I've been trying to implement quick sort and have been having a lot of problems. I even copied a lot of implementations and accepted answers from the net and they ALL crash on odd sized array/vector if you run it enough times (each time I run I run quick sort against random numbers to be sorted... rather than pretend my code works just cuz it can sort one particular set of numbers).
Here is my code and also prints to help debug the error.
template <typename T>
void quickSortMidPivot(vector<T>&vec, size_t left, size_t right)
{
mcount++;
if(right - left < 1)
return;
//crash all the time
//if(left >= right)
// return;
size_t l = left;
size_t r = right;
T pivot = vec[left + ((right-left)/2)];
cout << endl << "PivotValue:" << pivot << endl;
while (l <= r)
{
while (vec[l] < pivot)
l++;
while (vec[r] > pivot)
r--;
if (l <= r) {
cout << endl << "swap:" << vec[l] << "&" << vec[r] << endl;
std::swap(vec[l], vec[r]);
l++;
r--;
for (int i =left; i<=right; i++)
cout << vec[i] << " ";
}
}
cout << endl << "left:" << left << " r:" << r << endl;
cout << "l:" << l << " right:" << right << endl;
if(left < r)
quickSortMidPivot(vec, left, r);
if(l < right)
quickSortMidPivot(vec, l, right);
}
//in main
quickSortMidPivot(dsVector, 0, dsVector.size() - 1);
mcount is a global just so that I can count number of recursive calls. Help figure out most effective implementation...
Here is some debug info.
When run on even sized vector.
Test values are (PRE-SORTING):
8 4 6 5 2 4 1 2
PivotValue:5
swap:8&2
2 4 6 5 2 4 1 8
swap:6&1
2 4 1 5 2 4 6 8
swap:5&4
2 4 1 4 2 5 6 8
left:0 r:4
l:5 right:7
PivotValue:1
swap:2&1
1 4 2 4 2
left:0 r:0
l:1 right:4
PivotValue:2
swap:4&2
2 2 4 4
swap:2&2
2 2 4 4
left:1 r:1
l:3 right:4
PivotValue:4
swap:4&4
4 4
left:3 r:3
l:4 right:4
PivotValue:6
swap:6&6
5 6 8
left:5 r:5
l:7 right:7
# Recursions:5 0
Data Sorted.
Sorted test values are (POST-SORTING):
1 2 2 4 4 5 6 8
Here is case with odd sized array (9). Works 90% of time.
Test values are (PRE-SORTING):
7 7 5 6 5 8 9 5 8
PivotValue:5
swap:7&5
5 7 5 6 5 8 9 7 8
swap:7&5
5 5 5 6 7 8 9 7 8
swap:5&5
5 5 5 6 7 8 9 7 8
left:0 r:1
l:3 right:8
PivotValue:5
swap:5&5
5 5
left:0 r:0
l:1 right:1
PivotValue:8
swap:8&8
6 7 8 9 7 8
swap:9&7
6 7 8 7 9 8
left:3 r:6
l:7 right:8
PivotValue:7
swap:7&7
6 7 8 7
left:3 r:4
l:5 right:6
PivotValue:6
swap:6&6
6 7
left:3 r:2
l:4 right:4
PivotValue:8
swap:8&7
7 8
left:5 r:5
l:6 right:6
PivotValue:9
swap:9&8
8 9
left:7 r:7
l:8 right:8
# Recursions:7 0
Data Sorted.
Sorted test values are (POST-SORTING):
5 5 5 6 7 7 8 8 9
Here is print output for when odd sized (9) vector input causes crash.
