I need help with writing a backtracking recursion code in c++ that reach the min steps from 1 number to other number. you can only use +1 or *2
for example getting 23 from 10 by shortest is: ((10+1)*2)+1 has 3 steps,
or getting 65 from 12: (((12+1+1+1+1)*2)*2)+1 has 7 steps
TNX.
Here is some help, the fundamental recursion function:
void recursive(unsigned int count)
{
// Need to determine the end of recursion.
if (count == 0)
{
return;
}
// Call the function again, which is recursion.
recursive(count - 1);
}
Build upon this model to figure out how to implement your requirements.
Hint: With recursion, one step at a time. On a piece of paper, write down each step performed, along with all the values, until the operation is finished. Look at the last step to determine how to stop the recursion.
int goal=23;
int result=1e9; ## set the count to infinity to optimize it
void solve(int curNumber,int step){
if(goal == curNumber) // Check if we reached the goal number
result=min(result,step); // take the min of the solution, and number of steps we took to reach the goal
curNumber+=1 // add 1 to current Number
solve(curNumber,step+1);
curNumber-=1; // backtrack on the add 1 solution
curNumber*=2; // choose the other path *2
solve(curNumber, step+1);
}
Related
I was solving this question which is pretty interesting in leetcode and I found a solution for this.Though this iterative approach works,I want to know the logical thinking behind this solution.
Question:
There is only one character 'A' on the screen of a notepad. You can perform two operations on this notepad for each step:
Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).
Paste: You can paste the characters which are copied last time.
Given an integer n, return the minimum number of operations to get the character 'A' exactly n times on the screen.
Example testcases:
Input: n = 3
Output: 3
Explanation: Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.
My code:
class Solution {
public:
int minSteps(int n) {
int res=0;
for(int i=2;i<=n;i++)
{
while(n%i==0)
{
res+=i;
n=n/i;
}
}
return res;
}
};
The code is pretty simple,Just need to understand the intuition behind it.
Thanks
I am learning recursion in my current class and the idea is a little tricky for me. From my understanding, when we build a function it will run as many times until our "base case" is satisfied. What I am wondering is how this looks and is returned on the stack. For an example I wrote the following function for a simple program to count how many times a digit shows up in an integer.
What does this look and work in a stack frame view? I don't completely understand how the returning works. I appreciate the help!
int count_digits(int n, int digit) {
// Base case: When n is a single digit.
if (n / 10 == 0) {
// Check if n is the same as the digit.
// When recursion hits the base case it will end the recursion.
if (n == digit) {
return 1;
} else {
return 0;
}
} else {
if (n % 10 == digit) {
return (1 + count_digits(n / 10, digit));
} else {
return (count_digits(n / 10, digit));
}
}
}
What does this look and work in a stack frame view? I don't completely understand how the returning works. I appreciate the help!
Let's try to build the solution bottom-up.
If you called the function - int count_digits(int n, int digit) as count_digits(4, 4) what would happen ?
This is the base case of your solution so it is very easy to see what is the return value. Your function would return 1.
Now, let's add one more digit and call the function like- count_digits(42, 4). What would happen ?
Your function will check the last digit which is 2 and compare with 4 since they are not equal so it will call the function count_digits(4, 4) and whatever is the returned value, will be returned as the result of count_digits(42, 4).
Now, let's add one more digit and call the function like - count_digits(424, 4). What would happen ?
Your function will check the last digit which is 4 and compare with 4 since they are equal so it will call the function count_digits(42, 4) and whatever is the returned value, will be returned by adding 1 to it. Since, number of 4s in 424 is 1 + number of 4s in 42. The result of count_digits(42,4) will be calculated exactly like it was done previously.
The recursive function builds up the solution in a top-down manner. If there are n digits initially, then your answer is (0 or 1 depending on the last digit) + answer with n-1 digits. And this process repeats recursively. So, your recursive code, reduces the problems by one digit at a time and it depends on the result of the immediate sub-problem.
You can use the C++ tutor at pythontutor.com website for step by step visualization of the stack frame. http://pythontutor.com/cpp.html#mode=edit
You can also try with smaller inputs and add some debug output to help you track and see how recursion works.
