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Posts posted by bonanova


I get fewer ...
Spoiler47308.
To avoid double counting, I restricted a <= b <= c.
I let a be in [1, 100] and b in [a, 100]. I then counted the integers c in [ sqrt {(a^{2})+(b^{2})}, a+b 1 ].
I also noticed 52 right triangles (Pythagorean triples) .This is a plot of possible obtuse triangles for each value of a between 1 and 100.
When a = 27 the number reaches a maximum of 1166.

13 hours ago, plasmid said:Flipping the same coin twice would be indistinguishable from not flipping it at all to your buddy. So you would just be exposing yourself to more nyaning without accomplishing anything.
Naively.I might want the first n1 coins to be heads if the valuable one was tails in the nth position. I'd have you keep flipping them until they showed heads. But /// I watched the video, so ... nah.

I can ask you to flip a coin for a 2^{nd} time?

17 hours ago, Pickett said:Approach
Let's start by identifying the 7 different groups of people involved and label them
 Total people that had only Wine = W
 Total people that had only Beer = B
 Total people that had only Water = H
 Total people that had Wine and Beer = WB
 Total people that had Wine and Water = WH
 Total people that had Beer and Water = BH
 Total people that had Wine, Beer, and Water = WBH
Now, given that, we can write our system of equations based on the problem statement:
EQUATION 1 (total cost of all drinks):
5*(WINE) + 2*(BEER) + 1*(WATER) = 293
5*(W+WB+WH+WBH) + 2*(B+WB+BH+WBH) + (H+WH+BH+WBH) = 293
Simplifies to:
5W + 7WB + 6WH + 8WBH + 2B + 3BH + H = 293EQUATION 2 (total glasses used):
W + B + H + 2WB + 2WH + 2BH + 3WBH = 106EQUATION 3 (total nonwater drinkers):
W + B + WB = 18EQUATION 4 (total wine drinkers):
W + WB + WH + WBH = 39EQUATION 5 (total nonalcohol drinkers):
H = 9EQUATION 6 (ending assumption that wineonly was half the number of beer/water drinkers):
BH = 2WSo, now that we have our 6 equations, we can start doing our algebra:
Plug equation 5 into equation 1 and 2 to simplify slightly:
1) 5W + 7WB + 6WH + 8WBH + 2B + 3BH = 284
2) W + 2WB + 2WH + 3WBH + B + 2BH = 97Plug equation 6 into equations 1 and 2:
1) 11W + 2B + 7WB + 6WH + 8WBH = 284
2) 5W + B + 2WB + 2WH + 3WBH = 97Reorder equation 1 and 2:
1) 2(W+B) + 9W + 7WB + 6WH + 8WBH = 284
2) (W+B) + 4W + 2WB + 2WH + 3WBH = 97Plug equation 3 (W+B = 18WB) into equations 1 and 2:
1) 9W + 5WB + 6WH + 8WBH = 248
2) 4W + WB + 2WH + 3WBH = 79Plug equation 4 (W=39WBWHWBH) into equation 1 and 2:
1) 4WB + 3WH + WBH = 103
2) 3WB + 2WH + WBH = 77Solve for one of the remaining variables (WB):
WB = (103  3WH  WBH) / 4Plug into equation 2:
WH = 1 + WBHAlright, so we now have 1 equation with 2 unknowns, where one of the unknowns is what we're trying to determine! So, we can list out all of the possible values of WBH and WH...which happen to be:
WBH = 1, WH = 2
WBH = 2, WH = 3
...
WBH = 18, WH = 19Anything above WBH = 18 will cause equation 4 to not work...so we can stop there with those 18 possibilities. Now it's just a matter of substituting in each of those values into our equations to find the lowest one that works. I'll spare the gory details, but WBH < 11 results in negative values for other variables at some point...so let's just show trying WBH = 11:
WBH = 11, WH = 12:
Substitute those values into our 4 equations and simplify:
 5W + 2B + 7WB + 3BH = 124
 W + B + 2WB + 2BH = 40
 W + B + WB = 18
 W + WB + = 16
Now substitute 4 into 3 to solve for B:
B = 2...so we know B = 2 and W = 16WB
Plug those two into equations 1 and 2 and simplify
 2WB + 3BH = 40
 WB + 2BH = 22
Solve (WB = 222BH)
You find BH = 4...so given BH = 4, WBH = 11, WH = 12, B = 2, and H = 9, we can find the other two values of WB = 14 and W = 2...at which point we can verify all of the original equations hold true. so therefore, WBH (total number of people who drank all three drinks) is at least 11 given the assumptions we stated.@Pickett It may not impact your solution, but the OP did not intend to say that all of us drank something.
"Class of drinkers" was not meant to preclude anyone from drinking nothing.
Meaning there are eight "classes" of drinkers.Sorry for that. I have this thing about insisting that zero is a number, rather than a denial.
Also, I get a different answer. I'll check my analysis against yours to see why.

