I looked at http://rosettacode.org/wiki/Knapsack_problem/0-1 to do the basic knapsack dynamic programming problem, and I got a working solution (knapsack1()), but when I tried a different solution (knapsack2()) I feel like I am off-by-one somewhere because I am not getting the correct value.
The code in question is the method knapsack2, at the bottom. I'm sure the issue is small, but I feel absurd because I can't find the problem. The correct answer should be 1030 and not 880.
(I know that an array of size [n][S] should be enough, as shown in knapsack1(), which works, but knapsack2() doesn't oddly.)
Thanks in advance for anyone who looks through this.
/**
* http://rosettacode.org/wiki/Knapsack_problem/0-1
*/
public class Knapsack {
public static void main(String[] args) {
int n = 22;
int S = 400;
int s[] = new int[22];
int v[] = new int[22];
int i = 0;
s[i] = 9;
v[i] = 150;
i++;
s[i] = 13;
v[i] = 35;
i++;
s[i] = 153;
v[i] = 200;
i++;
s[i] = 50;
v[i] = 160;
i++;
s[i] = 15;
v[i] = 60;
i++;
s[i] = 68;
v[i] = 45;
i++;
s[i] = 27;
v[i] = 60;
i++;
s[i] = 39;
v[i] = 40;
i++;
s[i] = 23;
v[i] = 30;
i++;
s[i] = 52;
v[i] = 10;
i++;
s[i] = 11;
v[i] = 70;
i++;
s[i] = 32;
v[i] = 30;
i++;
s[i] = 24;
v[i] = 15;
i++;
s[i] = 48;
v[i] = 10;
i++;
s[i] = 73;
v[i] = 40;
i++;
s[i] = 42;
v[i] = 70;
i++;
s[i] = 43;
v[i] = 75;
i++;
s[i] = 22;
v[i] = 80;
i++;
s[i] = 7;
v[i] = 20;
i++;
s[i] = 18;
v[i] = 12;
i++;
s[i] = 4;
v[i] = 50;
i++;
s[i] = 30;
v[i] = 10;
System.out.println("--Given items with these values--");
System.out.println("[item, weight (dag), value]");
System.out.println("map 9 150");
System.out.println("compass 13 35");
System.out.println("water 153 200");
System.out.println("sandwich 50 160");
System.out.println("glucose 15 60");
System.out.println("tin 68 45");
System.out.println("banana 27 60");
System.out.println("apple 39 40");
System.out.println("cheese 23 30");
System.out.println("beer 52 10");
System.out.println("suntan cream 11 70");
System.out.println("camera 32 30");
System.out.println("T-shirt 24 15");
System.out.println("trousers 48 10");
System.out.println("umbrella 73 40");
System.out.println("waterproof trousers 42 70");
System.out.println("waterproof overclothes 43 75");
System.out.println("note-case 22 80");
System.out.println("sunglasses 7 20");
System.out.println("towel 18 12");
System.out.println("socks 4 50");
System.out.println("book 30 10");
System.out.println("--The max value you can achieve in your knapsack--");
System.out.println(knapsack2(n, S, s, v));
// proper value should be 1030, from website
}
/**
* #param n items
* #param S bag size/capacity
* #param s array where s[i] is size/weight of item at index i
* #param v array where v[i] is value of item at index i
* #return best/max value possible
*/
#SuppressWarnings("unused")
private static int knapsack1(int n, int S, int s[], int v[]) {
int dp[][] = new int[n][S];
for (int i=n-1; i >= 0; i--) { // # items being left out + 1
for (int j=0; j < S; j++) { // size/weight + 1
if (i == n-1) {
dp[i][j] = 0;
} else {
int choices[] = {0,0};
choices[0] = dp[i+1][j];
if (j >= s[i])
choices[1] = v[i] + dp[i+1][j-s[i]];
dp[i][j] = max(choices);
}
}
}
return dp[0][S-1];
}
private static int max(int choices[]) {
if (choices[0] > choices[1])
return choices[0];
else
return choices[1];
}
/**
* #param n items
* #param S bag size/capacity
* #param s array where s[i] is size/weight of item at index i
* #param v array where v[i] is value of item at index i
* #return best/max value possible
*/
// #SuppressWarnings("unused")
private static int knapsack2(int n, int S, int s[], int v[]) {
int dp[][] = new int[n][S]; // dp[i][j] holds max value using i items and not exceeding weight j+1
for (int i=0; i < n; i++) { // # items. don't need to reach n because we assume we can fit at most n-1 items.
for (int j=0; j < S; j++) { // size/weight+1
if (i == 0) {
dp[i][j] = 0;
} else {
int choices[] = {0,0};
choices[0] = dp[i-1][j];
if (j >= s[i])
choices[1] = v[i] + dp[i-1][j-s[i]];
dp[i][j] = max(choices);
}
}
}
return dp[n-1][S-1]; // TODO: WHY IS THIS 880 AND NOT 1030?
