Related
I have a string in the form of "AsKcQsJd" that represents 4 cards from a deck of playing cards. The uppercase value represnts the card value (in this case, Ace, King, Queen, and Jack) and the lowercase value represents the suit (in this case, spade, club, spade, diamond).
Say I have another value that tells me what suit I'm looking for. So in this case, I have:
$hand = 'AsKcQsJd';
$suit = 's';
How can I write a regular expression that checks if the hand has an Ace in it, followed by the suit, so in this case 'As' and also any other card that has the suit? Or in 'poker terms', I'm trying to determine if the hand has the 'ace high flush draw' for the suit defined as $suit.
To further explain, I need to check if any combination of the following two cards exist:
AsKs, AsQs, AsJs, AsTs,As9s,As8s,As7s,As6s,As5s,As4s,As3s,As2s
With the added complexity that these cards could occur anywhere in the hand. For example, the string could have As at the front and Ks at the end. That's why I think a regular expression is the best method for determining if the two coexist in the string.
You might use two lookaheads, one for As, and one for [^A]s, like this:
(?=.*As)(?=.*[^A]s)
https://regex101.com/r/8hkWTv/1
$suit = 's';
$re = '/(?=.*A' . $suit . ')(?=.*[^A]' . $suit . ')/';
print($re); // /(?=.*As)(?=.*[^A]s)/
print(preg_match($re, 'AsKcQsJd')); // 1
print(preg_match($re, 'AdKcQsJd')); // 0
print(preg_match($re, 'KsKcQsJd')); // 0
I'm not sure regex is the best solution but if that's your cup of tea you can do it pretty easily with alternation like this:
As.*s|s.*As
Or better yet - to capture the actual cards giving you a match:
(As).*(.s)|(.s).*(As)
These basically say - the hand has a spade followed by an ace of spades OR has ace of spades followed by any other spade. https://regex101.com/r/pdwHPQ/1
That said, I'd probably consider building a simple class to parse the hand and give you more flexibility when it comes to answering questions about what cards are present. Whether or not this is worth it really depends a lot on your app. Here's an idea:
$hand = 'AsKh4c5c9h2s';
$cards = new Cards($hand);
$spades = $cards->getCardsBySuit('s');
if (in_array('As',array_keys($spades)) && count($spades) > 1) {
// hand has ace high flush draw
echo 'yep';
}
class Cards {
private $cards = '';
public function __construct($hand) {
foreach (str_split($hand,2) as $card) {
$this->cards[$card] = [
'rank' => substr($card,0,1),
'suit' => substr($card,1,1)
];
}
}
public function getCardsBySuit($suit) {
$response = [];
foreach ($this->cards as $k => $card) {
if ($card['suit'] == $suit) {
$response[$k] = $card;
}
}
return $response;
}
}
EDIT 1 -since posting I have learnt that the underlying question is about how to find the CARTESIAN PRODUCT (now go google), but not only because I don't want every perm, I want to find the cartesian products that use the same subarray Key never more than once per permuation AND my 'extra' question then is more about how to minimise the workload that a cartesian product would require (accepting a small error rate, I have to say)-
Imagine... I have four cooks and four recipes, each cook has a score for each recipe and today I'd like each cook to make one dish (but no dish should be made twice) and the decision should be based on the best (highest total scores) permutation for all four (so maybe a cook won't make his personal best).
I have put the data into a multi-dimensional array as such
array(
array (1,2,3,4),
array (35,0,0,0),
array (36,33,1,1),
array (20,20,5,3)
)
it has the same number of valuepairs in each sub array as the number of sub-arrays (if that helps any)
in reality the number of sub-arrays would reach a maximum of 8 (max perms therefore =8!, approx 40,000 not 8^8 because many combinations are not allowed)
the choice of having the data in this format is flexible if that helps
I am trying to create a second array that would output the best (ie HIGHEST value) possible combination of the sub-arrays as per KEYs where only ONE of each subarray can be used
--so here each subarray[0][1][2][3] would be used once per permutation
and each subarrayKey [0][1][2][3] would be used once per permutaion, in my actual problem I'm using associated arrays, but that is extra to this issue.--
So the example would create an array as such
newArray (35,33,5,4) // note that [2][0] was not used
IDEALLY I would prefer to not produce the ALL perms but rather, SOMEHOW, discard many combinations that would clearly not be best fit.
