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Backtracking: Why We Replaced External Feedback With a Lightweight Classifier

The basics of backtracking has proven itself to be a universal algorithmic technique that can be applied while solving all kinds of computational problems, from the most primitive (such as counting all possible paths between two vertices) to the most complex ones (such as finding all Hamiltonian paths or solving the N Queen problem). Backtracking According to the study on , it can be defined as an improvement to the brute force approach. In other words, the basic idea behind the backtracking technique is the search for a solution to a particular problem among all the options available. fundamental backtracking The backtracking algorithm owes its success and effectiveness to the . When the algorithm initiates its extensive search for solutions, it promptly applies the abounding function, making it possible to determine whether a proposed solution meets the constraints set. In case the answer is positive, the algorithm continues its search. If this is not the case, the algorithm removes the branch and simply returns to the previous level. depth-first search method In general, there are three common types of problems that backtracking solves: problems - finding a feasible solution; Decision problems - finding the best solution possible; Optimization problems - finding all the feasible solutions. Enumeration Now, let's move on from theory to practice and have a look at by the backtracking techniques. Noteworthy is that both of the problems were taken from real job interviews, and they are included in the list. 2 algo problems solved Leetcode Top Interview Questions Subsets Problem description Given an integer array of unique elements, return . The solution set must not contain duplicate subsets. Return the solution in any order. nums all possible subsets (the power set) Input:  nums = [0,1,2] Output: [[],[0],[1],[0,1],[2],[0,2],[1,2],[0,1,2]] Idea It is common knowledge that the of a set is considered to be the set of all subsets of S, including both the empty set and itself. The power set of is graphically illustrated below: power set S S {0,1,2} The most direct way is to: compute all the subsets that do not contain a particular element (which could be any single element); U compute all the subsets that do contain the required element. V After it, each subset set must appear either in or in , providing us with the final result of . The construction is recursive, and the is when the input set is empty, in which case we return an empty list. U V U⋃V base case Example Now, it’s the right moment to study the example: Let = [0,1,2]. Pick any element, e.g., 0. nums First, we recursively compute all subsets of {1,2}. To do this, we select 1. This leaves us with {2}. Now, we pick 2 and get to the base case. -> {} Returning from the recursive calls: 3. So, the set of subsets of {2} is {} union with {2}, i.e., {{}, {2}}.
2. The set of subsets of {1, 2} is then {{}, {2}} union with {{1}, {1, 2}}, i.e., {{}, {2}, {1}, {1, 2}}. The set of subsets of {0,1,2} is then {{}, {2}, {1}, {1, 2}} union with {{0}, {0, 2}, {0, 1}, {0, 1, 2}}, i.e., {{}, {2}, {1}, {1, 2), {0}, {0, 2}, {0, 1}, {0, 1, 2}}. Solution class Solution {
  public List<List<Integer>> subsets(int[] nums) {
    List<List<Integer>> ans = new LinkedList<>();
    backtrack(nums, 0, new LinkedList<>(), ans);
    return ans;
  }

  void backtrack(int[] nums, int i, List<Integer> cur, List<List<Integer>> ans){
    ans.add(new ArrayList<>(cur));
    for (int j = i; j<nums.length; j++){
      cur.add(nums[j]);
      backtrack(nums, j + 1, cur, ans);
      cur.remove(cur.size() - 1);
    }
  }
} Performance all the subsets are generated and further copied to the output list using . The number of recursive calls, , satisfies the recurrence , which solves to . Since we spend time within a call, the time complexity is . The time complexity: O(N×2^N) C(n) C(n) = 2C(n - 1) C(n) = (2^N) O(N) O(N×2^N) is calculated by the following formula: . We are using the space to maintain , and are modifying in-place with backtracking. The space complexity O(N) O(N) cur cur Permutations Problem description Given an array of distinct integers, return . The answer can be returned in any order. nums all the possible permutations Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]] Idea It is generally known that every permutation of begins with one of . The basic idea here is to generate all permutations starting with , then all permutations starting with , and so on. Computing all permutations starting with means computing all permutations of , which suggests the use of recursion. A A[0], A[1], ..., A[n-1] A[0] A[1] A[0] A[1:n-1] To compute all permutations starting with , we swap with and compute all permutations of the updated . As we go on, we restore the original state before embarking on the computation of all permutations that start with , and so on. A[1] A[0] A[1] A[1:n-1] A[2] Example For instance, for the array (1, 2, 3), we would first generate all permutations starting with 1. This implies generating all permutations of (2, 3). Next, we find all permutations of (2, 3) starting with 2. Given that (3) is an array of length 1, it has the only possible permutation. The implication of this is that (2, 3) has a single permutation starting with 2. After that, we look for permutations of (2, 3) starting with 3, which can be executed by swapping 2 and 3. Consequently, we will find out that there is a single permutation of (2, 3) starting with 3, namely, (3, 2). Hence, there are two permutations starting with 1: (1,2,3) and (1,3,2). To find all permutations starting with 2, we swap 1 with 2 and get two permutations again: (2,1,3) and (2,3,1). The last two permutations we add are (3,2,1) and (3,1,2). In total, there are six permutations: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,2,1) and (3,1,2). Solution class Solution {
  List<List<Integer>> result;

