I've been asked to write a permutation function that uses recursion. The only parameter of the function should be the string that I should find all the permutations of. The function should return a vector with all possible permutations. I know I can use next_permutation in STL Algorithms, but I've been asked not to.
I have the base case set up, and I know I need a for loop, but I'm not quite sure where to go from there. Can someone point me in the right direction?
vector <string> getPerm(string str)
{
vector<string> v;
if(w.length() <= 1)
{
v.push_back(str);
return v;
}
else
{
for(int i = 0; i < str.size(); i++)
{
//Some code
}
}
}
Any help would be appreciated.
Imagine you already have the result of the previous iteration of your function, with returns all the permutations of the first n-1 elements of your string.
vector<string>& v_prev = getPerm(str.substr(0, str.length()-1));
Use this in the
//Some code
part of your code.
Another tip: use the 0-length string as the stop-condition of your recursion. You can construct the 1-lenght permutations recursively ;)
Here is the entire solution:
vector<string> getPerm(string str)
{
vector<string> v;
if (str.empty())
{
v.push_back(string());
return v;
}
else
{
vector<string>& v_prev = getPerm(str.substr(0, str.length()-1));
for(int i = 0; i < v_prev.size(); i++)
{
for (int j = 0; j < v_prev[i].length() + 1; j++)
{
string p = v_prev[i];
p.insert(j, str.substr(str.length() - 1, 1));
v.push_back(p);
}
}
return v;
}
}
Think about these permutations of the string "123"
123
132
213
231
312
321
And think about these permutations of "12"
12
21
Can you see how you might construct the permutations of a n letter string if you know the permutations of all the n-1 letter substrings. That type of solution would be recursive.
For each element x in yourArray
Make a copy of yourArray without element x. Call this new array newArray.
Find all of the permutations of newArray
add element x to the beginning of each of those permutations
Implementing what just Ken Bloom wrote:
vector <string> getPerm(string str)
{
vector<string> v;
if(str.length() <= 1)
{
v.push_back(str);
return v;
}
else
{
for(int i = 0; i < str.size(); i++){
vector<string> perms = getPerm(str.substr(0,i)+str.substr(i+1));
for(int j = 0; j < perms.size(); j++){
v.push_back(str[i] + perms[j]);
}
}
}
}
Try something like this:
permut(s) :
if s.length=0 : exit;
else :
for i=0 to s.length :
front:=s[i];
remove(s,i);
s2 := front + permut(s);
print s2, NEWLINE;
Related
I tried to write an algorithm to guess correctly in the game "Masterminds",
it works the average number of guesses is 6, but it takes a lot of time to calculate the best guess.
I used the idea of Knuth the algorithm works as follows:
Create the set S of 1296 possible codes (1111, 1112 ... 6665, 6666).
Start with initial guess 1122 (Knuth gives examples showing that other first guesses such as 1123, 1234 do not win in five tries on
every code).
Play the guess to get a response of colored and white pegs.
If the response is four colored pegs, the game is won, the algorithm terminates.
Otherwise, remove from S any code that would not give the same response if the current guess were the code.
In my code step 2 is to take random number.
I used vector<string> for this.
AllPoss is the vector full of strings, I guess is the last guess that was used. answer is the count of bulls and cows looks like "x,y" (where x and y are numbers)
void bullpgia::SmartGuesser::remove(string guess, string answer)
{
for (auto i= AllPoss.begin();i != AllPoss.end();i++){
string token = *i;
if (calculateBullAndPgia(token, guess) != answer)
AllPoss.erase(i--);
}
}
this is the part it take a lot of time to calculate is there any way of improvement?
