I have a code that is supposed to reduce a variable, by one in each iteration, in a loop. The problem is that it does not, it remains the same. Here is the code I have:
bool IsRed(int speed, int distance, int time)
{
if ((18 * distance) % (10 * speed * time) >= (5 * distance * time)) {return true;}
return false;
}
std::vector<std::pair<int, int> > DistanceTime = {{ 300,10 }, { 1500,10 }, { 3000,10 }};
int maxSpeed = 90 * 0.277778;
int traficLights = 3;
for (int i = 0; i < traficLights; i++)
{
for (auto j = DistanceTime.begin(); j != DistanceTime.end(); ++j)
{
int distance = j->first;
int time = j->second;
if (IsRed(maxSpeed, distance, time) == true)
{
maxSpeed--; //should get reduced
i = 0;
j = DistanceTime.begin();
}
//debug line
std::cout << maxSpeed * 3.6 << std::endl; //didn't change
}
}
The variable maxSpeed should get reduced every time the IsRed() function returns "true". Then the first and second for loop should re-initialize and test the new reduced speed again. At the end, the maxSpeed that returns "false", for all the pairs of the vector, in the IsRed() function, should be printed (did not do it in the code above).
Have I messed up the for loops? Any help would be very welcome!
In order for any code within:
if (IsRed(maxSpeed, distance, time) == true)
{
//stuff
}
to be executed, your function IsRed(...) needs to return true, and with the inputs you provided it doesn't.
you also don't need to specify == true in your if statement you can simply write:
if (IsRed(maxSpeed, distance, time))
{
//stuff
}
because of the return type of the function IsRed.
Related
I am trying to use find_if in a function with Lambda Expressions. I would like to get the last iterator of the vector, but I am not sure how to write it.
I can get the first iterator with this code. In this code, struct has an index and score. A vector in the code save structs and if an index is multiplies of five, the score is higher than usual. In this code, the first iterator's index is five and the last iterator's index is 25.
struct Recorder
{
Recorder(int idx, float score)
{
this->idx = idx;
this->score = score;
}
int idx;
float score;
};
int main(int argc, char** argv)
{
std::vector<std::shared_ptr<Recorder>> recorders;
for (int i = 0; i < 30; ++i)
{
float score = 0.3;
bool isLost = false;
if (i % 5 == 0 && i != 0) score = 0.8;
std::shared_ptr<Recorder> poseRecord = std::make_shared<Recorder>(i, score);
recorders.push_back(poseRecord);
}
auto record = std::find_if(recorders.begin(), recorders.end(),
[](const std::shared_ptr<Recorder>& r) {
return (r->score >= 0.8f);
});
std::cout << "idx : " << (*record)->idx << std::endl;
return 0;
}
How can I get the last iterate in this code? (here (*record)->idx should be 25).
You can use reverse iterators to start the search from the other end. Also, as Fureeish and Ted pointed out in the comments, it's more efficient to pass the shared pointer to the lambda function as reference.
auto record = std::find_if(recorders.rbegin(), recorders.rend(),
[&](const std::shared_ptr<Recorder> & r) {
return (r->isLost == false && r->score >= 0.8f);
});
I'm trying to create a array of prime numbers done by calculation. As a project to learn coding. Ultimately to build my own math library so this is something I can add onto at a variety of levels as I learn to code c++.
The following is code that works great for printing prime numbers to the screen based on the search range, but my totalPrimes iterator is stuck at 1. So each time it places the last prime found in the PrimeNumbers[1] position.
Any advice would be awesome.
#include <iostream>
#include <array>
std::array<long, 10000000> PrimeNumbers={0};
void isPrime(long x);
int main() {
for (long i = 1; i < 10; i++) {
isPrime(i);
}
for(int h = 0; h < 10; h++) {
std::cout << "\nSecond Prime is : " << PrimeNumbers[h];
}
}
void isPrime(long x) {
int count(0), totalPrimes(0);
for (long a = 1; a < x; a++) {
if ((x % a) == 0) {
count += 1;
}
}
if (count == 1) {
++totalPrimes;
std::cout << '\n' << x << " is a Prime number";
PrimeNumbers[totalPrimes] = x;
}
}
You're initializing totalPrimes to 0 every time the function runs. You would need to have totalPrimes as a global variable, or better yet (because global variables can become problematic), set it equal to the first available member of PrimeNumbers before you do anything else in that function.