  public List<List<Integer>> permute(int[] nums) {
    Arrays.sort(nums);
    result = new ArrayList<>();
    permute(0, nums);
    return result;
   }

   private void permute(int i, int[] nums){
     if (i == nums.length - 1){
       copyToResult(nums);
     }
     for (int j = i; j<nums.length; j++){
       swap(i,j,nums);
       permute(i+1,nums);
       swap(j,i,nums);
     }
   }

  private void copyToResult(int[] nums){
    List<Integer> list = new ArrayList<>();
    for (int num: nums){
      list.add(num);
    }
    result.add(list);
  }

  private void swap(int i, int j, int[] nums){
    int t = nums[i];
    nums[i] = nums[j];
    nums[j] = t;
  }
} Performance The time complexity is , since there are permutations, and we spend time to store each one. Having studied this example, we can observe that backtracking only visits each state once, which is why its time complexity should be similar to the graph traversal of O(|V|+|E|). 0(n×n!) n! 0(n) Conclusion Summing up the above, backtracking is definitely a perfect algorithmic technique for solving both subset and permutation problems. With the help of its depth-first search method, the backtracking algorithm will inevitably find the required solution in strict compliance with all given constraints. Since 1950, when the term “ ” was coined by D. H. Lehmer, an American mathematician, backtracking managed to handle a wide range of practice problems, starting from printing all paths from a given source to a destination and finishing with finding the shortest safe route in a path with landmines. In this regard, the next time you deal with either permutations or subsets, I strongly recommend that you use the backtracking algo pattern. backtracking List of references Aziz, Adnan; Prakash, Amit; Lee, Tsung-Hsien (2012). “Elements of Programming Interviews in Java” Gurari, Eitan (1999). “CIS 680: DATA STRUCTURES: Chapter 19: Backtracking Algorithms” “Introduction to Backtracking - Data Structure and Algorithm Tutorials” - GeeksforGeeks Upadhyay, Soni (2023). “What is Backtracking Algorithm? Types, Examples & its Application” Zibbu, Shirsh (2018). “Sudoku and Backtracking”

Backtracking LeetCode Pattern: Permutations vs Subsets in Java

in this blog post, we are going to take a look at what is backtracking and how to implement it using ruby First thing first what is backtracking? according to Backtracking is a general algorithm for finding all (or some) solutions to some computational problem notably constrain satisfaction problems that incrementally builds candidates to the solutions, and abandons a candidate ("backtracks") as soon as it determines that the candidate cannot possibly be completed to a valid solution. wikipedia To understand what backtracking is let's take a problem and try to solve it using backtracking. we will be given a tow input where the first number is the desired sum and the remaining is an array of numbers. Determine if the array or any of its sub-arrays can be summed to get the exact sum analyzing the above problem here are the steps we want to follow to solve the problem Implement the recursive function to find all the possible sub-arrays recursively. Follow backtracking strategy and implement required base-cases to process each recursive level. subtract  the first element of every sub-array to the sum and return true if the sum is 0 and return false if the sum is less than 0 or the array is empty. since our steps seem complete let's take a sample case sum== sum < arr.empty?
exact_sum?(sum - arr[ ], arr[ ,arr.length]) puts exact_sum?( , [ , , , , ]); def exact_sum? (sum, arr) return true if 0 return false if 0 || 0 1 end 12 1 2 3 4 5 # expected result => true # actual resut => false looking at the input you can see that if you remove 3 from the array and try to sum it you will get 12 but why do we get false? well the answer is that our above still recursive so we need to make it backtracking and the way to do that is by checking both cases where we include the first element and when we don't By doing that here's how the flow of our algorithm should look like let's include a variable K representing our sum and see how it would work let's implement that in our previous algorithm sum== sum < arr.empty? exact_sum?(sum - arr[ ], arr[ ,arr.length]) exact_sum?(sum, arr[ ,arr.length]) puts exact_sum?( , [ , , , , ]); def exact_sum? (sum, arr) return true if 0 return false if 0 || # now we are checking both cases 0 1 || 1 end 12 1 2 3 4 5 # => true by adding a or statement the algorithm will be checking both cases and will become a backtracking algorithm but this still a little bit confusing let's print the values to fully understand the flow of our algorithm [ , , , , ] [ , , , ] [ , , ] [ , ] [ ] [ ] [ , ] [ ] # the algorithm start 1 2 3 4 5 12 # check for the first condition : 12-1 is not less than 0 so we subtract it from the sum 2 3 4 5 11 # check again 11-2 > 0 is true so we proceed 3 4 5 9 # same thing here 4 5 6 # here 2-5 will return false so we go back to the parent call([4,5]6) and try to exclude the first case without touching the sum 5 2 # we still get false because the array will be empty and the sum will be 1 so go back to the ancestor call([3,4,5]9)and exclude the first element without touching the sum 5 6 # 9-4 > 0 is true so we proceed 4 5 9 # 5-5 is 0 so we return true 5 5 true And Voila!!! just with those few lines of code we have implemented a backtracking algorithm. Thank you for reading If you have any question or suggestion contact me on Twitter

