My array is 1D m in length. say m = 16
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
The way I actually interpret the array is n x n = m
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
I require to read the array in this manner due to the way my physical environment is set up
0 4 8 12 13 9 5 1 2 6 10 14 15 11 7 3
What I came up with works but I really don't think it is the best way to do this:
bool isFlipped = true;
int x = 0; x < m; x++
if(isFlipped)
newLine[x] = line[((n-1)-x%n)*n + x/n)]
else
newLine[x] = line[x%n*n +x/n]
if(x != 0 && x % n == 0)
isFlipped = !isFlipped
This gives me the required result but I really think there is a way to get rid of this boolean by purely using a math formula. I am stuffing this into a 8kb microcontroller and I need to conserve as much space as I can because I will have some bluetooth communication and more math going into it later on.
Edit:
Thanks to a user I got to a one line solution-ish. (the below would replace the lines in the for-loop)
c=x/n
newLine[x] = line[((c+1)%2)*((x%n)*n+c) + (c%2)*((n-1)-2*(x%n))*n ];
You should be able to utilize the fact that odd columns in the n*n matrix are read from down up, and even columns are read from up down.
A number at index x in newLine is located in column number c=floor(x/n) in the n*n matrix. c%2 is 0 for even columns and 1 for odd columns. So something like this should work:
int c = x/n;
newLine[x] = line[(x%n)*n + (c%2)*((n-1)-2*(x%n))*n + c];
Related
powers = c(c(1:10), seq(from = 12, to=20, by=2));
While going through WGCNA i came across this code which i am not able to understand, can anybody explain me the meaning of that piece of code
The code will create a vector of numbers stored in powers.
Specifically: 1:10 creates the numbers 1 2 3 4 5 6 7 8 9 10 (can read as 1 through 10) and seq(from = 12, to = 20, by = 2) creates a sequence of every other number from 12 to 20, i.e. 12 14 16 18 20.
Powers will contain the following 15 numbers: 1 2 3 4 5 6 7 8 9 10 12 14 16 18 20
I am not familiar with the WGCNApackage or if powers is an argument to a function, but this is what powers contains.
I have an adjancecy matrix stored in CSR format. Eg
xadj = 0 2 5 8 11 13 16 20 24 28 31 33 36 39 42 44
adjncy = 1 5 0 2 6 1 3 7 2 4 8 3 9 0 6 10 1 5 7 11 2 6 8 12 3 7 9 13 4 8 14 5 11 6 10 12 7 11 13 8 12 14 9 13
I am now paritioning said graph using METIS. This gives me the partition vector part of the graph. Basically a list that tells me in which partition each vertex is. Is there an efficient way to build the new adjacency matrix for this partitioning such that I can partition the new graph again? Eg a function rebuildAdjacency(xadj, adjncy, part). If possible reusing xadj and adjncy.
I'm assuming that what you mean by "rebuild" is removing the edges between vertices that have been assigned different partitions? If so, the (probably) best you can do is iterate your CSR list, generate a new CSR list, and skip all edges that are between partitions.
In pseudocode (actually, more or less Python):
new_xadj = []
new_adjcy = []
for row in range(0, n):
row_index = xadj[row]
next_row_index = xadj[row+1]
# New row index for the row we are currently building
new_xadj.append(len(new_adjcy))
for col in adjncy[row_index:next_row_index]:
if partition[row] != partition[col]:
pass # Not in the same partition
else:
# Put the row->col edge into the new CSR list
new_adjcy.append(col)
# Last entry in the row index field is the number of entries
new_xadj.append(len(new_adjcy))
I don't think that you can do this very efficiently re-using the old xadj and adjcy fields. However, if you are doing this recursively, you can save memory allocation / deallocation by having exacyly two copies of xadj and adjc, and alternating between them.
