W4 2013: Fill & Download for Free

There is an indicator like a fuel meter in your console. Check it and please refer to car manual for more details about W4

Wow, this post is long!Here is the tl,dr; version:We start with the 21 candidates.We show a simple lemma that can give us some lower bound for the number of questions we need.We design a simple recursive algorithm to check all possible ways of asking the questions. We use the above lemma to prune almost all branches of the search.Within the simple algorithm, we use a solver of linear programs to check whether an answer is still feasible.The smallest number of questions is indeed 6.Jaimal Ichharam explains nicely how we can reduce the number of possible answers to 21. In fact, it is impossible to reduce it any further: for each of those 21 options it is possible to construct a set of 8 weights such that it is the (only) answer.Now, the number 21 can be used to prove that we need at least 4 questions in the worst case. This is not an optimal lower bound (as we shall see later), but the proof is very simple. The argument is the same information-theoretic one that can be used to prove that searching in a sorted array is [math]\Omega(\log n)[/math] and that comparison-based sort is [math]\Omega(n\log n)[/math]: in order to distinguish among many possibilities, you need enough information, and for that you need to ask enough questions.Lemma: Any deterministic algorithm that solves the "8 bags problem" and always makes at most [math]q[/math] questions can distinguish between at most [math]2^{q+1}-1[/math] different candidates.Proof: The number is the maximal number of different outputs our program can produce. And if there are more candidates than possible outputs, some candidate is never output, and therefore the program cannot be correct.Why is that the maximum number of possible outputs? We'll prove that by using induction. If [math]q=1[/math], we get to ask 1 question, and then we can give 3 different answers: one in the case the left side of the scale was lighter, one if it was heavier, and one if they were equal. (In the last case, the answer would be that this is the split we are looking for.)In the general case with [math]q[/math] questions, after the program asks the first question, there are three branches. If the two pans were equal we know the answer and terminate. In each of the other two cases we have [math]q-1[/math] questions left and thus (by the induction hypothesis) we can give at most [math]2^q - 1[/math] different outputs. Thus the total number of different outputs for a program that asks [math]q[/math] questions is at most [math]1+2(2^q-1)[/math], q.e.d.(Note that a very similar lemma can be used in any other problem where our questions can receive the answers <, =, >, and where getting the answer = uniquely determines the correct answer. Do you see the similarity to binary search?)Corollary: If we ask at most 3 questions, we can have at most 15 different outputs. Therefore we need at least 4 questions to distinguish among 21 candidates.Even though the lemma does not give a tight lower bound, it is still important: we will use it later to prune our search.Our ultimate goal will be to implement a brute force search that will consider all orders in which we can ask the questions. The schema of the algorithm is very simple:is_solvable(number_of_questions, past):  count the answers that are consistent with the past  if there is at most one, return True  if we are down to 0 questions left, return False  for each possible answer A:  p1 = past + [ when weighing A left pan is lighter ]  p2 = past + [ when weighing A left pan is heavier ]  if is_solvable(number_of_questions-1, p1)  and is_solvable(number_of_questions-1, p2): return True  return False That is it. In each possible situations, consider all possible (meaningful) questions you could ask next. The current situation is solvable in [math]q[/math] questions if and only if you can now ask a question such that regardless of the answer you get you can guarantee solving the new state in [math]q-1[/math] questions.One additional optimization of this algorithm is our use of the above lemma. Instead of "if we are down to 0 questions left (and have more than 1 candidate), return False" we can have the more general check "if the number of candidates is too large for the number of questions, return False".The only "building block" we now need is the check whether a given candidate is consistent with a given set of answers we already have. So let's take a look at that.What do we have at the beginning? Eight unknown weights: [math]0\leq w_1<w_2<\cdots<w_8[/math]. Now, each time we ask a question and don't find the correct split, we get a new information of the following form:[math][/math][math] w_1 [/math][math][/math][math]+ w_3 [/math][math][/math][math]+ w_4 [/math][math][/math][math]+ w_8 < w_2 [/math][math][/math][math]+ w_5 [/math][math][/math][math]+ w_6 [/math][math][/math][math]+ w_7[/math]Now, what does it mean when we say that an answer is still feasible? It means that the above system of inequalities is consistent with equality in that particular case.