An Introduction to Backtracking in Ruby

Using Sudoku to explore backtracking Sudoku is a number-placement where the objective is to fill a square grid of size ‘n’ with numbers between 1 to ‘n’. The numbers must be placed so that each column, each row, and each of the sub-grids (if any) contains all of the numbers from 1 to ‘n’. Sudoku puzzle The most common Sudoku puzzles use a 9x9 grid. The grids are partially filled (with hints) to ensure a solution can be reached. And here’s the solution. Notice how each row, each column and each sub-grid have all numbers from 1 to 9. Some puzzles may even have multiple solutions. The solution to above Sudoku puzzle. One row, column and sub-grid have been highlighted. Aside: solving a Sudoku puzzle Sudoku is a logic-based puzzle. Needless to say, solving one requires a series of logical moves and might require a bit of guesswork. Since this isn’t an article to explore how to solve a Sudoku puzzle, I’ll just a leave a link to one that helped me getting started: . kristanix.com/sudokuepic/sudoku-solving-techniques Backtracking Backtracking is an algorithm for finding all (or some) of the solutions to a problem that incrementally builds candidates to the solution(s). As soon as it determines that a candidate cannot possibly lead to a valid solution, it abandons the candidate. Backtracking is all about choices and consequences. Abandoning a candidate typically results in visiting a previous stage of the problem-solving-process. This is what it means to “backtrack” — visit a previous stage and explore new possibilities from thereon. Usually, apart from the original problem and the end goal, we also have a set of constraints that the solution must satisfy. The simplest (read ‘dumbest’) implementations often use little to no “logic” or “insight” to the problem. Instead, they frantically try to find a solution by guesswork. A backtracking algorithm can be thought of as a . In this tree, the root node is the original problem, each node is a candidate and the leaf nodes are possible solution candidates. We traverse the tree depth-first from root to a leaf node to solve the problem. tree of possibilities Tree of Possibilities for a typical backtracking algorithm The tree diagram also shows 2 groups — and . Unexplored Possible Candidates Impossible Candidates UPC marks nodes that were never explored. Some of them could have been viable candidates, leading to another solution. Since we never explored them, we can never know. Problems where multiple solutions are acceptable won’t have this group. IC groups are the obvious ones. It contains nodes which have a failed candidate node as one of their ancestor nodes. None of the nodes in this group are candidate nodes and none of the leaf nodes are solution nodes. Sudoku & Backtracking We will now create a Sudoku solver using backtracking by our problem, goal and constraints in a step-by-step algorithm. encoding Problem Given a, possibly, partially filled grid of size ‘n’, completely fill the grid with number between 1 and ‘n’. Goal Goal is defined for verifying the solution. Once the goal is reached, searching terminates. A fully filled grid is a solution if: Each row has all numbers form 1 to ‘n’. Each column has all numbers form 1 to ‘n’. Each sub-grid (if any) has all numbers form 1 to ‘n’. Constraints Constraints are defined for verifying each candidate. A candidate is valid if: Each row has unique numbers form 1 to ’n’ or empty spaces. Each column has unique numbers form 1 to ‘n’ or empty spaces. Each sub-grid (if any) has unique numbers form 1 to ‘n’ or empty spaces. Termination conditions Typically, backtracking algorithms have termination conditions other than reaching goal. These help with failures in solving the problem and special cases of the problem itself. There are no empty spots left to fill and the candidate still doesn’t qualify as a the solution. There are no empty spots to begin with, i.e., the grid is already fully filled. Step-by-step algorithm Here’s how our code will “guess” at each step, all