I have spent time working on almost every impossible problem there is.

This is a lie, but it is rooted in a kernel of truth. I remember my very first impossible problem. When I was eight or nine, my dad gave me a word problem about firemen and houses or some bullshit, but what it boiled down to was:

Draw three points wherever you want on a paper. Label them A, B, C.

Draw three other points, also wherever you want. Label them 1, 2, 3.

Try to draw lines so that A, B, and C each have three lines going out to 1, 2, and 3—and have none of them cross.

And I spent days drawing nonsense like that (thanks Google Images), even after he tried to convey with a straightforward topological proof why it was impossible. But I wasn't convinced until I tried and tried and tried. It was especially painful because as it turns out, you can get all but one last little line to connect.

I'm older now, and have some modest understanding of math, so one might think I no longer fall for traps like this. Not so. I still have to try. But at least now I know on a rational level that it's a lost cause.

The most recent example is a couple of nights ago when I was reading about the subset sum problem. It's about as easy to explain as the problem above; in layman's terms, it's "Figure out how to quickly look through a bunch of positive and negative numbers and see if you can pick some out that add up to 0."

In a slightly more rigorous sense, it's a problem known to be NP-complete (you would think that might tip me off). Given an arbitrary set of integers, determine whether or not there is any subset thereof that sums to N, where N is some given integer (but classically 0.) The trick, of course, is that you have to do it in polynomial time; the most efficient algorithm offered on Wikipedia requires O(2^(N//2)).

After thinking on it, my essentially random approach was to mod everything. The logic went like this.

You have N numbers. A naive approach would be to check the sum of every possible combination of your numbers, which is a combinatoric nightmare on the order of O(2^N * N). No good.

Well, if we're shooting for 0, then we can divide our set into positives and negatives. Surely this is an improvement; now instead of dealing with the power set of N—that is, the number of subsets in a set of size N, which is 2^N—we've made some progress, since we won't waste time trying sets that are all positive numbers, for example. But it turns out this by itself buys us very little.

Then I go, hey, what if you take every integer modulo 2? Or, in other words, go through and mark down which numbers are even and which are odd.

Why? Because we know that our potential sum is going to be even (i.e. 0). We already knew that we'd have to take one or more numbers from each the positives and the negatives. Put those truths together, and it means we know that any group that might work will have a certain property: if the sum of our positive numbers is odd, then the sum of our negatives will have to be odd too, or else the sum of both together would be odd, which would mean not zero. Likewise, if our positives sum is even, the negs have to be even too. Or,

 \forall_P {P \subseteq S, P_i \geq 0}, \sum\P \equiv x \ ("mod"\ 2) \leftrightarrow \forall_Q {Q \subseteq S, Q_i < 0}, \sum\Q \equiv x \ ("mod"\ 2) .

This is surely good, since on average, half of all possible combinations won't meet that constraint and thus can be discarded out of hand.

But wait, it gets better. Why not take everything modulo 3? The same principle applies, which is that our overall sum must equal 0 \ ("mod"\ 3). So if the sum of the positives ("mod"\ 3) is 1, we know we only have to check those groups of negatives which sum to 2 \ ("mod"\ 3). And now it seems like we're getting a threefold rather than twofold improvement on our deal.

But wait, it gets better. Why not throw everything so far into a big pot and see what happens? It's not too hard to show that if you take mod 2, note which possible combinations remain, and then do the same thing with mod 3, you'll generally multiplicatively reduce your potential viable combinations. Then continue this with a handful of ascending integers (maybe primes?) and with almost no work at all, you'll all but perfectly sieve out any invalid combinations.

And so far as I could tell, this is all true. The insidious problem, and this is where I stopped working on this, is that the long cold years have taught me that things are never easy, particularly extremely difficult mathematical things. I have absolutely no doubt that if I sat down and did the math carefully, I would find that I was using up at least as many operations in logistical overhead as I was saving in the sieving process.

That's how these deceptive problems work. You feel like you're making progress, but all you've really done is paint a horse like a zebra; the nature of the thing itself remains unchanged.

Primes are especially brutal about that. Watch out for primes.