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Can't call it a problem on Stack Overflow apparently, however I am currently trying to understand how to integrate constraints in the form of item groups within the Knapsack problem. My math skills are proving to be fairly limiting in this situation, however I am very motivated to both make this work as intended as well as figure out what each aspect does (in that order since things make more sense when they work).
With that said, I have found an absolutely beautiful implementation at Rosetta Code and cleaned up the variable names some to help myself better understand this from a very basic perspective.
Unfortunately I am having an incredibly difficult time figuring out how I can apply this logic to include item groups. My purpose is for building fantasy teams, supplying my own value & weight (points/salary) per player but without groups (positions in my case) I am unable to do so.
Would anyone be able to point me in the right direction for this? I'm reviewing code examples from other languages and additional descriptions of the problem as a whole, however I would like to get the groups implemented by whatever means possible.
<?php
function knapSolveFast2($itemWeight, $itemValue, $i, $availWeight, &$memoItems, &$pickedItems)
{
global $numcalls;
$numcalls++;
// Return memo if we have one
if (isset($memoItems[$i][$availWeight]))
{
return array( $memoItems[$i][$availWeight], $memoItems['picked'][$i][$availWeight] );
}
else
{
// At end of decision branch
if ($i == 0)
{
if ($itemWeight[$i] <= $availWeight)
{ // Will this item fit?
$memoItems[$i][$availWeight] = $itemValue[$i]; // Memo this item
$memoItems['picked'][$i][$availWeight] = array($i); // and the picked item
return array($itemValue[$i],array($i)); // Return the value of this item and add it to the picked list
}
else
{
// Won't fit
$memoItems[$i][$availWeight] = 0; // Memo zero
$memoItems['picked'][$i][$availWeight] = array(); // and a blank array entry...
return array(0,array()); // Return nothing
}
}
// Not at end of decision branch..
// Get the result of the next branch (without this one)
list ($without_i,$without_PI) = knapSolveFast2($itemWeight, $itemValue, $i-1, $availWeight,$memoItems,$pickedItems);
if ($itemWeight[$i] > $availWeight)
{ // Does it return too many?
$memoItems[$i][$availWeight] = $without_i; // Memo without including this one
$memoItems['picked'][$i][$availWeight] = array(); // and a blank array entry...
return array($without_i,array()); // and return it
}
else
{
// Get the result of the next branch (WITH this one picked, so available weight is reduced)
list ($with_i,$with_PI) = knapSolveFast2($itemWeight, $itemValue, ($i-1), ($availWeight - $itemWeight[$i]),$memoItems,$pickedItems);
$with_i += $itemValue[$i]; // ..and add the value of this one..
// Get the greater of WITH or WITHOUT
if ($with_i > $without_i)
{
$res = $with_i;
$picked = $with_PI;
array_push($picked,$i);
}
else
{
$res = $without_i;
$picked = $without_PI;
}
$memoItems[$i][$availWeight] = $res; // Store it in the memo
$memoItems['picked'][$i][$availWeight] = $picked; // and store the picked item
return array ($res,$picked); // and then return it
}
}
}
$items = array("map","compass","water","sandwich","glucose","tin","banana","apple","cheese","beer","suntan cream","camera","t-shirt","trousers","umbrella","waterproof trousers","waterproof overclothes","note-case","sunglasses","towel","socks","book");
$weight = array(9,13,153,50,15,68,27,39,23,52,11,32,24,48,73,42,43,22,7,18,4,30);
$value = array(150,35,200,160,60,45,60,40,30,10,70,30,15,10,40,70,75,80,20,12,50,10);
## Initialize
$numcalls = 0;
$memoItems = array();
$selectedItems = array();
## Solve
list ($m4, $selectedItems) = knapSolveFast2($weight, $value, sizeof($value)-1, 400, $memoItems, $selectedItems);
# Display Result
echo "<b>Items:</b><br>" . join(", ", $items) . "<br>";
echo "<b>Max Value Found:</b><br>$m4 (in $numcalls calls)<br>";
echo "<b>Array Indices:</b><br>". join(",", $selectedItems) . "<br>";
echo "<b>Chosen Items:</b><br>";
echo "<table border cellspacing=0>";
echo "<tr><td>Item</td><td>Value</td><td>Weight</td></tr>";
$totalValue = 0;
$totalWeight = 0;
foreach($selectedItems as $key)
{
$totalValue += $value[$key];
$totalWeight += $weight[$key];
echo "<tr><td>" . $items[$key] . "</td><td>" . $value[$key] . "</td><td>".$weight[$key] . "</td></tr>";
}
echo "<tr><td align=right><b>Totals</b></td><td>$totalValue</td><td>$totalWeight</td></tr>";
echo "</table><hr>";
?>
That knapsack program is traditional, but I think that it obscures what's going on. Let me show you how the DP can be derived more straightforwardly from a brute force solution.
