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How about 12345679*9 = 111111111.
If you have a simple calculator and want to impress people ask them their favorite number between 1 and 9. If they say 3 then type 12345679*3*9 and you get 3.33333333 and the little e symbol. It's a great trick.
-Ben
:( I am sad because you were correct about the problem though my answer is slightly different from yours....But generally, the same, yes, same...:)Quote:
Originally posted by bmacri
How about 12345679*9 = 111111111.
If you have a simple calculator and want to impress people ask them their favorite number between 1 and 9. If they say 3 then type 12345679*3*9 and you get 3.33333333 and the little e symbol. It's a great trick.
-Ben
Mr Ben your up now, you have any question ?:)
I hate to burst your bubble so to speak, but the post before was in fact the thousandth. (What did you expect, "I hate to burst your bubble, so I won't"? :rolleyes: )Quote:
Originally posted by hometown
I just wanted to declare one important thing to all people who are reading this thread that THIS POST OF MINE IS EXACTLY 1000th.:) Really nice....
So Could you give me the answer to the question above?
One of these proofs are not correct. Which one?
Proof 1:
Assume a=b where a,b are nonzero integers.
a^2= ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
(a + b)(a - b)/(a - b) = b(a - b)/(a - b)
a + b = b
but since a=b then
2b = b
2 = 1
Proof 2:
Posting on CodeGuru takes both time and money (for the computer, electricity, ...) so
Posting = Time * Money
We all know time is money so
Posting = Money * Money = Money^2
But money is the root of all evil so
Posting = (sqrt(Evil))^2
Which becomes
Posting = Evil.
So posting on CodeGuru is Evil.
One of these is not true but which one????
-Ben
What are you crazy ;).Quote:
Originally posted by bmacri
(a + b)(a - b)/(a - b) = b(a - b)/(a - b)
A bit sketchy, but if you say it's true...Quote:
Posting on CodeGuru takes both time and money (for the computer, electricity, ...) so
Posting = Time * Money
We all know time is money so
Posting = Money * Money = Money^2
But money is the root of all evil so
Posting = (sqrt(Evil))^2
Which becomes
Posting = Evil.
So posting on CodeGuru is Evil.
Well I couldn't come up with much else. At least I thought this was some what humorous. Oh well. I do have an interesting coding question. It's long and not quite on the topic so if someone else wants to ask a better question that fine. But here goes.
My boss goes on a golfing trip every year with about a dozen friends. They play usually seven rounds of golf over a few days. At the beginning of each round they form into groups for that round. If possible they want to play with people they haven't played with before or at least not that much with. One thought was to write a program that came up with the pairings based on a few rules. Most of this has been coded but one piece is missing. How can we get all the possible ways of pairing the people up without any duplication? Here's an example of what I mean.
We have 12 people and want to form groups of 4. An array of
1,2,3,4,5,6,7,8,9,10,11,12 means person 1,2,3,4 play in one group, 5,6,7,8 play in another and 9,10,11,12 play in the last. The next round could look like 1,2,5,10,3,4,6,11,7,8,9,12 which means 1,2,5,10 play in the first group, 3,4,6,11 play in the next and 7,8,9,12 play in the last.
We have ways of evaluating each round and finding the best combination to use for that round. The problem is coming up with all possible combinations for that round WITHOUT duplicates. Does anyone have code or thoughts on how to do this? The kicker is everything needs to be variable. There may be 14 people instead of 12. The groups might be 3 people instead of 4. So nothing can be hard coded.
Any help would be great.
-Ben
I've not given this much thought but one brutal way of doing it would be to do a genetic search. Each round try a bunch of random groupings and evaluate the fitness of each one by penalizing combinations that include players who have played before. Then continue with the genetic search until you have come up with some good pairings at each level. It wouldn't be too hard to code this or use someones existing code. It would also be fairly easy to make it work for different sample and grouping sizes.Quote:
Originally posted by bmacri
Well I couldn't come up with much else. At least I thought this was some what humorous. Oh well. I do have an interesting coding question. It's long and not quite on the topic so if someone else wants to ask a better question that fine. But here goes.
