I’m trying to understand Heap’s algorithm from the Wikipedia page and I’m trying to compare the picture with the algorithm and I can’t seem to figure it out
this picture is from the wikipedia page
why would it switch the #1 and #2 first, shouldn’t it switch the #1 and #4 first?
I’m using java but this is just the code copied from Wikipedia, I understand that there is switching involved for the code in general
if k = 1 then
output(A)
else
// Generate permutations with kth unaltered
// Initially k == length(A)
generate(k - 1, A)
// Generate permutations for kth swapped with each k-1 initial
for i := 0; i < k-1; i += 1 do
// Swap choice dependent on parity of k (even or odd)
if k is even then
swap(A[i], A[k-1]) // zero-indexed, the kth is at k-1
else
swap(A[0], A[k-1])
end if
generate(k - 1, A)
end for
end if
k is initially 4 so wouldn’t it switch A[0] and A[3] first?
Sorry in advance if this is a stupid question...
When k > 1, the very first thing it does is recurse. So, the calls go:
generate(4,A) calls
generate(3,A) calls
generate(2,A) calls
generate(1,A) which prints A
Now we do the processing for k==2.
swap(0,1)
generate(1,A) which prints the new A
etc.
Related
I was reading book about competitive programming and was encountered to problem where we have to count all possible paths in the n*n matrix.
Now the conditions are :
`
1. All cells must be visited for once (cells must not be unvisited or visited more than once)
2. Path should start from (1,1) and end at (n,n)
3. Possible moves are right, left, up, down from current cell
4. You cannot go out of the grid
Now this my code for the problem :
typedef long long ll;
ll path_count(ll n,vector<vector<bool>>& done,ll r,ll c){
ll count=0;
done[r][c] = true;
if(r==(n-1) && c==(n-1)){
for(ll i=0;i<n;i++){
for(ll j=0;j<n;j++) if(!done[i][j]) {
done[r][c]=false;
return 0;
}
}
count++;
}
else {
if((r+1)<n && !done[r+1][c]) count+=path_count(n,done,r+1,c);
if((r-1)>=0 && !done[r-1][c]) count+=path_count(n,done,r-1,c);
if((c+1)<n && !done[r][c+1]) count+=path_count(n,done,r,c+1);
if((c-1)>=0 && !done[r][c-1]) count+=path_count(n,done,r,c-1);
}
done[r][c] = false;
return count;
}
Here if we define recurrence relation then it can be like: T(n) = 4T(n-1)+n2
Is this recurrence relation true? I don't think so because if we use masters theorem then it would give us result as O(4n*n2) and I don't think it can be of this order.
The reason, why I am telling, is this because when I use it for 7*7 matrix it takes around 110.09 seconds and I don't think for n=7 O(4n*n2) should take that much time.
If we calculate it for n=7 the approx instructions can be 47*77 = 802816 ~ 106. For such amount of instruction it should not take that much time. So here I conclude that my recurrene relation is false.
This code generates output as 111712 for 7 and it is same as the book's output. So code is right.
So what is the correct time complexity??
No, the complexity is not O(4^n * n^2).
Consider the 4^n in your notation. This means, going to a depth of at most n - or 7 in your case, and having 4 choices at each level. But this is not the case. In the 8th, level you still have multiple choices where to go next. In fact, you are branching until you find the path, which is of depth n^2.
So, a non tight bound will give us O(4^(n^2) * n^2). This bound however is far from being tight, as it assumes you have 4 valid choices from each of your recursive calls. This is not the case.
I am not sure how much tighter it can be, but a first attempt will drop it to O(3^(n^2) * n^2), since you cannot go from the node you came from. This bound is still far from optimal.
I was recently asked an interview question about testing the validity of a Sudoku board. A basic answer involves for loops. Essentially:
for(int x = 0; x != 9; ++x)
for(int y = 0; y != 9; ++y)
// ...
Do this nested for loops to check the rows. Do it again to check the columns. Do one more for the sub-squares but that one is more funky because we're dividing the suoku board into sub-boards so we end end up more than two nested loops, maybe three or four.
I was later asked the complexity of this code. Frankly, as far as I'm concerned, all the cells of the board are visited exactly three times so O(3n). To me, the fact that we have nested loops doesn't mean this code is automatically O(n^2) or even O(n^highest-nesting-level-of-loops). But I have suspicion that that's the answer the interviewer expected...
Posed another way, what is the complexity of these two pieces of code:
for(int i = 0; i != n; ++i)
// ...
and:
for(int i = 0; i != sqrt(n); ++i)
for(int j = 0; j != sqrt(n); ++j)
// ...
