I want to use this question to improve a bit in my general understanding of how computer works, since I'll probably never have the chance to study in a profound and deep manner. Sorry in advance if the question is silly and not useful in general, but I prefer to learn in this way.
I am learning c++, I found online a code that implements the Newton-Raphson method for finding the root of a function. The code is pretty simple, as you can see from it, at the beginning it asks for the tolerance required, and if I give a "decent" number it works fine. If instead, when it asks for the tolerance I write something like 1e-600, the program break down immediately and the output is Enter starting value x: Failed to converge after 100 iterations .
The output of failed convergence should be a consequence of running the loop for more than 100 iterations, but this isn't the case since the loop doesn't even start. It looks like the program knows already it won't reach that level of tolerance.
Why does this happen? How can the program write that output even if it didn't try the loop for 100 times?
Edit: It seems that everything meaningless (too small numbers, words) I write when it asks for tolerance produces a pnew=0.25 and then the code runs 100 times and fails.
The code is the following:
#include <iostream>
#include <cmath>
using namespace std;
#define N 100 // Maximum number of iterations
int main() {
double p, pnew;
double f, dfdx;
double tol;
int i;
cout << "Enter tolerance: ";
cin >> tol;
cout << "Enter starting value x: ";
cin >> pnew;
// Main Loop
for(i=0; i < N; i++){
p = pnew;
//Evaluate the function and its derivative
f = 4*p - cos(p);
dfdx= 4 + sin(p);
// The Newton-Raphson step
pnew = p - f/dfdx;
// Check for convergence and quit if done
if(abs(p-pnew) < tol){
cout << "Root is " << pnew << " to within " << tol << "\n";
return 0;
}
}
// We reach this point only if the iteration failed to converge
cerr << "Failed to converge after " << N << " iterations.\n";
return 1;
}
1e-600 is not representable by most implementations of double. std::cin will fail to convert your input to double and fall into a failed state. This means that, unless you clear the error state, any future std::cin also automatically fails without waiting for user input.
From cppreference (since c++17) :
If extraction fails, zero is written to value and failbit is set. If extraction results in the value too large or too small to fit in value, std::numeric_limits<T>::max() or std::numeric_limits<T>::min() is written and failbit flag is set.
As mentioned, 1e-600 is not a valid double value. However, there's more to it than being outside of the range. What's likely happening is that 1 is scanned into tol, and then some portion of e-600 is being scanned into pnew, and that's why it ends immediately, instead of asking for input for pnew.
Like François said, you cannot exeed 2^64 when you work on an 64bit machine (with corresponding OS) and 2^32 on a 32bit machine, you can use SSE which are 4 32 bytes data used for floating point representation. In your program the function fails at every iteration and skips your test with "if" and so never returns before ending the loop.
Related
#include <iostream>
#include <iomanip>
#include <math.h>
using namespace std;
int main() {
int t;
double n;
cin>>t;
while(t--)
{
cin>>n;
double x;
for(int i=1;i<=10000;i++)
{
x=n*i;
if(x==ceilf(x))
{
cout<<i<<endl;
break;
}
}
}
return 0;
}
For I/P:
3
5
2.98
3.16
O/P:
1
If my code is:
#include <iostream>
#include <iomanip>
#include <math.h>
using namespace std;
int main() {
int t;
double n;
cin>>t;
while(t--)
{
cin>>n;
double x;
for(int i=1;i<=10000;i++)
{
x=n*i;
cout<<"";//only this statement is added;
if(x==ceilf(x))
{
cout<<i<<endl;
break;
}
}
}
return 0;
}
For the same input O/P is:
1
50
25
The only extra line added in 2nd code is: cout<<"";
Can anyone please help in finding why there is such a difference in output just because of the cout statement added in the 2nd code?
Well this is a veritable Heisenbug. I've tried to strip your code down to a minimal replicating example, and ended up with this (http://ideone.com/mFgs0S):
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
float n;
cin >> n; // this input is needed to reproduce, but the value doesn't matter
n = 2.98; // overwrite the input value
cout << ""; // comment this out => y = z = 149
float x = n * 50; // 149
float y = ceilf(x); // 150
cout << ""; // comment this out => y = z = 150
float z = ceilf(x); // 149
cout << "x:" << x << " y:" << y << " z:" << z << endl;
}
The behaviour of ceilf appears to depend on the particular sequence of iostream operations that occur around it. Unfortunately I don't have the means to debug in any more detail at the moment, but maybe this will help someone else to figure out what's going on. Regardless, it seems almost certain that it's a bug in gcc-4.9.2 and gcc-5.1. (You can check on ideone that you don't get this behaviour in gcc-4.3.2.)
