So the task I have to solve is to calculate the binomial coefficient for 100>=n>k>=1 and then say how many solutions for n and k are over an under barrier of 123456789.
I have no problem in my formula of calculating the binomial coefficient but for high numbers n & k -> 100 the datatypes of c get to small to calculated this.
Do you have any suggestions how I can bypass this problem with overflowing the datatypes.
I thought about dividing by the under barrier straight away so the numbers don't get too big in the first place and I have to just check if the result is >=1 but i couldn't make it work.
Say your task is to determine how many binomial coefficients C(n, k) for 1 ≤ k < n ≤ 8 exceed a limit of m = 18. You can do this by using the recurrence C(n, k) = C(n − 1, k) + C(n − 1, k − 1) that can visualized in Pascal's triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 (20) 15 6 1
1 7 (21 35 35 21) 7 1
1 8 (28 56 70 56 28) 8 1
Start at the top and work your way down. Up to n = 5, everything is below the limit of 18. On the next line, the 20 exceeds the limit. From now on, more and more coefficients are beyond 18.
The triangle is symmetric and strictly increasing in the first half of each row. You only need to find the first element that exceeds the limit on each line in order to know how many items to count.
You don't have to store the whole triangle. It is enough to keey the last and current line. Alternatively, you can use the algorithm detailed [in this article][ot] to work your way from left to right on each row. Since you just want to count the coefficients that exceed a limit and don't care about their values, the regular integer types should be sufficient.
First, you'll need a type that can handle the result. The larget number you need to handle is C(100,50) = 100,891,344,545,564,193,334,812,497,256. This number requires 97 bits of precision, so your normal data types won't do the trick. A quad precision IEEE float would do the trick if your environment provides it. Otherwise, you'll need some form of high/arbitrary precision library.
Then, to keep the numbers within this size, you'll want cancel common terms in the numerator and the denominator. And you'll want to calculate the result using ( a / c ) * ( b / d ) * ... instead of ( a * b * ... ) / ( c * d * ... ).
I struggle with this simple problem: I want to create some random poll numbers. I have 4 variables I need to fill with data (actually an array of integer). These numbers should represent a random percentage. All percentages added will be 100% . Sounds simple.
But I think it isn't that easy. My first attempt was to generate a random number between 10 and base (base = 100), and substract the number from the base. Did this 3 times, and the last value was assigned the base. Is there a more elegant way to do that?
My question in a few words:
How can I fill this array with random values, which will be 100 when added together?
int values[4];
You need to write your code to emulate what you are simulating.
So if you have four choices, generate a sample size of random number (0..1 * 4) and then sum all the 0's, 1's, 2's, and 3's (remember 4 won't be picked). Then divide the counts by the sample size.
for (each sample) {
poll = random(choices);
survey[poll] += 1;
}
It's easy to use a computer to simulate things, simple simulations are very fast.
Keep in mind that you are working with integers, and integers don't divide nicely without converting them to floats or doubles. If you are missing a few percentage points, odds are it has to do with your integers dividing with remainders.
What you have here is a problem of partitioning the number 100 into 4 random integers. This is called partitioning in number theory.
This problem has been addressed here.
The solution presented there does essentially the following:
If computes, how many partitions of an integer n there are in O(n^2) time. This produces a table of size O(n^2) which can then be used to generate the kth partition of n, for any integer k, in O(n) time.
In your case, n = 100, and k = 4.
Generate x1 in range <0..1>, subtract it from 1, then generate x2 in range <0..1-x1> and so on. Last value should not be randomed, but in your case equal 1-x1-x2-x3.
I don't think this is a whole lot prettier than what it sounds like you've already done, but it does work. (The only advantage is it's scalable if you want more than 4 elements).
Make sure you #include <stdlib.h>
int prev_sum = 0, j = 0;
for(j = 0; j < 3; ++j)
{
values[j] = rand() % (100-prev_sum);
prev_sum += values[j];
}
values[3] = 100 - prev_sum;
It takes some work to get a truly unbiased solution to the "random partition" problem. But it's first necessary to understand what "unbiased" means in this context.
