Im using c++ and is using insertion sort
Where in the insertion sort algoithm should we put a counter to monitor number of item movements and number of item comparison. I have included my setup below
void InsertionSort::insertion_sort()
{
int key,i,count = 0;
for(int j=1;j<10;j++)
{
key=Arr1[j];
i=j-1;
while(Arr1[i]>key && i>=0)
{
Arr1[i+1]=Arr1[i];
i--;
numberOfItemMovements++;
}
Arr1[i+1]=key;
}
}
}
as you can see, i cant seem to figure out where comparison counter should be put, although the item movement counter is good and work as expected. thanks
A way of getting this to work is using numberOfComparisons in the loop.
while(++numberOfComparisons && Arr1[i]>key && i>=0)
{
Arr1[i+1]=Arr1[i];
numberOfItemMovements++;
i--;
}
1) Do you understand why ++numberOfComparisons can be used in the while loop but numberOfComparisons++ would fail?
2) You have a problem where Arr1[-1] could be evaluated. Can you figure out where this is happening and how you can change your code to fix it?
The movement counter is indeed correct.
In order to implement the comparison counter, you need to restructure your code a bit. In particular, you need to be able to distinguish between Arr1[i]>key and i>=0, since one of them entails a comparison while the other doesn't.
Since this looks like homework, I'll leave the rest for you to figure out.
Related
I was just wondering about something that popped into my mind while writing some code.
for (int i = 0; i < num_bits; i++) {
if (bits.at(i) == 0) {
}
else if (bits.at(i) == 1) {
}
}
In this code, bits is a string and num_bits is the length of the string.
In this case, would the program run string.at(i) at both the if and the `else if``, or would it run it once and then store it somewhere and use it at both of the statements? I don't know if the question was clear enough, but thanks for any answer.
Think about it. How would the engine know that every call to that function would produce the same result?
It wil run the function whenever you call it, so for this example 2 times. You can declare it at the top of the for loop or use a foreach if you need to do more heavy operations.
I'm trying to implement a quick sort that sorts numbers and words based on the number value. I can't seem to figure out how to fix the following code to work right.
if (high!=low&& high>low)//compares hashes and finds the number in the middle. swaps hashes and corresponding words
{
long one=hash[low];
long two=hash[high];
long three = hash[high/2];
if((one<=two&&one>=three)||(one<=three&&one>=two))
{
swap(hash[low], hash[high]);
swap(copyOfWords[low], copyOfWords[high]);
}
else if((three<=one&&three>=two)||(three<=two&&three>=one))
{
swap(hash[high/2], hash[high]);
swap(copyOfWords[high/2], copyOfWords[high]);
}
else
{
}
int i=low;
int j=high-1;
while(i!=j&&i<j)
{
while(hash[i]<hash[high]&&i<j)// find higher numbers and lower numbers then the middlle and swaps them
{
i++;
}
while(hash[j]>hash[high]&&i<j)
{
j--;
}
if(i==j||i>j)
{
}
else
{
swap(hash[i],hash[j]);
swap(copyOfWords[i],copyOfWords[j]);
i++;
j--;
}
}
swap(hash[i],hash[high]);
swap(copyOfWords[i], copyOfWords[high]);
quickSort(low, j-1);//recursive
quickSort(j+1, high);
}
}
I know the values in hash and copyOfWords are correct because when I use shell sort, it sorts them the right way. for example if there are two words, copyOfWOrds[0]="1994," and copyOfWords[1]="a" then hash[0]=549456039 and hash[1]=197000000, but the sort puts them a 1994, a instead of a 1994,. It causes more problems with more elements. Any help would be appreciated. Thanks
Why don't you go to the quick sort wiki page and see how it's done?
Your code tries to do unnecessary stuffs and eventually trips over its own feet. Keep it simple and it will work.
And Btw Quicksort works very well on arrays, so it's a shame to make one version where the array is hard-coded.