You're probably getting an issue with floating point representations - which is to say that computers cannot perfectly represent all fractions. So while you see 50, the result is probably something closer to 50.00000000001. This is a pretty common problem you'll run across when dealing with doubles and floats.
A common way to deal with it is to define a very small constant (in mathematical terms this is Epsilon, a number which is simply "small enough")
const double EPSILON = 0.000000001;
And then your comparison will change from
if (x==ceilf(x))
to something like
double difference = fabs(x - ceilf(x));
if (difference < EPSILON)
This will smooth out those tiny inaccuracies in your doubles.
"Comparing for equality
Floating point math is not exact. Simple values like 0.2 cannot be precisely represented using binary floating point numbers, and the limited precision of floating point numbers means that slight changes in the order of operations can change the result. Different compilers and CPU architectures store temporary results at different precisions, so results will differ depending on the details of your environment. If you do a calculation and then compare the results against some expected value it is highly unlikely that you will get exactly the result you intended.
In other words, if you do a calculation and then do this comparison:
if (result == expectedResult)
then it is unlikely that the comparison will be true. If the comparison is true then it is probably unstable – tiny changes in the input values, compiler, or CPU may change the result and make the comparison be false."
From http://www.cygnus-software.com/papers/comparingfloats/Comparing%20floating%20point%20numbers.htm
Hope this answers your question.
Also you had a problem with
if(x==ceilf(x))
ceilf() returns a float value and x you have declared as a double.
Refer to problems in floating point comparison as to why that wont work.
change x to float and the program runs fine,
I made a plain try on my laptop and even online compilers.
g++ (4.9.2-10) gave the desired output (3 outputs), along with online compiler at geeksforgeeks.org. However, ideone, codechef did not gave the right output.
All I can infer is that online compilers name their compiler as "C++(gcc)" and give wrong output. While, geeksforgeeks.org, which names the compiler as "C++" runs perfectly, along with g++ (as tested on Linux).
So, we could arrive at a hypothesis that they use gcc to compile C++ code as a method suggested at this link. :)
I'm a student, and my Operating Systems class project has a little snag, which is admittedly a bit superfluous to the assignment specifications itself:
While I can push 1 million deques into my deque of deques, I cannot push ~10 million or more.
Now, in the actual program, there is lots of stuff going on, and the only thing already asked on Stack Overflow with even the slightest relevance had exactly that, only slight relevance. https://stackoverflow.com/a/11308962/3407808
Since that answer had focused on "other functions corrupting the heap", I isolated the code into a new project and ran that separately, and found everything to fail in exactly the same ways.
Here's the code itself, stripped down and renamed for the sake of space.
#include <iostream>
#include <string>
#include <sstream>
#include <deque>
using namespace std;
class cat
{
cat();
};
bool number_range(int lower, int upper, double value)
{
while(true)
{
if(value >= lower && value <= upper)
{
return true;
}
else
{
cin.clear();
cerr << "Value not between " << lower << " and " << upper << ".\n";
return false;
}
}
}
double get_double(char *message, int lower, int upper)
{
double out;
string in;
while(true) {
cout << message << " ";
getline(cin,in);
stringstream ss(in); //convert input to stream for conversion to double
if(ss >> out && !(ss >> in))
{
if (number_range(lower, upper, out))
{
return out;
}
}
//(ss >> out) checks for valid conversion to double
//!(ss >> in) checks for unconverted input and rejects it
cin.clear();
cerr << "Value not between " << lower << " and " << upper << ".\n";
}
}
int main()
{
int dq_amount = 0;
deque<deque <cat> > dq_array;
deque<cat> dq;
do {
dq_amount = get_double("INPUT # OF DEQUES: ", 0, 99999999);
for (int i = 0; i < number_of_printers; i++)
{
dq_array.push_back(dq);
}
} while (!number_range(0, 99999999, dq_amount));
}
In case that's a little obfuscated, the design (just in case it's related to the error) is that my program asks for you to input an integer value. It takes your input and verifies that it can be read as an integer, and then further parses it to ensure it is within certain numerical bounds. Once it's found within bounds, I push deques of myClass into a deque of deques of myClass, for the amount of times equal to the user's input.