One line of reasoning is based on the intuition of a random coin toss. An unbiased coin will come up heads as often as it comes up tails, so we might think that we could produce an unbiased partition of 100 tosses into two parts (head-count and tail-count) by tossing the unbiased coin 100 times and counting. That's the essence of Edwin Buck's proposal, modified to produce a four-partition instead of a two-partition.
However, what we'll find is that many partitions never show up. There are 101 two-partitions of 100 -- {0, 100}, {1, 99} … {100, 0} but the coin sampling solution finds less than half of them in 10,000 tries. As might be expected, the partition {50, 50} is the most common (7.8%), while all of the partitions from {0, 100} to {39, 61} in total achieved less than 1.7% (and, in the trial I did, the partitions from {0, 100} to {31, 69} didn't show up at all.) [Note 1]
So that doesn't seem like a unbiased sample of possible partitions. An unbiased sample of partitions would return every partition with equal probability.
So another temptation would be to select the size of the first part of the partition from all the possible sizes, and then the size of the second part from whatever is left, and so on until we've reached one less than the size of the partition at which point anything left is in the last part. However, this will turn out to be biased as well, because the first part is much more likely to be large than any other part.
Finally, we could enumerate all the possible partitions, and then choose one of them at random. That will obviously be unbiased, but unfortunately there are a lot of possible partitions. For the case of 4-partitions of 100, for example, there are 176,581 possibilities. Perhaps that is feasible in this case, but it doesn't seem like it will lead to a general solution.
For a better algorithm, we can start with the observation that a partition
{p1, p2, p3, p4}
could be rewritten without bias as a cumulative distribution function (CDF):
{p1, p1+p2, p1+p2+p3, p1+p2+p3+p4}
where the last term is just the desired sum, in this case 100.
That is still a collection of four integers in the range [0, 100]; however, it is guaranteed to be in increasing order.
It's not easy to generate a random sorted sequence of four numbers ending in 100, but it is trivial to generate three random integers no greater than 100, sort them, and then find adjacent differences. And that leads to an almost unbiased solution, which is probably close enough for most practical purposes, particularly since the implementation is almost trivial:
(Python)
def random_partition(n, k):
d = sorted(randrange(n+1) for i in range(k-1))
return [b - a for a, b in zip([0] + d, d + [n])]
Unfortunately, this is still biased because of the sort. The unsorted list is selected without bias from the universe of possible lists, but the sortation step is not a simple one-to-one match: lists with repeated elements have fewer permutations than lists without repeated elements, so the probability of a particular sorted list without repeats is much higher than the probability of a sorted list with repeats.
As n grows large with respect to k, the number of lists with repeats declines rapidly. (These correspond to final partitions in which one or more of the parts is 0.) In the asymptote, where we are selecting from a continuum and collisions have probability 0, the algorithm is unbiased. Even in the case of n=100, k=4, the bias is probably ignorable for many practical applications. Increasing n to 1000 or 10000 (and then scaling the resulting random partition) would reduce the bias.
There are fast algorithms which can produce unbiased integer partitions, but they are typically either hard to understand or slow. The slow one, which takes time(n), is similar to reservoir sampling; for a faster algorithm, see the work of Jeffrey Vitter.
Notes
Here's the quick-and-dirty Python + shell test:
$ python -c '
from random import randrange
n = 2
for i in range(10000):
d = n * [0]
for j in range(100):
d[randrange(n)] += 1
print(' '.join(str(f) for f in d))
' | sort -n | uniq -c
1 32 68
2 34 66
5 35 65
15 36 64
45 37 63
40 38 62
66 39 61
110 40 60
154 41 59
219 42 58
309 43 57
385 44 56
462 45 55
610 46 54
648 47 53
717 48 52
749 49 51
779 50 50
788 51 49
723 52 48
695 53 47
591 54 46
498 55 45
366 56 44
318 57 43
234 58 42
174 59 41
118 60 40
66 61 39
45 62 38
22 63 37
21 64 36
15 65 35
2 66 34
4 67 33
2 68 32
1 70 30
1 71 29
You can brute force it by, creating a calculation function that adds up the numbers in your array. If they do not equal 100 then regenerate the random values in array, do calculation again.