In Python (sorry; this is my scripting language of choice), a brute force solution could look like this. First, there's a function for generating all subsets with breadth-first search (this is important).
def all_subsets(S): # brute force
subsets_so_far = [()]
for x in S:
new_subsets = [subset + (x,) for subset in subsets_so_far]
subsets_so_far.extend(new_subsets)
return subsets_so_far
Then there's a function that returns True if the solution is valid (within budget and with a proper position breakdown) – call it is_valid_solution – and a function that, given a solution, returns the total player value (total_player_value). Assuming that players is the list of available players, the optimal solution is this.
max(filter(is_valid_solution, all_subsets(players)), key=total_player_value)
Now, for a DP, we add a function cull to all_subsets.
def best_subsets(S): # DP
subsets_so_far = [()]
for x in S:
new_subsets = [subset + (x,) for subset in subsets_so_far]
subsets_so_far.extend(new_subsets)
subsets_so_far = cull(subsets_so_far) ### This is new.
return subsets_so_far
What cull does is to throw away the partial solutions that are clearly not going to be missed in our search for an optimal solution. If the partial solution is already over budget, or if it already has too many players at one position, then it can safely be discarded. Let is_valid_partial_solution be a function that tests these conditions (it probably looks a lot like is_valid_solution). So far we have this.
def cull(subsets): # INCOMPLETE!
return filter(is_valid_partial_solution, subsets)
The other important test is that some partial solutions are just better than others. If two partial solutions have the same position breakdown (e.g., two forwards and a center) and cost the same, then we only need to keep the more valuable one. Let cost_and_position_breakdown take a solution and produce a string that encodes the specified attributes.
def cull(subsets):
best_subset = {} # empty dictionary/map
for subset in filter(is_valid_partial_solution, subsets):
key = cost_and_position_breakdown(subset)
if (key not in best_subset or
total_value(subset) > total_value(best_subset[key])):
best_subset[key] = subset
return best_subset.values()
That's it. There's a lot of optimization to be done here (e.g., throw away partial solutions for which there's a cheaper and more valuable partial solution; modify the data structures so that we aren't always computing the value and position breakdown from scratch and to reduce the storage costs), but it can be tackled incrementally.
One potential small advantage with regard to composing recursive functions in PHP is that variables are passed by value (meaning a copy is made) rather than reference, which can save a step or two.
