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I'm trying to implement the Join Five game. It is a game where, given a grid and a starting configuration of dots, you have to add dots in free crossings, so that each dot that you add forms a 5-dot line with those already in the grid. Two lines may only have 1 dot in common (they may cross or touch end to end)
My game grid is an int array that contains 0 or 1. 1 if there is a dot, 0 if there isn't.
I'm doing kinda well in the implementation, but I'd like to display all the possibles moves.
I made a very long and ugly function that is available here : https://pastebin.com/tw9RdNgi (it was way too long for my post i'm sorry)
here is a code snippet :
if(jeu->plat[i][j] == 0) // if we're on a empty spot
{
for(k = 0; k < lineSize; k++) // for each direction
{
//NORTH
if(jeu->plat[i-1-k][j] == 1) // if there is a dot north
{
n++; // we count it
}
else
{
break; //we change direction
}
} //
This code repeats itself 7 other times changing directions and if n or any other variable reaches 4 we count the x and y as a possible move.
And it's not even treating all the cases, if the available spot is between 2 and 2 dots it will not count it. same for 3 and 1 and 1 and 3.
But I don't think the way I started doing it is the best one. I'm pretty sure there is an easier and more optimized way but i can't figure it out.
So my question is: could somebody help me figure out how to find all the possible 5-dot alignments, or tell me if there is a better way of doing it?
Ok, the problem is more difficult than it appears, and a lot of code is required. Everything would have been simpler if you posted all of the necessary code to run it, that is a Minimal, Complete, and Verifiable Example. Anyway, I resorted to putting together a structure for the problem which allows to test it.
The piece which answers your question is the following one:
typedef struct board {
int side_;
char **dots_;
} board;
void board_set_possible_moves(board *b)
{
/* Directions
012
7 3
654 */
static int dr[8] = { -1,-1,-1, 0, 1, 1, 1, 0 };
static int dc[8] = { -1, 0, 1, 1, 1, 0,-1,-1 };
int side_ = b->side_;
char **dots_ = b->dots_;
for (int r = 0; r < side_; ++r) {
for (int c = 0; c < side_; ++c) {
// The place already has a dot
if (dots_[r][c] == 1)
continue;
// Count up to 4 dots in the 8 directions from current position
int ndots[8] = { 0 };
for (int d = 0; d < 8; ++d) {
for (int i = 1; i <= 4; ++i) {
int nr = r + dr[d] * i;
int nc = c + dc[d] * i;
if (nr < 0 || nc < 0 || nr >= side_ || nc >= side_ || dots_[nr][nc] != 1)
break;
++ndots[d];
}
}
// Decide if the position is a valid one
for (int d = 0; d < 4; ++d) {
if (ndots[d] + ndots[d + 4] >= 4)
dots_[r][c] = 2;
}
}
}
}
Note that I defined a square board with a pointer to pointers to chars, one per place. If there is a 0 in one of the places, then there is no dot and the place is not a valid move; if there is a 1, then there is a dot; if there is a 2, then the place has no dot, but it is a valid move. Valid here means that there are at least 4 dots aligned with the current one.
You can model the directions with a number from 0 to 7 (start from NW, move clockwise). Each direction has an associated movement expressed as dr and dc. Moving in every direction I count how many dots are there (up to 4, and stopping as soon as I find a non dot), and later I can sum opposite directions to obtain the total number of aligned points.
Of course these move are not necessarily valid, because we are missing the definition of lines already drawn and so we cannot check for them.
Here you can find a test for the function.