Keep track of a position along with your PrimeNumbers array.
size_t nLastPos=0;
...
for(size_t x = 0; 1000 > x; ++x)
{
if(isPrime(x))
{
PrimeNumbers[nLastPos++] = x;
}
}
for(size_t i = 0; nLastPos > n; ++n)
{/* print out number PrimeNumbers[n] */ }
It looks like you're having some trouble with variable scoping. The reason for your problem (as I noted in the comment) is that totalPrimes is local, so you're creating a new integer variable and setting it to 0 every time the function is called.
However, you've made PrimeNumbers global and are having the isPrime function modify it, which doesn't look like good practice.
All of this can be fixed with a little restructuring to make the code nicer:
#include <iostream>
#include <array>
bool isPrime(long x);
int main() {
std::array<long, 10000000> PrimeNumbers={0};
int totalPrimes = 0;
for (long i = 1; i < 10; i++) {
if (isPrime(i)) {
std::cout << '\n' << i << " is a Prime number";
PrimeNumbers[totalPrimes++] = i;
}
}
for(int h = 0; h < 10; h++) {
std::cout << h << " Prime is : " << PrimeNumbers[h] << std::endl;
}
}
bool isPrime(long x) {
int count(0);
for (long a = 1; a < x; a++) {
if ((x % a) == 0) {
count += 1;
}
}
return count == 1;
}
Your program can be re-structured little bit to make it easier to follow and debug.
Don't put things in isPrime other than the logic to decide whether a number is prime. Make sure it returns a bool. This will make the function a bit simpler and easier to debug.
Use the return value of isPrime in the calling function to perform other bookkeeping tasks.
The logic you have used to check whether a number is prime is incorrect. That needs to be fixed.
Here's an updated version of your posted code.
#include <iostream>
#include <array>
#include <cmath>
std::array<long, 10000000> PrimeNumbers={0};
bool isPrime(long x);
int main()
{
int totalPrimes = 0;
for (long i = 1; i < 10; i++)
{
if ( isPrime(i) )
{
std::cout << i << " is a Prime number" << std::endl;
PrimeNumbers[totalPrimes] = i;
++totalPrimes;
}
}
}
bool isPrime(long x) {
// 1, 2, and 3 are primes.
if ( x <= 3 )
{
return true;
}
// Even numbers are not primes.
if ( x % 2 == 0 )
{
return false;
}
// Check the rest.
long end = (long)std::sqrt(x);
for (long a = 3; a < end; a += 2) {
if ((x % a) == 0)
{
return false;
}
}
return true;
}
and its output:
1 is a Prime number
2 is a Prime number
3 is a Prime number
5 is a Prime number
7 is a Prime number
9 is a Prime number
Everybody is talking about how your totalPrimes variable is reset each time the function is called, and this is obviously true. You could return the value from the function and increment it from main, you could use global variables having the variable being defined outside of the function so that it's not reset each time inside the function or you could use
A static variable!
Take a look at this simple case. I have a function called up_two which increases the value of by two each time the function is called. The static variable int value has a memory of each time the function up_two() is called which increments value by two each time. If I were to use a just an integer it would always reset the value and have it be zero, which is what I initially defined it to be.
The advantage of using a static variable is that I can count how many times a function has been called, and I can keep my counter specific to a particular function.
#include <iostream>
using namespace std;
void up_two();
int main()
{
for(int i = 0; i < 10; i++)
{
up_two();
}
return 0;
}
void up_two()
{
static int value = 0;
cout << value << endl;
value += 2;
}
This program doesn't solve the particular problem that you want to solve, but if you figure out how the static variable is working, it should make your workflow easier.
The magic line here is this:
static int value = 0;
With it like this my program prints the following:
0
2
4
6
8
10
12
14
16
18
Without the static declaration, you just get 10 lines of zeroes
which is troublesome.
Hope that helps you optimize your program the way you want it to be.
I have a bunch of maths manipulations that have thresholds, yet no matter what I change, the if statements always return true. No compile errors, can't get the debugger to work. This is a function, the X Y and Z arrays are all correct (I printed them to check earlier), the maths is right at least for the blade distance check, yet that always returns true. I ran the same code (rewritten obviously) through matlab and that returns true or false depending on my data, so clearly its something with the way I've written this that's wrong. Also is there any way to slimline this?