to creating the list i used :
void bullpgia::SmartGuesser::All() {
/**
* creates a pool of all the possibilities strings
* we then delete the ones we dont need
* #param length is the length of the word we need to guess
*/
for(int i=0;i<pow(10,length);i++){
stringstream ss;
ss << setw(length) << setfill('0') << i;
string s = ss.str();
AllPoss.push_back(s);
}
}
the function calculateBullAndPgia(string , string) is:
string calculateBullAndPgia(const string &choice, const string &guess) {
string temp = choice;
string temp2 = guess;
unsigned int bull = 0;
unsigned int pgia = 0;
for (int i = 0; i < temp.length(); i++) {
if (temp[i] == temp2[i]) {
bull++;
temp[i] = 'a';
temp2[i] = 'z';
}
}
for (int i = 0; i < temp.length(); i++) {
for (int j = 0; j < temp2.length(); j++) {
if (i != j && temp[i] == temp2[j]) {
pgia++;
temp[i] = 'a';
temp2[j] = 'z';
}
}
}
return to_string(bull) + "," + to_string(pgia);
}
Erasing a single element in the middle of a vector is O(n). My guess is that you wind up doing it O(n) times per call to SmartGuesser::remove. Then you loop over that so you probably have a O(n^3) algorithm. You instead could use std::remove_if, which is O(n), to move all the to-be-erased elements to the end of the vector where they can be cheaply erased.:
AllPoss.erase(std::remove_if(AllPos.begin(), AllPos.end(), [&](const std::string& token, const std::string& guess) { return calculateBullAndPgia(token, guess) != answer; }), AllPos.end());
I have created a function that creates all the possible solutions for a game that I am creating... Maybe some of you know the bullcow game.
First I created a function that creates a combination of numbers of max four integers and the combination can't have any repeating number in it... like...
'1234' is a solution but not '1223' because the '2' is repeating in the number. In total there is 5040 numbers between '0123' and '9999' that haven't repeating numbers.
Here is my function:
std::vector <std::array<unsigned, 4>> HittaAllaLosningar(){
std::vector <std::array<unsigned, 4>> Losningar;
for (unsigned i = 0; i < 10; i++) {
for (unsigned j = 0; j < 10; j++) {
for (unsigned k = 0; k < 10; k++) {
for (unsigned l = 0; l < 10; l++) {
if (i != j && i != k && i != l && j != k && j != l && k != l) {
Losningar.push_back({i,j,k,l});
}
}
}
}
}
return Losningar;
}
Now let's say I have the number '1234' and that is not the solution I am trying to find, I want to remove the solution '1234' from the array since that isn't a solution... how do I do that? have been trying to find for hours and can't find it. I have tried vector.erase but I get errors about unsigned and stuff... also its worth to mention the guesses are in strings.
What I am trying to do is, to take a string that I get from my program and if it isn't a solution I want to remove it from the vector if it exists in the vector.
Here is the code that creates the guess:
std::string Gissning(){
int random = RandomGen();
int a = 0;
int b = 0;
int c = 0;
int d = 0;
for (unsigned i = random-1; i < random; i++) {
for (unsigned j = 0; j < 4; j++) {
if (j == 0) {
a = v[i][j];
}
if (j == 1) {
b = v[i][j];
}
if (j == 2) {
c = v[i][j];
}
if (j == 3) {
d = v[i][j];
}
}
std::cout << std::endl;
AntalTry++;
}
std::ostringstream test;
test << a << b << c << d;
funka = test.str();
return funka;
}
The randomgen function is just a function so I can get a random number and then I go in the loop so I can take the element of the vector and then I get the integers of the array.
Thank you very much for taking your time to help me, I am very grateful!
You need to find the position of the element to erase.
std::array<unsigned, 4> needle{1, 2, 3, 4};
auto it = std::find(Losningar.begin(), Losningar.end(), needle);
if (it != Losningar.end()) { Losningar.erase(it); }
If you want to remove all the values that match, or you don't like checking against end, you can use std::remove and the two iterator overload of erase. This is known as the "erase-remove" idiom.
std::array<unsigned, 4> needle{1, 2, 3, 4};
Losningar.erase(std::remove(Losningar.begin(), Losningar.end(), needle), Losningar.end());
To erase from a vector you just need to use erase and give it an iterator, like so:
std::vector<std::array<unsigned, 4>> vec;
vec.push_back({1,2,3,4});
vec.push_back({4,3,2,1});
auto it = vec.begin(); //Get an iterator to first elements
it++; //Increment iterator, it now points at second element
it = vec.erase(it); // This erases the {4,3,2,1} array
After you erase the element, it is invalid because the element it was pointing to has been deleted. Ti continue to use the iterator you can take the return value from the erase function, a valid iterator to the next element after the one erased, in this the case end iterator.