2 hours ago, rocdocmac said:Does it necessarily mean that all 9 of the teetotalers drank water or is that an assumption? What if some didn't drink at all that day and therefore never touched one of the 106 glasses?
 It means that all of the teetotalers drank neither wine nor beer. Let's say also that the others did consume alcohol.
 One class of friends may have drunk nothing, and perforce not used a glass.

Thinking. Meantime I'll call your 10 minutes of cat flight and raise you 25 minutes of Roomfull of Teeth, as they explore new uses for the human voice, which you might actually enjoy.

A bunch of friends went to the sports bar and got a group rate on the drinks: $5/glass for wine, $2/glass for beer, and $1/glass for water. When we left, the waiter asked me to sort out the bill. There was enough uncertainty in what people remembered that I could not be precise. So we happily just threw in enough to cover the bill, which came to $293 and we went home.
But it got me thinking. None of us had multiple glasses of the same beverage. The waiter said 106 glasses were used, once each. 18 of us did not drink water. 39 people had wine. I was certain that 9 of us were teetotalers. If I had known the sizes of just three classes of drinkers I could have figured out the bill, as it was, I could not.
But it did occur to me that if those who drank beer and water but not wine were as many as possible, and if those who drank only wine were half as many as that, I could say the smallest number of us who drank all three. Can you?

Let there be n red cards and n black cards.
SpoilerWhat is the smallest n that permits a winning strategy?

On 2/22/2018 at 9:18 AM, harey said:Maybe an idea. There is an infinite number of numbers between 0 and 1. I have the opportunity to discard an infinite number of numbers between 0 and 1.
How much are you willing to bet that nothing remains between 0 and 1?
I just reread this, and my reply wasn't totally responsive. But let's see whether Charlie even makes us address this question. Remember, Charlie doesn't get any more coins to deal with than Albert did. If we believe Albert's box (can) get emptied, we can't say discarding Charlie's coins is an impossible task. We simply have to show that none of Charlie's coins is able to "hide" indefinitely, in hopes of hearing the clock strike while still being in the box.
Charlie's coins are not inscribed, so we have to treat them collectively. We can't trace the destiny of any individual coin, as we could in Albert's case. Can we still show they are all eventually discarded, and not face the task of depleting an infinite set along the way? We claim that we can do that.
We can refer to each of Charlie's coins as "one of the coins that entered Charlie's box at event n that was also not discarded at event n." The n coins in his box at any such time are indistinguishably equivalent. Their "fate" is countably infinite events with nonzero probability of being discarded. We have shown in a preceding post that each of Charlie's coins has a survival probability of k/n where k is the event when the coin entered his box and n is the current event number. k is fixed. n becomes infinite. The probability of the coin that entered Charlie's box at event k still being in his box after midnight is limit (n>inf) k/n = 0. Since this holds for each k it holds for each coin that entered Charlie's box.
Using other words, every coin that is in Charlie's box at midnight is not a coin that entered Charlie's box. A contradiction.
Also, at no event time did Charlie have an infinite number of coins.

7 hours ago, Molly Mae said:It takes 8 moves. All of the second player's moves are forced, but he usually has two ways to respond. In either case, though, the second player can't prolong the game any longer than move 8 (that I can find).
1. f6, f3
2. d6, g3
3. d4, g7
4. e4, g4
5. g6, d3
6. e6, c6
7. e3, resignsOne of 8. e2 and 8. e7 is unstoppable.
Spoiler4 in a row with one end unblocked, or 3 in a row with both ends unblocked, force the next move. A forced win must do this on each move. Nice.

Pencils down. Discussion time over.
For the believers in the utility of clever play, choose a convenient number of cards (e.g. pick a small number and enumerate the cases) and quantify the advantage that can be gained.
For the naysayers, give a convincing argument that all the factors already discussed (permission granted to peek at spoilers) exactly balance each other out.

@plainglazed (first) and @plasmid (with a formula) both have it.
Both posts show how to get the answer, with the formula garnering "best answer" designation. Nice job both.Spoilerplainglazed's construction of successively halving the group implies logarithmic dependence on n, while plasmid produces the n^{th} partial sum of the harmonic series. These differ only by the addition of the Euler constant (.577...). Because that's so close to 1/2 it may just represent the dilemma of whether or not, in plainglazed's method, to halve the final car.
Answers given in the posts differ, because I changed n from 100 to 1024 at one point.