Perhaps you could better clarify what you are looking for by including a sample input and output. Here's an example that makes combinations from given groups - I'm not sure if that's your intention... I made the section accessing the partial result allow combinations with less value to be considered if their weight is lower - all of this can be changed to prune in the specific ways you would like.
function make_teams($players, $position_limits, $weights, $values, $max_weight){
$player_counts = array_map(function($x){
return count($x);
}, $players);
$positions = array_map(function($x){
$positions[] = [];
},$position_limits);
$num_positions = count($positions);
$combinations = [];
$hash = [];
$stack = [[$positions,0,0,0,0,0]];
while (!empty($stack)){
$params = array_pop($stack);
$positions = $params[0];
$i = $params[1];
$j = $params[2];
$len = $params[3];
$weight = $params[4];
$value = $params[5];
// too heavy
if ($weight > $max_weight){
continue;
// the variable, $positions, is accumulating so you can access the partial result
} else if ($j == 0 && $i > 0){
// remember weight and value after each position is chosen
if (!isset($hash[$i])){
$hash[$i] = [$weight,$value];
// end thread if current value is lower for similar weight
} else if ($weight >= $hash[$i][0] && $value < $hash[$i][1]){
continue;
// remember better weight and value
} else if ($weight <= $hash[$i][0] && $value > $hash[$i][1]){
$hash[$i] = [$weight,$value];
}
}
// all positions have been filled
if ($i == $num_positions){
$positions[] = $weight;
$positions[] = $value;
if (!empty($combinations)){
$last = &$combinations[count($combinations) - 1];
if ($weight < $last[$num_positions] && $value > $last[$num_positions + 1]){
$last = $positions;
} else {
$combinations[] = $positions;
}
} else {
$combinations[] = $positions;
}
// current position is filled
} else if (count($positions[$i]) == $position_limits[$i]){
$stack[] = [$positions,$i + 1,0,$len,$weight,$value];
// otherwise create two new threads: one with player $j added to
// position $i, the other thread skipping player $j
} else {
if ($j < $player_counts[$i] - 1){
$stack[] = [$positions,$i,$j + 1,$len,$weight,$value];
}
if ($j < $player_counts[$i]){
$positions[$i][] = $players[$i][$j];
$stack[] = [$positions,$i,$j + 1,$len + 1
,$weight + $weights[$i][$j],$value + $values[$i][$j]];
}
}
}
return $combinations;
}
Output:
$players = [[1,2],[3,4,5],[6,7]];
$position_limits = [1,2,1];
$weights = [[2000000,1000000],[10000000,1000500,12000000],[5000000,1234567]];
$values = [[33,5],[78,23,10],[11,101]];
$max_weight = 20000000;
echo json_encode(make_teams($players, $position_limits, $weights, $values, $max_weight));
/*
[[[1],[3,4],[7],14235067,235],[[2],[3,4],[7],13235067,207]]
*/
I have this array which links numbers to letters at the moment like this:
1-26 = A-Z
But there is more, 27=AA and 28=AB etc...
so basically when I do this:
var_dump($array[2]); //shows B
var_dump($array[29]); //shows AC
Now this array I made myself but it's becoming way too long. Is there a way to actually get this going on till lets say 32? I know there is chr but I dont think I can use this.
Is there an easier way to actually get this without using this way too long of an array?
It's slower calculating it this way, but you can take advantage of the fact that PHP lets you increment letters in the same way as numbers, Perl style:
function excelColumnRange($number) {
$character = 'A';
while ($number > 1) {
++$character;
--$number;
}
return $character;
}
var_dump(excelColumnRange(2));
var_dump(excelColumnRange(29));
here is the code which you are looking for :
<?php
$start = "A";
$max = 50;
$result = array();
for($i=1; $i<=$max; $i++) {
$result[$i] = $start++;
}
print_r($result);
?>
Ref: http://www.xpertdeveloper.com/2011/01/php-strings-unusual-behaviour/
This should work for you:
Even without any loops. First I calculate how many times the alphabet (26) goes into the number. With this I define how many times it has to str_repleat() A. Then I simply subtract this number and calculate the number in the alphabet with the number which is left.
<?php
function numberToLetter($number) {
$fullSets = (($num = floor(($number-1) / 26)) < 0 ? 0 : $num);
return str_repeat("A", $fullSets) . (($v = ($number-$fullSets*26)) > 0 ? chr($v+64) : "");
}
echo numberToLetter(53);
?>
output:
AAA
Pattern search within a string.
for eg.