I have a vector holding 10 items (all of the same class for simplicity call it 'a'). What I want to do is to check that 'A' isn't either a) hiding the walls or b) hiding another 'A'. I have a collisions function that does this.
The idea is simply to have this looping class go though and move 'A' to the next position, if that potion is causing a collision then it needs to give itself a new random position on the screen. Because the screen is small, there is a good chance that the element will be put onto of another one (or on top of the wall etc). The logic of the code works well in my head - but debugging the code the object just gets stuck in the loop, and stay in the same position. 'A' is supposed to move about the screen, but it stays still!
When I comment out the Do while loop, and move the 'MoveObject()' Function up the code works perfectly the 'A's are moving about the screen. It is just when I try and add the extra functionality to it is when it doesn't work.
void Board::Loop(void){
//Display the postion of that Element.
for (unsigned int i = 0; i <= 10; ++i){
do {
if (checkCollisions(i)==true){
moveObject(i);
}
else{
objects[i]->ResetPostion();
}
}
while (checkCollisions(i) == false);
objects[i]->SetPosition(objects[i]->getXDir(),objects[i]->getYDir());
}
}
The class below is the collision detection. This I will expand later.
bool Board::checkCollisions(int index){
char boundry = map[objects[index]->getXDir()][objects[index]->getYDir()];
//There has been no collisions - therefore don't change anything
if(boundry == SYMBOL_EMPTY){
return false;
}
else{
return true;
}
}
Any help would be much appreciated. I will buy you a virtual beer :-)
Thanks
Edit:
ResetPostion -> this will give the element A a random position on the screen
moveObject -> this will look at the direction of the object and adjust the x and Y cord's appropriately.
I guess you need: do { ...
... } while (checkCollisions(i));
Also, if you have 10 elements, then i = 0; i < 10; i++
And btw. don't write if (something == true), simply if (something) or if (!something)
for (unsigned int i = 0; i <= 10; ++i){
is wrong because that's a loop for eleven items, use
for (unsigned int i = 0; i < 10; ++i){
instead.
You don't define what 'doesn't work' means, so that's all the help I can give for now.
There seems to be a lot of confusion here over basic language structure and logic flow. Writing a few very simple test apps that exercise different language features will probably help you a lot. (So will a step-thru debugger, if you have one)
do/while() is a fairly advanced feature that some people spend whole careers never using, see: do...while vs while
I recommend getting a solid foundation with while and if/else before even using for. Your first look at do should be when you've just finished a while or for loop and realize you could save a mountain of duplicate initialization code if you just changed the order of execution a bit. (Personally I don't even use do for that any more, I just use an iterator with while(true)/break since it lets me pre and post code all within a single loop)
I think this simplifies what you're trying to accomplish:
void Board::Loop(void) {
//Display the postion of that Element.
for (unsigned int i = 0; i < 10; ++i) {
while(IsGoingToCollide(i)) //check is first, do while doesn't make sense
objects[i]->ResetPosition();
moveObject(i); //same as ->SetPosition(XDir, YDir)?
//either explain difference or remove one or the other
}
}
This function name seems ambiguous to me:
bool Board::checkCollisions(int index) {
I'd recommend changing it to:
// returns true if moving to next position (based on inertia) will
// cause overlap with any other object's or structure's current location
bool Board::IsGoingToCollide(int index) {
In contrast checkCollisions() could also mean:
// returns true if there is no overlap between this object's
// current location and any other object's or structure's current location
bool Board::DidntCollide(int index) {
Final note: Double check that ->ResetPosition() puts things inside the boundaries.
Hello I am trying to get enemies to move left and right as if they are sliding backwards and forwards I know this can be done with the following code:
slide += slide_incr;
if(abs(slide)>30) slide_incr = -slide_incr;
However this is of no use to me as I need to set a boolean so I can cycle through the frames for when the enemy is going right or going left.