15 hours ago, harey said:If the blue are excluded because every room houses a green man, the same way you cannot accept even a single new guest.
Precisely.
If every room houses a green man, new customers have no place to sleep.
This does not work: { g1 g2 g3 ... gi .... b1 } <=> { r1 r2 r3 ... ri .... rinf+1 }
Do we want another guest? We can't assign gi to ri.
This does work: { b1 g1 g2 g3 ... gi .... } <=> { r1 r2 r3 r4 ... ri+1 .... } gi is no longer in ri.


4 hours ago, aiemdao said:The way it's set up, he can say Stop just once.

7 hours ago, aiemdao said:Is O win if we get XOOOOOX?
Yes. That’s a win for white

OK so I was a kid once, back in the 40s and 50s, and we had this game called Pegity.
It had a board with a 16x16 array of holes, and in turn each of 24 players inserted a peg of his own color into a hole. The object was to get 5 pegs of your color in a row: vertically, horizontally or diagonally. Not every game was won: similar to tictactoe, you could run out of holes. But with only two players, there was usually a win.
For simplicity let's shrink to a 9x9 board, mark the holes like in chess (A1, E7, etc.) and say that after O and X have each made three moves we have this position, with O to move:
1 2 3 4 5 6 7 8 9
A + + + + + + + + +
B + + + + + + + + +
C + + + + + + + + +
D + + + + + + + + +
E + + + + O + + + +
F + + + O + O + + +
G + + + + X + + + +
H + + + X + X + + +
I + + + + + + + + +With best play by both players, how soon can O win?
Use chess notation (naming the holes, in two columns) to list the moves. 
@CaptainEd shuffles a standard deck of playing cards and begins dealing the cards face up on the table. At any time @plainglazed can say Stop and bet $1 that the next card will be red. If he does not interrupt, the bet will automatically be placed on the last card.
What is the best strategy? How much better than 50% can @plainglazed do?

On 2/23/2018 at 4:52 PM, plainglazed said:I think I see the pattern here and am encouraged by the beauty of binary symmetry so here goes:
For integers i and the number of prisoners n equal to 2^i1, the probability of success p is (2^nm)/2^n or n/2^i where m is the number of strings needed to memorize and is 2^(ni).
i n=2^i1 m=2^(ni) p=(2^nm)/2^n or n/2^i
1 1 1 1/2
2 3 2 3/4
3 7 16 7/8
4 15 2,048 15/16
5 31 67,108,864 31/32
.
.
.That is amazing. And you've shown that 1 fewer than a power of 2 is the optimal value for n.
Can you reduce it to only q memorized strings, and the success rate to (q1)/q, where q is the highest power of 2 less than or equal to n?
(Purposely leaving this unspoilered, cuz this is a tough problem.)

On 2/23/2018 at 10:44 AM, CaptainEd said:I see, I didn’t pay attention. I really want to count only segments that are surrounded by shorter segments.
Yup. And a bit of possibly helpful thinking:
SpoilerWhile (see previous post for definitions) small, medium and large intervals (segments) are equal in (expected) number, their combined lengths of course will not be equal: the expected length of the large intervals is clearly greater than 1/3. But it's not intuitively clear whether it reaches 2/3.
Edit: The math gets a little demanding here. Integrals and stuff. 
On 2/24/2018 at 3:42 AM, harey said:Brilliant, surprisingly interesting and leading to a major annoyance.
Yeah, I hate annoyances, too...
SpoilerThis discussion has been a challenge to sharpen my thinking, and my words. And it has been fun. Thanks for that.
Meantime, I recalled the problem of proving that the rationals, { p/q } for integers p and q, are countable. By definition { integers } are countable. And so { p } and { q } individually are countable. But can we establish that { all pairs of p and q } are countable? This was doubted at one time, but in fact they are: { pairs of p and q } can be matched 11 with { integers }. However, care must be taken in constructing the proof.
One failed proof first counts { 1/q }, then counts { 2/q }, then { 3/q }, and so on. But there is a problem with this. Once { 1/q } has been counted, we're done counting! Every element of { 1/q } matches to an integer, but every integer points to an element of { 1/q }. Ouch. That leaves { 2/q }, { 3/q }, .... uncounted. Matching two infinite indexes to the integers is a little tricky. It can be done, but it takes a bit of cleverness. If you haven't seen that proof, you can read it here. Fun to recall. Now back to the Hotel.
Let's combine the infinite sets { blue men } and { green men } into the set { men } and try to prove that { men } is countable. We need to find an integer for each man, and a man for each integer. That is, we need to find a bijection between { men } and { integers }: But wait. If we think of { rooms } as { integers } this is precisely our room assignment task: to find a bijection between { men } and { rooms }. And we should be careful in creating it, recalling the failed attempt regarding rational numbers. Some schemes may succeed, but clearly not every scheme.
In fact, one way to fail is to first assign rooms to the green men. That fails because it only serves only to create a bijection of { rooms } with { green men }. Every green man gets a room, true, but (gulp) also, every room houses a green man. Oy vey! The blue contingent got excluded. Yes  we see them now, trudging down the road to the Holiday Inn. Want their room numbers? Call HI.
But we can house { men } at HH, interleaved. And also prove { men } is countable: just read their room numbers.
That's the crux, I think, of all the matters we've discussed.

