Sorry for the long title.
Wanted it to be as descriptive as possible.
Disclaimer : Could find some "find the differences" code here and elsewhere on Stackoverflow, but not quite the functionality I was looking for.
I'll be using these terminoligy later on:
'userguess' : a word that will be entered by the user
'solution' : the secret word that needs to be guessed.
A word guessing game where:
Pitfall Scenario : - Let's say the 'solution' is 'aabbc' and the user guesses 'abaac'
In the above scenario this would result in : (green)a(/green)(red)b(/red)(red)a(/red)(black)a(/black)(green)c(/green)
Notice how the last "a" is black cause 'userguess' has 3 a's but 'solution' only has 2
Code is working more or less, but I've got a feeling it can be 10 times more lean and mean.
I'm filling up 2 new Arrays (one for solution and one for userguess) as I go along to prevent the pitfall (see above) from messing things up.
function checkWord($toCheck) {
global $solution; // $solution word is defined outside the scope of this function
$goodpos = array(); // array that saves the indexes of all the right letters in the RIGHT position
$badpos = array(); // array that saves the indexes of all the right letters in the WRONG position
$newToCheck = array(); // array that changes overtime to help us with the Pitfall (see above)
$newSolution = array();// array that changes overtime to help us with the Pitfall (see above)
// check for all the right letters in the RIGHT position in entire string first
for ($i = 0, $j = strlen($toCheck); $i < $j; $i++) {
if ($toCheck[$i] == $solution[$i]) {
$goodpos[] = $i;
$newSolution[$i] = "*"; // RIGHT letters in RIGHT position are 'deleted' from solution
} else {
$newToCheck[] = $toCheck[$i];
$newSolution[$i] = $solution[$i];
}
}
// go over the NEW word to check for letters that are not in the right position but show up elsewhere in the word
for ($i = 0, $j = count($newSolution); $i <= $j; $i++) {
if (!(in_array($newToCheck[$i], $newSolution))) {
$badpos[] = $i;
$newSolution[$i] = "*";
}
}
// use the two helper arrays above 'goodpos' and 'badpos' to color the characters
for ($i = 0, $j = strlen($toCheck), $k = 0; $i < $j; $i++) {
if (in_array($i,$goodpos)) {
$colored .= "<span class='green'>";
$colored .= $toCheck[$i];
$colored .= "</span>";
} else if (in_array($i,$badpos)) {
$colored .= "<span class='red'>";
$colored .= $toCheck[$i];
$colored .= "</span>";
} else {
$colored .= $toCheck[$i];
}
}
// output to user
$output = '<div id="feedbackHash">';
$output .= '<h2>Solution was : ' . $solution . '</h2>';
$output .= '<h2>Color corrected: ' . $colored . '</h2>';
$output .= 'Correct letters in the right position : ' . count($goodpos) . '<br>';
$output .= 'Correct letters in the wrong position : ' . count($badpos) . '<br>';
$output .= '</div>';
return $output;
} // checkWord
Nice question. I'd probably do it slightly differently to you :) (I guess that's what you were hoping for!)
You can find my complete solution function here http://ideone.com/8ojAG - but I'm going to break it down step by step too.
Firstly, please try and avoid using global
. There's no reason why you can't define your function as:
function checkWord($toCheck, $solution) {
You can pass the solution in and avoid potential nasties later on.
I'd start by splitting both the user guess, and the solution into arrays, and have another array to store my output in.
$toCheck = str_split($toCheck, 1);
$solution = str_split($solution, 1);
$out = array();
At each stage of the process, I'd remove the characters that have been identified as correct or incorrect from the users guess or the solution, so I don't need to flag them in any way, and the remaining stages of the function run more efficiently.
So to check for matches.
foreach ($toCheck as $pos => $char) {
if ($char == $solution[$pos]) {
$out[$pos] = "<span class=\"green\">$char</span>";
unset($toCheck[$pos], $solution[$pos]);
}
}
So for your example guess/solution, $out
now contains a green 'a' at position 0, and a green c at position 4. Both the guess and the solution no longer have these indices, and will not be checked again.
A similar process for checking letters that are present, but in the wrong place.
foreach ($toCheck as $pos => $char) {
if (false !== $solPos = array_search($char, $solution)) {
$out[$pos] = "<span class=\"red\">$char</span>";
unset($toCheck[$pos], $solution[$solPos]);
}
}
In this case we are searching for the guessed letter in the solution, and removing it if it is found. We don't need to count the number of occurrences because the letters are removed as we go.
Finally the only letters remaining in the users guess, are ones that are not present at all in the solution, and since we maintained the numbered indices throughout, we can simply merge the leftover letters back in.
$out += $toCheck;
Almost there. $out
has everything we need, but it's not in the correct order. Even though the indices are numeric, they are not ordered. We finish up with:
ksort($out);
return implode($out);
The result from this is:
"<span class="green">a</span><span class="red">b</span><span class="red">a</span>a<span class="green">c</span>"
Here try this, See In Action
Example output:
<?php
echo checkWord('aabbc','abaac').PHP_EOL;
echo checkWord('funday','sunday').PHP_EOL;
echo checkWord('flipper','ripple').PHP_EOL;
echo checkWord('monkey','kenney').PHP_EOL;
function checkWord($guess, $solution){
$arr1 = str_split($solution);
$arr2 = str_split($guess);
$arr1_c = array_count_values($arr1);
$arr2_c = array_count_values($arr2);
$out = '';
foreach($arr2 as $key=>$value){
$arr1_c[$value]=(isset($arr1_c[$value])?$arr1_c[$value]-1:0);
$arr2_c[$value]=(isset($arr2_c[$value])?$arr2_c[$value]-1:0);
if(isset($arr2[$key]) && isset($arr1[$key]) && $arr1[$key] == $arr2[$key]){
$out .='<span style="color:green;">'.$arr2[$key].'</span>';
}elseif(in_array($value,$arr1) && $arr2_c[$value] >= 0 && $arr1_c[$value] >= 0){
$out .='<span style="color:red;">'.$arr2[$key].'</span>';
}else{
$out .='<span style="color:black;">'.$arr2[$key].'</span>';
}
}
return $out;
}
?>
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With