(Technical note: To be completely correct, our check for feasibility should find such values of weights that not only is there equality in our case, but at the same time there must be inequalities in all other cases. However, this is trivially true: whenever we have weights such that more than one split would balance the scales, we can apply some small random perturbation to the weights so that there would only be equality in the one case we want. Details are left as an exercise.)In order to do the above checks I use an LP solver (lp_solve in particular). Checking whether a system of inequalities and equalities has a solution is pretty close to linear programming. The only technicality we need to deal with are the sharp inequalities. Luckily, in our case we can simply turn them into less-than-or-equal inequalities by simply requiring the difference to be at least 1. (Note that the weights can always be scaled.)Here is a sample linear program:min: w8; w1+1 <= w2; w2+1 <= w3; w3+1 <= w4; w4+1 <= w5;  w5+1 <= w6; w6+1 <= w7; w7+1 <= w8; w1+w3+w6+w8+1 <= w2+w4+w5+w7; w1+w2+w3+w8+1 <= w4+w5+w6+w7; w1+w2+w4+w8 = w3+w5+w6+w7; Without the last constraint this program is feasible. One possible assignment of weights is (0,2,3,4,5,6,7,8). With the last constraint the program stops being feasible, so the corresponding way of splitting the bags cannot be the solution any more after those two weighings.(Note that I chose minimizing the maximum weight as the goal of the LP. This is pretty much arbitrary, any positive linear combination of our variables would do, as we are only interested in the existence of a solution.)Code posted here: http://ideone.com/xnivHM(Runs for about 3 minutes on my laptop, verifies that there is no solution with fewer than 6 questions in the worst case, and finds one with 6. Keeping the past in sorted order and adding memoization brings runtime down under a minute.)Some simple modifications of the code above can give us an optimal strategy as well. Here's one that I found by making some arbitrary choices:The first three questions will always be the same: 1567 vs. 2348, 1458 vs. 2367, and 1368 vs. 2457. If one of them is the solution, we are done. Otherwise, we get one of 8 possible combinations of < and >. Each of them will correspond to at most 7 candidates. These are the feasible answers at this point:<<<: 1468 1568 1278 1378 1478 1578 1678  <<>: 1268 1278  <><: 1278 1378  <>>: 1238 1248 1348 1258 1358 1268 1278  ><<: 1467 1468 1278 1378 1478  ><>: 1267 1367 1467 1268 1278  >><: 1456 1457 1467 1278 1378  >>>: 1467 1258 1358 1268 1278  If 1368 < 2457, compare 1378 vs. 2456 as the fourth question, otherwise compare 1268 vs. 3457 as the fourth question. In most cases you'll be left with at most 3 feasible solutions, and that is trivial to solve in two questions. There are only two cases that are harder:: <<<<: 1468 1568 1478 1578 1678 <>>>: 1238 1248 1348 1258 1358 Both of these can again be solved in two questions, an optimal strategy is left as an exercise.

Before talking about a business, I'd like to comment on your tax refund. The fact that you have a $7000 refund and the fact that you talk about it like it is a financial windfall indicates that you do not have your financial house in order. That $7000 is your money. In effect, you gave the government an interest free loan of $583 per month in 2013. You can avoid this by increasing the number of allowances on your W4 at your place of employment. Do a bit of research and some simple math to figure out how many allowances results in zero refund or to pay a small amount at the end of the year. Wouldn't you rather use that money yourself than letting the government have use of it for free?Now on the cleaning business. My first recommendation would be to get a normal job. Anyone with a brush and some soap can start a cleaning business. This means two things. The pay sucks and there is no stability of income. So, how do some survive and grow? They have a customer base and have built that customer base over years of time and they have more than $7000 in the bank. You just don't walk into the hotel and ask for the cleaning contract off the street. You have to have a solid reputation and track record of good service. Start with small expectations. See if you can get some clients through networking or on craigslist. Build the customer base. Save your money to help grow the business. Better still, keep looking for a job.