4 hours ago, CaptainEd said:Thank you for the kind words, Bonanova. I fear my answer to this one is not so brilliant.
63 is the lowest I can get.
7 vertical cuts of length 8,
56 horizontal cuts of length 1
i get the same result if I divide the board into 4 quarters, divide those into quarters(16), and divided THOSE into quarters(64).
Can you think of a reason?

OK I think I just found your first blue man.
SpoilerIf a process can be reduced to an algorithm (a program) that involves no more than countably infinite steps, then the process can be carried out in finite time. At least in theory. But that's sufficient to at least discuss the consequences or effect of the process.
We can do a process with countably infinite steps in finite time by scheduling step_{i} at a time t_{i} = 1/2^{i} minutes before midnight. The steps are executed at times 1, 1/2, 1/4, 1/8 ...., 1/2^{i}, .... and the process completes in one minute. But what if we wanted to run a (countably) infinite number of such processes. Can we also do that?
Well, the universe will freeze in less than countably infinite minutes. But we can run each successive process in half the time of the previous one and finish everything in finite time. So, Yes, we can do that.
SpoilerLet process_{1} begin at 2 minutes before midnight and run its steps at these times thereafter { 0 1/2 1/4 1/8 ... } as before, and now end at 1 minute before midnight, at which time process_{2} begins. We run its steps at these times thereafter 1/2 x { 0 1/2 1/4 1/8 ... }, so that it ends at 1/2 minutes before midnight, at which time process_{3} begins, subsequently ending at 1/4 minutes before midnight. In general, process_{j} will begin at 1/2^{(}^{j}^{1)} minutes before midnight with steps executed at times 1/2^{j} x { 0 1/2 1/4 1/8 ... } thereafter, and compete at 1/2^{j} minutes before midnight. At midnight then, a countably infinite number of processes, each with a countably infinite number of steps will have completed. And as before the j^{th} step of the i^{th} process has a unique and unambiguous execution date. So now we can do processes of complexity O(N^{2}) on countably infinite sets. That is another way of saying that { Aleph_{0} }^{2} = { Aleph_{0} }. To get to a higher order of infinity you need 2^{{ }^{Aleph}^{0 }}.
Now lets revisit Hilbert's Hotel and write some algorithms:
We have 3 countably infinite sets: { r = rooms } { b = blue men } { g = green men } and a bunch of things to do. Let's list them.
 Place b_{i} in r_{i}. Blue men have all checked in.
 Evacuate all rooms. Sorry to bother you, but some green men want to check in.
 Place b_{i} in r_{2}_{i}. Check Blue men into evennumbered rooms

