bonanova

Previous post can be touched up a bit to form a proof. Binary representation of the reals can be thought of as selecting subsets of a countably infinite sequence of binary digits: where the"1"'s are the selectors. Cantor's diagonal proof assumes the reals [subsets] are countable and associates integers with the [assumed complete] list. Cantor's diagonals [2's complemented] are seen not to be in the list, so the list is not complete. The assumption of countability is thus negated. Briefly, countable infinite sets have cardinality Aleph0. The subsets of a countably infinite set [in this case the reals] have cardinality 2Aleph0 = Aleph1 and are not countable. e.g. there are 23 subsets of a set of 3 objects. In my distinction of the countability of the rationals and reals I showed a 1-1 correspondence of the rationals [having digits that terminate] with the reals that start with the binary representation of a rational and append Aleph0 binary digits that are freely specified as 0 or 1. Appending those binary digits defines a set of reals with cardinality 2Aleph0 = Aleph1. Thus each rational can be paired with an uncountable set of reals. The set of all reals is thus uncountable.

Both correct.

I was distracted by personal business last week and worked on this only through the first weighing. Agree the first weighing has to be 4 vs 4 -- each case, < = > gives 26 cases. I could not get beyond that, however. Particularly having two possible locations for the heavy balls after the first weighing, and then to limit the cases for each outcome to 9 and 3 for the next two. I'll look over final's approach, but it seems he's got it. Great puzzle and solution both! I also wondered about the easier cases of 10, 11 and 12 balls. First thought is that for 10 balls, your first weighing is 3 vs 3.

OK. The question is about topology, not boundaries. Think of a normal table in a normal room. But to make it somewhat rigorous ... assume that The floor is infinite in extent.The surface is continuous and differentiable.

Yeah, there wasn't a smiley for "tongue in cheek"

A nice counterintuitive puzzle asked the ratio of boys to girls in a village where the rule was to stop having children after the first son was born. If you haven't solved it yet, you might want to give it a try. Here's a follow-on. If all the families obeyed this rule, and continued having children until a son was born, then stopped, what is the average size of a family?

Kudos. That's correct. The process of changing p = abcd into s = uvwx has an interesting property. Start with any p [where abcd are not all the same] to get s = uvwx. Continue the process, this time using s. Repeated application of the process always leads to a final result of 6174. Which gives rise to two questions: How many repetitions are sufficient? What do the initials D.R.K. have to do with any of this?

An old favorite asks whether a termite can bore his way from a corner cell of a 3x3x3 Rubik's Cube [assumed to be wooden] to the center, passing once through the center of every cell. His path is always parallel to one of the cube edges. Forget for the sake of the puzzle that a true Rubik's Cube has no center cell. You may know the answer to this, or want to figure it out. But let's generalize to other orders of cubes and ask: For an nxnxn cube, for what values of n is it possible for a termite, starting from a corner, and passing once through each of the cells, either to end his journey at the center cell, or come back to his starting point? Enjoy.

mojail [with glib] and tpaxatb have it. Nice going. sands: we use a balance scale, comparing two balls or groups of balls. decal-last: How do you distinguish HHH from HHL in the heavy pan? overide and dvalDragon: close. [dD very close.] Gaspar Zoltan [gotta love that handle] rethink it.

DeeGee, who I think miscounted his own solution, rohit_bd and CaptainEd all have it. Kudos! BTW there is a formula that predicts the number of races needed for up to 11 horses. If you want to waste some more time, you can analyze these cases and look at the answers below. The formula breaks down for 12 horses, predicting 29 races are needed. But this is wrong: clearly we need 30 races for 12 horses.

One vote against being able to place the table so it does not wobble. Comments added in spoiler.

One vote in favor of being able to adjust the table so it does not wobble. Comments added in the spoiler.

One of the site's classic weighing problems has three pairs of balls colored red white and blue. In each pair, one ball is slightly heavier. All heavy balls weigh the same, and all light balls weigh the same. In just two weighings on a balance scale, one may identify the three heavy balls! How many weighings are required if the balls are all white?

It's said that milking stools have exactly three legs so they won't wobble when set on an uneven barn floor. Not so for tables, since home floors are assumed to be planar. Which gives rise to an interesting puzzle. Assume you have a table whose four legs are the same length. But the floor is warped. Three legs touch the floor; the 4th does not. Is it always the case that for any mildly warped floor your table can be positioned so that all four legs contact the floor? Prove your answer.

In a previous horse race puzzle we were asked to find the three fastest horses from a group of 25 horses. Five horses could be raced at a time, and the determination was to be made running the fewest heats. If you have not tried solving it yet, you might find it interesting. In this puzzle, we want to rank order the speeds of five horses, [call them A, B ,C, D, E] assuming the speeds are all different. Simple enough if there are five lanes on the race track, but alas now there are only two lanes. How many head-to-head heats are required to do the task? For 2 horses, a single heat is required. For 3 horses, three heats are required in worst case. For 4 horses, it's fairly easy to determine how many heats it takes. For 5 horses, it becomes a challenge. Enjoy!

An old problem asks how many queens may be placed on a standard chess board so that none of them attacks another. It has been asked at least once in this forum. Let's now permit some hostility. We ask the maximum number of queens that can be placed on a standard chess board such that each queen attacks exactly one other queen. Escalating, what is the maximum number where each queen attacks exactly two others? Three others? Four others? For each answer, show the placement: with a sketch, codebox or chess notation, where a1 is the lower left, and h8 is the upper right corner. Have fun, and please use spoilers.

phillip1882 and alizawi have N. Care to give the sequence? Probability is still up for grabs: phillip1882 is close, but not quite.

Three switches, numbered 1, 2 and 3, and wired in series, control a single light bulb, which is not lighted. For the bulb to light, all three switches must be On. The switches' On positions are not labeled. You flip the switches, one at a time, in any sequence you choose. After how many flips [call this number N] can you be guaranteed to light the bulb? What is the probability of lighting the bulb after each flip: 1, 2, ... N? What is N if there are S switches [instead of 3]?

Yup.