I am having a problem with i think is the index. i can't run the code, somehow the "while" is wrong and i don't know how. I am coping with a youtube video how to build it. but i think the video is outdated, is from 3 years ago. if somone get a ideia how to run it i would be grateful. (sorry bad English, not my formal language)
some phrases will be in portuguese but i think is not gonna damege it...
edit: i think is not a error. but its a red phrase is on the scrubber area
extends Node2D
# Grid Variaveis
export (int) var width;
export (int) var height;
export (int) var x_start;
export (int) var y_start;
export (int) var offset;
var possible_pieces = [
preload("res://cenario/abacate.tscn"),
preload("res://cenario/sprite2.tscn"),
preload("res://cenario/Node2D.tscn"),
preload("res://cenario/sprite 3.tscn"),
preload("res://cenario/tomato.tscn"),
preload("res://cenario/uva.tscn")
];
var all_pieces =[];
func _ready():
randomize();
all_pieces = make_2d_array();
spawn_pieces();
func make_2d_array():
var array = [];
for i in width:
array.append([]);
for j in height:
array[i].append(null)
return array;
func spawn_pieces():
for i in width:
for j in height:
#choose a random number and store it
var rand = floor(rand_range(0, possible_pieces.size()));
var piece = possible_pieces[rand].instance();
var loops = 0
while(match_at(i, j, piece[rand].color) && loops < 100):
rand = floor(rand_range(0, possible_pieces.size()));
loops += 1;
piece = possible_pieces[rand].instance();
#instance that piece from the array
add_child(piece);
piece.position = grid_to_pixel(i, j);
all_pieces[i][j] = piece;
func match_at(i, j, color):
if i > 1:
if all_pieces[i - 1][j] != null && all_pieces[i - 2][j] != null:
if all_pieces[i - 1][j].color == color && all_pieces[i - 2][j].color == color:
return true;
if j > 1:
if all_pieces[i][j - 1] != null && all_pieces[i][j - 2] != null:
if all_pieces[i][j - 1].color == color && all_pieces[i][j - 2].color == color:
return true;
pass;
func grid_to_pixel(column, row):
var new_x = x_start + offset * column;
var new_y = y_start + -offset * row;
return Vector2(new_x, new_y);
You have possible_pieces with a bunch of PackedScenes:
var possible_pieces = [
preload("res://cenario/abacate.tscn"),
preload("res://cenario/sprite2.tscn"),
preload("res://cenario/Node2D.tscn"),
preload("res://cenario/sprite 3.tscn"),
preload("res://cenario/tomato.tscn"),
preload("res://cenario/uva.tscn")
];
Then you pick one and instantiate it:
var rand = floor(rand_range(0, possible_pieces.size()));
var piece = possible_pieces[rand].instance();
Thus piece is a Node. So when you do this: piece[rand] you are trying to use index access on a Node and that does not work. That is the problem with this line:
while(match_at(i, j, piece[rand].color) && loops < 100):
I believe you want to do this instead:
while(match_at(i, j, piece.color) && loops < 100):
For future reference, specifying the types of your variables might help you avoid this kind of errors. See Static typing in GDScript. By the way, even though the term "Type Hints" stuck, they are not hints, they are actual type declarations.
The function match_atreturns null if it doesn't match any of the conditions.
Replace the pass; for return false; or create an else statement to do something if needed, and then the return false;.
As mentioned below by #Theraot i realized the pass; its not the problem
Related
I have been working on an exercise from google's dev tech guide. It is called Compression and Decompression you can check the following link to get the description of the problem Challenge Description.