Test values are (PRE-SORTING):
8 3 2 3 9 3 8 1 5
PivotValue:9
swap:9&5
8 3 2 3 5 3 8 1 9
left:0 r:7
l:8 right:8
PivotValue:3
swap:8&1
1 3 2 3 5 3 8 8
swap:3&3
1 3 2 3 5 3 8 8
swap:3&3
1 3 2 3 5 3 8 8
left:0 r:2
l:4 right:7
PivotValue:3
swap:3&2
1 2 3
left:0 r:1
l:2 right:2
PivotValue:1
swap:1&1
1 2
swap:2&0
1 0
swap:3&0
1 0
swap:3&1
1 0
swap:5&0
1 0
swap:3&1
1 0
swap:8&0
1 0
swap:8&0
1 0
swap:9&0
1 0
swap:7274596&0
1 0
swap:666050571&0
1 0
swap:369110150&0
1 0
swap:1&0
1 0
swap:1&0
1 0
swap:110&0
1 0
swap:649273354&0
1 0
swap:134229126&0
1 0
swap:3764640&0
1 0
swap:2293216&0
1 0
swap:8&0
1 0
swap:2&0
1 0
swap:649273354&0
1 0
swap:134229127&0
1 0
swap:3764672&0
1 0
swap:3764608&0
1 0
swap:3&0
1 0
swap:649273354&0
1 0
swap:134229127&0
1 0
swap:3764704&0
1 0
swap:3764640&0
1 0
swap:2&0
1 0
swap:649273354&0
1 0
swap:134229127&0
1 0
swap:3764736&0
1 0
swap:3764672&0
1 0
swap:3&0
1 0
swap:649273354&0
1 0
swap:134229127&0
1 0
swap:3764768&0
1 0
swap:3764704&0
1 0
swap:9&0
1 0
swap:649273354&0
1 0
swap:134229127&0
1 0
swap:3764800&0
1 0
swap:3764736&0
1 0
swap:3&0
1 0
swap:6619252&0
1 0
swap:649273354&0
1 0
swap:134229127&0
1 0
swap:3764832&0
1 0
swap:3764768&0
1 0
swap:8&0
1 0
swap:666050571&0
1 0
swap:402664583&0
1 0
swap:3765152&0
1 0
swap:3764800&0
1 0
swap:1&0
1 0
swap:900931609&0
1 0
swap:268446854&0
1 0
swap:2046&0
1 0
swap:2046&0
1 0
swap:649273354&0
1 0
swap:134229140&0
1 0
swap:2293216&0
1 0
swap:3764832&0
1 0
swap:5&0
1 0
swap:11399&0
1 0
swap:3735896&0
1 0
swap:3735896&0
1 0
swap:548610060&1
1 0
swap:50342980&0
1 0
swap:6356944&-1
1 0
swap:3735800&-2
1 0
swap:3735648&0
1 0
swap:3735648&-1
1 0
swap:3768320&0
1 0
swap:32768&1
1 0
Related
This is the complete code that I wrote:
( [numsin the function] and [arrin main] is the array that needs to be sorted, sizeis the amount of numbers in the array, minis the smallest number in the unsorted part)
#include <iostream>
#include <vector>
using namespace std;
void sort(vector <int> &nums, int size){
int min = 0;
for(int i=0;i<size;i++){
min = i;
for(int j=i+1;j<size;j++){
if(nums[j]<nums[min]){
min = j; //comparing
}
}
nums[i] = nums[min] + nums[i]; //swaping
nums[min] = nums[i] - nums[min];
nums[i] = nums[i] - nums[min];
}
}
int main(){
cout<<"\nEnter Numbers:\n";
vector <int> arr;
int num;
while(cin>>num){
arr.push_back(num);
}
sort(arr,arr.size());
cout<<"\nSorted:\n";
for(int i=0;i<arr.size();i++){
cout<<arr[i]<<" ";
}
}
I'm writing a code that simply sorts the given array. But after trying to debug and find solutions online, I can't figure out which part is wrong. These are some examples of my results:
Enter Numbers:
9 8 7 6 1 2 3 4 5 ^Z
Sorted:
1 2 3 4 5 6 0 0 0
Enter Numbers:
6 4 8 7 2 3 5 ^Z
Sorted:
2 3 4 5 0 7 0
Enter Numbers:
9 8 7 6 5 4 3 2 1 ^Z
Sorted:
1 2 3 4 0 0 0 0 0
This is the result when I added a for loop under the swapping part to show what every round has done to the array:
Enter Numbers:
9 8 7 6 1 2 3 4 5 ^Z
1 8 7 6 9 2 3 4 5
1 2 7 6 9 8 3 4 5
1 2 3 6 9 8 7 4 5
1 2 3 4 9 8 7 6 5
1 2 3 4 5 8 7 6 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 0 8 9
1 2 3 4 5 6 0 0 9
1 2 3 4 5 6 0 0 0
Sorted:
1 2 3 4 5 6 0 0 0
Enter Numbers:
6 4 8 7 2 3 5 ^Z
2 4 8 7 6 3 5
2 3 8 7 6 4 5
2 3 4 7 6 8 5
2 3 4 5 6 8 7
2 3 4 5 0 8 7
2 3 4 5 0 7 8
2 3 4 5 0 7 0
Sorted:
2 3 4 5 0 7 0
9 8 7 6 5 4 3 2 1 ^Z
1 8 7 6 5 4 3 2 9
1 2 7 6 5 4 3 8 9
1 2 3 6 5 4 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 0 6 7 8 9
1 2 3 4 0 0 7 8 9
1 2 3 4 0 0 0 8 9
1 2 3 4 0 0 0 0 9
1 2 3 4 0 0 0 0 0
Sorted:
1 2 3 4 0 0 0 0 0
Please help, thanks.