$string = "111111110000";
FindOut($string);
Function should return 0
function FindOut($str){
$items = str_split($str, 3);
print_r($items);
}
If I understand you correctly, your problem comes down to finding out whether a substring of 3 characters occurs in a string twice without overlapping. This will get you the first occurence's position if it does:
function findPattern($string, $minlen=3) {
$max = strlen($string)-$minlen;
for($i=0;$i<=$max;$i++) {
$pattern = substr($string,$i,$minlen);
if(substr_count($string,$pattern)>1)
return $i;
}
return false;
}
Or am I missing something here?
What you have here can conceptually be solved with a sliding window. For your example, you have a sliding window of size 3.
For each character in the string, you take the substring of the current character and the next two characters as the current pattern. You then slide the window up one position, and check if the remainder of the string has what the current pattern contains. If it does, you return the current index. If not, you repeat.
Example:
1010101101
|-|
So, pattern = 101. Now, we advance the sliding window by one character:
1010101101
|-|
And see if the rest of the string has 101, checking every combination of 3 characters.
Conceptually, this should be all you need to solve this problem.
Edit: I really don't like when people just ask for code, but since this seemed to be an interesting problem, here is my implementation of the above algorithm, which allows for the window size to vary (instead of being fixed at 3, the function is only briefly tested and omits obvious error checking):
function findPattern( $str, $window_size = 3) {
// Start the index at 0 (beginning of the string)
$i = 0;
// while( (the current pattern in the window) is not empty / false)
while( ($current_pattern = substr( $str, $i, $window_size)) != false) {
$possible_matches = array();
// Get the combination of all possible matches from the remainder of the string
for( $j = 0; $j < $window_size; $j++) {
$possible_matches = array_merge( $possible_matches, str_split( substr( $str, $i + 1 + $j), $window_size));
}
// If the current pattern is in the possible matches, we found a duplicate, return the index of the first occurrence
if( in_array( $current_pattern, $possible_matches)) {
return $i;
}
// Otherwise, increment $i and grab a new window
$i++;
}
// No duplicates were found, return -1
return -1;
}
It should be noted that this certainly isn't the most efficient algorithm or implementation, but it should help clarify the problem and give a straightforward example on how to solve it.
Looks like you more want to use a sub-string function to walk along and check every three characters and not just break it into 3
function fp($s, $len = 3){
$max = strlen($s) - $len; //borrowed from lafor as it was a terrible oversight by me
$parts = array();
for($i=0; $i < $max; $i++){
$three = substr($s, $i, $len);
if(array_key_exists("$three",$parts)){
return $parts["$three"];
//if we've already seen it before then this is the first duplicate, we can return it
}
else{
$parts["$three"] = i; //save the index of the starting position.
}
}
return false; //if we get this far then we didn't find any duplicate strings
}
Based on the str_split documentation, calling str_split on "1010101101" will result in:
Array(
[0] => 101
[1] => 010
[2] => 110
[3] => 1
}
None of these will match each other.
You need to look at each 3-long slice of the string (starting at index 0, then index 1, and so on).
I suggest looking at substr, which you can use like this:
substr($input_string, $index, $length)
And it will get you the section of $input_string starting at $index of length $length.
quick and dirty implementation of such pattern search:
function findPattern($string){
$matches = 0;
$substrStart = 0;
while($matches < 2 && $substrStart+ 3 < strlen($string) && $pattern = substr($string, $substrStart++, 3)){
$matches = substr_count($string,$pattern);
}
if($matches < 2){
return null;
}
return $substrStart-1;
I am trying to to mark some trends, so I have 1 as the lowest and 5 as the biggest value.
So for example,
I may have the following case:
5,4,5,5 (UP)
3,4, (UP)
4,3,3 (DOWN)
4,4,4,4, (FLAT - this is OK for all same numbers)
I am planning to have unlimited number of ordered values as input, an as an output I will just show an (UP), (DOWN), or (FLAT) image.