}
}
Both your knapsack functions have issues. Both won't pass the simple test - S=1 n=1 s[0]=1 v[0]=10. The answer should be 10 but your code will return 0.
You can see that the issue are in the lines if (i==(n-1)) (knapsack1) or if (i==0) (knapsack2).
Also, you will see issues because of condition if (j >= s[i]. Your j starts at 0 so, if S was 1, you'll never be able to the item with s[i] == 1 because j will never reach 1.
Hope this helps. Good luck.
Related
I want to find the sum of all the positive integers in the range [1, N] with a given digit sum d. For example, if n = 100 and d = 7, the answer will be 7 + 16 + 25 + 34 + 43 + 52 + 61 + 70 = 308.
Following code can be used to count the numbers in the range [1, N] with a given digit sum d.
cnt[i][0][s] denotes count of suffixes that can be formed starting from index i, whose digits add up to s.
cnt[i][1][s] count of suffixes that can be formed starting from index i, whose digits add up to s such that the formed suffix is not greater than corresponding suffix in input string
#include <bits/stdc++.h>
using namespace std;
typedef long long int i64;
i64 cnt[20][2][200];
void digit_sum_dp(string ss) {
int n = ss.size();
for (int i = 0; i < 20; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 200; k++) {
cnt[i][j][k] = 0;
}
}
}
cnt[n][0][0] = 1;
cnt[n][1][0] = 1;
for (int i = n - 1; i >= 0; i--) {
for (int tight = 0; tight < 2; tight++) {
for (int sum = 0; sum < 200; sum++) {
if (tight) {
for (int d = 0; d <= ss[i] - '0'; d++) {
if (d == ss[i] - '0') {
cnt[i][1][sum] += cnt[i + 1][1][sum - d];
} else {
cnt[i][1][sum] += cnt[i + 1][0][sum - d];
}
}
} else {
for (int d = 0; d < 10; d++) {
cnt[i][0][sum] += cnt[i + 1][0][sum - d];
}
}
}
}
}
return cnt[0][1][d];
}
int main() {
string str = "100";
int d = 7;
cout << digit_sum_dp(str, d) << "\n";
return 0;
}
I have tried to extend the code to find out the sum of numbers instead of the count of numbers. Following is a code snippet.
cnt[i][1][sum] += cnt[i + 1][1][sum - d];
tot[i][1][sum] += (d * cnt[i + 1][1][sum - d] + tot[i + 1][1][sum - d] * pow(10, i));
I am getting incorrect results for some of the inputs. I shall be grateful if someone can help me.
I am using 1D array to get the final answer, but I also need to get selected items. How to achieve that?
private static int UnboundedKnapsack(int capacity, int n, int[] itemValue, int[] itemWeight)
{
int[] dp = new int[capacity + 1];
for (int i = 0; i <= capacity; i++)
{
for (int j = 0; j < n; j++)
{
if (itemWeight[j] <= i)
{
dp[i] = Math.Max(dp[i], dp[i - itemWeight[j]] + itemValue[j]);
}
}
}
return dp[capacity];
}
Let's introduce a new path function that gives the optimal selcetions of items using the previously calculated dp array.
private static void path(int capacity, int n, int[] itemValue, int[] itemWeight, int[] dp){
if(capacity == 0) return; // here you handle when the function will end. I assume capacity should be empty at the last
int ans = 0, chosenItem;
for(int j = 0; j < n; j++){
int newAns = dp[capacity - itemWeight[j]] + itemValue[j];
if(newAns > ans){
ans = newAns;
chosenItem = j;
}
}
printf("%d ",chosenItem); // here you get the current item you need to select;
path(capacity - itemWeight[chosenItem], n, itemValue, itemWeight, dp);
}
I have a function with 3 variables that can be modified based on random chance. For example a random number is picked from a range of 100, if random number is in range of 0 to 9, A is modified (10%), B is 11-30 (20%), C is 31 to 35 (5%).
I have rewritten this function so that the 3 variables all have a chance to be modified in the same call (random number is generated 3 times). What number ranges would I give to each variable if I wanted the overall chance of each variable being selected to be the same as the previous behaviour?
