Imagine that squares in a TicTacToe grid are numbered in a linear fashion from 1 to 9. A player puts an X on the grid by calling a class method:
$game->putX(1, 1); (the method accepts only integers from 0 to 2).
How do I calculate the linear value of the field where X was placed (here the linear value is 5)?
Your help will be much appreciated.
It's actually just x*3 + y+1. Assuming the games state is saved in an array (indexed 1-9, according to your question), your code could look like this:
// the board: examples:
// x 0 1 2 0 0 -> 1
// y 1 1 -> 5
// 0 1 2 3 2 2 -> 9
// 1 4 5 6
// 2 7 8 9
putX ($x, $y) {
$this->state[$x*3+$y+1] = 'X';
}
Related
How would one get the integer value of the n-th value from n?
This is hard to phrase so I'll just use English. If I wanted the 3rd integer value from 1...
1 = 1
2 = 3
3 = 4 <- (looking to get 4 using 3)
4 = 8
5 = 16 <- (or 16 using 5)
...
I could just do a lookup table, but I'm sure there's a better solution.
$bitvalue = 5;
$intvalue = 2 ** ($bitvalue - 1);
// gives 16
echo $intvalue;
The ** operator is the power operator. So I'm using powers of 2.
I wrote a sudoku generator that creates numbers cell by cell and checks immediately after a cell has been created if it is valid (horizontally, vertically and in a 3x3 block).
Now my problem is that the algorithm always gets stuck at some point, as it won't find a valid number for the current cell. Sometimes closer to the end, sometimes already after writing 30 cells.
This is my function to check the cell, which should change the number depending on its validity:
private function checkCell($index)
{
while ($this->isValid($index) === false) {
$this->cell[$index]->setValue(rand(1, 9));
$this->counter++;
echo 'counter: ' . $this->counter;
echo PHP_EOL;
if ($this->counter > 1000) {
$this->display();
die();
}
}
}
isValid() checks if the cell is valid horizontally, vertically and in a block (this is currently not working, it just returns true).
The counter is for debugging purposes so I can see when it gets stuck.
Here is the function generating my cells:
private function fillCell($index)
{
$rand = rand(1, 9);
$this->cell[$index]->setValue($rand);
$this->checkCell($index);
}
What should be changed so the algorithm doesn't get stuck all the time?
The issue might be that the algorithm is a little too random. You end up creating a grid that is invalid and can not be completed further.
I would propose starting from a known valid grid and shuffle the cells randomly. If a cell can't be moved, we can simply skip it.
A fair warning to the reader, the following will contain pseudo code rather than working code.
A perfectly valid starting grid:
1 2 3 | 4 5 6 | 7 8 9
7 8 9 | 1 2 3 | 4 5 6
4 5 6 | 7 8 9 | 1 2 3
------|-------|------
9 1 2 | 3 4 5 | 6 7 8
6 7 8 | 9 1 2 | 3 4 5
3 4 5 | 6 7 8 | 9 1 2
------|-------|------
8 9 1 | 2 3 4 | 5 6 7
5 6 7 | 8 9 1 | 2 3 4
2 3 4 | 5 6 7 | 8 9 1
We can store this in a single dimension array, as you already appear to do.
We follow a simple logic:
We create a single dimension array containing the cells
$cells = array(
1,2,3,4,5,6,7,8,9,7,8,9,1,2,3,4,5,6,4,5,6,7,8,9,1,2,3,
9,1,2,3,4,5,6,7,8,6,7,8,9,1,2,3,4,5,3,4,5,6,7,8,9,1,2,
8,9,1,2,3,4,5,6,7,5,6,7,8,9,1,2,3,4,2,3,4,5,6,7,8,9,1,
);
We create another array containing numbers from 0 to 80 in a random order, which are the indexes of $cells
$indexes = range(0, 80);
shuffle($indexes);
We iterate over $indexes and use the value to select a random $cell in $cells
foreach($indexes as $index) {
$cell = $cells[$index];
}
For each $cell, we iterate over $cells. In each iteration, we create a temporary grid where we switch the value of the current cell with the value of the target cell. If the temporary grid is valid, we save the target index in an array of candidates
// pseudo code because that's a lot of work
$candidates = getCandidates($cell, $cells);
We randomly choose one of the candidates and switch the cells. If no candidate is available, we simply ignore this step
candidatesCount = count(candidates);
if(candidatesCount > 0) {
$candidate = $candidates[range(0, candidatesCount -1)];
// switch values
$cells[$index] = $cells[$candidate];
$cells[candidate] = $cell;
}
Repeat until $cells is processed
There are likely more efficient ways to proceed, but this logic can not get stuck.