bool Device::_SafeD(char _Type, float _Data[20][3]) {
bool S;
double x[20], y[20], z[20];
for (int i=0; i<20; i++) {
x[i] = _Data[i][0];
y[i] = _Data[i][1];
z[i] = _Data[i][2];
}
// Check angles for needle
if (_Type == 'n') {
for (int i=0; i<20; i++) {
float dot, moda, modb, c, angle;
dot = ((x[i]*x[i+1]) + (y[i]*y[i+1]) + (z[i]*z[i+1]));
moda = sqrt(pow(x[i],2)+pow(y[i],2)+pow(z[i],2));
modb = sqrt(pow(x[i+1],2)+(y[i+1],2)+(z[i+1],2));
c = dot/(moda*modb);
angle = acos(c);
if (angle > 45){
S = 0;
} else {
S = 1;
}
}
}
// Check distance for blade
if (_Type == 'b'){
for (int i=0; i<19; i++) {
float distance = (x[i+1]-x[i]) + (y[i+1]-y[i]) + (z[i+1]-z[i]);
cout << "distance " << distance << endl;
if (distance > 5.0) {
S = 0;
} else {
S = 1;
}
}
}
if (S == 0) {
return 0;
}
if(S == 1) {
return 1;
}
}
Cheers
The most likely error is that you are comparing angle to an angle in degree while the return value of acos is in radians.
if (angle > 45){
Convert the angle to degrees before comparing to 45 and you should be OK.
if (radians_to_degrees(angle) > 45){
where
double radians_to_degrees(double in)
{
return (in*180/M_PI);
}
Another option is to compute the equivalent of 45 degrees in radians and compare it with angle.
double const radian_45 = M_PI*45.0/180;
and use
if (angle > radian_45){
The other error is spelled out clearly in a comment by #Bob__:
OP uses two loops to check the angle and the distance, inside those loops a flag (S) is set to 0 or 1, but it's done for every index in the array, so it's overwritten. The return value of the entire function (in the provided code) depends only by the last two elements in the array.
I'm trying to implement NegaMax ai for Connect 4. The algorithm works well some of the time, and the ai can win. However, sometimes it completely fails to block opponent 3 in a rows, or doesn't take a winning shot when it has three in a row.
The evaluation function iterates through the grid (horizontally, vertically, diagonally up, diagonally down), and takes every set of four squares. It then checks within each of these sets and evaluates based on this.
I've based the function on the evaluation code provided here: http://blogs.skicelab.com/maurizio/connect-four.html
My function is as follows:
//All sets of four tiles are evaluated before this
//and values for the following variables are set.
if (redFoursInARow != 0)
{
redScore = INT_MAX;
}
else
{
redScore = (redThreesInARow * threeWeight) + (redTwosInARow * twoWeight);
}
int yellowScore = 0;
if (yellowFoursInARow != 0)
{
yellowScore = INT_MAX;
}
else
{
yellowScore = (yellowThreesInARow * threeWeight) + (yellowTwosInARow * twoWeight);
}
int finalScore = yellowScore - redScore;
return turn ? finalScore : -finalScore; //If this is an ai turn, return finalScore. Else return -finalScore.
My negamax function looks like this:
inline int NegaMax(char g[6][7], int depth, int &bestMove, int row, int col, bool aiTurn)
{
{
char c = CheckForWinner(g);
if ('E' != c || 0 == depth)
{
return EvaluatePosition(g, aiTurn);
}
}
int bestScore = INT_MIN;
for (int i = 0; i < 7; ++i)
{
if (CanMakeMove(g, i)) //If column i is not full...
{
{
//...then make a move in that column.
//Grid is a 2d char array.
//'E' = empty tile, 'Y' = yellow, 'R' = red.
char newPos[6][7];
memcpy(newPos, g, sizeof(char) * 6 * 7);
int newRow = GetNextEmptyInCol(g, i);
if (aiTurn)
{
UpdateGrid(newPos, i, 'Y');
}
else
{
UpdateGrid(newPos, i, 'R');
}
int newScore = 0; int newMove = 0;
newScore = NegaMax(newPos, depth - 1, newMove, newRow, i, !aiTurn);
newScore = -newScore;
if (newScore > bestScore)
{
bestMove = i;
bestScore = newScore;
}
}
}
}
return bestScore;
}
I'm aware that connect four has been solved are that there are definitely better ways to go about this, but any help or suggestions with fixing/improving this will be greatly appreciated. Thanks!
Edit: to clarify, the problem is with the second algorithm.