#include <stdio.h>
#include <stdlib.h>
board *board_init(board *b, int side) {
b->side_ = side;
b->dots_ = malloc(side * sizeof(char*));
b->dots_[0] = calloc(side*side, 1);
for (int r = 1; r < side; ++r) {
b->dots_[r] = b->dots_[r - 1] + side;
}
return b;
}
board *board_free(board *b) {
free(b->dots_[0]);
free(b->dots_);
return b;
}
void board_cross(board *b) {
board_init(b, 18);
for (int i = 0; i < 4; ++i) {
b->dots_[4][7 + i] = 1;
b->dots_[7][4 + i] = 1;
b->dots_[7][10 + i] = 1;
b->dots_[10][4 + i] = 1;
b->dots_[10][10 + i] = 1;
b->dots_[13][7 + i] = 1;
b->dots_[4 + i][7] = 1;
b->dots_[4 + i][10] = 1;
b->dots_[7 + i][4] = 1;
b->dots_[7 + i][13] = 1;
b->dots_[10 + i][7] = 1;
b->dots_[10 + i][10] = 1;
}
}
void board_print(const board *b, FILE *f)
{
int side_ = b->side_;
char **dots_ = b->dots_;
for (int r = 0; r < side_; ++r) {
for (int c = 0; c < side_; ++c) {
static char map[] = " oX";
fprintf(f, "%c%s", map[dots_[r][c]], c == side_ - 1 ? "" : " - ");
}
fprintf(f, "\n");
if (r < side_ - 1) {
for (int c = 0; c < side_; ++c) {
fprintf(f, "|%s", c == side_ - 1 ? "" : " ");
}
fprintf(f, "\n");
}
}
}
int main(void)
{
board b;
board_cross(&b);
board_set_possible_moves(&b);
board_print(&b, stdout);
board_free(&b);
return 0;
}
I am given a set of elements from, say, 10 to 21 (always sequential),
I generate arrays of the same size, where size is determined runtime.
Example of 3 generated arrays (arrays # is dynamic as well as # of elements in all arrays, where some elements can be 0s - not used):
A1 = [10, 11, 12, 13]
A2 = [14, 15, 16, 17]
A3 = [18, 19, 20, 21]
these generated arrays will be given to different processes to to do some computations on the elements. My aim is to balance the load for every process that will get an array. What I mean is:
With given example, there are
A1 = 46
A2 = 62
A3 = 78
potential iterations over elements given for each thread.
I want to rearrange initial arrays to give equal amount of work for each process, so for example:
A1 = [21, 11, 12, 13] = 57
A2 = [14, 15, 16, 17] = 62
A3 = [18, 19, 20, 10] = 67
(Not an equal distribution, but more fair than initial). Distributions can be different, as long as they approach some optimal distribution and are better than the worst (initial) case of 1st and last arrays. As I see it, different distributions can be achieved using different indexing [where the split of arrays is made {can be uneven}]
This works fine for given example, but there may be weird cases..
So, I see this as a reflection problem (due to the lack of knowledge of proper definition), where arrays should be seen with a diagonal through them, like:
10|111213
1415|1617
181920|21
And then an obvious substitution can be done..
I tried to implement like:
if(rest == 0)
payload_size = (upper-lower)/(processes-1);
else
payload_size = (upper-lower)/(processes-1) + 1;
//printf("payload size: %d\n", payload_size);
long payload[payload_size];
int m = 0;
int k = payload_size/2;
int added = 0; //track what been added so far (to skip over already added elements)
int added2 = 0; // same as 'added'
int p = 0;
for (i = lower; i <= upper; i=i+payload_size){
for(j = i; j<(i+payload_size); j++){
if(j <= upper){
if((j-i) > k){
if(added2 > j){
added = j;
payload[(j-i)] = j;
printf("1 adding data: %d at location: %d\n", payload[(j-i)], (j-i));
}else{
printf("else..\n");
}
}else{
if(added < upper - (m+1)){
payload[(j-i)] = upper - (p*payload_size) - (m++);
added2 = payload[(j-i)];
printf("2 adding data: %d at location: %d\n", payload[(j-i)], (j-i));
}else{
payload[(j-i)] = j;
printf("2.5 adding data: %d at location: %d\n", payload[(j-i)], (j-i));
}
}
}else{ payload[(j-i)] = '\0'; }
}
p++;
k=k/2;
//printf("send to proc: %d\n", ((i)/payload_size)%(processes-1)+1);
}
..but failed horribly.
You definitely can see the problem in the implementation, because it is poorly scalable, not complete, messy, badly written and so on, and on, and on, ...
So, I need help either with the implementation or with an idea of a better approach to do what I want to achieve, given the description.
P.S. I need the solution to be as 'in-liney' as possible (avoid loop nesting) - that is why I am using bunch of flags and global indexes.
