I made this algorithm, i was debugging it to see why it wasnt working, but then i started getting weird stuff while printing arrays at the end of each cycle to see where the problem first occurred.
At a first glance, it seemed my while cycles didn't take into consideration the last array value, but i dunno...
all info about algorithm and everything is in the source.
What i'd like to understand is, primarily, the answer to this question:
Why does the output change sometimes?? If i run the program, 60-70% of the time i get answer 14 (which should be wrong), but some other times i get weird stuff as the result...why??
how can i debug the code if i keep getting different results....plus, if i compile for release and not debug (running codeblocks under latest gcc available in debian sid here), i get most of the times 9 as result.
CODE:
#include <iostream>
#include <vector>
/*void print_array
{
std::cout<<" ( ";
for (int i = 0; i < n; i++) { std::cout<<array[i]<<" "; }
std::cout<<")"<<std::endl;
}*/
///this algorithm must take an array of elements and return the maximum achievable sum
///within any of the sub-arrays (or sub-segments) of the array (the sum must be composed of adjacent numbers within the array)
///it will squeeze the array ...(...positive numbers...)(...negative numbers...)(...positive numbers...)...
///into ...(positive number)(negative number)(positive number)...
///then it will 'remove' any negative numbers in case it would be convienent so that the sum between 2 positive numbers
///separated by 1 negative number would result in the highest achievable number, like this:
// -- (3,-4,4) if u do 'remove' the negative number in order to unite the positive ones, i will get 3-4+4=3. So it would
// be better not to remove the negative number, and let 4 be the highest number achievable, without any sums
// -- (3,-1,4) in this case removing -1 will result in 3-1+4=6, 6 is bigger than both 3 and 4, so it would be convienent to remove the
// negative number and sum all of the three up into one number
///so what this step does is shrink the array furthermore if it is possible to 'remove' any negatives in a smart way
///i also make it reiterate for as long as there is no more shrinking available, because if you think about it not always
///can the pc know if, after a shrinking has occured, there are more shrinkings to be done
///then, lastly, it will calculate which of the positive numbers left is highest, and it will choose that as remaining maximum sum :)
///expected result for the array of input, s[], would be (i think), 7
int main() {
const int n=4;
int s[n+1]={3,-2,4,-4,6};
int k[n+1]={0};
///PRINT ARRAY, FOR DEBUG
std::cout<<" ( ";
for (int i = 0; i <= n; i++) { std::cout<<k[i]<<" "; }
std::cout<<")"<<std::endl;
int i=0, j=0;
// step 1: compress negative and postive subsegments of array s[] into single numbers within array k[]
/*while (i<=n)
{
while (s[i]>=0)
{
k[j]+=s[i]; ++i;
}
++j;
while (s[i]<0)
{
k[j]+=s[i]; ++i;
}
++j;
}*/
while (i<=n)
{
while (s[i]>=0)
{
if (i>n) break;
k[j]+=s[i]; ++i;
}
++j;
while (s[i]<0)
{
if (i>n) break;
k[j]+=s[i]; ++i;
}
++j;
}
std::cout<<"STEP 1 : ";
///PRINT ARRAY, FOR DEBUG
std::cout<<" ( ";
for (int i = 0; i <= n; i++) { std::cout<<k[i]<<" "; }
std::cout<<")"<<std::endl;
j=0;
// step 2: remove negative numbers when handy
std::cout<<"checked WRONG! "<<unsigned(k[3])<<std::endl;
int p=1;
while (p!=0)
{
p=0;
while (j<=n)
{
std::cout<<"checked right! "<<unsigned(k[j+1])<<std::endl;
if (k[j]<=0) { ++j; continue;}
if ( k[j]>unsigned(k[j+1]) && k[j+2]>unsigned(k[j+1]) )
{
std::cout<<"checked right!"<<std::endl;
k[j+2]=k[j]+k[j+1]+k[j+2];
k[j]=0; k[j+1]=0;
++p;
}
j+=2;
}
}
std::cout<<"STEP 2 : ";
///PRINT ARRAY, FOR DEBUG
std::cout<<" ( ";
for (int i = 0; i <= n; i++) { std::cout<<k[i]<<" "; }
std::cout<<")"<<std::endl;
j=0; i=0; //i will now use "i" and "p" variables for completely different purposes, as not to waste memory
// i will be final value that algorithm needed to find
// p will be a value to put within i if it is the biggest number found yet, it will keep changing as i go through the array....