Any ideas on how I can achieve this?
Sorry if I am not descriptive enough.
Thank you all for you time.
Use least square fit to calculate the "slope" of the values.
function leastSquareFit(array $values) {
$x_sum = array_sum(array_keys($values));
$y_sum = array_sum($values);
$meanX = $x_sum / count($values);
$meanY = $y_sum / count($values);
// calculate sums
$mBase = $mDivisor = 0.0;
foreach($values as $i => $value) {
$mBase += ($i - $meanX) * ($value - $meanY);
$mDivisor += ($i - $meanX) * ($i - $meanX);
}
// calculate slope
$slope = $mBase / $mDivisor;
return $slope;
} // function leastSquareFit()
$trend = leastSquareFit(array(5,4,5,5));
(Untested)
If the slope is positive, the trend is upwards; if negative, it's downwards. Use your own judgement to decide what margin (positive or negative) is considered flat.
A little bit hard to answer based on the limited info you provide, but assuming that:
if there's no movement at all the trend is FLAT,
otherwise, the trend is the last direction of movement,
then this code should work:
$input = array();
$previousValue = false;
$trend = 'FLAT';
foreach( $input as $currentValue ) {
if( $previousValue !== false ) {
if( $currentValue > $previousValue ) {
$trend = 'UP';
} elseif( $currentValue < $previousValue ) {
$trend = 'DOWN';
}
}
$previousValue = $currentValue;
}
For your examples :
Calculate longest increasing subsequence, A
Calulate longest decreasing subsequence , B
Going by your logic, if length of A is larger than B , its an UP , else DOWN.
You will also need to keep track of all equals using one boolean variable to mark FLAT trend.
Query :
What trend would be :
3,4,5,4,3 ?
3,4,4,4,3 ?
1,2,3,4,4,3,2,2,1 ?
Then the logic might need some alterations depending upon what your requirements are .
I'm not sure if i understand your problem totally but I would put the values in an array and use a code like this (written in pseudocode):
int i = 0;
String trend = "FLAT":
while(i<length(array)) {
if(array(i)<array(i+1)) {
trend = "UP";
}
else if(array(i)>array(i+1) {
trend = "DOWN";
}
i++;
}
EDIT: this would obviously only display the trend of the latest alteration
one would also may count the number of times the trend is up or down and determine the overall trend by that values
echo foo(array(5,4,5,5)); // UP
echo foo(array(3,4)); // UP
echo foo(array(4,3,3)); // DOWN
echo foo(array(4,4,4,4)); // FLAT
function foo($seq)
{
if (count(array_unique($seq)) === 1)
return 'FLAT';
$trend = NULL;
$count = count($seq);
$prev = $seq[0];
for ($i = 1; $i < $count; $i++)
{
if ($prev < $seq[$i])
{
$trend = 'UP';
}
if ($prev > $seq[$i])
{
$trend = 'DOWN';
}
$prev = $seq[$i];
}
return $trend;
}
I used the code from #liquorvicar to determine Google search page rank trends, but added some extra trend values to make it more accurate:
nochange - no change
better (higher google position = lower number)
worse (lower google position = higher number)
I also added extra checks when the last value had no change, but taking in account the previous changes i.e.