Surely this can be done with extra loops and unnecessary iterations. I invite people that can and appreciate t̲h̲e̲ ̲a̲r̲t̲ ̲o̲f̲ ̲i̲n̲d̲e̲x̲i̲n̲g̲ when it comes to arrays.
I am sure there is a solution somewhere out there, but I just cannot make an appropriate Google query to find it.
Hint? I thought of using index % size_of_my_data to achieve this task..
P.S. Application: described here
Here is an O(n) solution I wrote using deque (double-ended queue, a deque is not necessary and a simple array can be used, but a deque makes the code clean because of popRight and popLeft). The code is Python, not pseudocode, but it should be pretty to understand (because it's Python).:
def balancingSumProblem(seqStart = None, seqStop = None, numberOfArrays = None):
from random import randint
from collections import deque
seq = deque(xrange(seqStart or randint(1, 10),
seqStop and seqStop + 1 or randint(11,30)))
arrays = [[] for _ in xrange(numberOfArrays or randint(1,6))]
print "# of elements: {}".format(len(seq))
print "# of arrays: {}".format(len(arrays))
averageNumElements = float(len(seq)) / len(arrays)
print "average number of elements per array: {}".format(averageNumElements)
oddIteration = True
try:
while seq:
for array in arrays:
if len(array) < averageNumElements and oddIteration:
array.append(seq.pop()) # pop() is like popright()
elif len(array) < averageNumElements:
array.append(seq.popleft())
oddIteration = not oddIteration
except IndexError:
pass
print arrays
print [sum(array) for array in arrays]
balancingSumProblem(10,21,3) # Given Example
print "\n---------\n"
balancingSumProblem() # Randomized Test
Basically, from iteration to iteration, it alternates between grabbing large elements and distributing them evenly in the arrays and grabbing small elements and distributing them evenly in the arrays. It goes from out to in (though you could go from in to out) and tries to use what should be the average number of elements per array to balance it out further.
It's not 100 percent accurate with all tests but it does a good job with most randomized tests. You can try running the code here: http://repl.it/cJg
With a simple sequence to assign, you can just iteratively add the min and max elements to each list in turn. There are some termination details to fix up, but that's the general idea. Applied to your example the output would look like:
john-schultzs-macbook-pro:~ jschultz$ ./a.out
10 21 13 18 = 62
11 20 14 17 = 62
12 19 15 16 = 62
A simple reflection assignment like this will be optimal when num_procs evenly divides num_elems. It will be sub-optimal, but still decent, when it doesn't:
#include <stdio.h>
int compute_dist(int lower, int upper, int num_procs)
{
if (lower > upper || num_procs <= 0)
return -1;
int num_elems = upper - lower + 1;
int num_elems_per_proc_floor = num_elems / num_procs;
int num_elems_per_proc_ceil = num_elems_per_proc_floor + (num_elems % num_procs != 0);
int procs[num_procs][num_elems_per_proc_ceil];
int i, j, sum;
// assign pairs of (lower, upper) to each process until we can't anymore
for (i = 0; i + 2 <= num_elems_per_proc_floor; i += 2)
for (j = 0; j < num_procs; ++j)
{
procs[j][i] = lower++;
procs[j][i+1] = upper--;
}
// handle left overs similarly to the above
// NOTE: actually you could use just this loop alone if you set i = 0 here, but the above loop is more understandable
for (; i < num_elems_per_proc_ceil; ++i)
for (j = 0; j < num_procs; ++j)
if (lower <= upper)
procs[j][i] = ((0 == i % 2) ? lower++ : upper--);
else
procs[j][i] = 0;
// print assignment results
for (j = 0; j < num_procs; ++j)
{
for (i = 0, sum = 0; i < num_elems_per_proc_ceil; ++i)
{
printf("%d ", procs[j][i]);
sum += procs[j][i];
}
