void f2(int n)
{
if (n<=1)
return;
g2(n, n/3);
}
void g2(int n, int m)
{
int i=1;
while (m < n) {
m += i;
i++;
}
f2(n/2);
}
I tried alot to calculate the time complexity and got it wrong, I would really appreciate it if someone could help me on how to approach these programs. (The answer is O(sqrt(n)).
The following explanation can be simplified, but I tried to be as much scrupulous as possible.
Sum of arithmetic progression
First of all lets talk about complixity of the following loop (note the m=0):
int m=0;
int i=1;
while (m < n) {
m += i;
i++;
}
Invariant of the loop is: after ith iteration m == 1+2+...+i == (1+i)*i/2. So the loop stops when the following condition is met:
which is equavalent to
Big O of the left and right parts are equal and both equal to O(i), so O(i)=O(sqrt(n)) is the complexity of the loop.
Complexity of the loop inside g2
The loop inside the g2 is equavalent to the following loop:
int n_modified = n - m;
m = 0;
int i=1;
while (m < n_modified) {
m += i;
i++;
}
which complexity is O(sqrt(n-m)) as we've shown in the previous section.
Complexity of the f2
Now lets get overall formula for complexity of the f2 function. Its complexity is essentially the same as complexity of the g2(n, n/3) call. It consists of two parts: complexity of the g2's loop and complexitiy of the recursion. That is, the formula is
This can be simplified and estimated (factoring and sum of geometric progression):
which gives us the final answer: the complexity of f2 is O(sqrt(n)).
public void function2(long input) {
long s = 0;
for (long i = 1; i < input * input; i++){
for(long j = 1; j < i * i; j++){
s++;
}
}
}
l'm pretty certain that the time complexity of this function is n^3, however if someone could provide a line by line explanation of this, that would be great.
First of all, you need to define what n is if you write something like O(n^3), otherwise it doesn't make any sense. Let's say n is the value (as opposed to e.g. the bit-length) of input, so n = input.
The outer loop has k iterations, where k = n^2. The inner loop has 1^2, 2^2, 3^2, ... up to k^2 iterations, so summing up everything you get O(k^3) iterations (since the sum of the p-th powers of the first m integers is always O(m^(p+1))).
Hence the overall time complexity is O(n^6).
I am trying to find the tightest upper bound for the following upper bound. However, I am not able to get the correct answer. The algorithm is as follows:
public staticintrecursiveloopy(int n){
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
System.out.println("Hello.");
}
} if(n <= 2) {
return 1;
} else if(n % 2 == 0) {
return(staticintrecursiveloopy(n+1));
} else{
return(staticintrecursiveloopy(n-2));
}
}
I tried to draw out the recursion tree for this. I know that for each run of the algorithm the time complexity will be O(n2) plus the time taken for each of the recursive calls. Also, the recursion tree will have n levels. I then calculated the total time taken for each level:
For the first level, the time taken will be n2. For the second level, since there are two recursive calls, the time taken will be 2n2. For the third level, the time taken will be 4n 2 and so on until n becomes <= 2.
Thus, the time complexity should be n2 * (1 + 2 + 4 + .... + 2n). 1 + 2 + 4 + .... + 2n is a geometric sequence and its sum is equal to 2n - 1.Thus, the total time complexity should be O(2nn2). However, the answer says O(n3). What am I doing wrong?
Consider the below fragment
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
System.out.println("Hello.");
}
}
This doesn't need any introduction and is O(n2)
Now consider the below fragment
if(n <= 2) {
return 1;
} else if(n % 2 == 0) {
return(staticintrecursiveloopy(n+1));
} else {
return(staticintrecursiveloopy(n-2));
}
How many times do you think this fragment will be executed?
If n%2 == 0 then the method staticintrecursiveloopy will be executed 1 extra time. Otherwise it goes about decresing it by 2, thus it'll be executed n/2 times (or (n+1)/2 if you include the other condition).
