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How can I find a largest prime number which is less than n, where n ≤ 10¹⁸ ?
Please help me find an Efficient Algorithm.
for(j=n;j>=2;j--) {
if(j%2 == 1) {
double m = double(j);
double a = (m-1)/6.0;
double b = (m-5)/6.0;
if( a-(int)a == 0 || b-(int)b == 0 ) {
printf("%llu\n",j);
break;
}
}
}
I used this approach but this is not efficient to solve for n>10^10;
How to optimize this..
Edit:
Solution: Use Primality test on each j.
Miller Rabin, Fermat's Test.
I don't think this question should be so quickly dismissed, as efficiency is not so easy to determine for numbers in this range. First of all, given the average prime gap is ln(p), it makes sense to work down from the given (n). i.e., ln(10^18) ~ 41.44), so you would expect around 41 iterations on average working down from (n). e.g., testing: (n), (n - 2), (n - 4), ...
Given this average gap, the next step is to decide whether you wish to use a naive test - check for divisibility by primes <= floor(sqrt(n)). With n <= (10^18), you would need to test against primes <= (10^9). There are ~ 50 million primes in this range. If you are willing to precompute and tabulate these values (all of which fit in 32 bits), you have a reasonable test for 64-bit values n <= 10^18. In this case, is a 200MB prime table an acceptable approach? 20 years ago, it might not have been. Today, it's not an issue.
Combining a prime table with sieving and/or Pocklington's test might improve efficiency. Alternatively, if memory is more constrained, a deterministic variant of the Miller-Rabin test with bases: 2, 325, 9375, 28178, 450775, 9780504, 1795265022 (SPRP set). Most composites fail immediately with an SPRP-2 test.
The point is - you have a decision to make between algorithmic complexity, both theoretical and in terms of implementation difficulty, and a balance with space/time trade-offs.
Related
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I have come across a problem about the determination of triangle, it says:
Given a sorted integer array(length n), determinate whether you could
build a triangle by choosing three integers from the array, the
answer is "yes" or "no".
A naive solution is by scanning all the possibilities but it turn out to be O(n^3), seems
it will be C(n, 3) possibilities.
Assuming that the integers represent side lengths and array(0) > 0,
bool IsTriangle(int[] aray, int start) {
if(array.length - start <= 2) return false;
return (array(start+2) < array(start+1) + array(start+0))
|| IsTriangle(array,start+1);
}
This works because the list of integers is sorted; thus the RHS will always be larger using any subsequent elements of the array, and the LHS will be smaller using any previous elements of the array, and thus can satisfy the triangle inequality only if the selected three consecutive elements satisfy it. This is of course O(n) and can easily be converted to a (less elegant but more performant) iterative solution.
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I'm struggling with a dynamic programming problem for a couple of days. It goes like this:
John's working day is divided in N time slots, every slot i having associated a gain G[i] which he can receive is he works in that time slot. If he decides to work in the time interval [i, j] his total reward would be R[i,j]=G[i+1]+...+G[j] as the first slot is for warming up. Everyday he has to work exactly T slots - he can chose a subset of T slots from the available N total slots. He wants to maximize his profit by choosing a set of disjunct intervals [a1,b1], [a2,b2], ...[ak,bk] with 1 <= a1 <= b1 < a2 <= b2 <...< ak <= bk and Sum[i=1, k](bi-ai+1)=T.
Example: N=7, T=5 and the gain vector {3,9,1,1,7,5,4}. The optimal solution is selecting the intervals [1,2] and [4,6] with a total profit of 9+12=21.
DP solution:
int f[i][j][0..1];
let f[i][j][0] denotes the maximal gain for the first i time slots and using j time slots, and the i-th time slot is not used.
let f[i][j][1] denotes the maximal gain for the first i time slots and using j time slots, and the i-th time slot is used.
obviously,f[i][j][k] can determine f[i+1][j][k] or f[i+1][j+1][k]. details below:
f[i+1][j+1][1]=max(f[i+1][j+1][1],f[i][j][0],f[i][j][1]+G[i+1]);
f[i+1][j][0]=max(f[i+1][j][0],f[i][j][0],f[i][j][1]);
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I want to generate a very large pseudorandom permutation p : [0,n-1] -> [0,n-1], and then compute m specific values p[i], where m << n. Is it possible to do this in O(m) time? The motivation is a large parallel computation where each processor only needs to see a small piece of the permutation, but the permutation must be consistent between processors.
Note that in order to help in the parallel case, different processes computing disjoint sets of i values shouldn't accidentally produce p[i] == p[j] for i != j.
EDIT: There is a much more clever algorithm based on block ciphers that I think Geoff will write up.