printf(" = %d\n", sum);
}
return 0;
}
int main()
{
compute_dist(10, 21, 3);
return 0;
}
I have used this implementation, which I mentioned in this report (Implementation works for cases I've used for testing (1-15K) (1-30K) and (1-100K) datasets. I am not saying that it will be valid for all the cases):
int aFunction(long lower, long upper, int payload_size, int processes)
{
long result, i, j;
MPI_Status status;
long payload[payload_size];
int m = 0;
int k = (payload_size/2)+(payload_size%2)+1;
int lastAdded1 = 0;
int lastAdded2 = 0;
int p = 0;
int substituted = 0;
int allowUpdate = 1;
int s;
int times = 1;
int times2 = 0;
for (i = lower; i <= upper; i=i+payload_size){
for(j = i; j<(i+payload_size); j++){
if(j <= upper){
if(k != 0){
if((j-i) >= k){
payload[(j-i)] = j- (m);
lastAdded2 = payload[(j-i)];
}else{
payload[(j-i)] = upper - (p*payload_size) - (m++) + (p*payload_size);
if(allowUpdate){
lastAdded1 = payload[(j-i)];
allowUpdate = 0;
}
}
}else{
int n;
int from = lastAdded1 > lastAdded2 ? lastAdded2 : lastAdded1;
from = from + 1;
int to = lastAdded1 > lastAdded2 ? lastAdded1 : lastAdded2;
int tempFrom = (to-from)/payload_size + ((to-from)%payload_size>0 ? 1 : 0);
for(s = 0; s < tempFrom; s++){
int restIndex = -1;
for(n = from; n < from+payload_size; n++){
restIndex = restIndex + 1;
payload[restIndex] = '\0';
if(n < to && n >= from){
payload[restIndex] = n;
}else{
payload[restIndex] = '\0';
}
}
from = from + payload_size;
}
return 0;
}
}else{ payload[(j-i)] = '\0'; }
}
p++;
k=(k/2)+(k%2)+1;
allowUpdate = 1;
}
return 0;
}
So I need to fill in the code for the program to work for the question:
Array testGrades contains NUM_VALS test scores. Write a for loop that sets sumExtra to the total extra credit received. Full credit is 100, so anything over 100 is extra credit. Ex: If testGrades = {101, 83, 107, 90}, then sumExtra = 8, because 1 + 0 + 7 + 0 is 8.
#include <stdio.h>
int main(void) {
const int NUM_VALS = 4;
int testGrades[NUM_VALS];
int i = 0;
int sumExtra = -9999; // Initialize to 0 before your for loop
testGrades[0] = 101;
testGrades[1] = 83;
testGrades[2] = 107;
testGrades[3] = 90;
// STUDENT CODE GOES HERE
return 0;
}
So far all I have is:
for (i=0;i<NUM_VALS;++i) {
if (testGrades[i] > 100) {
sumExtra = testGrades[i] - 100;
}
}
I dont know how to find the sum of the array of the values over 100.
Firstly, initialise sumExtra to 0. Then just change:
sumExtra = testGrades[i] - 100;
to
sumExtra += testGrades[i] - 100;
because the sumExtra for a particular index i is testGrades[i]-100, and you want to find the total of the sumExtra, and hence keeping adding this to the sumExtra variable.
You are missing the if statement that checks if the test score is above 100. Here is the code that works:
sumExtra = 0;
for (i = 0; i < NUM_VALS; ++i) {
if (testGrades[i] >= 101) {
sumExtra += (testGrades[i] - 100);
}
}
cout << "sumExtra: " << sumExtra << endl;
Initialize to 0 before your for loop
int sumExtra = 0;
sumExtra = testGrades[i]-100;
sumExtra += testGrades[i]-100;
I have seen this question for other languages but not for AS3... and I'm having a hard time understanding it...
I need to generate 3 numbers, randomly, from 0 to 2, but they cannot repeat (as in 000, 001, 222, 212 etc) and they cannot be in the correct order (0,1,2)...
Im using
for (var u: int = 0; u < 3; u++)
{
mcCor = new CorDaCarta();
mcCor.x = larguraTrio + (mcCor.width + 5) * (u % 3);
mcCor.y = alturaTrio + (mcCor.height + 5) * (Math.floor(u / 3));
mcCor.gotoAndStop((Math.random() * (2 - u + 1) + u) | 0); // random w/ repeats
//mcCor.gotoAndStop(Math.floor(Math.random() * (2 - u + 1) + u)); // random w/ repeats
//mcCor.gotoAndStop((Math.random() * 3) | 0); // crap....