the way to the final solution: Make a list of all the empty spots. Select a spot and place a number, between 1 and ‘n’, in it and validate the candidate grid. If any of the constraints fails, . Otherwise, check if the goal is reached. abandon candidate and repeat step 2 with the next number If a solution is found, stop searching. Otherwise, repeat steps 2 to 4. Pen-paper demo . Now let’s try all this in practice with a simple 3x3 grid. Groovy A 3x3 Sudoku puzzle We start off by listing all the empty spots. If we label each cell in the grid with a pair of numbers and mark the first cell , then our empty spots will be at locations: (x,y) (1,1) (1,2) (2,2) (2,3) (3,1) (3,2) We now select the first spot to work with. Since this is a 3x3 grid, we have numbers 1 to 3 at our disposal and no sub-grids to worry about (sub-grids are only a bother for grids with squared sides, like 4, 9, 16 etc.). (1,2) Let’s place number in this spot and see if it fits. 1 Filling ‘1’ in spot (1, 2) It does. Great. We can now select the next spot on the list and do the same thing again. This time however, it fails. We already have a in this row. This means that we must — which is . (2,2) 1 abandon candidate and repeat step 2 with the next number 2 Filling 2 in spot (2, 2) One more spot is filled. Also, it might not look like it, but we did just perform backtracking on a single spot. We abandoned a candidate solution ( at spot ), visited a previous stage (empty spot ) and explored a new candidate solution (number at spot ). Huzzah! 1 (2,2) (2,2) 2 (2,2) When we move on to spot , we have another problem. As you can see, we are all out of options. None of the possible numbers fit in. This means that we must now . Only this time, we must visit spot first to fix spot . (2,3) abandon candidate and repeat step 2 with the next number (2,2) (2,3) Failure to fill spot (2, 3) We need to fill number in spot and that will resolve the issue. 3 (2,2) Backtracking to spot (2, 2) and then re-visiting spot (2, 3) We now repeat this process until with either reach the goal or we hit one of the termination conditions. Since this was a demo problem, it should be obvious that we’d arrive at the solution without any further complications. Solved 3x3 Sudoku puzzle However, consider the same grid with one small change. Replacing the in cell with a renders the grid unsolvable. Similarly, removing hints from cells and allows for multiple solutions. But since this algorithm has a single goal, it stops after the first solution is reached. 1 (3,3) 2 (2,1) (3,3) Unsolvable 3x3 puzzle (left) and Multi-solution 3x3 puzzle (right) Here’s what the looks like: tree of possibilities Tree of Possibilities for 3x3 Sudoku Implementation Enough talk. Let’s code … Disclaimer (for python-istas): The code you’re about to see is not pythonic.IMHO, it is borderline ugly. The purpose of this snippet is to explain convert the steps shown above into simple, meaningful code and not to boast the elegance of python. What next? This article aims to strengthen the concept of backtracking by drawing connections with a popular game of logic. I should tell you that Sudoku is an ‘ . As complex as it may sound, it really isn’t (not for the scope of this article): NP-complete combinatorial problem’ means that while we can verify a solution, it is rather to arrive at one. NP-complete easily difficult means that to arrive at a solution, we must perform selection and arrangement of some data — combination and permutation. Combinatorial Sample problems We can try to solve some other problems by basing our approach on our current understanding. The problem is a close enough relative to the Sudoku solver. Two more problems that rank similar are and . Binary Watch Combination Sum III Permutation Sequence There are that you can try on. However, trying to mold them into the Sudoku solver pattern might not always be trivial. It’s best to try to formulate your approach in the following pattern: a lot more problems Problem Goal Constraints Termination conditions Step-by-step algorithm