I decided to write a quick Java app to answer my question:
import java.util.*;
import java.lang.*;
class Main {
public static void main(String args[])
{
int a = 0;
int b = 0;
int c = 0;
System.out.println("old: ");
for (int i=0; i<10000; i++) {
int rand = random();
if (rand < 10) {
a++;
continue;
}
if (rand >= 10 && rand < 20) {
b++;
continue;
}
if (rand >= 20 && rand < 30) {
c++;
continue;
}
}
System.out.println("a= " + a);
System.out.println("b= " + b);
System.out.println("c= " + c);
int total = a+b+c;
System.out.println("total " + total);
a = 0;
b = 0;
c = 0;
System.out.println("new: ");
for (int i=0; i<10000; i++) {
int rand = random();
if (rand < 10) {
a++;
}
rand = random();
if (rand >= 10 && rand < 20) {
b++;
}
rand = random();
if (rand >= 20 && rand < 30) {
c++;
}
}
System.out.println("a= " + a);
System.out.println("b= " + b);
System.out.println("c= " + c);
total = a+b+c;
System.out.println("total " + total);
}
public static int random() {
return (int)(Math.random() * 100);
}
}
and the results between "new" and "old" were very similar:
old:
a= 1000
b= 1005
c= 1001
total 3006
new:
a= 949
b= 1023
c= 999
total 2971
So as it turns out, no difference at all.
I'm posting this although much has already been posted about this question. I didn't want to post as an answer since it's not working. The answer to this post (Finding the rank of the Given string in list of all possible permutations with Duplicates) did not work for me.
So I tried this (which is a compilation of code I've plagiarized and my attempt to deal with repetitions). The non-repeating cases work fine. BOOKKEEPER generates 83863, not the desired 10743.
(The factorial function and letter counter array 'repeats' are working correctly. I didn't post to save space.)
while (pointer != length)
{
if (sortedWordChars[pointer] != wordArray[pointer])
{
// Swap the current character with the one after that
char temp = sortedWordChars[pointer];
sortedWordChars[pointer] = sortedWordChars[next];
sortedWordChars[next] = temp;
next++;
//For each position check how many characters left have duplicates,
//and use the logic that if you need to permute n things and if 'a' things
//are similar the number of permutations is n!/a!
int ct = repeats[(sortedWordChars[pointer]-64)];
// Increment the rank
if (ct>1) { //repeats?
System.out.println("repeating " + (sortedWordChars[pointer]-64));
//In case of repetition of any character use: (n-1)!/(times)!
//e.g. if there is 1 character which is repeating twice,
//x* (n-1)!/2!
int dividend = getFactorialIter(length - pointer - 1);
int divisor = getFactorialIter(ct);
int quo = dividend/divisor;
rank += quo;
} else {
rank += getFactorialIter(length - pointer - 1);
}
} else
{
pointer++;
next = pointer + 1;
}
}
Note: this answer is for 1-based rankings, as specified implicitly by example. Here's some Python that works at least for the two examples provided. The key fact is that suffixperms * ctr[y] // ctr[x] is the number of permutations whose first letter is y of the length-(i + 1) suffix of perm.
from collections import Counter
def rankperm(perm):
rank = 1
suffixperms = 1
ctr = Counter()
for i in range(len(perm)):
x = perm[((len(perm) - 1) - i)]
ctr[x] += 1
for y in ctr:
if (y < x):
rank += ((suffixperms * ctr[y]) // ctr[x])
suffixperms = ((suffixperms * (i + 1)) // ctr[x])
return rank
print(rankperm('QUESTION'))
print(rankperm('BOOKKEEPER'))
Java version:
public static long rankPerm(String perm) {
long rank = 1;
long suffixPermCount = 1;
java.util.Map<Character, Integer> charCounts =
new java.util.HashMap<Character, Integer>();
for (int i = perm.length() - 1; i > -1; i--) {
char x = perm.charAt(i);
int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
charCounts.put(x, xCount);
for (java.util.Map.Entry<Character, Integer> e : charCounts.entrySet()) {
if (e.getKey() < x) {
rank += suffixPermCount * e.getValue() / xCount;
}
}
suffixPermCount *= perm.length() - i;
suffixPermCount /= xCount;
}
return rank;
}
Unranking permutations:
from collections import Counter
def unrankperm(letters, rank):
ctr = Counter()
permcount = 1
for i in range(len(letters)):
x = letters[i]
ctr[x] += 1
permcount = (permcount * (i + 1)) // ctr[x]
# ctr is the histogram of letters
# permcount is the number of distinct perms of letters
perm = []
for i in range(len(letters)):
for x in sorted(ctr.keys()):
# suffixcount is the number of distinct perms that begin with x
suffixcount = permcount * ctr[x] // (len(letters) - i)
if rank <= suffixcount:
perm.append(x)
permcount = suffixcount
ctr[x] -= 1
if ctr[x] == 0:
del ctr[x]
break
rank -= suffixcount
return ''.join(perm)
If we use mathematics, the complexity will come down and will be able to find rank quicker. This will be particularly helpful for large strings.