Any ideas for how to start? I would accept pseudo code.
For an example on SO about Cartesian Product, see PHP 2D Array output all combinations
EDIT 2
for more on making cartesian products more efficient, and maybe why it has to be case specific if you want to see if you can cut corners (with risk) Efficient Cartesian Product algorithm
Apologies, but this is going to be more of a logic layout than code...
It's not quite clear to me whether the array(1,2,3,4) are the scores for the first dish or for the first cook, but I would probably use an array such that
$array[$cook_id][$dish_number] = $score;
asort() each array so that $array[$cook_id] = array($lowest_scored_dish,...,$highest);
Consider a weighted preference for a particular cook to make a dish to be the difference between the score of the best dish and another.
As a very simple example, cooks a,b,c and dishes 0,1,2
$array['a'] = array(0=>100, 1=>50, 2=>0); // cook a prefers 0 over 1 with weight 50, over 2 with weight 100
$array['b'] = array(0=>100, 1=>100, 2=>50); // cook b prefers 0,1 over 2 with weight 50
$array['c'] = array(0=>50, 1=>50, 2=>100); // cook c prefers 2 with weight 50
After asort():
$array['a'] = array(0=>100, 1=>50, 2=>0);
$array['b'] = array(0=>100, 1=>100, 2=>50);
$array['c'] = array(2=>100, 0=>50, 1=>50);
Start with cook 'a' who prefers dish 0 over his next best dish by 50 points (weight). Cook 'b' also prefers dih 0, but with a weight of 0 over the next dish. Therefore it's likely (though not yet certain that cook 'a' should make dish 0.
Consider dish 0 to be reserved and move on to cook 'b'. Excluding dish 0, cook 'b' prefers dish 1. No other cook prefers dish 1, so cook 'b' is assigned dish 1.
Cook 'c' gets dish 2 by default.
This is a VERY convenient example where each cook gets to cook something that's a personal max, but I hope it's illustrative of some logic that would work out.
Let's make it less convenient:
$array['a'] = array(0=>75, 1=>50, 2=>0);
$array['b'] = array(0=>100, 1=>50, 2=>50);
$array['c'] = array(0=>100, 1=>25, 2=>25);
Start again with cook 'a' and see that 0 is preferred, but this time with weight 25. Cook 'b' prefers with a weight of 50 and cook 'c' prefers with a weight of 75. Cook 'c' wins dish 0.
Going back to the list of available cooks, 'a' prefers 1 with a weight of 50, but 'b' prefers it with weight 0. 'a' gets dish 1 and 'b' gets dish 2.
This still doesn't take care of all complexities, but it's a step in the right direction. Sometimes the assumption made for the first cook/dish combination will be wrong.
WAY less convenient:
$array['a'] = array(0=>200, 1=>148, 2=>148, 3=>0);
$array['b'] = array(0=>200, 1=>149, 2=>0, 3=>0);
$array['c'] = array(0=>200, 1=>150, 2=>147, 3=>147);
$array['d'] = array(0=>69, 1=>18, 2=>16, 3=>15);
'a' gets 0 since that's the max and no one else who prefers 0 has a higher weight
'b' wins 1 with a weight of 149
'd' wins 2 since 'c' doesn't have a preference from the available options
'c' gets 3
score: 200+149+147+16 = 512
While that's a good guess that's gathered without checking every permutation, it may be wrong. From here, ask, "If one cook traded with any one other cook, would the total increase?"
The answer is YES, a(0)+d(2) = 200+16 = 216, but a(2)+d(0) = 148+69 = 217.
I'll leave it to you to write the code for the "best guess" using the weighted approach, but after that, here's a good start for you:
// a totally uneducated guess...