This code has been working for the past few weeks that I've being making this project, but my upper bound had always been 9999, and I decided to standardize it with most of the other inputs in my program, which is an appreciably large 99,999,999. Trying to run this code with 9999 as the user input works fine, even with 99999999 as the upper bound. The issue is a runtime error that happens if the user input is 9999999+.
Is there any particular, clear reason why this doesn't work?
Oh, right, the error message itself from Code::Blocks 13.12:
terminate called after throwing an instance of 'std::bad_alloc'
what(): std::bad_alloc
This application has requested the Runtime to terminate it in an unusual way.
Please contact the application's supprt team for more information.
Process returned 3 (0x3) execution time : 12.559 s
Press any key to continue.
I had screenshots, but I need to be 10+ reputation in order to put images into my questions.
This looks like address space exhaustion.
If you are compiling for a 32-bit target, you will generally be limited to 2 GiB of user-mode accessible address space per process, or maybe 3 GiB on some platforms. (The remainder is reserved for kernel-mode mappings shared between processes)
If you are running on a 64-bit platform and build a 64-bit binary, you should be able to do substantially more new/alloc() calls, but be advised you may start hitting swap.
Alternatively, you might be hitting a resource quota even if you are building a 64-bit binary. On Linux you can check ulimit -d to see if you have a per-process memory limit.
I am trying to test a series of libraries for matrix-vector computations. For that I just make a large loop and inside I call the routine I want to time. Very simple. However I sometimes see that when I increase the level of optimization for the compiler the time drops to zero no matter how large the loop is. See the example below where I try to time a C macro to compute cross products. What is the compiler doing? how can I avoid it but to allow maximum optimization for floating point arithmetics? Thank you in advance
The example below was compiled using g++ 4.7.2 on a computer with an i5 intel processor.
Using optimization level 1 (-O1) it takes 0.35 seconds. For level two or higher it drops down to zero. Remember, I want to time this so I want the computations to actually happen even if, for this simple test, unnecessary.
#include<iostream>
using namespace std;
typedef double Vector[3];
#define VecCross(A,assign_op,B,dummy_op,C) \
( A[0] assign_op (B[1] * C[2]) - (B[2] * C[1]), \
A[1] assign_op (B[2] * C[0]) - (B[0] * C[2]), \
A[2] assign_op (B[0] * C[1]) - (B[1] * C[0]) \
)
double get_time(){
return clock()/(double)CLOCKS_PER_SEC;
}
int main()
{
unsigned long n = 1000000000u;
double start;
{//C macro cross product
Vector u = {1,0,0};
Vector v = {1,1,0};
Vector w = {1.2,1.2,1.2};
start = get_time();
for(unsigned long i=0;i<n;i++){
VecCross (w, =, u, X, v);
}
cout << "C macro cross product: " << get_time()-start << endl;
}
return 0;
}
Ask yourself, what does your program actually do, in terms of what is visible to the end-user?
It displays the result of a calculation: get_time()-start. The contents of your loop have no bearing on the outcome of that calculation, because you never actually use the variables being modified inside the loop.
Therefore, the compiler optimises out the entire loop since it is irrelevant.
One solution is to output the final state of the variables being modified in the loop, as part of your cout statement, thus forcing the compiler to compute the loop. However, a smart compiler could also figure out that the loop always calculates the same thing, and it can simply insert the result directly into your cout statement, because there's no need to actually calculate it at run-time. As a workaround to this, you could for example require that one of the inputs to the loop be provided at run-time (e.g. read it in from a file, command line argument, cin, etc.).
For more (and possibly better) solutions, check out this duplicate thread: Force compiler to not optimize side-effect-less statements
#include<iostream>
using namespace std;
int main(){
int i,x,max=0;
cin>>x;
int a[x];
for(i=0;i<x;i++){
cin>>a[i];
if(max<a[i]){
max=a[i];}
}
int b[max+1];
for(i=0;i<max+1;i++){
b[i]=-1;
}
for(i=0;i<x;i++){
if(b[a[i]]==-1){
b[a[i]]=1;
}
else{
b[a[i]]++;
}
}
i=0;
while(i<=max){
while(b[i]>0&&b[i]!=-1){
cout<<i<<endl;
b[i]--;
}
i++;
}
return 0;
}
Guys I tried indexing method for sorting and codechef shows tle .. complexity of this problem is not o(n) but closer to it ... the question has a time limit of 5 sec and source limit is 50000 bytes..
Any help on how to improve the performance either by faster i/o or code computations ...
