Is it possible to find prime factors of factorial without actually calculating the factorial?
My point here is to find prime factors of factorial not of a big number. Your algorithm should skip the step of having to calculate the factorial and derive prime factors from n! where n <= 4000.
Calculating the factorial and finding it's prime divisors is pretty easy, but my program crashes when the input is greater than n=22. Therfore I thought it would be pretty convinent to do the whole process without having to calculate the factorial.
function decomp(n){
var primeFactors = [];
var fact = 1;
for (var i = 2; i <= n; i++) {
fact = fact * i;
}
while (fact % 2 === 0) {
primeFactors.push(2);
fact = fact/2;
}
var sqrtFact = Math.sqrt(fact);
for (var i = 2; i <= sqrtFact; i++) {
while (fact % i === 0) {
primeFactors.push(i);
fact = fact/i;
}
}
return primeFactors;
}
I don't expect any code nor links, exemplifactions and a brief outline is enough.
Let's consider an example: 10! = 2^8 * 3^4 * 5^2 * 7^1. I computed that by computing the factors of each number from 2 to 10:
2: 2
3: 3
4: 2,2
5: 5
6: 2,3
7: 7
8: 2,2,2
9: 3,3
10: 2,5
Then I just counted each factor. There are eight 2's (1 in 2, 2 in 4, 1 in 6, 3 in 8, and 1 in 10), four 3's (1 in 3, 1 in 6, and 2 in 9), two 5's (1 in 5, and 1 in 10), and one 7 (in 7).
In terms of writing a program, just keep an array of counters (it only needs to be as large as the square root of the largest factorial you want to factor) and, for each number from 2 to the factorial, add the count of its factors to the array of counters.
Does that help?
Related
Combinations without repetitions look like this, when the number of elements to choose from (n) is 5 and elements chosen (r) is 3:
0 1 2
0 1 3
0 1 4
0 2 3
0 2 4
0 3 4
1 2 3
1 2 4
1 3 4
2 3 4
As n and r grows the amount of combinations gets large pretty quickly. For (n,r) = (200,4) the number of combinations is 64684950.
It is easy to iterate the list with r nested for-loops, where the initial iterating value of each for loop is greater than the current iterating value of the for loop in which it is nested, as in this jsfiddle example:
https://dotnetfiddle.net/wHWK5o
What I would like is a function that calculates only one combination based on its index. Something like this:
tuple combination(i,n,r) {
return [combination with index i, when the number of elements to choose from is n and elements chosen is r]
Does anyone know if this is doable?
You would first need to impose some sort of ordering on the set of all combinations available for a given n and r, such that a linear index makes sense. I suggest we agree to keep our combinations in increasing order (or, at least, the indices of the individual elements), as in your example. How then can we go from a linear index to a combination?
Let us first build some intuition for the problem. Suppose we have n = 5 (e.g. the set {0, 1, 2, 3, 4}) and r = 3. How many unique combinations are there in this case? The answer is of course 5-choose-3, which evaluates to 10. Since we will sort our combinations in increasing order, consider for a minute how many combinations remain once we have exhausted all those starting with 0. This must be 4-choose-3, or 4 in total. In such a case, if we are looking for the combination at index 7 initially, this implies we must subtract 10 - 4 = 6 and search for the combination at index 1 in the set {1, 2, 3, 4}. This process continues until we find a new index that is smaller than this offset.
Once this process concludes, we know the first digit. Then we only need to determine the remaining r - 1 digits! The algorithm thus takes shape as follows (in Python, but this should not be too difficult to translate),
from math import factorial
def choose(n, k):
return factorial(n) // (factorial(k) * factorial(n - k))
def combination_at_idx(idx, elems, r):
if len(elems) == r:
# We are looking for r elements in a list of size r - thus, we need
# each element.
return elems
if len(elems) == 0 or len(elems) < r:
return []
combinations = choose(len(elems), r) # total number of combinations
remains = choose(len(elems) - 1, r) # combinations after selection
offset = combinations - remains
if idx >= offset: # combination does not start with first element
return combination_at_idx(idx - offset, elems[1:], r)