There are two common algorithms for generating permutations. Knuth's shuffle is inherently sequential so not a nice choice for parallelism. The other is random selection with retry any time repetition is encountered. Random selection is clearly equivalent when applied in any order, thus I propose the following simple algorithm:
Randomly sample candidate p[i] in [0,n-1] for each i in Needed (in parallel).
Remove all non-collided entries from Needed, as well as (optionally) some deterministic choice from the collisions (e.g., keep p[i] if i < {j | p[j] = p[i]}).
Repeat from step 1 with new (smaller) set Needed.
Since we haven't lost entropy in this process, the result is essentially equivalent to sequential random sampling in some different order, starting with the locations i that did not collide (we just didn't know that order in advance). Note that if we used the computed value in a comparison, for example, we would have introduced bias.
An example very low strength version:
Generate 2k = O(1) random integers a_i,b_i in [0,n-1], with a_i relatively prime to n.
Pick a weak permutation wp : [0,n-1] -> [0,n-1], say w(i) = i with all the but the high bit flipped.
p[i] = b_0 + a_0 * wp(b_1 + a_1 * wp(... i ...))
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We have N books ( N<=200 ). All of them must be translated by K people (K<=100).
Every man can translate D books starting from index S to index S+D-1, 0<=D<=N.
Every man is paid c_1 dolars per page for the first book which he translates, c_2 for the second...
c_i for the book i.
0<=c_i<10000
The books must be translated in the order they are given.
input:
first row: 2 numbers N and K
second row: N numbers - number of pages per every book (<=10 000)
third row: N numbers - c_1, c_2, ... c_N; c_i is the price for translating a book by a man who has translated i-1 books;
output:
minimum price which must be paid for the translation of all the books.
Example:
Input:
6 3
50 100 60 5 6 30
1 2 3 4 5 6
Output: 339
(the first man translated the first book +50*1
the scond man translated the second,third,forth and fifth books:
+100*1+60*2+5*3+6*4
the third man translates the last book
+30*1
=339)
Can someone help me with this homework? I know i must be using dynamic programming to solve it.
Some clues: make a function F(BookNumber, ManNumber, NumOfBooksHeHaveTranslated).
Start with F(1, 1, 0). It is clear that
F(1, 1, 0) = Pages[1] * C[1] + Min(F(2, 1, 1), F(2, 2, 0))
i.e. we have to choose the best variant from - continue with the same translator, or use the next one. Elaborate this function for common case F(B, M, N), make recursive solution, check it for small inputs, transform recursion into DP (the methods are described in algo books)
The actual choice your algorithm needs to do is at which points in the sequence of books you change from one translator to the next one. That's the only decision point for the problem, and everything else follows from it. You can translate this to a recursive / dynamic programming problem by observing e.g. that the cost of translating N books by K translators is equivalent to the cost of translating x first books by one translator and N - x by the remaining K - 1 translators; or, equivalent to the cost of translating x first books by n translators and N - x by the remaining K - n translators. This is a subdivision / recursion step that you can use in a dynamic programming solution.
I hope no-one will post actually code to do this; it is homework.
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Good evening I was wondering if someone could please provide me with a simple pseudocode example of a deterministic algorithm... I will greatly appreciate it and surely give you points!!. thanks
To me, "deterministic" could mean many things:
Given the same input, produces the same output every time.
Given the same input, takes the same amount of time/memory/resources every time it is run.
Problems of complexity class P that can be solved in polynomial time by a deterministic computer, as opposed to problems of complexity class NP which can be only solved in polynomial time using a non-deterministic computer.
Which of these do you mean?
The most simple deterministic algorithm is this random number generator.
def random():
return 4 #chosen by fair dice roll, guaranteed to be random
It gives the same output every time, exhibits known O(1) time and resource usage, and executes in PTIME on any computer.
Do you really mean DETERMINISTIC and not NONdeterministic, I mean pretty much anything you see in any tutorial / guide / start book is deterministic, e.g.
for i from 1 to 9
print i
will always print 123456789
A deterministic algorithm is simply an algorithm that has a predefined output. For instance if you are sorting elements that are strictly ordered(no equal elements) the output is well defined and so the algorithm is deterministic.
In fact most of the computer algorithms are deterministic. Undeterminism usually apears when you have some parallelization or some equal elements that are only equal according to some non-full criteria.
Here's pseudo code for a deterministic algorithm that checks whether a given number is odd:
function is_odd(n):
if n mod 2 = 1
then return true
else return false
deterministic algorithm is an algorithm which, in informal terms, behaves predictably. Given a particular input, it will always produce the same output
public struct Point {
public int x;
public int y;
//other methods
public override int GetHashCode() {
return x ^ y;
}
}
Point P=new Point();
p.x=6;
p.y=3;
int res= p.GetHashCode();