// step 3: check which positive number is bigger: IT IS THE MAX ACHIEVABLE SUM!!
while (j<=n)
{
if(k[j]<=0) { ++j; continue; }
p=k[j]; if (p>i) { std::swap(p,i); }
j+=2;
}
std::cout<<std::endl<<"MAX ACHIEVABLE SUM WITHIN SUBSEGMENTS OF ARRAY : "<<i<<std::endl;
return 0;
}
might there be problems because im not using vectors??
Thanks for your help!
EDIT: i found both my algorithm bugs!
one is the one mentioned by user m24p, found in step 1 of the algorithm, which i fixed with a kinda-ugly get-around which ill get to cleaning up later...
the other is found in step2. it seems that in the while expression check, where i check something against unsigned values of the array, what is really checked is that something agains unsigned values of some weird numbers.
i tested it, with simple cout output:
IF i do unsigned(k[anyindexofk]) and the value contained in that spot is a positive number, i get the positive number of course which is unsigned
IF that number is negative though, the value won't be simply unsigned, but look very different, like i stepped over the array or something...i get this number "4294967292" when im instead expecting -2 to return as 2 or -4 to be 4.
(that number is for -4, -2 gives 4294967294)
I edited the sources with my new stuff, thanks for the help!
EDIT 2: nvm i resolved with std::abs() using cmath libs of c++
would there have been any other ways without using abs?
In your code, you have:
while (s[i]>=0)
{
k[j]+=s[i]; ++i;
}
Where s is initialized like so
int s[n+1]={3,-2,4,-4,6};
This is one obvious bug. Your while loop will overstep the array and hit garbage data that may or may not be zeroed out. Nothing stops i from being bigger than n+1. Clean up your code so that you don't overstep arrays, and then try debugging it. Also, your question is needs to be much more specific for me to feel comfortable answering your question, but fixing bugs like the one I pointed out should make it easier to stop running into inconsistent, undefined behavior and start focusing on your algorithm. I would love to answer the question but I just can't parse what you're specifically asking or what's going wrong.
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This is a C++ program to search a number and also display its index position if found in an array.
Program works fine. I just wanted to ask why we used k=1 in if statement because if we don't use it, then program won't work. Please explain the significance of k=1 in 2nd for loop in if statement.
#include<iostream>
using namespace std;
int main(){
int n;
cout<<"Size of Array: ";
cin>>n;
int arr[n];
cout<<"Type Elements of Array: ";
for(int i=0; i<n; i++)
{
cin>>arr[i];
}
int k;
cout<<"Enter No to be found: ";
cin>>k;
for(int i=0; i<n; i++)
{
if(arr[i]==k)
{
k=1;
cout<<"No is found at index position: "<<i<<endl;;
break;
}
else
{
cout<<"No is not found";
break;
}
}
return 0;
}
Looks like you are trying to do a linear search, there are some points I would like to clarify..
Never use int arr[n] using cin by user, it may work in yours, but not all compilers support it. Always allocate memory to the array statically like arr[10] or arr[20]
You are changing k which is the number that is to be found, I cannot understand why are you changing k . This is the number that is to be searched, you are mistakenly interpreting k as a flag variable, but it is NOT. Never try to change the input variable from the user.
Your program will only work, if the number you are searching is at 0th index, because, after that you are literally breaking from the loop in both if as well as else condition.
Here is a link to a better linear search code in c++, I hope it will clear your concepts C++ Linear Search
The problem is not with the k=1 statement. It lies in your logic of the else block.
Consider this:
n = 5
arr = [12, 2, 3, 14, 5]
k = 3
So initially i in the for loop starts from 0
It checks for equality:
Is arr[i] == k ?
arr[0] == 3 ?
12 == 3?
This is false, so your program goes to the else block, where it prints that the number was not found and breaks from the for loop. So essentially you are not checking all numbers, just the first one.
PS: Try and find a working logic for this on your own, it will be a good exercise and improve your understanding if you are getting started with programming (Hint: There are multiple ways, either through a flag variable or using the loop variable, etc). When your program doesn't work as you wanted it to, perform a dry run like I did on top.
You should go for the dynamic allocation of the array. Secondly, there is a mess with your if-else block. It checks only the first number in the array and then proceeds to the else portion. Your program can work fine but you should make sure that it searches the complete array for the number to be searched. You can do it by just removing the else part:
for(int i=0;i<n;i++)
{
if(arr[i]==k)
{
cout<<"No is found at index position: "<<i<<endl;;
return 0;
}
}
cout<<"No is not found.";
return 0;
I'm new too c++ and I had to design a program that determines the first four triangular square numbers and the output is exactly how I want it to be, but it wont quit after its printed the first four. I can't figure out what it could be. I can't CTRL C because I will get points taken off. What is the issue here?