$picks = array(0=>'a', 1=>'b', 2=>'c', 3=>'d');
do {
$best_change = false;
$best_change_weight = 0;
foreach ($picks as $dish1 => $cook1) {
foreach ($picks as $dish2 => $cook2) {
if (($array[$cook1][$dish1] + $array[$cook2][$dish2]) <
($array[$cook1][$dish2] + $array[$cook2][$dish1]))
{
$old_score = $array[$cook1][$dish1] + $array[$cook2][$dish2];
$new_score = $array[$cook1][$dish2] + $array[$cook2][$dish1];
if (($new_score - $old_score) > $best_change_weight) {
$best_change_weight = $new_score - $old_score;
$best_change = $dish2;
}
}
}
if ($best_change !== false) {
$cook2 = $picks[$best_change];
$picks[$dish1] = $cook2;
$picks[$dish2] = $cook1;
break;
}
}
} while ($best_change !== false);
I can't find a counter example to show that this doesn't work, but I'm suspicious of the case where
($array[$cook1][$dish1] + $array[$cook2][$dish2])
==
($array[$cook1][$dish2] + $array[$cook2][$dish1])
Maybe someone else will follow up with an answer to this "What if?"
Given this matrix, where the items in brackets are the "picks"
[a1] a2 a3
b1 [b2] b3
c1 c2 [c3]
If a1 + b2 == a2 + b1, then 'a' and 'b' will not switch dishes. The case I'm not 100% sure about is if there exists a matrix such that this is a better choice:
a1 [a2] a3
b1 b2 [b3]
[c1] c2 c3
Getting from the first state to the second requires two switches, the first of which seems arbitrary since it doesn't change the total. But, only by going through this arbitrary change can the last switch be made.
I tried to find an example 3x3 such that based on the "weighted preference" model I wrote about above, the first would be selected, but also such that the real optimum selection is given by the second. I wasn't able to find an example, but that doesn't mean that it doesn't exist. I don't feel like doing more matrix algebra right now, but maybe someone will pick up where I left off. Heck, maybe the case doesn't exist, but I thought I should point out the concern.
If it does work and you start with the correct pick, the above code will only loop through 64 times (8x8) for 8 cooks/dishes. If the pick is not correct and the first cook has a change, then it will go up to 72. If the 8th cook has a change, it's up to 128. It's possible that the do-while will loop several times, but I doubt it will get up near the CPU cycles required to sum all of the 40k combinations.
I may have a starting point for you with this algorithm that tries to choose cooks based on their ratio of max score to sum of scores (thus trying to choose chefs who are really good at one recipe but bad at the rest of the recipes to do that recipe)
$cooks = array(
array(1,2,3,4),
array(35,0,0,0),
array(36,33,1,1),
array(20,20,5,3)
);
$results = array();
while (count($cooks)) {
$curResult = array(
'cookId' => -1,
'recipe' => -1,
'score' => -1,
'ratio' => -1
);
foreach ($cooks as $cookId => $scores) {
$max = max($scores);
$ratio = $max / array_sum($scores);
if ($ratio > $curResult['ratio']) {
$curResult['cookId'] = $cookId;
$curResult['ratio'] = $ratio;
foreach ($scores as $recipe => $score) {
if ($score == $max) {
$curResult['recipe'] = $recipe;
$curResult['score'] = $score;
}
}
}
}
$results[$curResult['recipe']] = $curResult['score'];
unset($cooks[$curResult['cookId']]);
foreach ($cooks as &$cook) {
unset($cook[$curResult['recipe']]);
}
}
For the dataset provided, it does find what seems to be the optimum answer (35,33,5,4). However, it is still not perfect, for example, with the array:
$cooks = array(
array(1,2,3,4),
array(35,0,33,0),
array(36,33,1,1),
array(20,20,5,3)
);
The ideal answer would be (20,33,33,4), however this algorithm would return (35,33,5,4).
But since the question was asking for ideas of where to start, I guess this at least might suffice as something to start from :P
Try this
$mainArr = array(
array (1,2,3,4) ,
array (35,0,0,0) ,
array (36,33,1,1) ,
array (20,20,5,3)
);
$i = 0;
foreach( $mainArr as $subArray )
{
foreach( $subArray as $key => $value)
{
$newArr[$key][$i]=$value;
$i++;
}
}
$finalArr = array();
foreach( $newArr as $newSubArray )
{
$finalArr[] = max($newSubArray);
}
print_r( $finalArr );
OK here is a solution that allows you to find the best permutation of one cook to one recipe and no cook works twice and no recipe is made twice.