Your general intuition is correct. Let's clarify a bit about Big-O notation:
Big-O gives you an upper bound for the worst-case (time) complexity for your algorithm, in relation to n - the size of your input. In essence, it is a measurement of how the amount of work changes in relation to the size of the input.
When you say something like
all the cells of the board are visited exactly three times so O(3n).
you are implying that n (the size of your input) is the the number of cells in the board and therefore visiting all cells three times would indeed be an O(3n) (which is O(n)) operation. If this is the case you would be correct.
However usually when referring to Sudoku problems (or problems involving a grid in general), n is taken to be the number of cells in each row/column (an n x n board). In this case, the runtime complexity would be O(3n²) (which is indeed equal to O(n²)).
In the future, it is perfectly valid to ask your interviewer what n is.
As for the question in the title (Is a nested for loop automatically O(n^2)?) the short answer is no.
Consider this example:
for(int i = 0 ; i < n ; i++) {
for(int j = 0 ; j < n ; j * 2) {
... // some constant time operation
}
}
The outer loops makes n iterations while the inner loop makes log2(n) iterations - therefore the time complexity will be O(nlogn).
In your examples, in the first one you have a single for-loop making n iterations, therefore a complexity of (at least) O(n) (the operation is performed an order of n times).
In the second one you two nested for loops, each making sqrt(n) iterations, therefore a total runtime complexity of (at least) O(n) as well. The second function isn't automatically O(n^2) simply because it contains a nested loop. The amount of operations being made is still of the same order (n) therefore these two examples have the same complexity - since we assume n is the same for both examples.
This is the most crucial point to sail home. To compare between the performance of two algorithms, you must be using the same input to make the comparison. In your sudoku problem you could have defined n in a few different ways, and the way you did would directly affect the complexity calculation of the problem - even if the amount of work is all the same.
*NOTE - this is unrelated to your question, but in the future avoid using != in loop conditions. In your second example, if log(n) is not a whole number, the loop could run forever, depending on the language and how it is defined. It is therefore recommended to use < instead.
It depends on how you define the so-called N.
If the size of the board is N-by-N, then yes, the complexity is O(N^2).
But if you say, the total number of grids is N (i.e., the board id sqrt(N)-by-sqrt(N)), then the complexity is O(N), or 3O(N) if you mind the constant.
I'm interest in building an SVM multi class classifier, so I am currently implementing
Sequential minimal optimization SMO.
My implementation is based on the pseudo code in
`Fast Training of Support Vector Machines using Sequential Minimal Optimization" by John C. Platt
I observed that for certain training examples. The Smo may diverge and run into an infinite loop
The following loop in the main routine
numChanged = 0;
examineAll = 1;
while (numChanged > 0 || examineAll >0) {…}
may run into an infinite loop.
Is there there clue or criterion to prevent the smo algorithm routine from running into an infinite loop?
I would like to thank you for your answer.
Regards
You can add a max iteration condition if you want:
while ((numChanged > 0 || examineAll) && iter < MaxIter)
but for most cases it shouldn't run into an infinite loop, this is the full Platt's pseudocode:
while (numChanged > 0 || examineAll)
{
numChanged = 0;
// Adding curly brackets for better readability
if (examineAll)
{
loop I over all training examples
numChanged += examineExample(I);
}
else
{
loop I over examples where alpha is not 0 & not C
numChanged += examineExample(I);
}
if (examineAll == 1)
{
examineAll = 0;
}
else
{
examineAll = 1;
}
}
Notice that what it is doing is performing an iteration to examine the example and the next one do the same just to those examples where alpha is not 0 or C. If nothing changes after the "examine all" iteration, the while loop condition will be false hence stopping the loop.
So, for that to be in a infinite loop there must be a corner case (probably a numerical error) that introduce oscillations making examples to change during the examine all phase but not changing in the "examine only alpha == 0 and C".
Usually if the data is normalized in [-1,1] or [0,1] and the parameters of the algorithm have reasonable values, those corner cases would be rare. In any case, if you want to be extra careful you can put the max-iter safety net.
So I recently had this as an interview question and I was wondering what the optimal solution would be. Code is in Objective-c.
Say we have a very large data set, and we want to get a random sample
of items from it for testing a new tool. Rather than worry about the
specifics of accessing things, let's assume the system provides these
things:
// Return a random number from the set 0, 1, 2, ..., n-2, n-1.
int Rand(int n);
// Interface to implementations other people write.