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
//Prints name line
cout<<"*********** BY: ********"<<endl;
//Initializing
const int HOW_MANY=4;
int num=1;
int tsn=0;
int z=1;
int x=0;
//How many TSN it will find and be printed
while (x<=HOW_MANY)
{
//
int sum=0;
for (int y=0;y<512;y++)
{
sum+=y;
tsn=pow(num,2);
//Tests if the numbers are TSN
if ((sum==tsn) || (num+1)/sqrt(num)==sqrt(num))
{
//Prints 1-HOW_MANY TSN and what they are
cout<<"Square Triangular Number "<< z <<" is: "<< tsn <<endl;
z++;
x++;
}
}
num++;
}
return 0;
}
If x = 0 then instead of while (x<=HOW_MANY) you need write while (x<HOW_MANY).
x begins at 0. Every time you find and print a number it gets incremented. You'll continue this, so long as x<=HOW_MANY.
You say your program finds 4 numbers but keeps running. After 4 hits, x will be 4. Is 4 <= 4? The answer is yes, so your program keeps running.
Either change the condition to x < HOW_MANY, or initialize x to 1.
EDIT
Did a little leg work, it turns out the sum of all the numbers in the range [1,512] is 131328. The 5th square triangle number is 1413721.
This means after you find the fourth triangle number, you will never sum high enough to find the next one. This will result in the infinite loop you're seeing.
The answer above is still the correct fix, but this is the reason you end up with an infinite loop.
for should be used for iteration and while should be used for condition testing.
The problem, as has been noted, is that your x condition variable is never being incremented to get you out of the outer loop. That's a logic error that can be avoided by using the appropriate control structure for the job.
this is my first post here and I would be very happy if you could help me.
The task is - Create an array from 6 input numbers, then put the duplicated numbers in another array and then output the array with the repeated numbers.
Do you have any ideas? I'm still a newbie and need some help. Thanks in advance guys !!
EDIT:
I'm not sure if I'm on the right way, that's why I didn't post what I've done yet. But this is it:
#include <iostream>
using namespace std;
int main()
{
int a[6];
int b[6];
int i,z;
for (i=0; i<6; i++){
cin>>a[i];
}
for (z=0; z<6; z++){
if (a[0]==a[1]) b[z]=a[0];
if (a[0]==a[2]) b[z]=a[0];
if (a[0]==a[3]) b[z]=a[0];
if (a[0]==a[4]) b[z]=a[0];
if (a[0]==a[5]) b[z]=a[0];
if (a[1]==a[2]) b[z]=a[1];
if (a[1]==a[3]) b[z]=a[1];
if (a[1]==a[4]) b[z]=a[1];
if (a[1]==a[5]) b[z]=a[1];
if (a[2]==a[3]) b[z]=a[2];
if (a[2]==a[4]) b[z]=a[2];
if (a[2]==a[5]) b[z]=a[2];
if (a[3]==a[4]) b[z]=a[3];
if (a[3]==a[5]) b[z]=a[3];
if (a[4]==a[5]) b[z]=a[4];
else b[z]=0; cout << b[z];
}
return 0;
}
To give you a better understanding of how to solve this, i'll try to show you what is going on via an example.
Lets say you have just entered the 6 numbers you request via cin, and your a[] variable now looks like this in memory:
a[] = { 5, 2, 6, 2, 1, 6 };
The duplicates here are 2 and the 6. (pretty obvious for us humans) :-)
You start to compare the first 2 values in memory: a[0]==a[1], then the first with the third: a[0]==a[2] and so on. If one of these match, you know the value of a[0] has at least one duplicate in memory.
Whenever that happens, you would like to do something that that information. Store it somewhere (like your b[] array) or just output it directly with cout << a[0].
You are now finished with checking a[0] and can continue with a[1] in the same manor, except you do not have to compare with a[0] because you did that in the previous step. Looking at your code, it seems you already understand that you can skip that.
Lets say you really need to store the duplicates. It would help to keep track of how many duplicates you have found.
Pseudo code:
duplicates = 0;
if (a[0] has a duplicate) { b[duplicates] = a[0]; duplicates++; }
if (a[1] has a duplicate) { b[duplicates] = a[1]; duplicates++; }
// etc...
"has a duplicate" would be like the code you had earlier, like: a[0]==a[1] || a[0]==a[2] || a[0]==[3] and so on.