worsenochange (no change, previouse was worse - lower number)
betternochange (no change, previouse was better - lower number)
I used these values to display a range of trend icons:
$_trendIndicator="<img title="trend" width="16" src="/include/main/images/trend-'. $this->getTrend($_positions). '-icon.png">";
private function getTrend($_positions)
{
// calculate trend based on last value
//
$_previousValue = false;
$_trend = 'nochange';
foreach( $_positions as $_currentValue ) {
if( $_previousValue !== false ) {
if( $_currentValue > $_previousValue ) {
$_trend = 'better';
} elseif( $_currentValue < $_previousValue ) {
$_trend = 'worse';
}
if ($_trend==='worse' && ($_previousValue == $_currentValue)) {$_trend = 'worsenochange';}
if ($_trend==='better' && ($_previousValue == $_currentValue)) {$_trend = 'betternochange';}
}
$_previousValue = $_currentValue;
}
return $_trend;
}
There's a blog post comment on codinghorror.com by Paul Jungwirth which includes a little programming task:
You have the numbers 123456789, in that order. Between each number, you must insert either nothing, a plus sign, or a multiplication sign, so that the resulting expression equals 2001. Write a program that prints all solutions. (There are two.)
Bored, I thought, I'd have a go, but I'll be damned if I can get a result for 2001. I think the code below is sound and I reckon that there are zero solutions that result in 2001. According to my code, there are two solutions for 2002. Am I right or am I wrong?
/**
* Take the numbers 123456789 and form expressions by inserting one of ''
* (empty string), '+' or '*' between each number.
* Find (2) solutions such that the expression evaluates to the number 2001
*/
$input = array(1,2,3,4,5,6,7,8,9);
// an array of strings representing 8 digit, base 3 numbers
$ops = array();
$numOps = sizeof($input)-1; // always 8
$mask = str_repeat('0', $numOps); // mask of 8 zeros for padding
// generate the ops array
$limit = pow(3, $numOps) -1;
for ($i = 0; $i <= $limit; $i++) {
$s = (string) $i;
$s = base_convert($s, 10, 3);
$ops[] = substr($mask, 0, $numOps - strlen($s)) . $s;
}
// for each element in the ops array, generate an expression by inserting
// '', '*' or '+' between the numbers in $input. e.g. element 11111111 will
// result in 1+2+3+4+5+6+7+8+9
$limit = sizeof($ops);
$stringResult = null;
$numericResult = null;
for ($i = 0; $i < $limit; $i++) {
$l = $numOps;
$stringResult = '';
$numericResult = 0;
for ($j = 0; $j <= $l; $j++) {
$stringResult .= (string) $input[$j];
switch (substr($ops[$i], $j, 1)) {
case '0':
break;
case '1':
$stringResult .= '+';
break;
case '2':
$stringResult .= '*';
break;
default :
}
}
// evaluate the expression
// split the expression into smaller ones to be added together
$temp = explode('+', $stringResult);
$additionElems = array();
foreach ($temp as $subExpressions)
{
// split each of those into ones to be multiplied together
$multplicationElems = explode('*', $subExpressions);
$working = 1;
foreach ($multplicationElems as $operand) {
$working *= $operand;
}
$additionElems[] = $working;
}
$numericResult = 0;
foreach($additionElems as $operand)
{
$numericResult += $operand;
}
if ($numericResult == 2001) {
echo "{$stringResult}\n";
}
}
Further down the same page you linked to.... =)
"Paul Jungwirth wrote:
You have the numbers 123456789, in
that order. Between each number, you
must insert either nothing, a plus
sign, or a multiplication sign, so
that the resulting expression equals
2001. Write a program that prints all solutions. (There are two.)
I think you meant 2002, not 2001. :)
(Just correcting for anyone else like
me who obsessively tries to solve
little "practice" problems like this
one, and then hit Google when their
result doesn't match the stated
answer. ;) Damn, some of those Perl
examples are ugly.)"
The number is 2002.
Recursive solution takes eleven lines of JavaScript (excluding string expression evaluation, which is a standard JavaScript function, however it would probably take another ten or so lines of code to roll your own for this specific scenario):
function combine (digit,exp) {
if (digit > 9) {
if (eval(exp) == 2002) alert(exp+'=2002');
return;
}
combine(digit+1,exp+'+'+digit);
combine(digit+1,exp+'*'+digit);
combine(digit+1,exp+digit);
return;
}
combine(2,'1');