I have a bit of C++ code that samples cards from a 52 card deck, which works just fine:
void sample_allcards(int table[5], int holes[], int players) {
int temp[5 + 2 * players];
bool try_again;
int c, n, i;
for (i = 0; i < 5 + 2 * players; i++) {
try_again = true;
while (try_again == true) {
try_again = false;
c = fast_rand52();
// reject collisions
for (n = 0; n < i + 1; n++) {
try_again = (temp[n] == c) || try_again;
}
temp[i] = c;
}
}
copy_cards(table, temp, 5);
copy_cards(holes, temp + 5, 2 * players);
}
I am implementing code to sample the hole cards according to a known distribution (stored as a 2d table). My code for this looks like:
void sample_allcards_weighted(double weights[][HOLE_CARDS], int table[5], int holes[], int players) {
// weights are distribution over hole cards
int temp[5 + 2 * players];
int n, i;
// table cards
for (i = 0; i < 5; i++) {
bool try_again = true;
while (try_again == true) {
try_again = false;
int c = fast_rand52();
// reject collisions
for (n = 0; n < i + 1; n++) {
try_again = (temp[n] == c) || try_again;
}
temp[i] = c;
}
}
for (int player = 0; player < players; player++) {
// hole cards according to distribution
i = 5 + 2 * player;
bool try_again = true;
while (try_again == true) {
try_again = false;
// weighted-sample c1 and c2 at once
// h is a number < 1325
int h = weighted_randi(&weights[player][0], HOLE_CARDS);
// i2h uses h and sets temp[i] to the 2 cards implied by h
i2h(&temp[i], h);
// reject collisions
for (n = 0; n < i; n++) {
try_again = (temp[n] == temp[i]) || (temp[n] == temp[i+1]) || try_again;
}
}
}
copy_cards(table, temp, 5);
copy_cards(holes, temp + 5, 2 * players);
}
My problem? The weighted sampling algorithm is a factor of 10 slower. Speed is very important for my application.
Is there a way to improve the speed of my algorithm to something more reasonable? Am I doing something wrong in my implementation?
Thanks.
edit: I was asked about this function, which I should have posted, since it is key
inline int weighted_randi(double *w, int num_choices) {
double r = fast_randd();
double threshold = 0;
int n;
for (n = 0; n < num_choices; n++) {
threshold += *w;
if (r <= threshold) return n;
w++;
}
// shouldn't get this far
cerr << n << "\t" << threshold << "\t" << r << endl;
assert(n < num_choices);
return -1;
}
...and i2h() is basically just an array lookup.
Your reject collisions are turning an O(n) algorithm into (I think) an O(n^2) operation.
There are two ways to select cards from a deck: shuffle and pop, or pick sets until the elements of the set are unique; you are doing the latter which requires a considerable amount of backtracking.
I didn't look at the details of the code, just a quick scan.
you could gain some speed by replacing the all the loops that check if a card is taken with a bit mask, eg for a pool of 52 cards, we prevent collisions like so:
DWORD dwMask[2] = {0}; //64 bits
//...
int nCard;
while(true)
{
nCard = rand_52();
if(!(dwMask[nCard >> 5] & 1 << (nCard & 31)))
{
dwMask[nCard >> 5] |= 1 << (nCard & 31);
break;
}
}
//...
My guess would be the memcpy(1326*sizeof(double)) within the retry-loop. It doesn't seem to change, so should it be copied each time?
Rather than tell you what the problem is, let me suggest how you can find it. Either 1) single-step it in the IDE, or 2) randomly halt it to see what it's doing.
That said, sampling by rejection, as you are doing, can take an unreasonably long time if you are rejecting most samples.
Your inner "try_again" for loop should stop as soon as it sets try_again to true - there's no point in doing more work after you know you need to try again.
for (n = 0; n < i && !try_again; n++) {
try_again = (temp[n] == temp[i]) || (temp[n] == temp[i+1]);
}
Answering the second question about picking from a weighted set also has an algorithmic replacement that should be less time complex. This is based on the principle of that which is pre-computed does not need to be re-computed.
In an ordinary selection, you have an integral number of bins which makes picking a bin an O(1) operation. Your weighted_randi function has bins of real length, thus selection in your current version operates in O(n) time. Since you don't say (but do imply) that the vector of weights w is constant, I'll assume that it is.
You aren't interested in the width of the bins, per se, you are interested in the locations of their edges that you re-compute on every call to weighted_randi using the variable threshold. If the constancy of w is true, pre-computing a list of edges (that is, the value of threshold for all *w) is your O(n) step which need only be done once. If you put the results in a (naturally) ordered list, a binary search on all future calls yields an O(log n) time complexity with an increase in space needed of only sizeof w / sizeof w[0].