In your example you have just 6 values, so it is not much work to write all the compare statements yourself. If you needed to do this with many more numbers, it would take you ages to write it, and is prone to little mistakes like typo's. Using a for loop would work for few and many numbers:
Pseude code:
duplicates = 0;
for (z = 0 to 6) {
for (y = z+1 to 6) {
if (a[z]==a[y]) {
b[duplicates] = a[z];
duplicates++;
break; // We know it is a duplicate, continue with the next value
}
}
}
But even this is not perfect. If one number occurs more than 2 times in memory, this will store the same duplicate value multiple times.
I have a programming assignment to write a program in C++ that finds all primes less than n (user input). One half of the assignment involves the Sieve of Eratosthenes. My code is working (read: assignment is complete), but before I edited the output, it was unconditionally printing out n-3, n-2, and n-1 as primes even if they were not prime. I'm not sure why this is happening. I'd appreciate a bit of feedback and ideas as to why the program is acting the way it is. Here is the unaltered code:
Please note that I am using a ListNode class and a LinkedList class, both of which are fully functional. EDIT: partial main added; notice the second item in the for loop is size-3. If it's left at size, the program outputs 3 extra non-primes.
int main()
{
for(int i = 0; i<my_list.size()-3; i++)
{
if(marked[i]==true)
cout<<my_list[i]<<"\n";
}
}
void eratosthenes(int item)
{
bool run=true;
int p=2, count=0;
for(int i=2; i<=item; i++)
{
my_list.append(i); // Entire list is filled with integers from 2 to n
marked.append(true); // Entire list is filled with true entries
}
while(run==true&&(2*p)<item)
{
count = 0;
int i = (2*p);
do {
marked[i-2]=false; // marked values are false and not prime
i+=p;
} while(i<item-2);
for(int i=0; i<item-2; i++) // i starts at 0 and increments by 1
{ // each time through the loop
if(my_list[i]>p)
{
if(marked[i]==true) // If a value stored in a node is true
{ // (prime), it becomes the new p.
p=my_list[i]; // The loop is then broken.
break;
}
}
}
for(int j=1; j<item-2; j++)
{
if(marked[j]==false)
{
count=1;
}
}
if(count==0)
run=false;
}
Complete method
void Eratosthenes(int upperBound)
{
bool Prime[upperBound];
for(int i = 0;i<upperBound;i++)
Prime[i]=true;
for (int i = 2; i <= sqrt(upperBound); i++)
{
if (Prime[i])
{
for (int j = i * 2; j < upperBound; j += i)
Prime[j] = false;
}
}
for(int i=2;i<upperBound;i++)
{
if(Prime[i]==true)
cout<<i<<" ";
}
}
From your code:
do{
marked[i-2]=false;//marked values are false and not prime
i+=p;
}while(i<item-2);
This loop is responsible for going through all numbers i that are integer multiples of the prime number p and marking them not prime, as I understand. Why are you stopping on the condition i < item - 2? This would be fine if i were your index for the my_list and marked lists, but in this case it's not; it's the actual number you're marking not prime. I suspect this is why you're getting numbers near your limit (item) that are marked as prime—your loop here exits before i ever gets to those numbers!
By the way, you could do this as a for loop instead, which would be easier to read. The for loop has the meaning "go through each element in a set" (whether that's consecutive integers, or every nth integer, or elements in an array/list/deque, etc.), so a programmer reading your code knows that immediately and doesn't have to figure it out from your while loop.
// mark every multiple of the current prime as not prime
for(int i = 2*p; i < item - 2; i += p)
{
marked[i-2] = false;
}
(This is the same as your original code, no fixes applied).
Some general comments to improve your algorithm/code:
Try using more descriptive variable names. Your use of i two times to mean different things is confusing, and in general single letters don't mean much as to what the variable represents (although sometimes they're sufficient, e.g. a for loop where i is the index of a list/array).
Also, you're looping over your list a lot more than you need to. The minimum a sieve of Eratosthenes algorithm needs is two nested for loops (not including initializing a list/array to all true).
One example of where you're doing more work than necessary is that you're looping starting from index 0 to find the next p to use—instead of just remembering where your current p is and starting from there. You wouldn't even need to check my_list[i] > p in that case, since you know you'd be beyond it to start off. Also, your last loop could break; early and avoid continuing on after it finds a non-prime (and I'm not sure what the point of it is).