My boss goes on a golfing trip every year with about a dozen friends. They play usually seven rounds of golf over a few days. At the beginning of each round they form into groups for that round. If possible they want to play with people they haven't played with before or at least not that much with. One thought was to write a program that came up with the pairings based on a few rules. Most of this has been coded but one piece is missing. How can we get all the possible ways of pairing the people up without any duplication? Here's an example of what I mean.
We have 12 people and want to form groups of 4. An array of
1,2,3,4,5,6,7,8,9,10,11,12 means person 1,2,3,4 play in one group, 5,6,7,8 play in another and 9,10,11,12 play in the last. The next round could look like 1,2,5,10,3,4,6,11,7,8,9,12 which means 1,2,5,10 play in the first group, 3,4,6,11 play in the next and 7,8,9,12 play in the last.
We have ways of evaluating each round and finding the best combination to use for that round. The problem is coming up with all possible combinations for that round WITHOUT duplicates. Does anyone have code or thoughts on how to do this? The kicker is everything needs to be variable. There may be 14 people instead of 12. The groups might be 3 people instead of 4. So nothing can be hard coded.
Any help would be great.
-Ben
Thanks for the thoughts Tom. We actually tried that but we were not coming up with the best possible pairings. I wrote some code that did it brute force ( every possible combination including duplicates) and was hardcoded for the given values(12 people groups of 4). With this I was able to get a nice break down. When we tried the random test method the results were never as good. For example with 12 people and groups of four everyone played with each other at least once and nobody played more then 3 times with someone else. When we tried the random way we could get everyone to play at least once but many people payed four times with other people. We could never get as good of results than the brute force method. The brute force isn't that bad if we just found a way not to test duplicates. If I remember right there are only 1225 different possiblities for each round. My boss came close with a recursive function but it never seemed to work right.
-Ben
Something's wrong with that number, I think.Quote:
Originally posted by bmacri
For example with 12 people and groups of four [...] If I remember right there are only 1225 different possiblities for each round.
Arrange the numbers into sets, like so:
[1234][5678][9(10)(11)(12)]
There are 12! possibilities, including duplicates.
There is 1 distinct possibility for each 4!4!4! sets.
The sets are not distinct, and there are 3! possible orders of the sets.
Therefore, there should be 12!/(4!4!4!3!) = 5775 distinct possibilities.
Either that, or I'm way too tired to do math right now.
Well that isn't exactly correct, I think. Order doesn't matter and each person can only be used once. Here's a small example of what I mean. Say 4 people two groups. The possiblities are
1,2,3,4 1,2 play and 3,4 play
1,3,2,4 1,3 play and 2,4 play
1,4,2,3 1,4 play and 2,3 play
and that's it. Everything else is a duplicate. But according to your way of calculating it there should be 6 different values 4!/(2!2!) = 6.
Here's how I got to the 1225 number. If you have 8 people and want groups of 4, Comb(8,4) = 70 combinations but it's only 35 different arrays. Each array uses two combinations. So and here's the part I can't explain 35*35 = 1225. It makes sense in my head but I can't seem to exaplin it at all. Here's a try though.
With each 35 posibilites there are 4 numbers let that form the last group. Once those 35 are written down you can swap a number from the last four and form 35 more possibilities. This can only be done 35 times and get different combinations. More than that you get duplicates. That seems to be the case. Are you more confused now? I know I am. I'm pretty sure it's only 1225 different possibilites.
-Ben
I know I made a mistake in my last post (I will go back and edit it now), but I still don't understand your reasoning. Maybe I'm missing a part of the problem.
A surprising result. I've gotten excellent results from a genetic search on a problem with 10^10 possible combinations. If the method is done right, it should work. On smaller problems it usually finds the best solution.Quote:
Originally posted by bmacri
Thanks for the thoughts Tom. We actually tried that but we were not coming up with the best possible pairings.
-Ben
I think you guys are right about 5775 different possibilities. Looks like I'm too tired to do math.
Now can you think of some code that would produce only these possibilites but yet be flexable to handle runtime changes to the number of people in a group and the total number of people? I know it's asking a lot but we still haven't been able to do it.