# We now know the first element of the combination, but *not* yet the next
# r - 1 elements. These need to be computed as well, again recursively.
return [elems[0]] + combination_at_idx(idx, elems[1:], r - 1)
Test-driving this with your initial input,
N = 5
R = 3
for idx in range(choose(N, R)):
print(idx, combination_at_idx(idx, list(range(N)), R))
I find,
0 [0, 1, 2]
1 [0, 1, 3]
2 [0, 1, 4]
3 [0, 2, 3]
4 [0, 2, 4]
5 [0, 3, 4]
6 [1, 2, 3]
7 [1, 2, 4]
8 [1, 3, 4]
9 [2, 3, 4]
Where the linear index is zero-based.
Start with the first element of the result. The value of that element depends on the number of combinations you can get with smaller elements. For each such smaller first element, the number of combinations with first element k is n − k − 1 choose r − 1, with potentially some of-by-one corrections. So you would sum over a bunch of binomial coefficients. Wolfram Alpha can help you compute such a sum, but the result still has a binomial coefficient in it. Solving for the largest k such that the sum doesn't exceed your given index i is a computation you can't do with something as simple as e.g. a square root. You need a loop to test possible values, e.g. like this:
def first_naive(i, n, r):
"""Find first element and index of first combination with that first element.
Returns a tuple of value and index.
Example: first_naive(8, 5, 3) returns (1, 6) because the combination with
index 8 is [1, 3, 4] so it starts with 1, and because the first combination
that starts with 1 is [1, 2, 3] which has index 6.
"""
s1 = 0
for k in range(n):
s2 = s1 + choose(n - k - 1, r - 1)
if i < s2:
return k, s1
s1 = s2
You can reduce the O(n) loop iterations to O(log n) steps using bisection, which is particularly relevant for large n. In that case I find it easier to think about numbering items from the end of your list. In the case of n = 5 and r = 3 you get choose(2, 2)=1 combinations starting with 2, choose(3,2)=3 combinations starting with 1 and choose(4,2)=6 combinations starting with 0. So in the general choose(n,r) binomial coefficient you increase the n with each step, and keep the r. Taking into account that sum(choose(k,r) for k in range(r,n+1)) can be simplified to choose(n+1,r+1), you can eventually come up with bisection conditions like the following:
def first_bisect(i, n, r):
nCr = choose(n, r)
k1 = r - 1
s1 = nCr
k2 = n
s2 = 0
while k2 - k1 > 1:
k3 = (k1 + k2) // 2
s3 = nCr - choose(k3, r)
if s3 <= i:
k2, s2 = k3, s3
else:
k1, s1 = k3, s3
return n - k2, s2
Once you know the first element to be k, you also know the index of the first combination with that same first element (also returned from my function above). You can use the difference between that first index and your actual index as input to a recursive call. The recursive call would be for r − 1 elements chosen from n − k − 1. And you'd add k + 1 to each element from the recursive call, since the top level returns values starting at 0 while the next element has to be greater than k in order to avoid duplication.
def combination(i, n, r):
"""Compute combination with a given index.
Equivalent to list(itertools.combinations(range(n), r))[i].
Each combination is represented as a tuple of ascending elements, and
combinations are ordered lexicograplically.
Args:
i: zero-based index of the combination
n: number of possible values, will be taken from range(n)
r: number of elements in result list
"""
if r == 0:
return []
k, ik = first_bisect(i, n, r)
return tuple([k] + [j + k + 1 for j in combination(i - ik, n - k - 1, r - 1)])
I've got a complete working example, including an implementation of choose, more detailed doc strings and tests for some basic assumptions.
I have a spreadsheet where I put numbers that represent number of verses on each paragraph of a book.
I manually distribute sequential paragraphs by number of verses, so in the spreadsheet I'll have something like this:
Verses Day
5 1
6 1
3 1
10 2
8 3
4 3
2 3
6 4
3 4
10 5
3 5
2 6
5 6
10 7
= 2,7080128015
By summing the total of verses for each day - in this case, 7 days - I get the standard deviation and try to reduce it for a better distribution of paragraphs.