Place g_{i} in r_{2}_{i}_{1}. Check Green men into odd rooms, so it's g b g b g b g b g b ....
All these are O(N) processes, which we know we can do.
But now we want to sort the occupants, using a type of bubble sort,
which is an O(N^{2}) complexity process. But hey, we just showed how to do that, as well.
 For all j run process_{j} consisting of
 For all i if b(r_{i}) and g(r_{i}_{+1}) then swap those room occupants
At this point the clock has struck midnight and the horrific job is done. There is now no blue man in a room that has a lower number than a room occupied by a green man.
What have we accomplished? We'd like to say that in a countably infinite linear array of rooms we have placed a countably infinite set of green men, followed by* a countably infinite set of blue men. We know that we have enough rooms, because our rooms accommodated these same men when they were initially interleaved, all of which is to say
SizeOf ( { rooms } ) = SizeOf ( { green men } ) + SizeOf ( { blue men } ) = Aleph_{0.}
So let's note here that Aleph_{0}, the cardinality (SizeOf) of the counting numbers, is unchanged by "multiplying it by," or even "raising it to," any finite number. Those words are in quotes because Aleph_{0} is not a number and it does not follow the rules of algebra. But now at least we should hear an alarm bell and see a flashing red light if we try to say things like inf+1 or 2xinf or the like. These are not numbers, either.
But can we say anything more about the room assignments? For example initially a blue man was sleeping in Room_{2}. And now we know (just by examining the doublyinfinite set of steps described in lines 5 and 6) that man is soundly asleep, hopefully, in a different room. Presumably he could open his door and read his room number. If we had a really long string with tin cans on the ends we could ask him what it is. How would he answer? Well since { rooms } = { green men } + { blue men } and he is the first blue man, he'd have to say "My room number is ... Aleph_{0} + 1." What a prestigious address  like the ultimate penthouse. Wow. There we have it. But what we have is meaningless.
So anyway I was wrong before. We can, by following a welldefined procedure, place all the blue men after* all the green men. But after we've done it, we've lost track, in conventional terms, of their room numbers. Other than to say "they're waaaaaay out there, unspeakably greatly removed from the hotel office." And so much for room service! I think we can say that the blue men originally in rooms 2 and 4 are now neighbors, but maybe we can't even say that. It would be like asserting that (inf+2)  (inf+1) = 1, where the terms on the left are meaningless. Maybe we can just say the blue room numbers are all infinitely large. But maybe we can do better than that, in a different way?
Spoiler* Except that followed by, and after, etc., imply this completion thing again.
We handled the completion of infinite events by clever scheduling. So let's take a page from that playbook so that I can do a better job of describing the new location of the first blue man.
Hilbert's hotel undergoes a redesign:
Room_{1} is now its most luxurious room with a colossal width of 1 mile. Room_{2} less grand, but certainly nothing to sneeze at, weighs in at a comfortable width of 1/2 mile. Room_{3} is "only" 1/4 mile wide. Similarly, each Room_{i} has a width of precisely 1/2^{i} miles. Thus a countably infinite number of HH rooms fits alongside a workable 2milelong parking lot. Let's paint this building a refreshing sea green, and into its rooms let's place all the { green men }. Immediately adjacent to that building lies another set of rooms, the first of which is, again, 1 mile in width. Next to it is a 1/2milewide room, followed, in turn, by rooms 1/4, 1/8, ... miles wide. They comprise a second 2milewide building, which we will paint a lovely azure blue and into whose rooms we now place all of the { blue men }.
If all the rooms are numbered consecutively we still do not know the number of the first blue man's room. But if we want to visit him we can look at a map and clearly locate it! We only have to walk 2 miles from the office. Or we can simply jog down to the blue building, and knock on its first door. Hi, mister first blue man, nice to see you.
How's that?

making obtuse triangles
in New Logic/Math Puzzles
Posted · Report reply
Case by case results, inviting algorithm falsification, while limiting lengths to 10.
Rows: values of a; Columns: values of b; Elements: triangle counts.
OBTUSE 10
0 0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 1 1 0
0 0 1 1 2 2 2 2 1 0
0 0 0 2 2 2 2 2 1 0
0 0 0 0 2 3 2 1 0 0
0 0 0 0 0 2 1 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
Row sums (obtuse triangles for 10 values of a)
0 8 11 11 8 3 1 0 0 0 <= showing no possible obtuse triangles when 2> a >7
Total triangles
42
More matrices, for maximum lengths of 29:
OBTUSE 2
0 0
0 0
OBTUSE 3
0 0 0
0 1 0
0 0 0
OBTUSE 4
0 0 0 0
0 1 1 0 <= meaning a=b=2 permits (only) one obtuse triangle: c=3
0 0 0 0
0 0 0 0
OBTUSE 5
0 0 0 0 0
0 1 1 1 0 <= meaning a=2 b=4 permits (only) one obtuse triangle: c=5
0 0 1 0 0 <= showing that a=3 b=4 c=5 (right)triangle was not counted
0 0 0 0 0
0 0 0 0 0
OBTUSE 6
0 0 0 0 0 0
0 1 1 1 1 0 <= counting (only) 223, 234, 245, 256 for a=2
0 0 1 1 1 0 <= indicating a=3 b=4 c=6 obtuse triangle (only) was counted; not c=5,7
0 0 0 1 0 0
0 0 0 0 0 0
0 0 0 0 0 0
OBTUSE 7
0 0 0 0 0 0 0
0 1 1 1 1 1 0
0 0 1 1 2 1 0
0 0 0 2 1 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
OBTUSE 8
0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 0
0 0 1 1 2 2 1 0
0 0 0 2 2 1 0 0
0 0 0 0 1 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
OBTUSE 9
0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 1 0
0 0 1 1 2 2 2 1 0
0 0 0 2 2 2 1 1 0
0 0 0 0 2 2 1 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0