Here is my code for the solution:
public static String decompressV2 (String string, int start, int times) {
String result = "";
for (int i = 0; i < times; i++) {
inner:
{
for (int j = start; j < string.length(); j++) {
if (isNumeric(string.substring(j, j + 1))) {
String num = string.substring(j, j + 1);
int times2 = Integer.parseInt(num);
String temp = decompressV2(string, j + 2, times2);
result = result + temp;
int next_j = find_next(string, j + 2);
j = next_j;
continue;
}
if (string.substring(j, j + 1).equals("]")) { // Si es un bracket cerrado
break inner;
}
result = result + string.substring(j,j+1);
}
}
}
return result;
}
public static int find_next(String string, int start) {
int count = 0;
for (int i = start; i < string.length(); i++) {
if (string.substring(i, i+1).equals("[")) {
count= count + 1;
}
if (string.substring(i, i +1).equals("]") && count> 0) {
count = count- 1;
continue;
}
if (string.substring(i, i +1).equals("]") && count== 0) {
return i;
}
}
return -111111;
}
I will explain a little bit about the inner workings of my approach. It is a basic solution involves use of simple recursion and loops.
So, let's start from the beggining with a simple decompression:
DevTech.decompressV2("2[3[a]b]", 0, 1);
As you can see, the 0 indicates that it has to iterate over the string at index 0, and the 1 indicates that the string has to be evaluated only once: 1[ 2[3[a]b] ]
The core here is that everytime you encounter a number you call the algorithm again(recursively) and continue where the string insides its brackets ends, that's the find_next function for.
When it finds a close brackets, the inner loop breaks, that's the way I choose to make the stop sign.
I think that would be the main idea behind the algorithm, if you read the code closely you'll get the full picture.
So here are some of my concerns about the way I've written the solution:
I could not find a more clean solution to tell the algorithm were to go next if it finds a number. So I kind of hardcoded it with the find_next function. Is there a way to do this more clean inside the decompress func ?
About performance, It wastes a lot of time by doing the same thing again, when you have a number bigger than 1 at the begging of a bracket.
I am relatively to programming so maybe this code also needs an improvement not in the idea, but in the ways It's written. So would be very grateful to get some suggestions.
This is the approach I figure out but I am sure there are a couple more, I could not think of anyone but It would be great if you could tell your ideas.
In the description it tells you some things that you should be awared of when developing the solutions. They are: handling non-repeated strings, handling repetitions inside, not doing the same job twice, not copying too much. Are these covered by my approach ?
And the last point It's about tets cases, I know that confidence is very important when developing solutions, and the best way to give confidence to an algorithm is test cases. I tried a few and they all worked as expected. But what techniques do you recommend for developing test cases. Are there any softwares?
So that would be all guys, I am new to the community so I am open to suggestions about the how to improve the quality of the question. Cheers!
Your solution involves a lot of string copying that really slows it down. Instead of returning strings that you concatenate, you should pass a StringBuilder into every call and append substrings onto that.
That means you can use your return value to indicate the position to continue scanning from.
You're also parsing repeated parts of the source string more than once.
My solution looks like this:
public static String decompress(String src)
{
StringBuilder dest = new StringBuilder();
_decomp2(dest, src, 0);
return dest.toString();
}
private static int _decomp2(StringBuilder dest, String src, int pos)
{
int num=0;
while(pos < src.length()) {
char c = src.charAt(pos++);
if (c == ']') {
break;
}
if (c>='0' && c<='9') {
num = num*10 + (c-'0');
} else if (c=='[') {
int startlen = dest.length();
pos = _decomp2(dest, src, pos);
if (num<1) {
// 0 repetitions -- delete it
dest.setLength(startlen);
} else {
// copy output num-1 times
int copyEnd = startlen + (num-1) * (dest.length()-startlen);
for (int i=startlen; i<copyEnd; ++i) {
dest.append(dest.charAt(i));
}
}
num=0;
} else {
// regular char
dest.append(c);
num=0;
}
}
return pos;
}
I would try to return a tuple that also contains the next index where decompression should continue from. Then we can have a recursion that concatenates the current part with the rest of the block in the current recursion depth.