It is however not very efficient to remove elements in the middle of a vector, due to how it works internally. If it's not important in what order the different solution are stored, a small trick can simplify and make your code faster. Let's say we have this.
std::vector<std::array<unsigned, 4>> vec;
vec.push_back({1,2,3,4});
vec.push_back({4,3,2,1});
vec.push_back({3,2,1,4});
To remove the middle one we then do
vec[1] = vec.back(); // Replace the value we want to delete
// with the value in the last element of the vector.
vec.pop_back(); //Remove the last element
This is quite simple if you have ready other functions:
using TestNumber = std::array<unsigned, 4>;
struct TestResult {
int bulls;
int cows;
}
// function which is used to calculate bulls and cows for given secred and guess
TestResult TestSecretGuess(const TestNumber& secret,
const TestNumber& guess)
{
// do it your self
… … …
return result;
}
void RemoveNotMatchingSolutions(const TestNumber& guess, TestResult result)
{
auto iter =
std::remove_if(possibleSolutions.begin(),
possibleSolutions.end(),
[&guess, result](const TestNumber& possibility)
{
return result == TestSecretGuess(possibility, guess);
});
possibleSolutions.erase(iter, possibleSolutions.end());
}
Disclaimer: it is possible to improve performance (you do not care about order of elements).
I'm trying to do string matching algorithm a brute force method. but The algorithm is not working correctly, I get an out of bound index error.
here is my algorithm
int main() {
string s = "NOBODY_NOTICED_HIM";
string pattern="NOT";
int index = 0;
for (int i = 0; i < s.size();)
{
for (int j = 0; j < pattern.size();)
{
if(s[index] == pattern[j])
{
j++;
i++;
}
else
{
index = i;
j = 0;
}
}
}
cout<<index<<endl;
return 0;
}
FIXED VERSION
I fixed the out of bound exception. I don't know if the algorithm will work with different strings
int main() {
string s = "NOBODY_NOTICED_HIM";
string pattern="NOT";
int index = 0;
int i = 0;
while( i < s.size())
{
i++;
for (int j = 0; j < pattern.size();)
{
if(s[index] == pattern[j])
{
index++;
j++;
cout<<"i is " <<i << " j is "<<j <<endl;
}
else
{
index = i;
break;
}
}
}
cout<<i<<endl;
return 0;
}
Because the inner for loop has a condition to loop while j is less than pattern.size() but you are also incrementing i inside the body. When i goes out of bounds of s.size() then index also goes out of bounds and you'd get an OutOfBounds error.
The brute force method has to test the pattern with every possible subsequence. The main condition is the length, which has to be the same. All subsequence from s are:
['NOB', 'OBO', 'BOD', 'ODY', 'DY_', 'Y_N', 'NO', 'NOT', 'OTI', 'TIC',
'ICE', 'CED', 'ED', 'D_H', '_HI', 'HIM']
There are many ways to do it, you can do it char by char, or by using string operations like taking a substring. Both are nice excercises for learning.
Starting at zero in the s string you take the first three chars, compare to the pattern, and if equal you give the answer. Otherwise you move on to the char starting at one, etc.
I'm trying to step through a given string with a for loop, replacing one character per iteration with a character from a vector[char].
Problem is that the replace inserts the entire vector-k instead of the character at place k and I cannot figure out what I've done wrong.
Any and all help is appreciated.