Your swapping logic doesn't handle the case where the smallest remaining element is the first unsorted element (i.e., i == min). Consider what each line does in this case:
nums[i] = nums[min] + nums[i]; //nums[i] will be doubled
nums[min] = nums[i] - nums[min]; // nums[min] is subtracted from itself, making it 0
nums[i] = nums[i] - nums[min]; // nums[i] is subtracted from itself gain, but 0-0 is still 0
The goal of avoiding a temporary isn't bad by itself, but you do have this nasty edge case. You'd either have to detect the edge case or just bite the bullet and deal with a temporary. You could also call std::swap, but that likely uses a temporary as well.
The advantage to using a temporary or std::swap is that this code would be easier to make generic for other types (especially via std::swap). In addition, std::swap can be specialized for types to avoid temporaries if possible, and if this is actually a bottleneck.
I was solving the puzzles in 2017 Advent of Code. It was necessary to fill in circular buffer using certain algorithm. For the buffer implementation I first used vector, and then I tried with deque. I am getting different results when printing values of the vector and the queue. Here's the code:
#include <iostream>
#include <vector>
void PrintBuffer(std::vector<int> a_CircularBuffer)
{
for (std::vector<int>::iterator it = a_CircularBuffer.begin(); it != a_CircularBuffer.end(); ++it) {
std::cout << *it << " ";
}
std::cout << std::endl;
}
int main()
{
std::vector<int> circularBuffer;
circularBuffer.reserve(20);
circularBuffer.push_back(0);
circularBuffer.push_back(1);
std::vector<int>::iterator currentPosition = circularBuffer.begin() + 1;
for (int i = 2; i < 20; ++i) {
int steps = 378;
if (steps >= i) {
steps = (steps % i);
}
if ((circularBuffer.end() - currentPosition) <= steps) {
currentPosition = circularBuffer.begin() + (((currentPosition - circularBuffer.begin()) + steps) % i);
circularBuffer.insert(currentPosition, i);
}
else {
currentPosition = currentPosition + steps;
circularBuffer.insert(currentPosition, i);
}
PrintBuffer(circularBuffer);
}
return 0;
}
This is the result when using vector:
0 2 1
0 3 2 1
0 3 2 4 1
0 5 3 2 4 1
0 6 5 3 2 4 1
0 7 6 5 3 2 4 1
0 7 6 8 5 3 2 4 1
0 7 6 9 8 5 3 2 4 1
0 10 7 6 9 8 5 3 2 4 1
0 10 7 6 9 11 8 5 3 2 4 1
0 10 7 6 9 11 8 5 3 2 4 12 1
0 10 7 6 9 11 8 5 3 2 4 12 13 1
0 10 7 6 9 11 8 5 3 2 4 12 14 13 1
15 0 10 7 6 9 11 8 5 3 2 4 12 14 13 1
...
and this is when using deque (just change "vector" to "deque" and comment out circularBuffer.reserve(20) line):
0 2 1
0 3 2 1
0 3 2 4 1
0 5 3 2 4 1
0 5 6 3 2 4 1
0 5 6 7 3 2 4 1
0 5 6 7 3 8 2 4 1
0 5 6 7 3 9 8 2 4 1
0 5 6 10 7 3 9 8 2 4 1
0 5 6 10 7 3 9 8 11 2 4 1
0 5 12 6 10 7 3 9 8 11 2 4 1
0 5 12 6 13 10 7 3 9 8 11 2 4 1
0 5 12 6 13 14 10 7 3 9 8 11 2 4 1
0 5 12 6 13 14 10 7 3 15 9 8 11 2 4 1
...
Why there are different results for vector and deque?
You get undefined behaviour when you insert an element causing reallocation, and then use the old iterator again.
Anything can happen.
Use index to store current position and it'll work the same way.