(more details can be found here)
Suggest to programmatically define the approach shown here (screenshot attached below) given below)
I would say David post (the accepted answer) is super cool. However, I would like to improve it further for speed. The inner loop is trying to find inverse order pairs, and for each such inverse order, it tries to contribute to the increment of rank. If we use an ordered map structure (binary search tree or BST) in that place, we can simply do an inorder traversal from the first node (left-bottom) until it reaches the current character in the BST, rather than traversal for the whole map(BST). In C++, std::map is a perfect one for BST implementation. The following code reduces the necessary iterations in loop and removes the if check.
long long rankofword(string s)
{
long long rank = 1;
long long suffixPermCount = 1;
map<char, int> m;
int size = s.size();
for (int i = size - 1; i > -1; i--)
{
char x = s[i];
m[x]++;
for (auto it = m.begin(); it != m.find(x); it++)
rank += suffixPermCount * it->second / m[x];
suffixPermCount *= (size - i);
suffixPermCount /= m[x];
}
return rank;
}
#Dvaid Einstat, this was really helpful. It took me a WHILE to figure out what you were doing as I am still learning my first language(C#). I translated it into C# and figured that I'd give that solution as well since this listing helped me so much!
Thanks!
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
namespace CsharpVersion
{
class Program
{
//Takes in the word and checks to make sure that the word
//is between 1 and 25 charaters inclusive and only
//letters are used
static string readWord(string prompt, int high)
{
Regex rgx = new Regex("^[a-zA-Z]+$");
string word;
string result;
do
{
Console.WriteLine(prompt);
word = Console.ReadLine();
} while (word == "" | word.Length > high | rgx.IsMatch(word) == false);
result = word.ToUpper();
return result;
}
//Creates a sorted dictionary containing distinct letters
//initialized with 0 frequency
static SortedDictionary<char,int> Counter(string word)
{
char[] wordArray = word.ToCharArray();
int len = word.Length;
SortedDictionary<char,int> count = new SortedDictionary<char,int>();
foreach(char c in word)
{
if(count.ContainsKey(c))
{
}
else
{
count.Add(c, 0);
}
}
return count;
}
//Creates a factorial function
static int Factorial(int n)
{
if (n <= 1)
{
return 1;
}
else
{
return n * Factorial(n - 1);
}
}
//Ranks the word input if there are no repeated charaters
//in the word
static Int64 rankWord(char[] wordArray)
{
int n = wordArray.Length;
Int64 rank = 1;
//loops through the array of letters
for (int i = 0; i < n-1; i++)
{
int x=0;
//loops all letters after i and compares them for factorial calculation
for (int j = i+1; j<n ; j++)
{
if (wordArray[i] > wordArray[j])
{
x++;
}
}
rank = rank + x * (Factorial(n - i - 1));
}
return rank;
}
//Ranks the word input if there are repeated charaters
//in the word
static Int64 rankPerm(String word)
{
Int64 rank = 1;
Int64 suffixPermCount = 1;
SortedDictionary<char, int> counter = Counter(word);
for (int i = word.Length - 1; i > -1; i--)
{
char x = Convert.ToChar(word.Substring(i,1));
int xCount;
if(counter[x] != 0)
{
xCount = counter[x] + 1;
}
else
{
xCount = 1;
}
counter[x] = xCount;
foreach (KeyValuePair<char,int> e in counter)
{
if (e.Key < x)
{
rank += suffixPermCount * e.Value / xCount;
}
}
suffixPermCount *= word.Length - i;
suffixPermCount /= xCount;
}
return rank;
}
static void Main(string[] args)
{
Console.WriteLine("Type Exit to end the program.");
string prompt = "Please enter a word using only letters:";
const int MAX_VALUE = 25;
Int64 rank = new Int64();
string theWord;
do
{
theWord = readWord(prompt, MAX_VALUE);
char[] wordLetters = theWord.ToCharArray();
Array.Sort(wordLetters);
bool duplicate = false;
for(int i = 0; i< theWord.Length - 1; i++)
{
if(wordLetters[i] < wordLetters[i+1])
{
duplicate = true;
}
}
if(duplicate)
{
SortedDictionary<char, int> counter = Counter(theWord);
rank = rankPerm(theWord);
Console.WriteLine("\n" + theWord + " = " + rank);
}
else
{
char[] letters = theWord.ToCharArray();
rank = rankWord(letters);
Console.WriteLine("\n" + theWord + " = " + rank);
}
} while (theWord != "EXIT");
Console.WriteLine("\nPress enter to escape..");
Console.Read();
}
}
}
If there are k distinct characters, the i^th character repeated n_i times, then the total number of permutations is given by
(n_1 + n_2 + ..+ n_k)!