Sudoku and Backtracking

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Are you ready to join majestic world of C and C++ of programming? Do you want to question your existence after few simple lines of C++? If your answer is "Yeh!", “Yep“ or “Why not?“ - welcome  to test your knowledge. You will be given multiple questions related to C or C++. Please find correct answers and explanations in the end of the story. Good luck! 1. The smallest program main; What will happen if you try to compile this program using C compiler? will not compile
will compile, will not link
will compile and will link 2. The fork #include <iostream>
#include <unistd.h>

int main() {
    for(auto i = 0; i < 1000; i++)
        std::cout << "Hello world!\n";

    fork();
} How many lines will this program print? 1000
less than 1000
more than 1000 3. All you need is indices #include <iostream>

int main()  {
    int array[] = { 1, 2, 3 };
    std::cout << (4, (1, 2)[array]) << std::endl;
} What will this program print? 1
2
3
4
will not compile
undefined 4. Regular expressions #include <regex>
#include <iostream>

int main() {
    std::regex  re("(.*|.*)*O");
    std::string str("0123456789");

    std::cout << std::regex_match(str, re);

    return 0;
} How long will it take for this regular expression to match this input string? 1 ms
1 sec
1 min
1 hour
1 year
forever 5. Moves and lambdas #include <iostream>

struct Foo {
    Foo() { std::cout << "Foo()\n"; }
    Foo(Foo&&) { std::cout << "Foo(Foo&&)\n"; }
    Foo(const Foo&) { std::cout << "Foo(const Foo&)\n"; }
};

int main() {
    Foo f;
    auto a = [f = std::move(f)]() {
        return std::move(f);
    };

    Foo f2(a());
    return 0;
} The last line to be printed by this program is… Foo()
Foo(Foo&&)
Foo(const Foo&) 6. X and bar #include <iostream>

int x = 0;
int bar(int(x));

int main() {
    std::cout << bar;
} What will  this program print? 0
1
0x0
will not compile
will not link 7. Constructors #include <iostream>

struct Foo  {
    Foo() { std::cout << "Foo()\n"; }
    Foo(const Foo&) { std::cout << "Foo(const Foo&)\n"; }
    Foo(int) { std::cout << "Foo(int)\n"; }
    Foo(int, int) { std::cout << "Foo(int, int)\n"; }
    Foo(const Foo&, int) { std::cout << "Foo(const Foo&, int)\n"; }
    Foo(int, const Foo&) { std::cout << "Foo(int, const Foo&)\n"; }
};

void f(Foo) {}

struct Bar {
    int i, j;

    Bar() {
        f(Foo(i, j));
        f(Foo(i));
        Foo(i, j);
        Foo(i);
        Foo(i, j);
    }
};

int main() { Bar(); } The last line to be printed by this program is… Foo(int, int)
Foo(const Foo&, int)
Foo(int, const Foo&)
Foo(int) Instead of conclusion I hope you will never find one of this peculiar snippets in the wild. Answers The smallest program
This is legal C code. It will compile and link successfully. It will crash if you try to run it. main; - is global variable.In C code you can omit a lot of things. For example you can omit the type of a global variable. By the default compiler will assume this type is an int. Also there is no name mangling  in C (unlike in C++), so when linking there is no way to distinguish variable main from function main.Thus compiler will compile valid code, and linker will find something named main in object file to link a program.

The fork
This is more POSIX feature rather than C or C++ feature. Implementations of IO operations use buffers for optimising performance. When you invoke fork, the OS will create copy-on-write duplicate of process memory, the IO-buffers will likely duplicate as well and buffered strings likely will be printed more than 1000 times.

All you need is indices
Answer is 3To understand this code lets take a closer look on how indices in C and C++ work: array[index], is the same as *(array + index), is the same as (index + array) and the same as index[array.The second clue is operator,. Its binary operator, it discards left argument and returns right argument.


Regular expressions
Its impossible to predict what will happen! The behaviour is up to implementation.Im my environment this program raises the exception The complexity of an attempted match against a regular expression exceeded a pre-set level.Other probable options are surprisingly long time or work as expected. It’s because there are two possible approaches to implement regular expressions.First - transform regular expressions to finite automata O(n**2)(n - length of pattern), match string O(m) (m - length of string). This approach doesn’t support backtracking.Second - greedy approach + DFS, supports backtracking but prone to exponential time complexity on certain patterns.

Moves and lambdas
Answer is Foo(const Foo&). Lambdas are immutable by the default, all values captured into lambda with [] are implicitly const. This unlocks idempotent behaviour for lambdas.When you move f you create const Foo&&. const Foo&& is a weird type, thus compiler just copies Foo There are two ways to fix this:

Create mutable lambda
auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    };



Declare constructor Foo(const Foo&&)




X and bar
The program will print 1.
int bar(int(x)); — is weird way to declare function, it is equal to int bar(int x);.If you confused with type cast, int bar((int(x)));- this is type cast.
Than we try to implicitly cast function address tobool, the result of such cast is always true.Functionbar() has never been used, thats allow us to dodge unreferenced symbol error while linking.