I have been trying to solve this problem.
http://www.spoj.com/problems/DIV/
for calcuating interger factors, I tried two ways
first: normal sqrt(i) iteration.
int divCount = 2;
for (int j = 2; j * j <= i ; ++j) {
if( i % j == 0) {
if( i / j == j )
divCount += 1;
else
divCount += 2;
}
}
second: Using prime factorization (primes - sieve)
for(int j = 0; copy != 1; ++j){
int count = 0;
while(copy % primes.get(j) == 0){
copy /= primes.get(j);
++count;
}
divCount *= ( count + 1);}
While the output is correct, I am getting TLE. Any more optimization can be done? Please help. Thanks
You're solving the problem from the wrong end. For any number
X = p1^a1 * p2^a2 * ... * pn^an // p1..pn are prime
d(X) = (a1 + 1)*(a2 + 1)* ... *(an + 1)
For instance
50 = 4 * 25 = 2^2 * 5^2
d(50) = (1 + 2) * (1 + 2) = 9
99 = 3^2 * 11^1
d(99) = (2 + 1) * (1 + 1) = 6
So far so good you need to generate all the numbers such that
X = p1^a1 * p2^a2 <= 1e6
such that
(a1 + 1) is prime
(a2 + 1) is prime
having a table of prime numbers from 1 to 1e6 it's a milliseconds task
It is possible to solve this problem without doing any factoring. All you need is a sieve.
Instead of a traditional Sieve of Eratosthenes that consists of bits (representing either prime or composite) arrange your sieve so each element of the array is a pointer to an initially-null list of factors. Then visit each element of the array, as you would with the Sieve of Eratosthenes. If the element is a non-null list, it is composite, so skip it. Otherwise, for each element and for each of its powers less than the limit, add the element to each multiple of the power. At the end of this process you will have a list of prime factors of the number. That wasn't very clear, so let me give an example for the numbers up to 20. Here's the array, initially empty:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
Now we sieve by 2, adding 2 to each of its multiples:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
2 2 2 2 2 2 2 2 2 2
Since we also sieve by powers, we add 2 to each multiple of 4:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2
And likewise, by each multiple of 8 and 16:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2
2 2
2
Now we're finished with 2, so we go to the next number, 3. The entry for 3 is null, so we sieve by 3 and its power 9:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2
2 2
2
3 3 3 3 3 3
3 3
Then we sieve by 5, 7, 11, 13, 17 and 19:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2
2 2
2
3 3 3 3 3 3
3 3
5 5 5 5
7 7
11
13
17
19
Now we have a list of all the prime factors of all the numbers less than the limit, computed by sieving rather than factoring. It's easy then to calculate the number of divisors by scanning the lists; count the number of occurrences of each factor in the list, add 1 to each total, and multiply the results. For instance, 12 has 2 factors of 2 and 1 factor of 3, so take (2+1) * (1+1) = 3 * 2 = 6, and indeed 12 has 6 factors: 1, 2, 3, 4, 6 and 12.
The final step is to check if the number of divisors has exactly two factors. That's easy: just look at the list of prime divisors and count them.
Thus, you have solved the problem without doing any factoring. That ought to be very fast, just a little bit slower than a traditional Sieve of Eratosthenes and very much faster than factoring each number to compute the number of divisors.
The only potential problem is space consumption for the lists of prime factors. But you shouldn't worry too much about that; the largest list will have only 19 factors (since the smallest factor is 2, and 2^20 is greater than your limit), and 78498 of the lists will have only a single factor (the primes less than a million).
Even though the above mentioned problem doesn't require calculating number of divisors, It still can be solved by calculating d(N) (divisors of N) within the time limit (0.07s).
The idea is to pretty simple. Keep track of smallest prime factor f(N) of every number. This can be done by standard prime sieve. Now, for every number i keep dividing it by f(i) and increment the count till i = 1. You now have set of prime counts for each number i.
int d[MAX], f[MAX];
void sieve() {
for (int i = 2; i < MAX; i++) {
if (!f[i]) {
f[i] = i;
for (int j = i * 2; j < MAX; j += i) {
if (!f[j]) f[j] = i;
}
}
d[i] = 1;
}
for (int i = 1; i < MAX; i++) {
int k = i;
while (k != 1) {
int s = 0, fk = f[k];
while (k % fk == 0) {
k /= fk; s++;
}
d[i] *= (s + 1);
}
}
}
Once, d(N) is figured out, rest of the problem becomes much simpler. Keeping a smallest prime factor of every number also helps to solve lots of other problems.
I have been searching for long but unable to find a solution for this.
My question is "Suppose you have n street lights(cannot be moved) and if you get any m from them then it should have atleast k working.Now in how many ways can this be done"
This seems to be a combination problem, but the problem here is "m" must be sequential.