//mcCor.gotoAndStop(Math.round(Math.random()*u)); // 1,1,1
//mcCor.gotoAndStop(u + 1); // 1,2,3
mcCor.buttonMode = true;
mcCor.addEventListener(MouseEvent.CLICK, cliquetrio);
mcExplic.addChild(mcCor);
trio.push(mcCor);
}
those are the codes i've been trying.... best one so far is the active one (without the //), but it still gives me duplicates (as 1,1,1) and still has a small chance to come 0,1,2....
BTW, what I want is to mcCor to gotoAndStop on frames 1, 2 or 3....without repeating, so THE USER can put it on the right order (1,2,3 or (u= 0,1,2), thats why I add + 1 sometimes there)
any thoughts?? =)
I've found that one way to ensure random, unique numbers is to store the possible numbers in an array, and then sort them using a "random" sort:
// store the numbers 0, 1, 2 in an array
var sortedNumbers:Array = [];
for(var i:int = 0; i < 3; i++)
{
sortedNumbers.push(i);
}
var unsortedNumbers:Array = sortedNumbers.slice(); // make a copy of the sorted numbers
trace(sortedNumbers); // 0,1,2
trace(unsortedNumbers); // 0,1,2
// randomly sort array until it no longer matches the sorted array
while(sortedNumbers.join() == unsortedNumbers.join())
{
unsortedNumbers.sort(function (a:int, b:int):int { return Math.random() > .5 ? -1 : 1; });
}
trace(unsortedNumbers); // [1,0,2], [2,1,0], [0,1,2], etc
for (var u: int = 0; u < 3; u++)
{
mcCor = new CorDaCarta();
mcCor.x = larguraTrio + (mcCor.width + 5) * (u % 3);
mcCor.y = alturaTrio + (mcCor.height + 5) * (Math.floor(u / 3));
// grab the corresponding value from the unsorted array
mcCor.gotoAndStop(unsortedNumbers[u] + 1);
mcCor.buttonMode = true;
mcCor.addEventListener(MouseEvent.CLICK, cliquetrio);
mcExplic.addChild(mcCor);
trio.push(mcCor);
}
Marcela is right. Approach with an Array is widely used for such task. Of course, you will need to check 0, 1, 2 sequence and this will be ugly, but in common code to get the random sequence of integers can look like this:
function getRandomSequence(min:int, max:int):Array
{
if (min > max) throw new Error("Max value should be greater than Min value!");
if (min == max) return [min];
var values:Array = [];
for (var i:int = min; i <= max; i++) values.push(i);
var result:Array = [];
while (values.length > 0) result = result.concat(values.splice(Math.floor(Math.random() * values.length), 1));
return result;
}
for (var i:uint = 0; i < 10; i++)
{
trace(getRandomSequence(1, 10));
}
You will get something like that:
2,9,3,8,10,6,5,1,4,7
6,1,2,4,8,9,5,10,7,3
3,9,10,6,8,2,5,4,1,7
7,6,1,4,3,8,9,2,10,5
4,6,7,1,3,2,9,10,8,5
3,10,5,9,1,7,2,4,8,6
1,7,9,6,10,3,4,5,2,8
4,10,8,9,3,2,6,1,7,5
1,7,8,9,10,6,4,3,2,5
7,5,4,2,8,6,10,3,9,1
I created this for you. It is working but it can be optimized...
Hope is good for you.
var arr : Array = [];
var r : int;
for (var i: int = 0; i < 3; i++){
r=rand(0,2);
if(i == 1){
if(arr[0] == r){
i--;
continue;
}
if(arr[0] == 0){
if(r==1){
i--;
continue;
}
}
}else if(i==2){
if(arr[0] == r || arr[1] == r){
i--;
continue;
}
}
arr[i] = r;
}
trace(arr);
for(var i=0;i<3;i++){
mcCor = new CorDaCarta();
mcCor.x = larguraTrio + (mcCor.width + 5) * (i % 3);
mcCor.y = alturaTrio + (mcCor.height + 5) * (Math.floor(i / 3));
mcCor.gotoAndStop(arr[i]);
mcCor.buttonMode = true;
mcCor.addEventListener(MouseEvent.CLICK, cliquetrio);
mcExplic.addChild(mcCor);
trio.push(mcCor);
}
function rand(min:int, max:int):int {
return Math.round(Math.random() * (max - min) + min);
}
try this...