------------------------------------------------
n_1! n_2! ... n_k!
which is the multinomial coefficient.
Now we can use this to compute the rank of a given permutation as follows:
Consider the first character(leftmost). say it was the r^th one in the sorted order of characters.
Now if you replace the first character by any of the 1,2,3,..,(r-1)^th character and consider all possible permutations, each of these permutations will precede the given permutation. The total number can be computed using the above formula.
Once you compute the number for the first character, fix the first character, and repeat the same with the second character and so on.
Here's the C++ implementation to your question
#include<iostream>
using namespace std;
int fact(int f) {
if (f == 0) return 1;
if (f <= 2) return f;
return (f * fact(f - 1));
}
int solve(string s,int n) {
int ans = 1;
int arr[26] = {0};
int len = n - 1;
for (int i = 0; i < n; i++) {
s[i] = toupper(s[i]);
arr[s[i] - 'A']++;
}
for(int i = 0; i < n; i++) {
int temp = 0;
int x = 1;
char c = s[i];
for(int j = 0; j < c - 'A'; j++) temp += arr[j];
for (int j = 0; j < 26; j++) x = (x * fact(arr[j]));
arr[c - 'A']--;
ans = ans + (temp * ((fact(len)) / x));
len--;
}
return ans;
}
int main() {
int i,n;
string s;
cin>>s;
n=s.size();
cout << solve(s,n);
return 0;
}
Java version of unrank for a String:
public static String unrankperm(String letters, int rank) {
Map<Character, Integer> charCounts = new java.util.HashMap<>();
int permcount = 1;
for(int i = 0; i < letters.length(); i++) {
char x = letters.charAt(i);
int xCount = charCounts.containsKey(x) ? charCounts.get(x) + 1 : 1;
charCounts.put(x, xCount);
permcount = (permcount * (i + 1)) / xCount;
}
// charCounts is the histogram of letters
// permcount is the number of distinct perms of letters
StringBuilder perm = new StringBuilder();
for(int i = 0; i < letters.length(); i++) {
List<Character> sorted = new ArrayList<>(charCounts.keySet());
Collections.sort(sorted);
for(Character x : sorted) {
// suffixcount is the number of distinct perms that begin with x
Integer frequency = charCounts.get(x);
int suffixcount = permcount * frequency / (letters.length() - i);
if (rank <= suffixcount) {
perm.append(x);
permcount = suffixcount;
if(frequency == 1) {
charCounts.remove(x);
} else {
charCounts.put(x, frequency - 1);
}
break;
}
rank -= suffixcount;
}
}
return perm.toString();
}
See also n-th-permutation-algorithm-for-use-in-brute-force-bin-packaging-parallelization.
I have written an image resizer using Lanczos re-sampling. I've taken the implementation straight from the directions on wikipedia. The results look good visually, but for some reason it does not match the result from Matlab's resize with Lanczos very well (in pixel error).
Does anybody see any errors? This is not my area of expertise at all...
Here is my filter (I'm using Lanczos3 by default):
double lanczos_size_ = 3.0;
inline double sinc(double x) {
double pi = 3.1415926;
x = (x * pi);
if (x < 0.01 && x > -0.01)
return 1.0 + x*x*(-1.0/6.0 + x*x*1.0/120.0);
return sin(x)/x;
}
inline double LanczosFilter(double x) {
if (std::abs(x) < lanczos_size_) {
double pi = 3.1415926;
return sinc(x)*sinc(x/lanczos_size_);
} else {
return 0.0;
}
}
And my code to resize the image:
Image Resize(Image& image, int new_rows, int new_cols) {
int old_cols = image.size().cols;
int old_rows = image.size().rows;
double col_ratio =
static_cast<double>(old_cols)/static_cast<double>(new_cols);
double row_ratio =
static_cast<double>(old_rows)/static_cast<double>(new_rows);
// Apply filter first in width, then in height.