I have an array in MATLAB containing elements such as
A=[12 13 14 15 30 31 32 33 58 59 60];
How can I identify breaks in values of data? For example, the above data exhibits breaks at elements 15 and 33. The elements are arranged in ascending order and have an increment of one. How can I identify the location of breaks of this pattern in an array? I have achieved this using a for and if statement (code below). Is there a better method to do so?
count=0;
for i=1:numel(A)-1
if(A(i+1)==A(i)+1)
continue;
else
count=count+1;
q(count)=i;
end
end
Good time to use diff and find those neighbouring differences that aren't equal to 1. However, this will return an array which is one less than the length of your input array because it finds pairwise differences up until the last element, so naturally there will be one less. As such, when you find the locations that aren't equal to 1, make sure you add 1 to the locations to account for this:
>> A=[12 13 14 15 30 31 32 33 58 59 60];
>> q = find(diff(A) ~= 1) + 1
q =
5 9
This tells us that locations 5 and 9 in your array is where the jump happens, and that's right for your example data.
However, if you want to find the locations before the jump happens, such as in your code, don't add 1 to the result:
>> q = find(diff(A) ~= 1)
q =
4 8
Divide an array into K number of parts such that each part is less than N. Also, maximize the sum of all parts. If such configuration is not possible return -1.
For ex-
Array - 5,3,2,7,8,9
K - 3
N - 71
Solution- 53 + 27 + 89 = 169
Can someone suggest an efficient algorithm for this ?
I have a Sorted array .Lets assume
{4,7,9,12,23,34,56,78} Given min and max I want to find elements in array between min and max in efficient way.
Cases:min=23 and max is 78 op:{23,34,56,78}
min =10 max is 65 op:{12,23,34,56}
min 0 and max is 100 op:{4,7,9,12,23,34,56,78}
Min 30 max= 300:{34,56,78}
Min =100 max=300 :{} //empty
I want to find efficient way to do this?I am not asking code any algorithm which i can use here like DP exponential search?
Since it's sorted, you can easily find the lowest element greater than or equal to the minimum desired, by using a binary search over the entire array.
A binary search basically reduces the serch space by half with each iteration. Given your first example of 10, you start as follows with the midpoint on the 12:
0 1 2 3 4 5 6 7 <- index
4 7 9 12 23 34 56 78
^^
Since the element you're looking at is higher than 10 and the next lowest is lesser, you've found it.
Then, you can use a similar binary search but only over that section from the element you just found to the end. This time you're looking for the highest element less than or equal to the maximum desired.
On the same example as previously mentioned, you start with:
3 4 5 6 7 <- index
12 23 34 56 78
^^
Since that's less than 65 and the following one is also, you need to increase the pointer to the halfway point of 34..78:
3 4 5 6 7 <- index
12 23 34 56 78
^^
And there you have it, because that number is less and the following number is more (than 65)
Then you have the start at stop indexes (3 and 6) for extracting the values.
0 1 2 3 4 5 6 7 <- index
4 7 9 ((12 23 34 56)) 78
-----------
The time complexity of the algorithm is O(log N). Though keep in mind that this really only becomes important when dealing with larger data sets. If your data sets do consist of only about eight elements, you may as well use a linear search since (1) it'll be easier to write; and (2) the time differential will be irrelevant.
I tend not to worry about time complexity unless the operations are really expensive, the data set size gets into the thousands, or I'm having to do it thousands of times a second.
Since it is sorted, this should do:
List<Integer> subarray = new ArrayList<Integer>();
for (int n : numbers) {
if (n >= MIN && n <= MAX) subarray.add(n);
}
It's O(n) as you only look at every number once.