So i slurped in a file with numbers in it like: 39.00 into a vector<std::string> and now I need to convert groups of numbers that look like 3 9 0 0 back into the form 39.00
heres a small sample.
3 4 5 0 1 2 5 0 3 4 0 0 3 4 9 0 7 0 0 0 8 0 0 0 9 0 0 0 6 5 0 0 3 9 3 2 1 1 2 0 0 0 2 5 0 0 3 0 9 0 4 0 0 0 5 5 5 0 2 0 0 0 2 5 0 0 3 0 9 0 4 0 0 0 5 5 5 0 2 2 3 0 0 0 2 4 0 0 4 5 0 0 6 7 0 0 6 5 5 0 5 6 9 0 8 7 0 0 4 3 5 0 5 6 9 8 5 5 4 0 3 3 6 2 0 0 3 4 5 0 1 2 5 0 3 4 0 0 3 4 9 0 7 0 0 0 8 0 0 0 9 0 0 0 6 5 0 0 3 9 0 3 2 1 1 2 0 0 0 2 5 0 0 3 0 9 0 4 0 0 0 5 5 5 0 2 0 0 0 2 5 0 0 3 0 9 0 4 0 0 0 5 5 5 0 2 2 3 0 0 0 2 4 0 0 4 5 0 0 6 7 0 0 6 5 5 0 5 6 9 0 8 7 0 0 4 3 5 0 5 6 9 8 5 5 4 0 3 3 6 2 0 0 3 4 5 0 1 2 5 0 3 4 0 0 3 4 9 0 7 0 0 0 8 0 0 0 9 0 0 0 6 5 0 0 3 9 0 0
transformed into 34.50 12.50 34.00....
My goal is to eventually find the average of all the floats.
Of course if there is a way to slurp a file while keeping formatting only using the standard library that would be cool too.
#include <iostream>
#include <string>
#include <fstream>
#include <streambuf>
#include <regex>
#include <vector>
#include <math.h>
void tableWriter(std::string);
float employeeAverage(std::string);
float employeeTotal(std::string);
float totalAverage(std::string);
void totalPayroll(std::string, std::vector<std::string>);
std::string getEmployeeName(std::string, std::string[]);
int main(int argc, const char * argv[]) {
try {
std::vector<std::string> regexContainer;
std::ifstream t("TheSales.txt");
std::string theSales;
t.seekg(0, std::ios::end);
theSales.reserve(t.tellg());
t.seekg(0, std::ios::beg);
theSales.assign((std::istreambuf_iterator<char>(t)),
std::istreambuf_iterator<char>());
//std::cout << theSales << std::endl;
totalPayroll(theSales, regexContainer);
std::cout << std::endl << regexContainer.empty() << std::endl;
return 0;
} catch (int w) {
std::cout << "Could not open file. Exiting Now." << std::endl; return 0;
}
}
void tableWriter(std::string){}
float employeeAverage(std::string){return 0.0;}
float employeeTotal(std::string){return 0.0;}
float totalAverage(std::string){return 0.0;}
void totalPayroll(std::string theSales, std::vector<std::string> regexContainer) {
std::string matches;
std::regex pattern ("\\d");
const std::sregex_token_iterator end;
for (std::sregex_token_iterator i(theSales.cbegin(), theSales.cend(), pattern);
i != end;
++i)
{
regexContainer.push_back(*i);
for (std::vector<std::string>::const_iterator i = regexContainer.begin(); i != regexContainer.end(); ++i)
std::cout << *i << ' ';
}
}
this is the data:
2.40 5.30 6.30 65.34 65.34
3.40 7.80 3.20 65.34 65.34
3.40 5.20 8.20 23.54 12.34
2.42 5.30 6.30 5.00 65.34
3.44 7.80 3.20 34.55 65.34
3.45 5.20 8.20 65.34 65.34
Functions such as fscanf are able to read from your file and return a correctly formed float number. This should be more efficient than trying to reconstruct them from a stream of char...
The problem is defined as follows:
You're given a square. The square is lined with flat flagstones size 1m x 1m. Grass surround the square. Flagstones may be at different height. It starts raining. Determine where puddles will be created and compute how much water will contain. Water doesn't flow through the corners. In any area of grass can soak any volume of water at any time.
Input:
width height
width*height non-negative numbers describing a height of each flagstone over grass level.
Output:
Volume of water from puddles.
width*height signs describing places where puddles will be created and places won't.