Note that there is a low probability that the shuffle will undo itself and produce the original grid. But it's still a valid grid.
You never want to make a backtracking algorithm that uses random numbers. It can end up running infinitely.
What you want to do is:
Find the first empty cell
Try all possible values from 1 to 9 in this cell. When all values are tried, go back.
For every value you try in the cell (at step 2), recursively call the backtracking algorithm. (go back to step 1)
If the function is called and there are no empty cells, evaluate the board. If everything is ok, you found the solution! If it's not ok, go back.
The evaluation means, check that you have all numbers from 1 to 9 exactly once on every line, every column, and every 3x3 square.
Example of how it might look like:
function back($pos) {
if ($pos >= 9*9) {
if (evaluate()) {
// we found a solution
// do soemthing with it
} else {
return;
}
}
$x = pos / 9;
$y = pos % 9;
if ($m[x][y] != 0) {
// we already have a value assigned for this position
back($pos+1);
return;
}
for ($v = 1; $v <= 9; $v++) {
$m[x][y] = $v;
back($pos+1);
}
$m[x][y] = 0; // clean up tested value before going back
}
back(1)
The above algorithm can be optimized by evaluating lines/columns at every step, instead of just once at the end. If the algorithm tries to place number x, but x is already found on the line/column, then we can just move on to try x+1 since we know x will create an invalid solution.
I want to make the calculation of personal number by date of birth.
The calculation is done in this manner:
Ex. 8 (day) +12 (month) + 1 + 9 + 7 + 1 (year) = 38 = 3 + 8 = 11 = 1 + 1 = 2
(the final number)
This final number must not be greater than nine.
So:
The first number comes is 38 greater than 9 and it should make 3 + 8
The second number comes is 11 greater than 9 it should make 1 + 1
The third number comes is 2 less than 9 so it is the final number.
Taking all these calculations should let out the number 2.
How can I get it with php calculation?
I suppose, you can split date to array. Then
$arr = array(8,12,1,9,7,1);
// sum array, split sum to array per digit untill more than 1 digit in sum
while (count($arr = str_split(array_sum($arr))) != 1) {}
echo $arr[0]; // 2
What I need is to create five random integer (say rand(1,5)). Then, I generate a score based on these numbers. For instance, if I get a result of 1,2,3,4,5 then that would equal a zero score, but if I got 1,1,3,4,5 that would be 1 as we have a pair. Similar to a poker kind of scoring, so five of the same number would be a "full house" thus resulting in the highest score.
How would I go about the scoring system, even if it is just the mathematical equation?
More detail:
1-5 will hold separate images and then will be fought against "The House" which will have identical code to the user to determine the winner. Here's some example draws and the score they would receive:
1,2,3,4,5 = score 0
1,1,2,3,4 = score 1 (1 pair)
1,1,2,2,4 = score 2 (2 pair)
1,1,1,3,4 = score 3 (3 of a kind)
1,1,1,1,5 = score 4 (4 of a kind)
1,1,1,3,3 = score 5 (full house)
1,1,1,1,1 = score 6 (5 of a kind)
The combination of numbers is irreverent if they score 6 and the house scores 6, it's a tie.
if (isset($_POST['play'])) {
$rand1 = rand(1, 5);
$rand2 = rand(1, 5);
$rand3 = rand(1, 5);
$rand4 = rand(1, 5);
$rand5 = rand(1, 5);
if ($_POST['bet'] <= $user_data['coins']) {
if ($_POST['bet'] < 999999999) {
if ($_POST['bet'] > 0.99) {
if ($user_data['coins'] >= 1) {
$array = array($rand1,$rand2,$rand3,$rand4,$rand5);
print_r(array_count_values($array));
echo $rand1.', '.$rand2.', '.$rand3.', '.$rand4.', '.$rand5;
Array( // Here I don't understand
1 => 3,//
2 => 1,//
3 => 1 //
);
}
}
}
}
}
This outputs ; Array ( [5] => 2 [4] => 2 [1] => 1 ) 5, 5, 4, 4, 1
Use array_count_value function for this.