Check this stackoverflow answer for understanding what a stack frame is - Explain the concept of a stack frame in a nutshell
Check this stackoverflow answer for understanding recursion -
Understanding recursion
If you would like more help, please let me know in comments.
how does line return((count-2)+(count-1)) works in below cpp program?
ans of the given code is -18 .how to know the ans without running the code
and out of two function count(n-2) and count(n-1) which one is called first and how is it decided?
#include <iostream>
using namespace std;
int count(int n);
int main() {
int n, m;
n = 4;
m = count(n);
cout << m;
}
int count(int n)
{
if (n<0)
{
return n;
}
else
{
return (count(n - 2) + count(n - 1));
}
}
There's no sequencing between the left-hand and right-hand side of the + operator. So which one is evaluated first is unknown (and left up to the compiler).
The only way to figure it out is to step thought he code line by line, statement by statement, expression by expression in a debugger.
However, since each recursive call is not depending on any side-effects they can be executed independently of each other, and therefore the order doesn't matter as the result will always be the same.
We can simply draw a binary tree to know the answer without compiling it. Just start breaking branch into two for count(n-1) and count(n-2) then add all leaves of trees.
Like if we took n as 4 it would be split as 3 and 2, Which would be two branch of 4. Similarly and recursively break nodes in to branch. 3 would be split into 1 and 2 so on. till leaves node is less then 0. In the end add value of all leaves to get the answers.
Hello I have a homework question I am stuck in..any hint or tips would be appreciated. the questions is:
Write a single recursive function in C++ that takes as argument a positive integer n then print n, n-1, n-2,...3,2,1,2,3,...n. How many recursive call does your algorithm makes? What is the worst case running time of your algorithm?
I am stuck in the first part. writing a recursive function that prints n, n-1, n-2,...3,2,1,2,3,...n
so far I have:
print(int n)
{
if (n==0)
return;
cout<<n<<" ";
print(n-1);
return;
}
but this only prints from n to 1
I am lost about how I would print from 2 to n using just one parameter and recursively single function.
I tried this: which gives the correct output but has a loop and has two parameters:
p and z has the same value.
void print(int p,int z)
{
if (p==0)
{
for(int x=2;x<=z; x++)
cout<<x<<" ";
return;
}
else
cout<<p<<" ";
print(p-1,z);
return;
}
any hint or tips is much appreciated thank you.
so it is working now, but I am having trouble understanding how (question in comment):
void print(int n)
{
if (n==1){
cout<<n;
return;
}
else
cout<< n;
print(n-1); // how does it get passed this? to the line below?
cout<<n; // print(n-1) takes it back to the top?
return;
}
The output you want is mirrored, so you can have this series of steps:
print num
recursive step on num-1
print num again
That's the recursive case. Now you need an appropriate base case upon which to stop the recursion, which shouldn't be difficult.
Given the pseudocode:
recursive_print(n):
if n == 1:
print 1
return
print n
recursive_print(n-1)
print n
(If you prefer, just look at your solution instead).
Let's trace it. A dot will mark where we're up to in terms of printing.
. recursive_print(3) // Haven't printed anything
3 . recursive_print(2) 3 // Print the 3
3 2 . recursive_print(1) 2 3 //Print 2
3 2 1 . 2 3 // Print 1
3 2 1 2 . 3 // Print 2
3 2 1 2 3 . // Print 3
Each unrolling of the function gives us 2 numbers on opposite sides and we build down to the "1", then go back and print the rest of the numbers.
The "unrolling" is shown in this picture:
If you strip away the functions and leave yourself with a sequence of commands, you'll get:
print 3
print 2
print 1
print 2
print 3
where each indentation signifies a different function.
Simple solution for this:
Def fxn(n):
if n <= n:
if n > 0:
print(n)
fxn(n - 1)
print(n)
Def main():
Number = 6
fxn(Number)
Main()
If you struggle understanding how this works:
Basically, each time you call a function in a recursive problem, it isn't a loop. It's as if you were on the woods leaving a trail behind. Each time you call a function inside a function, it does its thing, then it goes right back to were you called it.
In other words, whenever you call a recursive function, once the newer attempt is done, it will go right back to were it used to be.
In loops, once each step is done, it is done, but in recursive functions you can do a lot with a lot less.
Printing before the recurse call in my code is the descension step, and once its descension finished, it will progresively unfold step by step, printing the ascension value and going back to the former recurse.