Eg:
1 2 3 4 5 6 7 (Street lamps)
Let m=3
Then the valid sets are,
1 2 32 3 43 4 54 5 65 6 7Whereas,1 2 4 and so are invalid selections.
So every set must have atleast 2 working lights. I have figured how to find the minimum lamps required to satisfy the condition but how can I find the number of ways in it can be done ?
There should certainly some formula to do this but I am unable to find it.. :(
Should always be (n-m)+1.
E.g., 10 lights (n = 10), 5 in set (m = 5):
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
6 7 8 9 10
Gives (10-5)+1 = 6 sets.
The answer should always be m choose k for all values of n where n > m > k. I'll try to explain why;
Given, for example, the values m = 10, n = 4, k = 2, you can start by generating all possible permutations of 1s and 0s for sets of 4 lights, with exactly 2 lights on;
1100
0110
0011
1001
0101
1010
As you can see, there are 6 permutations, because 4 choose 2 = 6. You can choose any of these 6 permutations to be the first 4 lights. You then continue the sequence until you get n (in this case 10) lights, ensuring that you only ever add a zero if you must in order to keep the condition true of having 2 lights on for every 4. What you will find is that the sequence simply repeats; for example:
1100 -> next can be 1, so 11001
Next can still be 1 and meet the condition, so 110011.
The next must now be a zero, giving 1100110, and then again -> 11001100. This simply continues until the length is n : 1100110011. Given that the starting four can only be one of the above set, you will only get 6 different permutations.
Now, since the sequence will repeat exactly the same for any value of n, it means that the answer will always be m choose k.
For your example in your comment of 6,3,2, I can only find the following permutations:
011011
110110
101101
Which works, because 3 choose 2 = 3. If you can find more, then I guess I'm wrong and I've probably misunderstood again :D but from my understanding of this problem, I'm certain that the answer will always be m choose k.
int maxValue = m[0][0];
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
if ( m[i][j] >maxValue )
{
maxValue = m[i][j];
}
}
}
cout<<maxValue<<endl;
int sum = 0;
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
sum = sum + m[i][j];
}
}
cout<< sum <<endl;
For the above code if we draw a flow graph like this basic independent paths would be following six
Path 1: 1 2 3 10 11 12 13 19
Path 2: 1 2 3 10 11 12 13 14 15 18 13 19
Path 3: 1 2 3 10 11 12 13 14 15 16 17 15 18 13 19
Path 4: 1 2 3 4 5 9 3 10 11 12 13 19
Path 5: 1 2 3 4 5 6 8 5 9 3 10 11 12 13 14 15 16 17 15 18 13 19
Path 6: 1 2 3 4 5 6 7 8 5 9 3 10 11 12 13 14 15 16 17 15 18 13 19
So the question here is according to the given code path 2, 3, 4 can not be tested (Note the "N" in loops). So is it okay not to have a actual execution path as given in the basic set?...
or according to macabe complexity metric do we have to change the code given above. Because a tutor of mine said we have to change the code also he said that there are unstructured loops so we have to change the code. (I don't see an unstructured loop as well)
But my feeling is, if we change the code actual output may differ to expected output. So please someone explain this
1) McCabe's complexity can be calculated as the number of decision points + 1. In your case there are 5 decision points (nodes 3, 5, 6, 13 and 15) meaning that the McCabe complexity of the code fragment is 5+1 = 6. 6 is by no means too high in terms of McCabe complexity: one could, of course, still argue that it is too high given the functionality the implementation has to provide.
2) McCabe's complexity is related to testability of a method/procedure but not to testability of a specific path. Paths can be feasible (= there exist values of the variables that force the execution through this path) or not, but McCabe's complexity is happily unaware of such complications. If you really want to look into feasibility of paths keep in mind that the problem in general is undecidable but many practical data flow analysis algorithms are available.
3) if we change the code actual output may differ to expected output Of course, you cannot introduce an arbitrary change and hope that the results will be the same. However, and, this is probably what your tutor intended, there is a way of restructuring your code such that the output produced remains the same, and the McCabe's complexity goes down. Think, e.g., on whether you really need to separate the tasks of calculating the maximum and the sum.