Wrongly asked or am I stupid?

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; $i <= 开发者_JAVA百科$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');