Thanks for the code to calculate perm of arrays goes to o'reilly...
http://docstore.mik.ua/orelly/webprog/pcook/ch04_26.htm
CONSIDERATIONS:
The number of cooks and the number of recipes are the same.
Going above a 5 by 5 matrix as here will get very big very fast. (see part 2 to be posted shortly)
The logic:
A permutation of an array assigns a place as well as just being included (ie what a combination does), so why not then assign each key of such an array to a recipe, the permutation guarantees no cook is repeated and the keys guarantee no recipe is repeated.
Please let me know if there are improvements or errors in my thinking or my code but here it is!
<?php
function pc_next_permutation($p, $size) {
//this is from http://docstore.mik.ua/orelly/webprog/pcook/ch04_26.htm
// slide down the array looking for where we're smaller than the next guy
for ($i = $size - 1; $p[$i] >= $p[$i+1]; --$i) { }
// if this doesn't occur, we've finished our permutations
// the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
if ($i == -1) { return false; }
// slide down the array looking for a bigger number than what we found before
for ($j = $size; $p[$j] <= $p[$i]; --$j) { }
// swap them
$tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp;
// now reverse the elements in between by swapping the ends
for (++$i, $j = $size; $i < $j; ++$i, --$j) {
$tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp;
}
return $p;
}
$cooks[441] = array(340=>5,342=>43,343=>50,344=>9,345=>0);
$cooks[442] = array(340=>5,342=>-33,343=>-30,344=>29,345=>0);
$cooks[443] = array(340=>5,342=>3,343=>0,344=>9,345=>10,);
$cooks[444] = array(340=>25,342=>23,343=>20,344=>19,345=>20,);
$cooks[445] = array(340=>27,342=>27,343=>26,344=>39,345=>50,);
//a consideration: this solution requires that the number of cooks equal the number of recipes
foreach ($cooks as $cooksCode => $cooksProfile){
$arrayOfCooks[]=$cooksCode;
$arrayOfRecipes = (array_keys($cooksProfile));
}
echo "<br/> here is the array of the different cooks<br/>";
print_r($arrayOfCooks);
echo "<br/> here is the array of the different recipes<br/>";
print_r($arrayOfRecipes);
$set = $arrayOfCooks;
$size = count($set) - 1;
$perm = range(0, $size);
$j = 0;
do {
foreach ($perm as $i) { $perms[$j][] = $set[$i]; }
} while ($perm = pc_next_permutation($perm, $size) and ++$j);
echo "<br/> here are all the permutations of the cooks<br/>";
print_r($perms);
$bestCombo = 0;
foreach($perms as $perm){
$thisScore =0;
foreach($perm as $key =>$cook){
$recipe= $arrayOfRecipes[$key];
$cookScore =$cooks[$cook][$recipe];
$thisScore = $thisScore+$cookScore;
}
if ($thisScore>$bestCombo){
$bestCombo=$thisScore;
$bestArray= $perm;
}
}
echo "<br/> here is the very best array<br/>";
print_r ($bestArray);
echo "<br/> best recipe assignment value is:".$bestCombo."<br/><br/>";
?>
I am working on a Web Application that includes long listings of names. The client originally wanted to have the names split up into divs by letter so it is easy to jump to a particular name on the list.
Now, looking at the list, the client pointed out several letters that have only one or two names associated with them. He now wants to know if we can combine several consecutive letters if there are only a few names in each.
(Note that letters with no names are not displayed at all.)
What I do right now is have the database server return a sorted list, then keep a variable containing the current character. I run through the list of names, incrementing the character and printing the opening and closing div and ul tags as I get to each letter. I know how to adapt this code to combine some letters, however, the one thing I'm not sure about how to handle is whether a particular combination of letters is the best-possible one. In other words, say that I have:
A - 12 names
B - 2 names
C - 1 name
D - 1 name
E - 1 name
F - 23 names
I know how to end up with a group A-C and then have D by itself. What I'm looking for is an efficient way to realize that A should be by itself and then B-D should be together.
I am not really sure where to start looking at this.
If it makes any difference, this code will be used in a Kohana Framework module.
