Блог пользователя Sinux

Need Help in this generating subset code!
Автор Sinux, история, 2 года назад, По-английски 

void search(int k) {
  if (k == n) {
    // process subset
  } else {
     search(k+1);
     subset.push_back(k);
     search(k+1);
     subset.pop_back();
   }
}

I am having a really hard time understand this code. I especially don't understand how 'k' turns to 2 after 'k' reaches 3. I don't understand because there's no such code that does search(k-1) but it backtracks automatically. I think It's related to the thing backtracking. Someone plz help!!!

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2 года назад, скрыть # |
 
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It's building subset from $$$1$$$ to $$$n$$$. For each element, you can decide: to include it or not? Its like a DFS.

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This code generating all unique ({1,2,3} and {3,2,1} are same) subsets from 1 to n. There are two options, to add element or left it and transit with k+1, which generate all subset with size 1 till n.

About why k become 2 from 3, just because it go with first possible way. If this is too difficult to you, I recomment you to read this and watch here Good luck!

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it doesn't return to, it's another stack of recursions

this code works as follows:

image

either an element comes or doesn't, in the else part, it handles when we haven't visited every element, the first 2 lines of the else part are when it comes, and the second 2 lines are when it doesn't.
bcuz of the way recursion(in total stack memory) works, it starts drawing it from above, not equally everywhere, so it first finishes writing 123, then goes to the last time it got branched and solves there etc

it's just understanding recursions
imagine you say to your friend, hey go put this box there, then come back, the other box here, he goes to put that box there, but when he does he sees another friend who tells him the same thing, and he goes and puts the box in the place the other friend tells him, and since there isn't anyone telling him to go do another thing, he returns to the second friend and moves the other box, then he returns to you and moves the other box, why? cause you told it to, not now but before, it's the same

the function get's called on k = 2 here, he goes and does the job but another friend calls him on k = 3 now and he goes and does his job and another friend on k = 4, each friend is giving him 2 tasks, when he does the first task of the k = 4, no one tells him to go do something, so he returns and does the second tasks of k = 4 and then he goes to do the second task of k = 3 and goes to do the second task of k = 2, he remembers each task, and does them in a last in first out order, which causes this going back to action

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