Constructors
The  last line is Foo(const Foo&, int).
Foo(i) is variable declaration, the same as Foo i. Thus class member under the name of i is hidden in this scope. Are you ready to join majestic world of C and C++ of programming? Do you want to question your existence after few simple lines of C++? If your answer is "Yeh!", “Yep“ or “Why not?“ - welcome  to test your knowledge. You will be given multiple questions related to C or C++. Please find correct answers and explanations in the end of the story. Good luck! 1. The smallest program main; main; What will happen if you try to compile this program using C compiler? will not compile will compile, will not link will compile and will link will not compile will compile, will not link will compile and will link 2. The fork #include <iostream>
#include <unistd.h>

int main() {
    for(auto i = 0; i < 1000; i++)
        std::cout << "Hello world!\n";

    fork();
} #include <iostream>
#include <unistd.h>

int main() {
    for(auto i = 0; i < 1000; i++)
        std::cout << "Hello world!\n";

    fork();
} How many lines will this program print? 1000 less than 1000 more than 1000 1000 less than 1000 more than 1000 3. All you need is indices #include <iostream>

int main()  {
    int array[] = { 1, 2, 3 };
    std::cout << (4, (1, 2)[array]) << std::endl;
} #include <iostream>

int main()  {
    int array[] = { 1, 2, 3 };
    std::cout << (4, (1, 2)[array]) << std::endl;
} What will this program print? 1 2 3 4 will not compile undefined 1 2 3 4 will not compile undefined 4. Regular expressions #include <regex>
#include <iostream>

int main() {
    std::regex  re("(.*|.*)*O");
    std::string str("0123456789");

    std::cout << std::regex_match(str, re);

    return 0;
} #include <regex>
#include <iostream>

int main() {
    std::regex  re("(.*|.*)*O");
    std::string str("0123456789");

    std::cout << std::regex_match(str, re);

    return 0;
} How long will it take for this regular expression to match this input string? 1 ms 1 sec 1 min 1 hour 1 year forever 1 ms 1 sec 1 min 1 hour 1 year forever 5. Moves and lambdas #include <iostream>

struct Foo {
    Foo() { std::cout << "Foo()\n"; }
    Foo(Foo&&) { std::cout << "Foo(Foo&&)\n"; }
    Foo(const Foo&) { std::cout << "Foo(const Foo&)\n"; }
};

int main() {
    Foo f;
    auto a = [f = std::move(f)]() {
        return std::move(f);
    };

    Foo f2(a());
    return 0;
} #include <iostream>

struct Foo {
    Foo() { std::cout << "Foo()\n"; }
    Foo(Foo&&) { std::cout << "Foo(Foo&&)\n"; }
    Foo(const Foo&) { std::cout << "Foo(const Foo&)\n"; }
};

int main() {
    Foo f;
    auto a = [f = std::move(f)]() {
        return std::move(f);
    };

    Foo f2(a());
    return 0;
} The last line to be printed by this program is… Foo() Foo(Foo&&) Foo(const Foo&) Foo() Foo() Foo(Foo&&) Foo(Foo&&) Foo(const Foo&) Foo(const Foo&) 6. X and bar #include <iostream>

int x = 0;
int bar(int(x));

int main() {
    std::cout << bar;
} #include <iostream>

int x = 0;
int bar(int(x));

int main() {
    std::cout << bar;
} What will  this program print? 0 1 0x0 will not compile will not link 0 0 1 1 0x0 0x0 will not compile will not link 7. Constructors #include <iostream>

struct Foo  {
    Foo() { std::cout << "Foo()\n"; }
    Foo(const Foo&) { std::cout << "Foo(const Foo&)\n"; }
    Foo(int) { std::cout << "Foo(int)\n"; }
    Foo(int, int) { std::cout << "Foo(int, int)\n"; }
    Foo(const Foo&, int) { std::cout << "Foo(const Foo&, int)\n"; }
    Foo(int, const Foo&) { std::cout << "Foo(int, const Foo&)\n"; }
};

void f(Foo) {}

struct Bar {
    int i, j;