Nikola Mitev's second answer is a more efficient and more readable implementation of the sieve (but replace sqrt(upperBound) with upperBound/2 for it to work correctly; the reason for upperBound/2 should be pretty clear from the way the Sieve works), although he didn't really give much comment or explanation on it. The first loop is "go through every number up to upperBound"; inside it, "if the current number is a prime, go through all the multiples of that prime and mark them non-prime". After that innerloop executes, the outer loop continues, going through the next numbers—no need to start from the beginning, or even type out another for loop, to find the next prime.
EDIT: sqrt(upperBound) is correct. I wasn't thinking about it carefully enough.
Why don't you work with array of booleans for simplicity starting from index 2, and when you will print the result, you will print indices with value of true
as the title explains this is a program to find lcm of numbers between 1 to 20. i found an algorithm to do this, here's the link
http://www.cut-the-knot.org/Curriculum/Arithmetic/LCM.shtml
there is a java applet on the webpage that might explain the algorithm better
Problem: i wrote the code compiler shows no error but when i run the code the program goes berserk, i guess may be some infinite loopig but i can't figure it out for the life of me. i use turbo c++ 4.5 so basically if anyone can look at the code and help me out it would be great . thanks in advance
Algorithm:
say we need to find lcm of 2,6,8
first we find the least of the series and add to it the number above it, i.e the series become
4,6,8
now we find the least value again and add to it the intitial value in the column i.e 2
6,6,8
so the next iteration becomes
8,6,8
8,12,8
10,12,8
10,12,16
12,12,16
14,12,16
14,18,16
16,18,16
18,18,16
18,18,24
20,18,24
20,24,24
22,24,24
24,24,24
as you can see at one point all numbers become equal which is our lcm
#include<iostream.h>
/*function to check if all the elements of an array are equal*/
int equl(int a[20], int n)
{
int i=0;
while(n==1&&i<20)
{
if (a[i]==a[i+1])
n=1;
else
n=0;
i++;
}
return n;
}
/*function to calculate lcm and return that value to main function*/
int lcm()
{
int i,k,j,check=1,a[20],b[20];
/*loading both arrays with numbers from 1 to 20*/
for(i=0;i<20;i++)
{
a[i]=i+1;
b[i]=i+1;
}
check= equl(a,1);
/*actual implementation of the algorith*/
while(check==0)
{
k=a[0]; /*looks for the least value in the array*/
for(i=0;i<20;i++)
{
if(a[i+1]<k)
{
k=a[i+1]; /*find the least value*/
j=i+1; /*mark the position in array */
}
else
continue;
}
a[j]=k+b[j]; /*adding the least value with its corresponding number*/
check= equl(a,1);
}
return (a[0]);
/*at this point all numbers in the array must be same thus any value gives us the lcm*/
}
void main()
{
int l;
l=lcm();
cout<<l;
}
In this line:
a[j]=k+b[j];
You use j but it is unitialized so it's some huge value and you are outside of the array bounds and thus you get a segmentation fault.
You also have some weird things going on in your code. void main() and you use cout without either saying std::cout or using namespace std; or something similar. An odd practice.
Also don't you think you should pass the arrays as arguments if you're going to make lcm() a function? That is int lcm(int a[], int b[]);.
You might look into using a debugger also and improving your coding practices. I found this error within 30 seconds of pasting your code into the compiler with the help of the debugger.
Your loop condition is:
while(n==1&&i<20)
So your equl function will never return 1 because if n happens to be 1 then the loop will just keep going and never return a 1.
However, your program still does not appear to return the correct result. You can split the piece of your code that finds the minimum element and replace it with this for cleanliness:
int least(int a[], int size){
int minPos = 0;
for(int i=0; i<size ;i++){
if (a[i] < a[minPos] ){
minPos = i;
}
}
return minPos;
}
Then you can call it by saying j = least(a, 20);. I will leave further work on your program to you. Consider calling your variables something meaningful instead of i,j,k,a,b.
Your equl function is using array indices from 0-20, but the arrays only have 1-19
j in lcm() is uninitialized if the first element is the smallest. It should be set to 0 at the top of the while loop
In the following code, when i=19, you are accessing a[20], which is out of the bounds of the array. Should be for(i=0;i<19;i++)
for(i=0;i<20;i++) {
if(a[i+1]<k)
You are not actually using the std namespace for the cout. this should be std::cout<<l
Your are including iostream.h. The standard is iostream without the .h, this may not work on such an old compiler tho
instead of hard-coding 20 everywhere, you should use a #define. This is not an error, just a style thing.
The following code does nothing. This is the default behavior
else
continue;