I'm pretty certain your code is problematic because you are using cout << x << endl; in a loop that will print a huge number of lines.
I will be back with "difference" in a few minutes.
Edit: Not sure I can make much of a difference either way. Obviously, depending on compiler, it may vary greatly, but with my g++ -O2 and 100000 input numbers, it takes 0.16 - 0.18s to use endl; and 0.06 - 0.07s to use '\n' for the output.
Using printf isn't faster than cout, but scanf is a little faster than cin (0.04s +/- 0.05).
However, that is really related to sync_with_stdio. If we use cin.sync_with_stdio(false); then the results are the same for scanf and cin.
All measurements are made with a file as input and a file as output - it takes much longer to write to the shell, but that's because it's scrolling 100k lines of text past me.
(Your program will crash with eithr "large" inputs or with large number of inputs - if max is greater than about 1 million, the code will crash due to out of stack - on many systems, that may happen for lower values too)
Avoid cin and cout, and use the C IO functions scanf and printf instead. You'll find they can be up to 5x faster than the slow c++ functions.
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Today I'm working on making my program faster. This program scans in a 100,000 fake social security numbers, first names, last names and gpa's. My professor has started talking about pointer's, referencing and dereferencing and has said that using these can help speed a program up by passing addresses. My explanation probably sucks because I am not really understanding the topics in class, so while you guys help me out I will be reading in my book for chapter 9 on call by value and call by reference. Any help would be appreciated. Thanks!!!
#include<iostream>
#include<fstream>
#include<cstdlib>
#include<ctime>
using namespace std;
struct nameType{
string ssno;
string fName;
string lName;
double gpa;
};
int load(istream &in,nameType []);
void shellsort(nameType [],int);
void exchange(nameType &, nameType &);
void print(ostream &out,nameType [],int);
int main(void)
{
ifstream in;
ofstream out;
char infile[40],outfile[40];
nameType name[100000];
clock_t start, stop;
double secl=0;
double secs=0;
double secp=0;
double total=0;
int n;
cout << "Please enter the input data file name(NO SPACES): ";
cin >> infile;
in.open(infile);
if(in.fail()) {
cerr<<"problem input file\n"<<endl;
exit(1);
}
cout << "Please enter the output data file name(NO SPACES): ";
cin >> outfile;
out.open(outfile);
if(out.fail()) {
cerr<<"problem output file\n"<<endl;
exit(1);
}
start = clock();
n = load(in,name);
stop = clock();
secl = (double)(stop - start)/CLOCKS_PER_SEC;
cout << "Load Time: " << secl << endl;
start = clock();
shellsort(name,n);
stop = clock();
secs = (double)(stop - start)/CLOCKS_PER_SEC;
cout << "Sort Time: " << secs << endl;
start = clock();
print(out,name,n);
stop = clock();
secp = (double)(stop - start)/CLOCKS_PER_SEC;
cout << "Print Time: " << secp << endl;
total = secl + secs + secp;
cout << "Total Time: " << total << endl;
in.close();
out.close();
return 0;
}
int load(istream &in,nameType name[])
{
int n=0;
in >> name[n].ssno >> name[n].fName >> name[n].lName >> name[n].gpa;
while(!in.eof()){
n++;
in >> name[n].ssno >> name[n].fName >> name[n].lName >> name[n].gpa;
}
return n;
}
void shellsort(nameType name[],int n)
{
int gap = n/2;
bool passOk;
while(gap>0){
passOk=true;
for(int i=0; i<n-gap; i++){
if(name[i].lName>name[i+gap].lName){
exchange(name[i],name[i+gap]);
passOk=false;
}
else if(name[i].lName == name[i+gap].lName && name[i].fName > name[i+gap].fName){
exchange(name[i],name[i+gap]);
passOk=false;
}
}
if(passOk){
gap/=2;
}
}
}
void exchange(nameType &a, nameType &b)
{
nameType temp;
temp = a;
a = b;
b = temp;
}
void print(ostream &out,nameType name[],int n)
{
for(int i=0;i<n;i++){
out << name[i].ssno << " " << name[i].fName << " " << name[i].lName << " " << name[i].gpa << endl;
}
out << endl;
}
Exact assignment details--- I'm on bullet number 5---
Efficiency - time and space
Time and space are always issues to consider when writing programs.
In this assignment, you will be modifying the sort_v4.cpp program created in the first programming assignment. In that assignment, you were required to process an array of integers. Let's update the program with several items:
Build a structure containing an social security number, first name, last name, and a GPA.