The question is: what is the best way to find the least standard deviation?
I thought on using brute force to generate all possible combinations, but that is not a good idea if the number increases.
EDIT: The standard deviation is based on total number of verses of each day, which are identified sequentialy. Day 1 has total of 14 verses, day 2, 10 and so on.
1 14
2 10
3 14
4 9
5 13
6 7
7 10
= 2,7080128015
Since the total number of verses and the number of days is constant, you want to minimize
sum (avg verse count - verse count of day i)^2
i
avg verse count is a constant and simply the total number of verses divided by the number of days.
This problem can be solved with a dynamic program over the days. Let us build the partial solution function f(days, paragraph) that gives us the minimal sum of squares for distributing paragraphs 0 through paragraph over days days. We are interested in the last value of this function.
We can build the function incrementally. Calculating f(1, p) for any p is straight-forward since we just need to calculate the differences to the average and square. Then, for all other days, we can calculate
f(d, p) = min f(d - 1, i) + (avg verse count - sum verse count of paragraph j)^2
i<p j:i+1..p
That means, we check the solutions for one day less and fill up the current day with the paragraphs between the previous day's end paragraph and p. While we calculate this function, we keep a pointer to the chosen minimum element (as usual for a dynamic program). When we are done calculating the entire function, we just follow the pointers back to the start, which will give us the partitioning.
The algorithm has a running time of O(d * p^2), where d is the number of days and p is the number of paragraphs.
Example Code
Here is some example C# code that implements the above algorithm:
struct Entry
{
public double minCost;
public int predecessor;
}
public static void Main()
{
//input data
int[] versesPerParagraph = { 5, 6, 3, 10, 8, 4, 2, 6, 3, 10, 3, 2, 5, 10 };
int days = 7;
//calculate constants
double avgVerses = (double)versesPerParagraph.Sum() / days;
//set up DP table (f(d,p))
int paragraphs = versesPerParagraph.Length;
Entry[,] dp = new Entry[days, paragraphs];
//initialize table
int verseCount = 0;
for(int p = 0; p < paragraphs; ++p)
{
verseCount += versesPerParagraph[p];
double diff = avgVerses - verseCount;
dp[0, p].minCost = diff * diff;
dp[0, p].predecessor = -1;
}
//run dynamic program
for(int d = 1; d < days; ++d)
{
for(int p = d; p < paragraphs; ++p)
{
verseCount = 0;
dp[d, p].minCost = double.MaxValue;
for(int i = p; i >= d; --i)
{
verseCount += versesPerParagraph[i];
double diff = avgVerses - verseCount;
double cost = dp[d - 1, i - 1].minCost + diff * diff;
if(cost < dp[d, p].minCost)
{
dp[d, p].minCost = cost;
dp[d, p].predecessor = i - 1;
}
}
}
}
//reconstruct the partitioning
{
int p = paragraphs - 1;
for (int d = days - 1; d >= 0; --d)
{
int predecessor = dp[d, p].predecessor;
//calculate number of verses, just to show them
verseCount = 0;
for (int i = predecessor + 1; i <= p; ++i)
verseCount += versesPerParagraph[i];
Console.WriteLine($"Day {d} ranges from paragraph {predecessor + 1} to {p} and has {verseCount} verses.");
p = predecessor;
}
}
}
The output is:
Day 6 ranges from paragraph 13 to 13 and has 10 verses.
Day 5 ranges from paragraph 10 to 12 and has 10 verses.
Day 4 ranges from paragraph 9 to 9 and has 10 verses.
Day 3 ranges from paragraph 6 to 8 and has 11 verses.
Day 2 ranges from paragraph 4 to 5 and has 12 verses.
Day 1 ranges from paragraph 2 to 3 and has 13 verses.
Day 0 ranges from paragraph 0 to 1 and has 11 verses.