    Bar() {
        f(Foo(i, j));
        f(Foo(i));
        Foo(i, j);
        Foo(i);
        Foo(i, j);
    }
};

int main() { Bar(); } #include <iostream>

struct Foo  {
    Foo() { std::cout << "Foo()\n"; }
    Foo(const Foo&) { std::cout << "Foo(const Foo&)\n"; }
    Foo(int) { std::cout << "Foo(int)\n"; }
    Foo(int, int) { std::cout << "Foo(int, int)\n"; }
    Foo(const Foo&, int) { std::cout << "Foo(const Foo&, int)\n"; }
    Foo(int, const Foo&) { std::cout << "Foo(int, const Foo&)\n"; }
};

void f(Foo) {}

struct Bar {
    int i, j;

    Bar() {
        f(Foo(i, j));
        f(Foo(i));
        Foo(i, j);
        Foo(i);
        Foo(i, j);
    }
};

int main() { Bar(); } The last line to be printed by this program is… Foo(int, int) Foo(const Foo&, int) Foo(int, const Foo&) Foo(int) Foo(int, int) Foo(int, int) Foo(const Foo&, int) Foo(const Foo&, int) Foo(int, const Foo&) Foo(int, const Foo&) Foo(int) Foo(int) Instead of conclusion I hope you will never find one of this peculiar snippets in the wild. Answers The smallest program
This is legal C code. It will compile and link successfully. It will crash if you try to run it. main; - is global variable.In C code you can omit a lot of things. For example you can omit the type of a global variable. By the default compiler will assume this type is an int. Also there is no name mangling  in C (unlike in C++), so when linking there is no way to distinguish variable main from function main.Thus compiler will compile valid code, and linker will find something named main in object file to link a program. The fork
This is more POSIX feature rather than C or C++ feature. Implementations of IO operations use buffers for optimising performance. When you invoke fork, the OS will create copy-on-write duplicate of process memory, the IO-buffers will likely duplicate as well and buffered strings likely will be printed more than 1000 times. All you need is indices
Answer is 3To understand this code lets take a closer look on how indices in C and C++ work: array[index], is the same as *(array + index), is the same as (index + array) and the same as index[array.The second clue is operator,. Its binary operator, it discards left argument and returns right argument. Regular expressions
Its impossible to predict what will happen! The behaviour is up to implementation.Im my environment this program raises the exception The complexity of an attempted match against a regular expression exceeded a pre-set level.Other probable options are surprisingly long time or work as expected. It’s because there are two possible approaches to implement regular expressions.First - transform regular expressions to finite automata O(n**2)(n - length of pattern), match string O(m) (m - length of string). This approach doesn’t support backtracking.Second - greedy approach + DFS, supports backtracking but prone to exponential time complexity on certain patterns. Moves and lambdas
Answer is Foo(const Foo&). Lambdas are immutable by the default, all values captured into lambda with [] are implicitly const. This unlocks idempotent behaviour for lambdas.When you move f you create const Foo&&. const Foo&& is a weird type, thus compiler just copies Foo There are two ways to fix this:

Create mutable lambda
auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    };