The new sort technique that uses the gap concept as described in class that sorts the information in ascending order based on the last name (if last names are the same then the first names need to be checked). The sort technique is know as shell sort.
Improve the efficiency - time and space as described below.
Each struct element contained an ID, first name, last name and gpa. Suppose you had to process 100,000 students. The program is inefficient for two reasons
Space issue: if social security number takes up 12 characters, first and last take up 20 characters each, and the gpa as a double takes up 8 bytes then the main array of structure elements consumes (12+20+20+8)*100000 bytes of memory. This may be OK if we load up 100,000 names. But if the average number of students we process is <50,000, then there is a considerable amount of wasted memory.
Time issue: When you exchange two elements that are out of order, 60 bytes are moved around 3 times. Total of 180 bytes are moved in memory. Again, inefficient.
As the number of member variables in the struct increase the problem gets worse.
The focus of this assignment is to
Read a file containing the social security number, first, last, and a gpa into an array of structure elements. Process until EOF. Each line contains information about one student. Inside the struct, the elements may be define as char [] or strings. Do you think it will make a difference in performance if we use char [] vs strings? Make a prediction.
Dump the array into a file along with timing information.
Try out the new sort technique - sort structure elements based on two items - first and last name.
Time each function to see where the most time is being spent.
Attempt to improve the use of memory by using an array of pointers to the structs.
Attempt to improve the efficiency by making the exchange faster by swapping pointers instead of elements.
With the above list in mind, there will be three possible grades for the assignment. For a maximum grade of a C - 14/20 points, you must complete the first two bullets. For a maximum grade of a B - 16/20 points, you must complete the first 4 bullets. For an A, you must complete all bullets.
I would recommend to start the assignment by implementing what I call a non pointer version. Define a struct above the main function to hold the items. The main function should declare the array of struct elements. Call the load, sort, and print functions as we did in the first assignment making adjustments to accommodate the struct and the new sort technique where there are two items to consider before exchanging two elements. NOTE: I want to see an exchange function this time. This will give you a total of 4 functions. Make sure you follow the guidelines stated in the first programming assignment.
For all that are attempting the A program, make sure you have the basic program for a B completed and tested before continuing. Make a copy of that program and modify the program as follows:
Change the array of struct elements to an array of pointers to struct elements by placing a * in the definition.
Using new (or malloc), dynamically create space for one structure element just before you read in the values for the members of the struct.
When you exchange two elements, exchange two pointers to struct elements instead of the struct elements themselves.
Take a look at your times for the non-pointer version and the pointer version. Is there a significant difference?
Following is the information about timing one function.
#include<ctime>
//Create a couple variables of clock_t type.
clock_t start, stop;
start = clock();
load(x, n); //call a function to perform a task
stop = clock();
cout << "load time: " << (double)(stop - start)/CLOCKS_PER_SEC << endl;
The function clock() returns the number of cpu clock cycles that have occurred since the program started. Ideally, start in the above code should be 0. To be able to make some sense of how much time a function takes, you have to convert the elapsed time into seconds. This is accomplished by taking the difference between start and stop, typecasting, then dividing by the system defined CLOCKS_PER_SEC. On linus public (or an alien ware machine in the lab) is 1,000,000. Also see class notes on the topic. The following is an example of what should appear at the end of the sorted data.
load time: 0.05
sort time: 2.36
print time: 0.01
Total Run time: 2.42
The Penalty for missing deadline, 1 pt per day for a max of 7 days. Programs will not be accepted 7 days after the deadline.
If you look here: How are arrays passed?
Arrays are passed as pointers already, and your implementation of std::swap (which you called "exchange") is already passing by reference. So that is not an issue in your case. Has your professor complained about your program execution speed?
A place that takes a lot of execution time in a program is called a bottleneck.
A common bottleneck is file input and output. You can make your program faster by reducing or optimizing the bottleneck.
Reading many small pieces of data from a file takes more time than reading one large piece of data. For example, reading 10 lines of data with one request is more efficient that 10 requests to read one line of data.
Accessing memory is fast, faster than reading from a file.
The general pattern is:
1. Read a lot of data into memory.
2. Process the data.
3. Repeat at 1 as necessary.
So, you could use the "block" read command, istream::read into an array of bytes. Load a string with a line of data from the array of bytes and process the line. Repeat loading from the array until you run out and then reload from the file.