This partitioning gives a standard deviation of 1.15.
Right now I have a code that can find the number of combinations of a sum of a value using numbers greater than zero and less than the value.
I need to alter the value in order to expand the combinations so that they include more than just the value.
For example:
The number 10 yields the results:
[1, 2, 3, 4], [1, 2, 7],
[1, 3, 6], [1, 4, 5],
[1, 9], [2, 3, 5], [2, 8],
[3, 7], [4, 6]
But I need to expand this to including any number that collapses to 1 as well. Because in essence, I need 100 = n in that the sum of the individual numbers within the digits = n. So in this case 100 = 1 because 100 --> 1+0+0 = 1
Therefore the number 1999 will also be a valid combination to list for value = 100 because 1999 = 1+9+9+9 = 28, and 28 = 2+8 = 10, and 10 = 1+0 = 1
Now I realize that this will yield an infinite series of combinations, so I will need to set limits to the range I want to acquire data for. This is the current code I am using to find my combinations.
def a(lst, target, with_replacement=False):
def _a(idx, l, r, t, w):
if t == sum(l): r.append(l)
elif t < sum(l): return
for u in range(idx, len(lst)):
_a(u if w else (u + 1), l + [lst[u]], r, t, w)
return r
return _a(0, [], [], target, with_replacement)
for val in range(100,101):
s = range(1, val)
solutions = a(s, val)
print(solutions)
print('Value:', val, "Combinations", len(solutions))
You seem to have multiple issues.
To repeatedly add the decimal digits of an integer until you end with a single digit, you could use this code.
d = val
while d > 9:
d = sum(int(c) for c in str(d))
This acts in just the way you describe. However, there is an easier way. Repeatedly adding the decimal digits of a number is called casting out nines and results in the digital root of the number. This almost equals the remainder of the number when divided by nine, except that you want to get a result of 9 rather than 1. So easier and faster code is
d = val % 9
if d == 0:
d == 9
or perhaps the shorter but trickier
d = (val - 1) % 9 + 1
or the even-more-tricky
d = val % 9 or 9
To find all numbers that end up at 7 (for example, or any digit from 1 to 9) you just want all numbers with the remainder 7 when divided by 9. So start at 7 and keep adding 9 and you get all such values.
The approach you are using to find all partitions of 7 then arranging them into numbers is much more complicated and slower than necessary.
To find all numbers that end up at 16 (for example, or any integer greater than 9) your current approach may be best. It is difficult otherwise to avoid the numbers that directly add to 7 or to 25 without going through 16. If this is really what you mean, say so in your question and we can look at this situation further.
I want to solve a mathematical problem in a fastest possible way.
I have a set of natural numbers between 1 to n, for example {1,2,3,4,n=5} and I want to calculate a formula like this:
s = 1*2*3*4+1*2*3*5+1*2*4*5+1*3*4*5+2*3*4*5
as you can see, each element in the sum is a multiplications of n-1 numbers in the set. For example in (1*2*3*4), 5 is excluded and in (1*2*3*5), 4 is excluded. I know some of the multiplications are repeated, for example (1*2) is repeated in 3 of the multiplications. How can I solve this problem with least number of multiplications.
Sorry for bad English.
Thanks.
Here is a way that does not "cheat" by replacing multiplication with repeated addition or by using division. The idea is to replace your expression with
1*2*3*4 + 5*(1*2*3 + 4*(1*2 + 3*(1 + 2)))
This used 9 multiplications for the numbers 1 through 5. In general I think the multiplication count would be one less than the (n-1)th triangular number, n * (n - 1) / 2 - 1. Here is Python code that stores intermediate factorial values to reduce the number of multiplications to just 6, or in general 2 * n - 4, and the addition count to the same (but half of them are just adding 1):
def f(n):
fact = 1
term = 2
sum = 3
for j in range(2, n):
fact *= j
term = (j + 1) * sum
sum = fact + term
return sum
The only way to find which algorithm is the fastest is to code all of them in one language, and run each using a timer.