. - no puddle
# - puddle
Examples
Input:
8 8
0 0 0 0 0 1 0 0
0 1 1 1 0 1 0 0
0 1 0 2 1 2 4 5
0 1 1 2 0 2 4 5
0 3 3 3 3 3 3 4
0 3 0 1 2 0 3 4
0 3 3 3 3 3 3 0
0 0 0 0 0 0 0 0
Output:
11
........
........
..#.....
....#...
........
..####..
........
........
Input:
16 16
8 0 1 0 0 0 0 2 2 4 3 4 5 0 0 2
6 2 0 5 2 0 0 2 0 1 0 3 1 2 1 2
7 2 5 4 5 2 2 1 3 6 2 0 8 0 3 2
2 5 3 3 0 1 0 3 3 0 2 0 3 0 1 1
1 0 1 4 1 1 2 0 3 1 1 0 1 1 2 0
2 6 2 0 0 3 5 5 4 3 0 4 2 2 2 1
4 2 0 0 0 1 1 2 1 2 1 0 4 0 5 1
2 0 2 0 5 0 1 1 2 0 7 5 1 0 4 3
13 6 6 0 10 8 10 5 17 6 4 0 12 5 7 6
7 3 0 2 5 3 8 0 3 6 1 4 2 3 0 3
8 0 6 1 2 2 6 3 7 6 4 0 1 4 2 1
3 5 3 0 0 4 4 1 4 0 3 2 0 0 1 0
13 3 6 0 7 5 3 2 21 8 13 3 5 0 13 7
3 5 6 2 2 2 0 2 5 0 7 0 1 3 7 5
7 4 5 3 4 5 2 0 23 9 10 5 9 7 9 8
11 5 7 7 9 7 1 0 17 13 7 10 6 5 8 10
Output:
103
................
..#.....###.#...
.......#...#.#..
....###..#.#.#..
.#..##.#...#....
...##.....#.....
..#####.#..#.#..
.#.#.###.#..##..
...#.......#....
..#....#..#...#.
.#.#.......#....
...##..#.#..##..
.#.#.........#..
......#..#.##...
.#..............
................
I tried different ways. Floodfill from max value, then from min value, but it's not working for every input or require code complication. Any ideas?
I'm interesting algorithm with complexity O(n^2) or o(n^3).
Summary
I would be tempted to try and solve this using a disjoint-set data structure.
The algorithm would be to iterate over all heights in the map performing a floodfill operation at each height.
Details
For each height x (starting at 0)
Connect all flagstones of height x to their neighbours if the neighbour height is <= x (storing connected sets of flagstones in the disjoint set data structure)
Remove any sets that connected to the grass
Mark all flagstones of height x in still remaining sets as being puddles
Add the total count of flagstones in remaining sets to a total t
At the end t gives the total volume of water.
Worked Example
0 0 0 0 0 1 0 0
0 1 1 1 0 1 0 0
0 1 0 2 1 2 4 5
0 1 1 2 0 2 4 5
0 3 3 3 3 3 3 4
0 3 0 1 2 0 3 4
0 3 3 3 3 3 3 0
0 0 0 0 0 0 0 0
Connect all flagstones of height 0 into sets A,B,C,D,E,F
A A A A A 1 B B
A 1 1 1 A 1 B B
A 1 C 2 1 2 4 5
A 1 1 2 D 2 4 5
A 3 3 3 3 3 3 4
A 3 E 1 2 F 3 4
A 3 3 3 3 3 3 A
A A A A A A A A
Remove flagstones connecting to the grass, and mark remaining as puddles
1
1 1 1 1
1 C 2 1 2 4 5 #
1 1 2 D 2 4 5 #
3 3 3 3 3 3 4
3 E 1 2 F 3 4 # #
3 3 3 3 3 3
Count remaining set size t=4
Connect all of height 1
G
C C C G
C C 2 D 2 4 5 #
C C 2 D 2 4 5 #
3 3 3 3 3 3 4
3 E E 2 F 3 4 # #
3 3 3 3 3 3
Remove flagstones connecting to the grass, and mark remaining as puddles
2 2 4 5 #
2 2 4 5 #
3 3 3 3 3 3 4
3 E E 2 F 3 4 # # #
3 3 3 3 3 3
t=4+3=7
Connect all of height 2
A B 4 5 #
A B 4 5 #
3 3 3 3 3 3 4
3 E E E E 3 4 # # #
3 3 3 3 3 3
Remove flagstones connecting to the grass, and mark remaining as puddles
4 5 #
4 5 #
3 3 3 3 3 3 4
3 E E E E 3 4 # # # #
3 3 3 3 3 3
t=7+4=11
Connect all of height 3
4 5 #
4 5 #
E E E E E E 4
E E E E E E 4 # # # #
E E E E E E
Remove flagstones connecting to the grass, and mark remaining as puddles
4 5 #
4 5 #
4
4 # # # #
After doing this for heights 4 and 5 nothing will remain.