This is a basic piece of code I've created that generates a random two-coloured level:
var map:Array = new Array();
var mLength:int = 28;
var mHeight:int = 16;
var mArea:int = mLength * mHeight;
var tileWidth:int = 20;
var tileHeight:int = 20;
var tileX:int = 0 - tileWidth;
var tileY:int = 0;
var genTile:int;
var genDone:Boolean = false;
var waterChance:int = 10;
var grassChance:int = 33;
var tile:Sprite = new Sprite();
var waterTile:Sprite = new Sprite();
var waterHolder:Sprite = new Sprite();
var genericTileHolder:Sprite = new Sprite();
var hitting:Boolean = false;
var i:int = 0;
for (i = 0; i < mArea; i++) {
if (map[i - 27] == 1) {
waterChance * 8;
grassChance / 2;
}
if (map[i - 28] == 1) {
waterChance * 8;
grassChance / 2;
}
if (map[i - 29] == 1) {
waterChance * 8;
grassChance / 2;
}
if (map[i - 1] == 1) {
waterChance * 8;
grassChance / 2;
}
tileX += tileWidth;
if (tileX >= mLength * tileWidth) {
tileX = 0;
tileY += tileHeight;
}
genTile = (Math.round((Math.random()*(waterChance+grassChance))));
if (0 < genTile < waterChance) {
waterTile.graphics.beginFill(0x0033CC);
waterTile.graphics.drawRect(tileX,tileY,tileWidth,tileHeight);
waterHolder.addChildAt(waterTile, 0);
map.push("1");
}
if ((waterChance + 1) < genTile && genTile < (waterChance + grassChance)) {
tile.graphics.beginFill(0x216B18);
tile.graphics.drawRect(tileX,tileY,tileWidth,tileHeight);
genericTileHolder.addChildAt(tile, 0);
map.push("2");
}
grassChance = 33;
waterChance = 10;
}
stage.addChildAt(waterHolder, 0);
stage.addChildAt(genericTileHolder, 1);
The problem is two-fold. One, whenever the array generates, the length seems to have a random bit of variation- using trace(map.length) I get a lot of different lengths, from 750 to 780, when it should only be 400 at most.
Secondly, whenever I trace the level itself, using trace(map) I find that the first element seems to be set by default to 1. I can't figure out why it's doing this, as it should be 2 at least once after 20 tries.
You cant compare more than 2 items in actionscript:
if (0 < genTile < waterChance) {...
It will always return true (i think). Instead:
if ((0 < genTile) && (genTile < waterChance)) {...
Also, looks like each loop can do map.push("1") and map.push("2") - hence map is larger than mArea - i assume it should do one or the other?
Problem 2: first element of map is 1 by default
if (0 < genTile < waterChance)
{
waterTile.graphics.beginFill (0x0033CC); // Blue
waterTile.graphics.drawRect (tileX,tileY,tileWidth,tileHeight);
waterHolder.addChildAt (waterTile, 0);
map.push ("1");
}
Since < operator is left-associative, the expression will be evaluated as (0 < genTile) < waterChance.
Since genTile is never less than 0, (0 < genTile) is always false.
And since waterChance is also never less than 0, (false < waterChance)
which translates to (0 < waterChance) is always true.
That makes (0 < genTile) < waterChance always true.
Therefore the first element of map is always "1".
Problem 1: map.length varies
if ((waterChance + 1) < genTile && genTile < (waterChance + grassChance))
{
tile.graphics.beginFill (0x216B18); // Green
tile.graphics.drawRect (tileX,tileY,tileWidth,tileHeight);
genericTileHolder.addChildAt (tile, 0);
map.push ("2");
}
Since mArea equals 448, the water tiles will always total to 448 .
And since genTile is a random number, the grass tiles will always total to a random number.
Therefore map.length varies.
I hope that helps.