5).
If r was not altered in step 3 then we're finished.
Sort the list of larger values and use the compare-and-increment method.
Observations:
If all values are in the lower list, there will be no sort because there is nothing to sort.
If all values are in the upper list, there will be no sort because there is no occasion on which r is moved into the hazardous area.
As k approaches n, the maximum size of the upper (sorted) list grows.
For a given k, if more value appear in the upper list (the bigger the sort), the chance of getting a hit in the lower list shrinks, reducing the likelihood of needing to do the sort.
Refinement:
Obviously things get very sorty for large k, but in such cases the list has comparatively few holes into which r is allowed to settle. This could surely be exploited.
I might suggest something different if many random values with the same
list and limits were needed. I hope that the list of illegal values is not the
list of results of previous calls to this function, because if it is then you
wouldn't want any of this -- instead you would want a Fisher-Yates shuffle.
Rejection sampling would be the simplest if possible as described already. However, if you didn't want use that, you could convert the range and disallowed values to sets and find the difference. Then, you could choose a random value out of there.
Assuming you wanted the range to be in [1,n] but not in [i,j] and that you wanted them uniformly distributed.
In Python
total = range(1,n+1)
disallowed = range(i,j+1)
allowed = list( set(total) - set(disallowed) )
return allowed[random.randrange(len(allowed))]
(Note that this is not EXACTLY uniform since in all likeliness, max_rand%len(allowed) != 0 but this will in most practical applications be very close)
I assume that you know how to generate a random number in [1, n) and also your list is ordered like in the example above.
Let's say that you have a list with k elements. Make a map(O(logn)) structure, which will ensure speed if k goes higher. Put all elements from list in map, where element value will be the key and "good" value will be the value. Later on I'll explain about "good" value. So when we have the map then just find a random number in [1, n - k - p)(Later on I'll explain what is p) and if this number is in map then replace it with "good" value.
"GOOD" value -> Let's start from k-th element. It's good value is its own value + 1, because the very next element is "good" for us. Now let's look at (k-1)th element. We assume that its good value is again its own value + 1. If this value is equal to k-th element then the "good" value for (k-1)th element is k-th "good" value + 1. Also you will have to store the largest "good" value. If the largest value exceed n then p(from above) will be p = largest - n.
Of course I recommend you this only if k is big number otherwise #Timothy Shields' method is perfect.
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.