Here's JavaScript code. It takes some thought to encapsulate the order of operations that reflects the rules.
function f(s, i=0){
if (i == s.length)
return ['', i];
// We might start with a multiplier
let m = '';
while (!isNaN(s[i]))
m = m + s[i++];
// If we have a multiplier, we'll
// also have a nested expression
if (s[i] == '['){
let result = '';
const [word, nextIdx] = f(s, i + 1);
for (let j=0; j<Number(m); j++)
result = result + word;
const [rest, end] = f(s, nextIdx);
return [result + rest, end]
}
// Otherwise, we may have a word,
let word = '';
while (isNaN(s[i]) && s[i] != ']' && i < s.length)
word = word + s[i++];
// followed by either the end of an expression
// or another multiplier
const [rest, end] = s[i] == ']' ? ['', i + 1] : f(s, i);
return [word + rest, end];
}
var strs = [
'2[3[a]b]',
'10[a]',
'3[abc]4[ab]c',
'2[2[a]g2[r]]'
];
for (const s of strs){
console.log(s);
console.log(JSON.stringify(f(s)));
console.log('');
}
I am writing a Maximum Value Knapsack algorithm. It takes in a Knapsack object with Items that have a value and cost. I declare a 2D array for calculating the max value. For the base cases I have set the zeroth row values to 0 and zeroth column values to 0. I am running into trouble when I grab an item in the knapsack because when I want to grab the zeroth item, I am really grabbing the first item in the knapsack and am consequently getting the wrong values in the 2D array. Can someone check out my code and see what I am missing?
public static double MaximumKnapsack(Knapsack knapsack) {
int numItems = knapsack.getNumOfItems();
int budget = (int) knapsack.getBudget();
double[][] DP = new double[numItems+1][budget+1];
boolean taken = false;
for (int i = 0; i < numItems + 1; i++) {
for (int b = 0; b < budget + 1; b++) {
if (i == 0 || b == 0) {
DP[i][b] = 0;
}
else
{
Item item = knapsack.getItem(i);
if (item.getCost() > b) {
DP[i][b] = DP[i-1][b];
}
else
{
DP[i][b] = Math.max(DP[i-1][b-(int) item.getCost()] + item.getValue(),
DP[i-1][b]);
if (DP[i][b] == DP[i-1][b-(int) item.getCost()] + item.getValue() && item.getCost() != 0.0) {
taken = true;
}
}
}
}
taken = false;
}
return DP[numItems][budget];
}
I think the problem is in
Item item = knapsack.getItem(i);
beacuse your loop will start with i = 1. You should use:
Item item = knapsack.getItem(i-1);
On the webpage
http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463
It lists prices for a particular item in a game, I wanted to grab the "Current guide price:" of said item, and store it as a variable so I could output it in a google spreadsheet. I only want the number, currently it is "643.8k", but I am not sure how to grab specific text like that.
Since the number is in "k" form, that means I can't graph it, It would have to be something like 643,800 to make it graphable. I have a formula for it, and my second question would be to know if it's possible to use a formula on the number pulled, then store that as the final output?
-EDIT-
This is what I have so far and it's not working not sure why.
function pullRuneScape() {
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var number = page.match(/Current guide price:<\/th>\n(\d*)/)[1];
SpreadsheetApp.getActive().getSheetByName('RuneScape').appendRow([new Date(), number]);
}
Your regex is wrong. I tested this one successfully:
var number = page.match(/Current guide price:<\/th>\s*<td>([^<]*)<\/td>/m)[1];
What it does:
Current guide price:<\/th> find Current guide price: and closing td tag
\s*<td> allow whitespace between tags, find opening td tag
([^<]*) build a group and match everything except this char <
<\/td> match the closing td tag
/m match multiline
Use UrlFetch to get the page [1]. That'll return an HTTPResponse that you can read with GetBlob [2]. Once you have the text you can use regular expressions. In this case just search for 'Current guide price:' and then read the next row. As to remove the 'k' you can just replace with reg ex like this:
'123k'.replace(/k/g,'')
Will return just '123'.