#interface Dataset : NSObject
// YES when there is no more data.
- (BOOL)endOfData;
// Get the next element and move forward.
- (NSString*)getNext;
#end
// This function reads elements from |input| until the end, and
// returns an array of |k| randomly-selected elements.
- (NSArray*)getSamples:(unsigned)k from:(Dataset*)input
{
// Describe how this works.
}
Edit: So you are supposed to randomly select items from a given array. So if k = 5, then I would want to randomly select 5 elements from the dataset and return an array of those items. Each element in the dataset has to have an equal chance of getting selected.
This seems like a good time to use Reservoir Sampling. The following is an Objective-C adaptation for this use case:
NSMutableArray* result = [[NSMutableArray alloc] initWithCapacity:k];
int i,j;
for (i = 0; i < k; i++) {
[result setObject:[input getNext] atIndexedSubscript:i];
}
for (i = k; ![input endOfData]; i++) {
j = Rand(i);
NSString* next = [input getNext];
if (j < k) {
[result setObject:next atIndexedSubscript:j];
}
}
return result;
The code above is not the most efficient reservoir sampling algorithm because it generates a random number for every entry of the reservoir past the entry at index k. Slightly more complex algorithms exist under the general category "reservoir sampling". This is an interesting read on an algorithm named "Algorithm Z". I would be curious if people find newer literature on reservoir sampling, too, because this article was published in 1985.
Interessting question, but as there is no count or similar method in DataSet and you are not allowed to iterate more than once, i can only come up with this solution to get good random samples (no k > Datasize handling):
- (NSArray *)getSamples:(unsigned)k from:(Dataset*)input {
NSMutableArray *source = [[NSMutableArray alloc] init];
while(![input endOfData]) {
[source addObject:[input getNext]];
}
NSMutableArray *ret = [[NSMutableArray alloc] initWithCapacity:k];
int count = [source count];
while ([ret count] < k) {
int index = Rand(count);
[ret addObject:[source objectAtIndex:index]];
[source removeObjectAtIndex:index];
count--;
}
return ret;
}
This is not the answer I did in the interview but here is what I wish I had done:
Store pointer to first element in dataset
Loop over dataset to get count
Reset dataset to point at first element
Create NSMutableDictionary for storing random indexes
Do for loop from i=0 to i=k. Each iteration, generate a random value, check if value exists in dictionary. If it does, keep generating a random value until you get a fresh value.
Loop over dataset. If the current index is within the dictionary, add it to a the array of random subset values.
Return array of random subsets.
There are multiple ways to do this, the first way:
1. use input parameter k to dynamically allocate an array of numbers
unsigned * numsArray = (unsigned *)malloc(sizeof(unsigned) * k);
2. run a loop that gets k random numbers and stores them into the numsArray (must be careful here to check each new random to see if we have gotten it before, and if we have, get another random, etc...)
3. sort numsArray
4. run a loop beginning at the beginning of DataSet with your own incrementing counter dataCount and another counter numsCount both beginning at 0. whenever dataCount is equal to numsArray[numsCount], grab the current data object and add it to your newly created random list then increment numsCount.
5. The loop in step 4 can end when either numsCount > k or when dataCount reaches the end of the dataset.
6. The only other step that may need to be added here is before any of this to use the next command of the object type to count how large the dataset is to be able to bound your random numbers and check to make sure k is less than or equal to that.
The 2nd way to do this would be to run through the actual list MULTIPLE times.
// one must assume that once we get to the end, we can start over within the set again
1. run a while loop that checks for endOfData
a. count up a count variable that is initialized to 0
2. run a loop from 0 through k-1
a. generate a random number that you constrain to the list size
b. run a loop that moves through the dataset until it hits the rand element
c. compare that element with all other elements in your new list to make sure it isnt already in your new list
d. store the element into your new list
there may be ways to speed up the 2nd method by storing a current list location, that way if you generate a random that is past the current pointer you dont have to move through the whole list again to get back to element 0, then to the element you wish to retreive.
A potential 3rd way to do this might be to:
1. run a loop from 0 through k-1
a. generate a random
b. use the generated random as a skip count, move skip count objects through the list
c. store the current item from the list into your new list
Problem with this 3rd method is without knowing how big the list is, you dont know how to constrain the random skip count. Further, even if you did, chances are that it wouldnt truly look like a randomly grabbed subset that could easily reach the last element in the list as it would become statistically unlikely that you would ever reach the end element (i.e. not every element is given an equal shot of being select.)