UPDATE 2012-04-04:
Here is a clarification of what I need:
Say the minimum number of items I want in a group is 30. Now say that letter A has 25 items, letters B, C, and D, have 10 items each, and letter E has 32 items. I want to leave A alone because it will be better to combine B+C+D. The simple way to combine them is A+B, C+D+E - which is not what I want.
In other words, I need the best fit that comes closest to the minimum per group.
If a letter contains more than 10 names, or whatever reasonable limit you set, do not combine it with the next one. However, if you start combining letters, you might have it run until 15 or so names are collected if you want, as long as no individual letter has more than 10. That's not a universal solution, but it's how I'd solve it.
I came up with this function using PHP.
It groups letters that combined have over $ammount names in it.
function split_by_initials($names,$ammount,$tollerance = 0) {
$total = count($names);
foreach($names as $name) {
$filtered[$name[0]][] = $name;
}
$count = 0;
$key = '';
$temp = array();
foreach ($filtered as $initial => $split) {
$count += count($split);
$temp = array_merge($split,$temp);
$key .= $initial.'-';
if ($count >= $ammount || $count >= $ammount - $tollerance) {
$result[$key] = $temp;
$count = 0;
$key = '';
$temp = array();
}
}
return $result;
}
the 3rd parameter is used for when you want to limit the group to a single letter that doesn't have the ammount specified but is close enough.
Something like
i want to split in groups of 30
but a has 25
to so, if you set a tollerance of 5, A will be left alone and the other letters will be grouped.
I forgot to mention but it returns a multi dimensional array with the letters it contains as key then the names it contains.
Something like
Array
(
[A-B-C-] => Array
(
[0] => Bandice Bergen
[1] => Arey Lowell
[2] => Carmen Miranda
)
)
It is not exactly what you needed but i think it's close enough.
Using the jsfiddle that mrsherman put, I came up with something that could work: http://jsfiddle.net/F2Ahh/
Obviously that is to be used as a pseudocode, some techniques to make it more efficient could be applied. But that gets the job done.
Javascrip Version: enhanced version with sort and symbols grouping
function group_by_initials(names,ammount,tollerance) {
tolerance=tollerance||0;
total = names.length;
var filtered={}
var result={};
$.each(names,function(key,value){
val=value.trim();
var pattern = /[a-zA-Z0-9&_\.-]/
if(val[0].match(pattern)) {
intial=val[0];
}
else
{
intial='sym';
}
if(!(intial in filtered))
filtered[intial]=[];
filtered[intial].push(val);
})
var count = 0;
var key = '';
var temp = [];
$.each(Object.keys(filtered).sort(),function(ky,value){
count += filtered[value].length;
temp = temp.concat(filtered[value])
key += value+'-';
if (count >= ammount || count >= ammount - tollerance) {
key = key.substring(0, key.length - 1);
result[key] = temp;
count = 0;
key = '';
temp = [];
}
})
return result;
}
I already have the pieces mapped out in the array and it prints just fine
$board = array(
array('1','rs','1','rs','1','rs','1','rs'),
array('rs','1','rs','1','rs','1','rs','1'),
array('1','rs','1','rs','1','rs','1','rs'),
array('rs','bs','rs','bs','rs','bs','rs','bs'),
array('bs','rs','bs','rs','bs','rs','bs','rs'),
array('2','bs','2','bs','2','bs','2','bs'),
array('bs','2','bs','2','bs','2','bs','2'),
array('2','bs','2','bs','2','bs','2','bs')
);
1 = black pieces
2 = red pieces
rs = red square
bs = black square
this code parse the input of a player : example FROM F2 into (0,0) coordinates
function parseSquareFrom($square) {
if (strlen($square) != 2) {
return FALSE;
}
$coords = array(ord('A') - ord($square[0]), $square[1] - 1);
// Perform bounds-checking.