(alphabet is a const string a-z, FirstWord is the given string).
vector<char> VectorAlphabet;
for (int i=0; i<alphabet.length(); ++i)
{
VectorAlphabet.push_back(alphabet.at(i));
}
for (int i = 0; i < FirstWord.length(); ++i )
{
for (int k = 0; k < VectorAlphabet.size(); ++k)
{
string TempWord = FirstWord;
TempWord.replace(i, 1, &VectorAlphabet[k]);
if (CheckForValidWord(TempWord, WordSet))
{
if(CheckForDuplicateChain(TempWord, DuplicateWordSet))
{
DuplicateWordSet.insert(TempWord);
stack<string> TempStack = WordStack;
TempStack.push(TempWord);
WordQueue.push(TempStack);
}
}
}
}
e.g TempWord = tempword, then after TempWord.replace() on the first iteration it is abcde...zempWord. and not aempword. On the second to last iteration of the second for loop it is yzempword.
What have I missed?
Problem solved, thanks to Dieter Lücking.
Looking closer at the string.replace reference, I see that I tried to use a replace which takes strings as the input, and then the vector[char] is interpreted as a c-string, starting from the k-position.
By using the fill-version of replace the vector position is correctly used as a char instead.
New code is:
for (int i = 0; i < FirstWord.length(); ++i )
{
for (int k = 0; k < VectorAlphabet.size(); ++k)
{
string TempWord = WordStack.top();
// Change:
TempWord.replace(i, 1, 1, VectorAlphabet[k]);
if (CheckForValidWord(TempWord, WordSet))
{
if(CheckForDuplicateChain(TempWord, DuplicateWordSet))
{
DuplicateWordSet.insert(TempWord);
stack<string> TempStack = WordStack;
TempStack.push(TempWord);
WordQueue.push(TempStack);
}
}
}
}
I have googled this question and couldn't find an answer that worked with my code so i wrote this to get the frequency of the words the only issue is that i am getting the wrong number of occurrences of words apart form one that i think is a fluke. Also i am checking to see if a word has already been entered into the vector so i don't count the same word twice.
fileSize = textFile.size();
vector<wordFrequency> words (fileSize);
int index = 0;
for(int i = 0; i <= fileSize - 1; i++)
{
for(int j = 0; j < fileSize - 1; j++)
{
if(string::npos != textFile[i].find(textFile[j]) && words[i].Word != textFile[j])
{
words[j].Word = textFile[i];
words[j].Times = index++;
}
}
index = 0;
}
Any help would be appreciated.
Consider using a std::map<std::string,int> instead. The map class will handle ensuring that you don't have any duplicates.
Using an associative container:
typedef std::unordered_map<std::string, unsigned> WordFrequencies;
WordFrequencies count(std::vector<std::string> const& words) {
WordFrequencies wf;
for (std::string const& word: words) {
wf[word] += 1;
}
return wf;
}
It is hard to get simpler...
Note: you can replace unordered_map with map if you want the worlds sorted alphabetically, and you can write custom comparisons operations to treat them case-insensitively.
try this code instead if you do not want to use a map container..
struct wordFreq{
string word;
int count;
wordFreq(string str, int c):word(str),count(c){}
};
vector<wordFreq> words;
int ffind(vector<wordFreq>::iterator i, vector<wordFreq>::iterator j, string s)
{
for(;i<j;i++){
if((*i).word == s)
return 1;
}
return 0;
}
Code for finding the no of occurrences in a textfile vector is then:
for(int i=0; i< textfile.size();i++){
if(ffind(words.begin(),words.end(),textfile[i])) // Check whether word already checked for, if so move to the next one, i.e. avoid repetitions
continue;
words.push_back(wordFreq(textfile[i],1)); // Add the word to vector as it was not checked before and set its count to 1
for(int j = i+1;j<textfile.size();j++){ // find possible duplicates of textfile[i]
if(file[j] == (*(words.end()-1)).word)
(*(words.end()-1)).count++;
}
}