Thus the total number of times the method staticintrecursiveloopy will be executed is roughly n/2 which when expressed in terms of complexity becomes O(n).
And the method staticintrecursiveloopy calls a part with complexity O(n2) in each iteration, thus the total time complexity becomes
O(n) * O(n2) = O(n3).
Using only for or while statements, I'm trying to come up with a program to generate and print a table of the first 10 factorials. Here's my code:
for (count = 1; count<=10; ++count)
{
n = count;
while (n > 0){
count *= (count-1);
n -= 1;
}
NSLog(#" %2g %3g", count, factorial);
}
I don't understand why this is not working. It never gets out of the loop and goes on forever. What's the correction? Thank you!
The reason:
count *= (count-1);
Since count starts at 1, it will always be reset to 0, so the count <= 10 condition of the outer loop will always be true, hence the infinite looping.
And you're overcomplicating it anyway.
for (int i = 1; i <= 10; i++) {
int r = 1, n = i;
while (n)
r *= n--;
printf("%d! = %d\n", i, r);
}
In Math, n! is the same thing as Γ(n+1) (see: http://en.wikipedia.org/wiki/Gamma_function)
So just use:
-(float)factorial:(float)number1 {
return tgammaf(++number1);
}
This will even work for floats and negative numbers,
other solutions posted are long and extraneous and only work with
positive integers.
During the first loop iteration count is 1 and so also n is 1, then you enter the while and you set count to zero (count-1), and decrease n which becomes zero and you exit the while. So during the second loop iteration count will be zero. You keep decreasing count and it never gets increased, so you never exit the loop until a numeric overflow occurs.
You're doing it harder that what it is (and also inefficient) . Is enough that you keep multiplying n for count to get the factorial:
int n=1;
for (count = 1; count<=10; ++count)
{
n*= count;
NSLog(#"%d",n);
}
How many possible combinations of the variables a,b,c,d,e are possible if I know that:
a+b+c+d+e = 500
and that they are all integers and >= 0, so I know they are finite.
#Torlack, #Jason Cohen: Recursion is a bad idea here, because there are "overlapping subproblems." I.e., If you choose a as 1 and b as 2, then you have 3 variables left that should add up to 497; you arrive at the same subproblem by choosing a as 2 and b as 1. (The number of such coincidences explodes as the numbers grow.)
The traditional way to attack such a problem is dynamic programming: build a table bottom-up of the solutions to the sub-problems (starting with "how many combinations of 1 variable add up to 0?") then building up through iteration (the solution to "how many combinations of n variables add up to k?" is the sum of the solutions to "how many combinations of n-1 variables add up to j?" with 0 <= j <= k).
public static long getCombos( int n, int sum ) {
// tab[i][j] is how many combinations of (i+1) vars add up to j
long[][] tab = new long[n][sum+1];
// # of combos of 1 var for any sum is 1
for( int j=0; j < tab[0].length; ++j ) {
tab[0][j] = 1;
}
for( int i=1; i < tab.length; ++i ) {
for( int j=0; j < tab[i].length; ++j ) {
// # combos of (i+1) vars adding up to j is the sum of the #
// of combos of i vars adding up to k, for all 0 <= k <= j
// (choosing i vars forces the choice of the (i+1)st).
tab[i][j] = 0;
for( int k=0; k <= j; ++k ) {
tab[i][j] += tab[i-1][k];
}
}
}
return tab[n-1][sum];
}
$ time java Combos
2656615626
real 0m0.151s
user 0m0.120s
sys 0m0.012s
The answer to your question is 2656615626.
Here's the code that generates the answer:
public static long getNumCombinations( int summands, int sum )
{
if ( summands <= 1 )
return 1;
long combos = 0;
for ( int a = 0 ; a <= sum ; a++ )
combos += getNumCombinations( summands-1, sum-a );
return combos;
}
In your case, summands is 5 and sum is 500.
Note that this code is slow. If you need speed, cache the results from summand,sum pairs.