Image horiz_image(new_cols, old_rows);
for (int r = 0; r < old_rows; r++) {
for (int c = 0; c < new_cols; c++) {
// x is the new col in terms of the old col coordinates.
double x = static_cast<double>(c)*col_ratio;
// The old col corresponding to the closest new col.
int floor_x = static_cast<int>(x);
horiz_image[r][c] = 0.0;
double weight = 0.0;
// Add up terms across the filter.
for (int i = floor_x - lanczos_size_ + 1; i < floor_x + lanczos_size_; i++) {
if (i >= 0 && i < old_cols) {
double lanc_term = LanczosFilter(x - i);
horiz_image[r][c] += image[r][i]*lanc_term;
weight += lanc_term;
}
}
// Normalize the filter.
horiz_image[r][c] /= weight;
// Strap the pixel values to valid values.
horiz_image[r][c] = (horiz_image[r][c] > 1.0) ? 1.0 : horiz_image[r][c];
horiz_image[r][c] = (horiz_image[r][c] < 0.0) ? 0.0 : horiz_image[r][c];
}
}
// Now apply a vertical filter to the horiz image.
Image new_image(new_cols, new_rows);
for (int r = 0; r < new_rows; r++) {
double x = static_cast<double>(r)*row_ratio;
int floor_x = static_cast<int>(x);
for (int c = 0; c < new_cols; c++) {
new_image[r][c] = 0.0;
double weight = 0.0;
for (int i = floor_x - lanczos_size_ + 1; i < floor_x + lanczos_size_; i++) {
if (i >= 0 && i < old_rows) {
double lanc_term = LanczosFilter(x - i);
new_image[r][c] += horiz_image[i][c]*lanc_term;
weight += lanc_term;
}
}
new_image[r][c] /= weight;
new_image[r][c] = (new_image[r][c] > 1.0) ? 1.0 : new_image[r][c];
new_image[r][c] = (new_image[r][c] < 0.0) ? 0.0 : new_image[r][c];
}
}
return new_image;
}
Here is Lanczosh in one single loop. no errors.
Uses mentioned at top procedures.
void ResizeDD(
double* const pixelsSrc,
const int old_cols,
const int old_rows,
double* const pixelsTarget,
int const new_rows, int const new_cols)
{
double col_ratio =
static_cast<double>(old_cols) / static_cast<double>(new_cols);
double row_ratio =
static_cast<double>(old_rows) / static_cast<double>(new_rows);
// Now apply a filter to the image.
for (int r = 0; r < new_rows; ++r)
{
const double row_within = static_cast<double>(r)* row_ratio;
int floor_row = static_cast<int>(row_within);
for (int c = 0; c < new_cols; ++c)
{
// x is the new col in terms of the old col coordinates.
double col_within = static_cast<double>(c)* col_ratio;
// The old col corresponding to the closest new col.
int floor_col = static_cast<int>(col_within);
double& v_toSet = pixelsTarget[r * new_cols + c];
v_toSet = 0.0;
double weight = 0.0;
for (int i = floor_row - lanczos_size_ + 1; i <= floor_row + lanczos_size_; ++i)
{
for (int j = floor_col - lanczos_size_ + 1; j <= floor_col + lanczos_size_; ++j)
{
if (i >= 0 && i < old_rows && j >= 0 && j < old_cols)
{
const double lanc_term = LanczosFilter(row_within - i + col_within - j);
v_toSet += pixelsSrc[i * old_rows + j] * lanc_term;
weight += lanc_term;
}
}
}
v_toSet /= weight;
v_toSet = (v_toSet > 1.0) ? 1.0 : v_toSet;
v_toSet = (v_toSet < 0.0) ? 0.0 : v_toSet;
}
}
}
The line
for (int i = floor_x - lanczos_size_ + 1; i < floor_x + lanczos_size_; i++)
should be
for (int i = floor_x - lanczos_size_ + 1; i <= floor_x + lanczos_size_; i++)
Do not know but perhaps other mistakes linger too.
I think there is a mistake in your sinc function. Below the fraction bar you have to square pi and x. Additional you have to multiply the function with lanczos size
L(x) = **a***sin(pi*x)*sin(pi*x/a) * (pi**²**x**²**)^-1
Edit: My mistake, there is all right.