The following would be the most straightforward answer.
def f(n):
result = 0
nList = [i+1 for i in range(n)]
for i in range(len(nList)):
result += reduce(lambda x, y: x*y,(nList[:i]+nList[i+1:]))
return result
Walkthrough - use the reduce function to multiply all list's of length n-1 and add to the variable result.
If you just want to minimise the number of multiplications, you can replace all the multiplications by additions, like this:
// Compute 1*2*…*n
mult_all(n):
if n = 1
return 1
res = 0
// by adding 1*2*…*(n-1) an entirety of n times
for i = 1 to n do
res += mult_all(n-1)
return res
// Compute sum of 1*2*…*(i-1)*(i+1)*…*n
sum_of_mult_all_but_one(n):
if n = 1
return 0
// by computing 1*2*…*(n-1) + (sum 1*2*…*(i-1)*(i+1)*…*(n-1))*n
res = mult_all(n-1)
for i = 1 to n do
res += sum_of_mult_all_but_one(n-1)
return res
Here is an answer that would work with javascript. It is not the fastest way because it is not optimized, but it should work if you want to just find the answer.
function combo(n){
var mult = 1;
var sum = 0;
for (var i = 1; i <= n; i++){
mult = 1;
for (var j = 1; j<= n; j++){
if(j != i){
mult = mult*j;
}
}
sum += mult;
}
return (sum);
}
alert(combo(n));
I was asked on an interview to find all numbers only divisible by 3, 5 and 7. I purposed we can make check like
if (num%3==0 || num%5==0 || num%7==0)
return true
else
return false.
But in this case if we have 6 it will pass the test but its also divisible by 2 so this doesn't work. Can you purpose something?
I am using java. Find mean to check if some number is divisible only to this number
I would approach this by removing all of the factors of 3, 5, and 7 from the original number, and seeing what's left.
while(num % 3 == 0)
{
num = num / 3;
}
while(num % 5 == 0)
{
num = num / 5;
}
while(num % 7 == 0)
{
num = num / 7;
}
return (num == 1);
I won't give you a Java algorithm, as it should be fairly easy to implement.
You can just:
1. check if (n%3 == 0)
2. if it is, set n /= 3 and repeat step 1.
3. do the same for the number 5 and 7
4. now if n != 1, return false, else return true
In a Java algorithm:
// n is some random natural number
if (n == 1 || n == 0)
return false
while (!n%3)
{
n /= 3;
}
while (!n%5)
{
n /= 5;
}
while (!n%7)
{
n /= 7;
}
if (n == 1)
{
return true;
}
else
{
return false;
}
It's not the best syntax, I'm just giving an straight-forward implementation of the algorithm presented above.
We first note that 1 is a member of the set. Although it is not divisible by 3, 5 or 7, neither is it divisible by any number other than 3, 5 or 7, so we will say that 1 is in the set. This conforms to the mathematical definition of the set { x = 3i · 5j · 7k | i, j, k ≥ 0 }.
One method is to count from 1, adding 2 at each step, and checking if the number is divisible only by 3, 5 and 7. That's slow because it does a lot of work that immediately gets discarded, since there are many fewer numbers divisible only by 3, 5 and 7 than there are odd numbers.
A better approach is to generate the desired numbers directly, by induction. The number 1 is in the set, and for any x in the set, so are 3 x, 5 x and 7 x. So the algorithm to generate all numbers divisible only by 3, 5 and 7, in order, is:
1. Initialize a priority queue with the number 1.
2. Pop the smallest number in the priority queue, call it x.
3. Add 3x, 5x and 7x to the priority queue.
4. Output x as the next integer in the set.
5. If you want more output, go to Step 2.
6. Halt.
I implemented both algorithms; you can see them at http://ideone.com/YwnAQ8. The brute-force method takes a little over ten seconds to find the 203 members of the 3,5,7 set less than a million; the priority queue does the same calculation in a hundredth of a second, a thousand times faster. The priority queue implementation used there is explained at my blog. You can also see the set of 3,5,7 numbers at OEIS.