if ($coords[0] < 0 || $coords[0] > 7 || $coords[1] < 0 || $coords[1] > 7) {
return FALSE;
}
return $coords;
}
I have repeated the same function for the TO input ( to where the player wants to move the piece
my question is this next code a valid way to move with the functions above
$board[$coords1[0]-1][$coords1[1]+1] = $board[$coords[0]][$coords[1]];
$board[$coords[0]][$coords[1]] = 0;
//eating action
$board[$coords1[0]][$coords1[1]] = 0;
$board[$coords1[0]-2][$coords1[1]+2] = $board[$coords[0]][$coords[1]];
$way = ($_POST['way'] === 'up')? 1:-1;
$way = ($_POST['way'] === 'down')? -1:+1;
//if player is 'up' then the value of $way is 1 so
$board[$x+(-1*$way)][$y+(1*$way)] = $board[$coords[0]][$coords[1]]; // position 2,2 becomes 1,3
//if player is not 'up' then the value of $way is -1 so
$board[$x+(-1*$way)][$y+(1*$way)] = $board[$coords[0]][$coords[1]]; // position 2,2 becomes 3,1
I plan to have a function to update the tile movements in the screen so the piece moves as the it highlights the piece as is moving to the next square
this is using serialize into file to hold start positions, movements, kings and queens positions
I'm going to assume that $coords is a $_POST variable like a suggested on the first part of this question. If that's the case, the first part of your code is correct:
$board[$coords1[0]-1][$coords1[1]+1] = $board[$coords[0]][$coords[1]];
$board[$coords[0]][$coords[1]] = 0;
This moves a piece diagonally one step up-right in the board.
The second part on the other hand skips the actual 'eating' action. Unless, the location specified by the user is the one where the enemy piece is. In which case you code would work.
As for the bound-checking you do, you're not being bullet-proof since adding 2 in the eating move, while a piece is next to a border would result in your code trying to place the piece out of the board. So you could check bounds, depending on the move, so if the move is a standard diagonal one, you should check that the ending position is within the limits, not the starting one since you are assuming that piece is in a correct position before.
I think you have a few issues
$board[$coords1[0]-1][$coords1[1]+1] = $board[$coords[0]][$coords[1]];
if $coords[1] == 7 (which appears to be allowed) then $coords[1]+1 == 8 which is out of bounds. Let alone the coords +/- 2 we see later.
You probably don't want that 'bs'/'rs' in your storage, all of that can be inferred.
$getSquareColor = function($x,$y){return ($x+$y)/2 == 1 ? 'red' : 'black';} //or better yet, const or enum
function getSquareColor($x, $y) //alternatively
{
if ( ($x+$y) % 2 == 0) return 'black';
else return 'red';
}
ps - if you drop those dollar signs the function is just as good in javascript
I'm writing an algorithm in PHP to solve a given Sudoku puzzle. I've set up a somewhat object-oriented implementation with two classes: a Square class for each individual tile on the 9x9 board, and a Sudoku class, which has a matrix of Squares to represent the board.
The implementation of the algorithm I'm using is a sort of triple-tier approach. The first step, which will solve only the most basic puzzles (but is the most efficient), is to fill in any squares which can only take a single value based on the board's initial setup, and to adjust the constraints accordingly on the rest of the unsolved squares.
Usually, this process of "constant propagation" doesn't solve the board entirely, but it does solve a sizable chunk. The second tier will then kick in. This parses each unit (or 9 squares which must all have unique number assignments, e.g. a row or column) for the "possible" values of each unsolved square. This list of possible values is represented as a string in the Square class:
class Square {
private $name; // 00, 01, 02, ... , 86, 87, 88
private $peers; // All squares in same row, col, and box
private $number; // Assigned value (0 if not assigned)
private $possibles; // String of possible numbers (1-9)
public function __construct($name, $p = 0) {
$this->name = $name;
$this->setNumber($p);
if ($p == 0) {
$this->possibles = "123456789";
}
}
// ... other functions
Given a whole array of unsolved squares in a unit (as described in the second tier above), the second tier will concatenate all the strings of "possibles" into a single string. It will then search through that single string for any unique character values - values which do not repeat themselves. This will indicate that, within the unit of squares, there is only one square that can take on that particular value.
My question is: for implementing this second tier, how can I parse this string of all the possible values in a unit and easily detect the unique value(s)? I know I could create an array where each index is represented by the numbers 1-9, and I could increment the value at the corresponding index by 1 for each possible-value of that number that I find, then scan the array again for any values of 1, but this seems extremely inefficient, requiring two linear scans of an array for each unit, and in a Sudoku puzzle there are 27 units.