Arguably the FASTEST way to do this is method 1, where you generate the random numerics first, then traverse the list only once (yes its actually twice, once to get the size of the dataset list then again to grab the random elements)
We need a little probability theory. As others, I will ignore the case n < k. The probability that the n'th item will be selected into the set of size k is just C(n-1, k-1) / C(n, k) where C is the binomial coefficient. A bit of math says shows that this is just k/n. For the rest, note that the selection of the n'th item is independent of all other selections. In other words, "the past doesn't matter."
So an algorithm is:
S = set of up to k elements
n = 0
while not end of input
v = next value
n = n + 1
if |S| < k add v to S
else if random(0,1) >= k/n replace a randomly chosen element of S with v
I will let the coders code this one! It's pretty trivial. All you need is an array of size k and one pass over the data.
If you care about efficiency (as your tags suggest) and the number of items in the population is known, don't use reservior sampling. That would require you to loop through the entire data set and generate a random number for each.
Instead, pick five values ranges from 0 to n-1. In the unlikely case, there is a duplicate among the five indexes, replace the duplicate with another random value. Then use the five indexes to do a random-access lookup to the i-th element in the population.
This is simple. It uses a minimum number of calls the random number generator. And it accesses memory only for the relevant selections.
If you don't know the number of data elements in advance, you can loop-over the data once to get the population size and proceed as above.
If you aren't allow to iterate over the data more than once, use a chunked form of reservior sampling: 1) Choose the first five elements as the initial sample, each having a probability of 1/5th. 2) Read in a large chunk of data and choose five new samples from the new set (using only five calls to Rand). 3) Pairwise, decide whether to keep the new sample item or old sample element (with odds proportional the the probablities for each of the two sample groups). 4) Repeat until all the data has been read.
For example, assume there are 1000 data elements (but we don't know this in advance).
Choose the first five as the initial sample: current_sample = read(5); population=5.
Read a chunk of n datapoints (perhaps n=200 in this example):
subpop = read(200);
m = len(subpop);
new_sample = choose(5, subpop);
loop-over the two samples pairwise:
for (a, b) in (current_sample and new_sample): if random(0 to population + m) < population, then keep a, otherwise keep *b)
population += m
repeat
I am writing a battleship program. Right now I am testing a couple lines of code to see if it will place the boat going in the up direction. How my program is set up is that if, for example, the user clicks on the aircraft carrier button to set his aircraft carrier, the program should also set the ai's aircraft carrier. The boats are placed on a button array, called tlba. aifirstclicki is set by a random generator so that it will choose a random row. aifirstclickj chooses a random column, in conjunction the two pinpoint a spot on the button array (which is 10x10). I wrote the following code to try to make it so that if the program has an outofboundsexception error,or in other words if the program chooses a first spot that will eventually cause an outofbounds exception error because the for loop will keep adding spots until aiclickcount = 5, it should start over and pick a different spot until it finds a spot that will allow it to place all 5 spots. I keep getting stuck in an infinite loop though.
int aiclickcount = 0;
while (directiondecider == 0)
{//up
aifirstclicki = generator.nextInt(10);
aifirstclickj = generator.nextInt(10);
while (aifirstclicki != 3 &&
aifirstclicki != 2 &&
aifirstclicki != 1 &&
aifirstclicki != 0)
{
for(int k=0; k<shiplength; k++)
{
tlba[aifirstclicki - k][aifirstclickj].setBackground(Color.RED);
aistringarray[aifirstclicki - k][aifirstclickj] = "aircraftcarrier";
aioccupied2d[aifirstclicki - k][aifirstclickj] = true;
aiclickcount++;
}
if (aiclickcount == 5)
{
shipset = true;
break;
}
}
System.out.println(shipset);
}
Does anyone know what's wrong or have a different solution to my problem?
You never have aiclickcount == 5 if your shiplength is not 5. Put if into your for loop. You don't need the second while at all, you don't break out of it as well. Just generate number greater than 3 by nextInt(6) + 4.
Your code does not tell us, which value the variable shiplength has. If it's 0 the for-loop will never be entered thus aiclickcount will remain 0 and your break statement is never reached (under the premise that the random value of aifirstclicki is greater than 3).
Try to step through your code with a debugger and let it display the values for the variables to you to find out what's going on.
Your break; is only going to get you out of the second while loop, not the first as it only works on the inner-most loop that it is part of.
Java allows you to specify multi-level breaks, rather than having to complicate your loop conditions:
Breaking out of nested loops in Java