A preprocessing step to create lists of all locations with each height should mean that the algorithm is close to O(n^2).
This seems to be working nicely. The idea is it is a recursive function, that checks to see if there is an "outward flow" that will allow it to escape to the edge. If the values that do no have such an escape will puddle. I tested it on your two input files and it works quite nicely. I copied the output for these two files for you. Pardon my nasty use of global variables and what not, I figured it was the concept behind the algorithm that mattered, not good style :)
#include <fstream>
#include <iostream>
#include <vector>
using namespace std;
int SIZE_X;
int SIZE_Y;
bool **result;
int **INPUT;
bool flowToEdge(int x, int y, int value, bool* visited) {
if(x < 0 || x == SIZE_X || y < 0 || y == SIZE_Y) return true;
if(visited[(x * SIZE_X) + y]) return false;
if(value < INPUT[x][y]) return false;
visited[(x * SIZE_X) + y] = true;
bool left = false;
bool right = false;
bool up = false;
bool down = false;
left = flowToEdge(x-1, y, value, visited);
right = flowToEdge(x+1, y, value, visited);
up = flowToEdge(x, y+1, value, visited);
down = flowToEdge(x, y-1, value, visited);
return (left || up || down || right);
}
int main() {
ifstream myReadFile;
myReadFile.open("test.txt");
myReadFile >> SIZE_X;
myReadFile >> SIZE_Y;
INPUT = new int*[SIZE_X];
result = new bool*[SIZE_X];
for(int i = 0; i < SIZE_X; i++) {
INPUT[i] = new int[SIZE_Y];
result[i] = new bool[SIZE_Y];
for(int j = 0; j < SIZE_Y; j++) {
int someInt;
myReadFile >> someInt;
INPUT[i][j] = someInt;
result[i][j] = false;
}
}
for(int i = 0; i < SIZE_X; i++) {
for(int j = 0; j < SIZE_Y; j++) {
bool visited[SIZE_X][SIZE_Y];
for(int k = 0; k < SIZE_X; k++)//You can avoid this looping by using maps with pairs of coordinates instead
for(int l = 0; l < SIZE_Y; l++)
visited[k][l] = 0;
result[i][j] = flowToEdge(i,j, INPUT[i][j], &visited[0][0]);
}
}
for(int i = 0; i < SIZE_X; i++) {
cout << endl;
for(int j = 0; j < SIZE_Y; j++)
cout << result[i][j];
}
cout << endl;
}
The 16 by 16 file:
1111111111111111
1101111100010111
1111111011101011
1111000110101011
1011001011101111
1110011111011111
1100000101101011
1010100010110011
1110111111101111
1101101011011101
1010111111101111
1110011010110011
1010111111111011
1111110110100111
1011111111111111
1111111111111111
The 8 by 8 file
11111111
11111111
11011111
11110111
11111111
11000011
11111111
11111111
You could optimize this algorithm easily and considerably by doing several things. A: return true immediately upon finding a route would speed it up considerably. You could also connect it globally to the current set of results so that any given point would only have to find a flow point to an already known flow point, and not all the way to the edge.
The work involved, each n will have to exam each node. However, with optimizations, we should be able to get this much lower than n^2 for most cases, but it still an n^3 algorithm in the worst case... but creating this would be very difficult(with proper optimization logic... dynamic programming for the win!)
EDIT:
The modified code works for the following circumstances:
8 8
1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 1
1 0 1 1 1 1 0 1
1 0 1 0 0 1 0 1
1 0 1 1 0 1 0 1
1 0 1 1 0 1 0 1
1 0 0 0 0 1 0 1
1 1 1 1 1 1 1 1
And these are the results:
11111111
10000001
10111101
10100101
10110101
10110101
10000101
11111111
Now when we remove that 1 at the bottom we want to see no puddling.