Ive tried the follow code with no luck:
if(abs(eSlide)<=0)
{
eSlide += eSlide_incr;
}
if(abs(eSlide)>30)
{
eSlide_incr = -eSlide_incr;
}
Any ideas on how I can implement it?
Thanks
You want to hold a hysteresis state for if you're sliding forward or backward. You are also mixing up how to use the abs() function when bounds checking. Try something along the lines of:
eSlide += eSlide_incr;
if (abs(eSlide) >= 30) {
eSlide_incr = -eSlide_incr;
}
the first thing that stands out for me is that the contents of the block:
if (abs(eSlid) <= 0) {
eSlide += eSlide_incr;
}
will never ever run (the absolute value will always be greater than or equal to 0)
as for your boolean facing, that can be achieved with:
bool isSlidingRight = eSlide_incr > 0;
(note: this would still use the left animation set for values of 0)
Guys, I have the following code that is inside a big while loop that iterates over a tree. This is as fast as I can get this routine but I have to use a goto. I am not fundamentally against goto but if I can avoid them I would like to. (I am not trying to start a flame war, please.)
The constraints:
The current=current->child() is expensive (it's a shared_ptr) so I'd like to minimize the use of that operation at all cost.
After the operation current should be the last child it found.
cnt must count each child it encounters.
cnt++ will be replaced by some other operation (or several operations) and should only appear once :)
the code:
insideloopy:
cnt++;
if ( current->hasChild() )
{
current = current->child();
goto insideloopy;
}
Edit: Sorry guys, originally forgot to mention cnt++ should only appear once. It will be some kind of operation on the node, and should thus only be there one time. I'm also trying to avoid making that another function call.
Original answer
Assuming this is C or C++:
while (cnt++, current->hasChild())
{
current = current->child();
}
I'm not a big fan of the comma operator usually, but I don't like repeating myself either :)
Updated 'fun' answer
After learning that cnt++ is actually some multiline operation, this particular syntax would be less than ideal. Something more along the lines of your accepted answer would be better.
If you want to be really funky, this would also work:
do
{
cnt++;
} while (current->hasChild() && (current = current->child()));
Now I feel really dirty though, with my abusing the short circuiting on the && operator :)
Sane answer
Exercises in compactness aside and striving for readable code, I'm forced to conclude that one of the existing answers is best suited (I'm just including this for completeness' sake):
while (true)
{
cnt++;
if (!current->hasChild()) break;
current = current->child();
}
The while (true) will be optimized by the compiler into a regular infinite loop, so there is only one conditional statement (if you care about that).
The only thing going against this solution is if your node operation was a long piece of code. I don't mind infinite loops so much, as long as I can see where they terminate at a glance. Then again, if it were really long, it should be a function anyway.
cnt++;
while(current->hasChild())
{
cnt++;
current = current->child();
}
EDIT:
If you only want cnt++ to be in your code once:
while(true)
{
cnt++;
if(current->hasChild())
current = current->child();
else
break;
}
insideloopy:
cnt++;
if ( current->hasChild() )
{
current = current->child();
goto insideloopy;
}
I love infinite loops.
while (true) {
cnt++;
if (!current->hasChild()) break;
current = current->child();
}
Of course you can do it in many other ways (see other answers). do while, put the check in the while, etc. In my solution, I wanted to map nearly to what you are doing (an infinite goto, unless break)
You can use break to get out of the loop in the middle of the code:
while (true) {
cnt++;
if (!current->hasChild()) break;
current = current->child();
}
while (current->hasChild())
{
cnt++;
current = current->child();
}
Or am I missing something?
for(cnt++ ; current->hasChild() ; cnt++) {
current = current->child();
}
I'd investigate the possibility of making current->child() return NULL when it has no child if it doesn't already -- that seems the best possible result and leaving it undefined in this case seems error prone -- and then use:
for (; current; current = current->child())
{
cnt++;
}
No break statements:
notDone=true;
while(notDone){
cnt++;
if ( current->hasChild() ){
current = current->child();
} else {
notDone=false;
}
}