This is somewhat like what you have already ruled out as "extremely inefficient", but with builtin functions so it might be quite efficient:
$all_possibilities = "1234567891234";
$unique = array();
foreach (count_chars($all_possibilities, 1) as $c => $occurrences) {
if ($occurrences == 1)
$unique[] = chr($c);
}
print join("", $unique) . "\n";
Prints: "56789"
Consider using a binary number to represent your "possibles" instead, because binary operations like AND, OR, XOR tend to be much faster than string operations.
E.g. if "2" and "3" are possible for a square, use the binary number 000000110 to represent the possibilities for that square.
Here's how you could find uniques:
$seenonce = 0;
$seenmore = 0;
foreach(all_possibles_for_this_unit as $possibles) {
$seenmore |= ($possibles & $seenonce);
$seenonce |= $possibles;
}
$seenonce ^= $seenmore;
if ($seenonce) {
//something was seen once - now it must be located
}
I'm not sure if this method will actually work faster but it's worth looking into.
function singletonsInString($instring) {
$results = array();
for($i = 1; $i < 10; $i++) {
$first_pos = strpos($instring, str($i));
$last_pos = strrpos($instring, str($i));
if ( $first_pos !== FALSE and $first_pos == $last_pos )
$results[] = $i;
}
return $results;
}
That'll give you every singleton. Get the first and last positions of a number in that array, and if they match and aren't both FALSE (strict comparison in case there's a singleton right at the start) then there's only one such number in that array.
If you're super super worried about speed here, you can probably replace the interior of that loop with
$istr = str($i);
if ( ($first = strpos($instring, $istr)) !== FALSE
and $first == $strrpos($instring, $istr) ) $results[] = $i;
for a minimum number of computations. Well, assuming PHP's native strpos is the best way to go about these things, which as far as I know is not unreasonable.
The last time I fooled with Sudoku solving, I had a third class called "Run". A Run instance is created for each row, col and 3x3 square. Every square has three runs associated with it. The Run class contains the set of numbers not yet placed within the run. Solving the board then involves intersecting the sets at each square iteratively. This takes care of 80% of most medium boards and 60% of most hard boards. Once you've gone through the whole board with no changes, you can move on to higher level logic. Each time your higher level logic fills a square, you run through your squares again.
The nice thing about this setup is you can easily add variants to the solver. Say you use the variant where the two diagonals are also unique. You just add a 4th run to those 18 squares.
What I would do, is actually use binary bits for storing actual values as another answer suggested. That allows to do efficient checks and in general might lend Sudoku itself to more mathematical(=efficient and shorter) solution (just my impression, I have not researched this).
Basically, you represent the numbers in squares not with digits, but with powers of 2
"1" = 2^0 = 1 = 000000001
"2" = 2^1 = 2 = 000000010
"3" = 2^2 = 4 = 000000100
"4" = 2^3 = 8 = 000001000
... etc up to
"9" = 2^8 = 256= 100000000
this way, you can simply add contents' of single squares to find out what numbers are missing in a 3x3 or a row or any other subset of sudoku, like this:
// shows the possibles for 3x3 square number 1 (00-22)
$sum=0;
for ($i=0; $i< 3; $i++)
for ($j=0; $j < 3; $j++)
$sum += $square["${i}${j}"]->number
$possibles = $sum ^ 511 // ^ stands for bitwise XOR and 511 is binary 11111111
now the $possibles contains "1" in bit positions of digits that are possible in this square and you can do bitwise operations with the results for other squares to match them together, like this:
eg. let's say:
$possibles1 = 146 // is binary 100100101,
//indicating that this row or 3x3 square has place for "9", "6", "3" and "1"
$possibles2 = 7 // is binary 000000111, indicating it has place for "3", "2" and "1".
// so:
$possibles1 & $possibles2
// bitwise AND, will show binary 101 saying that "3" and "1" is unfilled in both bloces
$possibles1 | $possibles2
// bitwise OR will give that in total it is possible to use "9", "6", "3", "2" and "1" in those two squares together
Here is a way using only PHP built-in functions which should be pretty fast.
function getUniques($sNumbers)
{
return join(array_keys(array_count_values(str_split($sNumbers)),1));
}
echo getUniques("1234567891234"); // return 56789;