I'm assuming you want numbers >=0. If you want >0, replace the loop initialization with a = 1 and the loop condition with a < sum. I'm also assuming you want permutations (e.g. 1+2+3+4+5 plus 2+1+3+4+5 etc). You could change the for-loop if you wanted a >= b >= c >= d >= e.
I solved this problem for my dad a couple months ago...extend for your use. These tend to be one time problems so I didn't go for the most reusable...
a+b+c+d = sum
i = number of combinations
for (a=0;a<=sum;a++)
{
for (b = 0; b <= (sum - a); b++)
{
for (c = 0; c <= (sum - a - b); c++)
{
//d = sum - a - b - c;
i++
}
}
}
This would actually be a good question to ask on an interview as it is simple enough that you could write up on a white board, but complex enough that it might trip someone up if they don't think carefully enough about it. Also, you can also for two different answers which cause the implementation to be quite different.
Order Matters
If the order matters then any solution needs to allow for zero to appear for any of the variables; thus, the most straight forward solution would be as follows:
public class Combos {
public static void main() {
long counter = 0;
for (int a = 0; a <= 500; a++) {
for (int b = 0; b <= (500 - a); b++) {
for (int c = 0; c <= (500 - a - b); c++) {
for (int d = 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2656615626.
Order Does Not Matter
If the order does not matter then the solution is not that much harder as you just need to make sure that zero isn't possible unless sum has already been found.
public class Combos {
public static void main() {
long counter = 0;
for (int a = 1; a <= 500; a++) {
for (int b = (a != 500) ? 1 : 0; b <= (500 - a); b++) {
for (int c = (a + b != 500) ? 1 : 0; c <= (500 - a - b); c++) {
for (int d = (a + b + c != 500) ? 1 : 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2573155876.
One way of looking at the problem is as follows:
First, a can be any value from 0 to 500. Then if follows that b+c+d+e = 500-a. This reduces the problem by one variable. Recurse until done.
For example, if a is 500, then b+c+d+e=0 which means that for the case of a = 500, there is only one combination of values for b,c,d and e.
If a is 300, then b+c+d+e=200, which is in fact the same problem as the original problem, just reduced by one variable.
Note: As Chris points out, this is a horrible way of actually trying to solve the problem.
link text
If they are a real numbers then infinite ... otherwise it is a bit trickier.
(OK, for any computer representation of a real number there would be a finite count ... but it would be big!)
It has general formulae, if
a + b + c + d = N
Then number of non-negative integral solution will be C(N + number_of_variable - 1, N)
#Chris Conway answer is correct. I have tested with a simple code that is suitable for smaller sums.
long counter = 0;
int sum=25;
for (int a = 0; a <= sum; a++) {
for (int b = 0; b <= sum ; b++) {
for (int c = 0; c <= sum; c++) {
for (int d = 0; d <= sum; d++) {
for (int e = 0; e <= sum; e++) {
if ((a+b+c+d+e)==sum) counter=counter+1L;
}
}
}
}
}
System.out.println("counter e "+counter);
The answer in math is 504!/(500! * 4!).
Formally, for x1+x2+...xk=n, the number of combination of nonnegative number x1,...xk is the binomial coefficient: (k-1)-combination out of a set containing (n+k-1) elements.
The intuition is to choose (k-1) points from (n+k-1) points and use the number of points between two chosen points to represent a number in x1,..xk.
Sorry about the poor math edition for my fist time answering Stack Overflow.
Just a test for code block
Just a test for code block
Just a test for code block
Including negatives? Infinite.
Including only positives? In this case they wouldn't be called "integers", but "naturals", instead. In this case... I can't really solve this, I wish I could, but my math is too rusty. There is probably some crazy integral way to solve this. I can give some pointers for the math skilled around.
being x the end result,
the range of a would be from 0 to x,
the range of b would be from 0 to (x - a),
the range of c would be from 0 to (x - a - b),
and so forth until the e.
The answer is the sum of all those possibilities.
I am trying to find some more direct formula on Google, but I am really low on my Google-Fu today...