https://developers.google.com/apps-script/reference/url-fetch/
https://developers.google.com/apps-script/reference/url-fetch/http-response
Obviously, you are not getting anything because the regexp is wrong. I'm no regexp expert but I was able to extract the number using basic string manipulation
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var TD = "<td>";
var start = page.indexOf('Current guide price');
start = page.indexOf(TD, start);
var end = page.indexOf('</td>',start);
var number = page.substring (start + TD.length , end);
Logger.log(number);
Then, I wrote a function to convert k,m etc. to the corresponding multiplying factors.
function getMultiplyingFactor(symbol){
switch(symbol){
case 'k':
case 'K':
return 1000;
case 'm':
case 'M':
return 1000 * 1000;
case 'g':
case 'G':
return 1000 * 1000 * 1000;
default:
return 1;
}
}
Finally, tie the two together
function pullRuneScape() {
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var TD = "<td>";
var start = page.indexOf('Current guide price');
start = page.indexOf(TD, start);
var end = page.indexOf('</td>',start);
var number = page.substring (start + TD.length , end);
Logger.log(number);
var numericPart = number.substring(0, number.length -1);
var multiplierSymbol = number.substring(number.length -1 , number.length);
var multiplier = getMultiplyingFactor(multiplierSymbol);
var fullNumber = multiplier == 1 ? number : numericPart * multiplier;
Logger.log(fullNumber);
}
Certainly, not the optimal way of doing things but it works.
Basically I parse the html page as you did (with corrected regex) and split the string into number part and multiplicator (k = 1000). Finally I return the extracted number. This function can be used in Google Docs.
function pullRuneScape() {
var pageContent = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var matched = pageContent.match(/Current guide price:<.th>\n<td>(\d+\.*\d*)([k]{0,1})/);
var numberAsString = matched[1];
var multiplier = "";
if (matched.length == 3) {
multiplier = matched[2];
}
number = convertNumber(numberAsString, multiplier);
return number;
}
function convertNumber(numberAsString, multiplier) {
var number = Number(numberAsString);
if (multiplier == 'k') {
number *= 1000;
}
return number;
}
I have recently come across with this problem,
you have to find an integer from a sorted two dimensional array. But the two dim array is sorted in rows not in columns. I have solved the problem but still thinking that there may be some better approach. So I have come here to discuss with all of you. Your suggestions and improvement will help me to grow in coding. here is the code
int searchInteger = Int32.Parse(Console.ReadLine());
int cnt = 0;
for (int i = 0; i < x; i++)
{
if (intarry[i, 0] <= searchInteger && intarry[i,y-1] >= searchInteger)
{
if (intarry[i, 0] == searchInteger || intarry[i, y - 1] == searchInteger)
Console.WriteLine("string present {0} times" , ++cnt);
else
{
int[] array = new int[y];
int y1 = 0;
for (int k = 0; k < y; k++)
array[k] = intarry[i, y1++];
bool result;
if (result = binarySearch(array, searchInteger) == true)
{
Console.WriteLine("string present inside {0} times", ++ cnt);
Console.ReadLine();
}
}
}
}
Where searchInteger is the integer we have to find in the array. and binary search is the methiod which is returning boolean if the value is present in the single dimension array (in that single row).
please help, is it optimum or there are better solution than this.