It seems way harder than it is, but it is really easy once you grasp it.
The answer is simpler than you think.
A recursive call is no different from a regular call. The only difference is that the function called is also the caller, so you need to make sure you don't call it forever (you already did that). Let's think about a regular call. If you have the following code snippet:
statement1
f();
statement2
The statement1 is executed, then f is called and does it's thing and then, after f finishes, statement2 is executed.
Now, let's think about your problem. I can see that your hard work on this question from the second program you've written, but forget about it, the first one is very close to the answer.
Let's think about what your function does. It prints all numbers from n to 0 and then from 0 to n. At the first step, you want to print n, then all the numbers from n-1 to 0 and from 0 to n-1, and print another n. See where it's going?
So, you have to do something like this:
print(n)
call f(n-1)
print(n)
I hope my explanation is clear enough.
This is more of hack -- using the std::stream rather than recursion...
void rec(int n) {
if (n==1) { cout << 1; return; }
cout << n;
rec(n-1);
cout << n;
}
int main() {
rec(3);
}
prints
32123
I'm new to C++ and have to do a home assignment (sudoku). I'm stuck on a problem.
Problem is that to implement a search function which to solve a sudoku.
Instruction:
In order to find a solution recursive search is used as follows. Suppose that there is a
not yet assigned field with digits (d1....dn) (n > 1). Then we first try to
assign the field to d1, perform propagation, and then continue with search
recursively.
What can happen is that propagation results in failure (a field becomes
empty). In that case search fails and needs to try different digits for one of
the fields. As search is recursive, a next digit for the field considered last
is tried. If none of the digits lead to a solution, search fails again. This in
turn will lead to trying a different digit from the previous field, and so on.
Before a digit d is tried by assigning a field to
it, you have to create a new board being a copy of the current board (use
the copy constructor and allocate the board from the heap with new). Only
then perform the assignment on the copy. If the recursive call to search
returns unsuccessfully, a new board can be created for the next digit to be
tried.
I've tried:
// Search for a solution, returns NULL if no solution found
Board* Board::search(void) {
// create a copy of the cur. board
Board *copyBoard = new Board(*this);
Board b = *copyBoard;
for(int i = 0; i < 9; i++){
for(int j = 0; j < 9; j++){
// if the field has not been assigned, assign it with a digit
if(!b.fs[i][j].assigned()){
digit d = 1;
// try another digit if failed to assign (1 to 9)
while (d <=9 && b.assign(i, j, d) == false){
d++;
// if no digit matches, here is the problem, how to
// get back to the previous field and try another digit?
// and return null if there's no pervious field
if(d == 10){
...
return NULL;
}
}
}
}
return copyBoard;
}
Another problem is where to use the recursive call? Any tips? thx!
Complete instruction can been found here: http://www.kth.se/polopoly_fs/1.136980!/Menu/general/column-content/attachment/2-2.pdf
Code: http://www.kth.se/polopoly_fs/1.136981!/Menu/general/column-content/attachment/2-2.zip
There is no recursion in your code. You can't just visit each field once and try to assign a value to it. The problem is that you may be able to assign, say, 5 to field (3,4) and it may only be when you get to field (6,4) that it turns out there can't be a 5 at (3, 4). Eventually you need to back out of recursion until you come back to (3,4) and try another value there.
With recursion you might not use nested for loops to visit fields, but visit the next field with a recursive call. Either you manage to reach the last field, or you try all possibilities and then leave the function to get back to the previous field you visited.
Sidenote: definitely don't allocate dynamic memory for this task:
//Board *copyBoard = new Board(*this);
Board copyBoard(*this); //if you need a copy in the first place
Basically what you can try is something like this (pseudocode'ish)
bool searchSolution(Board board)
{
Square sq = getEmptySquare(board)
if(sq == null)
return true; // no more empty squares means we solved the puzzle
// otherwise brute force by trying all valid numbers
foreach (valid nr for sq)
{
board.doMove(nr)
// recurse
if(searchSolution(board))
return true
board.undoMove(nr) // backtrack if no solution found
}
// if we reach this point, no valid solution was found and the puzzle is unsolvable
return false;
}
The getEmptySquare(...) function could return a random empty square or the square with the least number of options left.
Using the latter will make the algorithm converge much faster.