$array = array(1,1,1,2,5);
print_r(array_count_values($array));
Array(
1 => 3,
2 => 1,
3 => 1
);
Here's the approach I would consider, building on #Lele's answer. Warning: this is a bit confusing, so sit down with a cup of tea for this one.
Build a set of five buckets, [1] to [5], and scan a player's numbers, so that the count for each number is stored in the corresponding bucket
Then count the numbers you are left with into a new bucket system, with each position representing the number of counts you have for something.
So, if your score is this:
1 1 2 2 4
Then your first buckets are:
2 2 0 1 0
That's because you have two ones, two twos, and one four. And your second buckets are:
1 2 0 0 0
That's because you have two two-counts, and one one-count. Here, you disregard the first position (since a one-count for something does not score anything) and score for the others. So, test for two twos, and score that two.
If you score is this:
5 5 5 5 1
Then your first buckets are:
1 0 0 0 4
That's one one and four fives. So your second buckets are:
1 0 0 1 0
Your lookup table for this could be:
x 1 0 0 0 -> one pair
x 2 0 0 0 -> two pairs
x 0 1 0 0 -> three of a kind
x 1 1 0 0 -> full house
x 0 0 1 0 -> four of a kind
x 0 0 0 1 -> five of a kind
The 'x' means that you don't match on this. So, your lookup table matches four numbers to a score.
I was rather interested in this problem, so I have written some code to do the above. You'll still need to do the lookup table, but that is relatively trivial, and will be good practice for you. Here is a demo, with comments (run code here):
<?php
function counting(array $array) {
// Input figures
print_r($array);
// Run the figures twice through the bucket-counter
$firstBuckets = bucketCounter($array);
$secondBuckets = bucketCounter($firstBuckets);
// Ignore counts of 1
array_shift($secondBuckets);
// Output, just need to do the lookup now
echo ' converts to ';
print_r($secondBuckets);
echo "<br />";
}
/**
* Bucket counter
*/
function bucketCounter(array $array) {
$result = array(0, 0, 0, 0, 0, );
foreach($array as $value) {
if ($value > 0) {
$result[$value - 1]++;
}
}
return $result;
}
// Try some demos here!
counting(array(1, 2, 3, 4, 5));
counting(array(1, 1, 2, 4, 2));
counting(array(1, 1, 1, 1, 1));
?>
The demos I've included seem to work, but do hunt for bugs!
If the range is quite small, you can use counting sort approach. For each number, provide a "bucket" to count how many times a number appear. Scan once to fill in the buckets. Then another scan, but this time against the bucket to get the highest value. That's your score.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Understanding PHP's & operator
I was just looking at array_filter() function doc and they had the following code to return odd numbers...
<?php
function odd($var)
{
// returns whether the input integer is odd
return($var & 1);
}
?>
Why does $var & 1 return odd number? how does that work?
& is bitwise and. It acts as a mask on the bits of $var. All odd numbers end with 1
no bit &1
1 001 1
2 010 0
3 011 1
4 100 0
5 101 1
6 110 0
7 111 1
You are using a bitwise function with always returns 1 when anded with an odd number.
A few examples:
11 = 3
01 = 1
----
01 = odd -- return 1 (true)
100 = 4
01 = 1
-----
000 = even -- return 0 (false)
One more:
10101 = 21
01 = 1
-------
00001 = odd -- return 1 (true)
That function return 1 if var is an odd number, 0 otherwise. "&" is the AND binary operator, so it considers the last binary digit of a number.
For example:
5 in binary is 101 -> 101 & 1 = 1 -> odd number.
8 in binary is 1000 -> 1000 & 1 = 0 -> even number.