Declare constructor Foo(const Foo&&) X and bar
The program will print 1.
int bar(int(x)); — is weird way to declare function, it is equal to int bar(int x);.If you confused with type cast, int bar((int(x)));- this is type cast.
Than we try to implicitly cast function address tobool, the result of such cast is always true.Functionbar() has never been used, thats allow us to dodge unreferenced symbol error while linking. Constructors
The  last line is Foo(const Foo&, int).
Foo(i) is variable declaration, the same as Foo i. Thus class member under the name of i is hidden in this scope. The smallest program This is legal C code. It will compile and link successfully. It will crash if you try to run it. main; - is global variable. In C code you can omit a lot of things. For example you can omit the type of a global variable. By the default compiler will assume this type is an int. Also there is no name mangling  in C (unlike in C++), so when linking there is no way to distinguish variable main from function main. Thus compiler will compile valid code, and linker will find something named main in object file to link a program. The smallest program The smallest program This is legal C code. It will compile and link successfully. It will crash if you try to run it. main; - is global variable. It will compile and link successfully. main; In C code you can omit a lot of things. For example you can omit the type of a global variable. By the default compiler will assume this type is an int . Also there is no name mangling in C (unlike in C++), so when linking there is no way to distinguish variable main from function main . int name mangling main main Thus compiler will compile valid code, and linker will find something named main in object file to link a program. main The fork This is more POSIX feature rather than C or C++ feature. Implementations of IO operations use buffers for optimising performance. When you invoke fork, the OS will create copy-on-write duplicate of process memory, the IO-buffers will likely duplicate as well and buffered strings likely will be printed more than 1000 times. The fork The fork This is more POSIX feature rather than C or C++ feature. Implementations of IO operations use buffers for optimising performance. When you invoke fork , the OS will create copy-on-write duplicate of process memory, the IO-buffers will likely duplicate as well and buffered strings likely will be printed more than 1000 times . fork likely will be printed more than 1000 times All you need is indices Answer is 3 To understand this code lets take a closer look on how indices in C and C++ work: array[index], is the same as *(array + index), is the same as (index + array) and the same as index[array. The second clue is operator,. Its binary operator, it discards left argument and returns right argument. All you need is indices All you need is indices Answer is 3 Answer is 3 To understand this code lets take a closer look on how indices in C and C++ work: array[index] , is the same as *(array + index) , is the same as (index + array) and the same as index[array . array[index] *(array + index) (index + array) index[array The second clue is operator , . Its binary operator, it discards left argument and returns right argument. , Regular expressions Its impossible to predict what will happen! The behaviour is up to implementation. Im my environment this program raises the exception The complexity of an attempted match against a regular expression exceeded a pre-set level. Other probable options are surprisingly long time or work as expected. It’s because there are two possible approaches to implement regular expressions. First - transform regular expressions to finite automata O(n**2)(n - length of pattern), match string O(m) (m - length of string). This approach doesn’t support backtracking. Second - greedy approach + DFS, supports backtracking but prone to exponential time complexity on certain patterns. Regular expressions Regular expressions Its impossible to predict what will happen! The behaviour is up to implementation. impossible to predict Im my environment this program raises the exception The complexity of an attempted match against a regular expression exceeded a pre-set level. The complexity of an attempted match against a regular expression exceeded a pre-set level. Other probable options are surprisingly long time or work as expected. It’s because there are two possible approaches to implement regular expressions. First - transform regular expressions to finite automata O(n**2) (n - length of pattern), match string O(m) (m - length of string). This approach doesn’t support backtracking. O(n**2) O(m) backtracking. Second - greedy approach + DFS, supports backtracking but prone to exponential time complexity on certain patterns. Moves and lambdas Answer is Foo(const Foo&). Lambdas are immutable by the default, all values captured into lambda with [] are implicitly const. This unlocks idempotent behaviour for lambdas. When you move f you create const Foo&&. const Foo&& is a weird type, thus compiler just copies Foo There are two ways to fix this: Create mutable lambda
auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    };



Declare constructor Foo(const Foo&&) Moves and lambdas Moves and lambdas Answer is Foo(const Foo&) . Lambdas are immutable by the default, all values captured into lambda with [] are implicitly const . This unlocks idempotent behaviour for lambdas. Foo(const Foo&) [] const idempotent When you move f you create const Foo&& . const Foo&& is a weird type, thus compiler just copies Foo f const Foo&& const Foo&& Foo There are two ways to fix this: Create mutable lambda
auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    }; Declare constructor Foo(const Foo&&) Create mutable lambda auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    }; Create mutable lambda auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    }; auto a = [f = std::move(f)]() mutable {
        return std::move(f);
    }; Declare constructor Foo(const Foo&&) Declare constructor Foo(const Foo&&) Foo(const Foo&&) X and bar The program will print 1. int bar(int(x)); — is weird way to declare function, it is equal to int bar(int x);. If you confused with type cast, int bar((int(x)));- this is type cast. Than we try to implicitly cast function address tobool, the result of such cast is always true. Functionbar() has never been used, thats allow us to dodge unreferenced symbol error while linking. X and bar X and bar The program will print 1 . The program will print 1 int bar(int(x)); — is weird way to declare function, it is equal to int bar(int x); . int bar(int(x)); int bar(int x); If you confused with type cast, int bar((int(x))); - this is type cast. int bar((int(x))); Than we try to implicitly cast function address to bool , the result of such cast is always true . bool true Function bar() has never been used, thats allow us to dodge unreferenced symbol error while linking. bar() Constructors The  last line is Foo(const Foo&, int). Foo(i) is variable declaration, the same as Foo i. Thus class member under the name of i is hidden in this scope. Constructors Constructors The  last line is Foo(const Foo&, int) . Foo(const Foo&, int) Foo(i) is variable declaration, the same as Foo i . Thus class member under the name of i is hidden in this scope. Foo(i) Foo i i

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