8 8
1 1 1 1 1 1 1 1
1 0 0 0 0 0 0 1
1 0 1 1 1 1 0 1
1 0 1 0 0 1 0 1
1 0 1 1 0 1 0 1
1 0 1 1 0 1 0 1
1 0 0 0 0 1 0 1
1 1 1 1 1 1 0 1
And these are the results
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
I'm looking for a J code to do the following.
Suppose I have a list of random integers (sorted),
2 3 4 5 7 21 45 49 61
I want to start with the first element and remove any multiples of the element in the list then move on to the next element cancel out its multiples, so on and so forth.
Thus the output
I'm looking at is 2 3 5 7 61. Basically a Sieve Of Eratosthenes. Would appreciate if someone could explain the code as well, since I'm learning J and find it difficult to get most codes :(
Regards,
babsdoc
It's not exactly what you ask but here is a more idiomatic (and much faster) version of the Sieve.
Basically, what you need is to check which number is a multiple of which. You can get this from the table of modulos: |/~
l =: 2 3 4 5 7 21 45 49 61
|/~ l
0 1 0 1 1 1 1 1 1
2 0 1 2 1 0 0 1 1
2 3 0 1 3 1 1 1 1
2 3 4 0 2 1 0 4 1
2 3 4 5 0 0 3 0 5
2 3 4 5 7 0 3 7 19
2 3 4 5 7 21 0 4 16
2 3 4 5 7 21 45 0 12
2 3 4 5 7 21 45 49 0
Every pair of multiples gives a 0 on the table. Now, we are not interested in the 0s that correspond to self-modulos (2 mod 2, 3 mod 3, etc; the 0s on the diagonal) so we have to remove them. One way to do this is to add 1s on their place, like so:
=/~ l
1 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1
(=/~l) + (|/~l)
1 1 0 1 1 1 1 1 1
2 1 1 2 1 0 0 1 1
2 3 1 1 3 1 1 1 1
2 3 4 1 2 1 0 4 1
2 3 4 5 1 0 3 0 5
2 3 4 5 7 1 3 7 19
2 3 4 5 7 21 1 4 16
2 3 4 5 7 21 45 1 12
2 3 4 5 7 21 45 49 1
This can be also written as (=/~ + |/~) l.
From this table we get the final list of numbers: every number whose column contains a 0, is excluded.
We build this list of exclusions simply by multiplying by column. If a column contains a 0, its product is 0 otherwise it's a positive number:
*/ (=/~ + |/~) l
256 2187 0 6250 14406 0 0 0 18240
Before doing the last step, we'll have to improve this a little. There is no reason to perform long multiplications since we are only interested in 0s and not-0s. So, when building the table, we'll keep only 0s and 1s by taking the "sign" of each number (this is the signum:*):
* (=/~ + |/~) l
1 1 0 1 1 1 1 1 1
1 1 1 1 1 0 0 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 0 1 1
1 1 1 1 1 0 1 0 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1
so,
*/ * (=/~ + |/~) l
1 1 0 1 1 0 0 0 1
From the list of exclusion, you just copy:# the numbers to your final list:
l #~ */ * (=/~ + |/~) l
2 3 5 7 61
or,
(]#~[:*/[:*=/~+|/~) l
2 3 5 7 61
Tacit iteration is usually done with the conjunction Power. When the test for completion needs to be something other than hitting a fixpoint, the Do While construction works well.
In this solution filterMultiplesOfHead is applied repeatedly until there are no more numbers not either applied or filtered. Numbers already applied are accumulated in a partial answer. When the list to be processed is empty the partial answer is the result, after stripping off the boxing used to segregate processed from unprocessed data.
filterMultiplesOfHead=: {. (((~: >.)# %~) # ]) }.
appendHead=: (>#[ , {.#>#])/
pass=: appendHead ; filterMultiplesOfHead#>#{:
prep=: a: , <
unfinished=: [: -. a: -: {:
sieve=: [: ; [: pass^:unfinished^:_ prep
sieve 2 3 4 5 7 21 45 49 61
2 3 5 7 61
prep 2 3 4 7 9 10
┌┬────────────┐
││2 3 4 7 9 10│
└┴────────────┘
appendHead prep 2 3 4 7 9 10
2
filterMultiplesOfHead 2 3 4 7 9 10
3 7 9
pass^:2 prep 2 3 4 7 9 10
┌───┬─┐
│2 3│7│
└───┴─┘
sieve 1-.~/:~~.>:?.$~100
2 3 7 11 29 31 41 53 67 73 83 95 97