Thanks
Provided you have declared the array intarry, x and y as follows:
int[,] intarry =
{
{0,7,2},
{3,4,5},
{6,7,8}
};
var y = intarry.GetUpperBound(0)+1;
var x = intarry.GetUpperBound(1)+1;
// intarry.Dump();
You can keep it as simple as:
int searchInteger = Int32.Parse(Console.ReadLine());
var cnt=0;
for(var r=0; r<y; r++)
{
for(var c=0; c<x; c++)
{
if (intarry[r, c].Equals(searchInteger))
{
cnt++;
Console.WriteLine(
"string present at position [{0},{1}]" , r, c);
} // if
} // for
} // for
Console.WriteLine("string present {0} times" , cnt);
This example assumes that you don't have any information whether the array is sorted or not (which means: if you don't know if it is sorted you have to go through every element and can't use binary search). Based on this example you can refine the performance, if you know more how the data in the array is structured:
if the rows are sorted ascending, you can replace the inner for loop by a binary search
if the entire array is sorted ascending and the data does not repeat, e.g.
int[,] intarry = {{0,1,2}, {3,4,5}, {6,7,8}};
then you can exit the loop as soon as the item is found. The easiest way to do this to create
a function and add a return statement to the inner for loop.
Let me explain.
I have to do some fuzzy matching for a company, so ATM I use a levenshtein distance calculator, and then calculate the percentage of similarity between the two terms. If the terms are more than 80% similar, Fuzzymatch returns "TRUE".
My problem is that I'm on an internship, and leaving soon. The people who will continue doing this do not know how to use excel with macros, and want me to implement what I did as best I can.
So my question is : however inefficient the function may be, is there ANY way to make a standard function in Excel that will calculate what I did before, without resorting to macros ?
Thanks.
If you came about this googling something like
levenshtein distance google sheets
I threw this together, with the code comment from milot-midia on this gist (https://gist.github.com/andrei-m/982927 - code under MIT license)
From Sheets in the header menu, Tools -> Script Editor
Name the project
The name of the function (not the project) will let you use the func
Paste the following code
function Levenshtein(a, b) {
if(a.length == 0) return b.length;
if(b.length == 0) return a.length;
// swap to save some memory O(min(a,b)) instead of O(a)
if(a.length > b.length) {
var tmp = a;
a = b;
b = tmp;
}
var row = [];
// init the row
for(var i = 0; i <= a.length; i++){
row[i] = i;
}
// fill in the rest
for(var i = 1; i <= b.length; i++){
var prev = i;
for(var j = 1; j <= a.length; j++){
var val;
if(b.charAt(i-1) == a.charAt(j-1)){
val = row[j-1]; // match
} else {
val = Math.min(row[j-1] + 1, // substitution
prev + 1, // insertion
row[j] + 1); // deletion
}
row[j - 1] = prev;
prev = val;
}
row[a.length] = prev;
}
return row[a.length];
}
You should be able to run it from a spreadsheet with
=Levenshtein(cell_1,cell_2)
While it can't be done in a single formula for any reasonably-sized strings, you can use formulas alone to compute the Levenshtein Distance between strings using a worksheet.
Here is an example that can handle strings up to 15 characters, it could be easily expanded for more:
https://docs.google.com/spreadsheet/ccc?key=0AkZy12yffb5YdFNybkNJaE5hTG9VYkNpdW5ZOWowSFE&usp=sharing
This isn't practical for anything other than ad-hoc comparisons, but it does do a decent job of showing how the algorithm works.
looking at the previous answers to calculating Levenshtein distance, I think it would be impossible to create it as a formula.
Take a look at the code here
Actually, I think I just found a workaround. I was adding it in the wrong part of the code...
Adding this line
} else if(b.charAt(i-1)==a.charAt(j) && b.charAt(i)==a.charAt(j-1)){
val = row[j-1]-0.33; //transposition
so it now reads
if(b.charAt(i-1) == a.charAt(j-1)){
val = row[j-1]; // match
} else if(b.charAt(i-1)==a.charAt(j) && b.charAt(i)==a.charAt(j-1)){
val = row[j-1]-0.33; //transposition
} else {
val = Math.min(row[j-1] + 1, // substitution
prev + 1, // insertion
row[j] + 1); // deletion
}
Seems to fix the problem. Now 'biulding' is 92% accurate and 'bilding' is 88%. (whereas with the original formula 'biulding' was only 75%... despite being closer to the correct spelling of building)