-Ben
The problem I think we were having was we were testing duplicates all the time(1,3,6,7,2,5,10,12,11,8,9,4 is the same as 3,6,7,1,2,5,10,12,11,8,9,4). Also if you don't get the best possibility on round one then that effects all the rounds following it. So we figured if there really are only so few unique possibilities then why not just test them all. It's very possible that there was something wrong in the code that wasn't getting truelly random test groups.Quote:
A surprising result. I've gotten excellent results from a genetic search on a problem with 10^10 possible combinations. If the method is done right, it should work. On smaller problems it usually finds the best solution.
-Ben
A nice way to see the 5775 result is to break the process of calculating up.
For group 1 you have 12 possible players to choose from and must pick 4. Order doesn't matter, so you use combinations.
(12 choose 4) = 12! / (4! 8!)
For group 2 you now have only 8 players left to pick from.
(8 choose 4) = 8! / (4! 4!)
And the final group, just to work it all out, leaves 4 to choose from (yes I know its 1)
(4 choose 4) = 4! / (4! 0!)
And, as solarflare points out, the grouping doesn't matter. So you have 3 groups and thus must divide out all of the 3! permutations of groups. Since the choices are dependent, we use multiplication, ie.
number of possibilities = 12! 8! 4! / (4! 8! 4! 4! 4! 0! 3!)
which can be simplified to solarflare's
12! / (4! 4! 4! 3!) = 5775
Now, concerning the code to enumerate the possibilities. A straightforward approach could be just to normal order the 5775 possibilities in a fashion like
1) arrange the players in each group from least number to greatest (ie, {11, 4, 7, 2} becomes {2, 4, 7, 11})
2) arrange the groups in order from least initial number to greatest
A full set is considered "less than" another if its very first member in this normal ordering is less than the very first member of the other. If equal move to comparing the next member (this is a standard dictionary ordering).
Now you can walk through the dictionary orderings and build each set by using a couple of rules:
1) all groups must be increasing.
2) the first group must start with 1 (some group has to have 1 in it, and it becomes the first group in the ordering).
3) the second group must begin with the lowest number not in the first group.
4) the second group must not have any members that are numbers in group 1.
5) the third group is completely determined by the first two groups (so really it does not need to be considered, if you want to save space).
So the problem gets reduced to enumerating all first groups that begin with 1 and increase and then for each first group built, enumerate all second groups that start with the necessary number. You only have 3 members to decide for each enumeration. The 3 members of the first group can be any increasing sequence from [2, 12]. The 3 members of the second group are restricted (so the second group enumeration requires knowing the data of the first group). That's all there is to it, and this is already almost pseudocode. Enumerating increasing sequences in order is a standard combinatorial algorithm (though, I forget its name), and you can apply the restrictions on group 2 by discarding those that aren't compatible with group 1. If this is c++, just toss the suckers in a vector and you're there!
and feel happy when seeing you back.... I like what you write and what you were/are thinking...though I couldnot read your mind much and sometimes some posts which wrote about me with negative points or there were some misunderstandings...
Amazing...!!!Your answer great !:)
Thanks for the help galathaea. I'll try to code it this weekend and I'll let you know how it works. Thanks to everyone for all their help. So any one got a fun question?
-Ben
You are coding and willnot have time to come back,Quote:
Originally posted by bmacri
Thanks for the help galathaea. I'll try to code it this weekend and I'll let you know how it works. Thanks to everyone for all their help. So any one got a fun question?
-Ben
even when there is a fun question you will not join..., right ?
I dont have a fun question,
Mr Ben, Will you also let me know if your program works fine Okay ? I just would like to hear how it will be...
Regards,:)
From Home to Mr. George Benard Shaw:
Very polite. Very cute. I love girls, Home. >:->Quote:
You are coding and willnot have time to come back,
even when there is a fun question you will not join..., right ?
I dont have a fun question
My question is : Can someone please send HomeTown to another forum etc.?Quote:
Mr Ben, Will you also let me know if your program works fine Okay ? I just would like to hear how it will be...
She makes me cringe by her politeness alone and it reminds me of.....ugh! :cool: :eek: :eek:
ok!
I have a riddle now. Here it is:
Pretend you have two strings which burn at random rates but which will always be completely burned in exactly one hour. (Thus, after 1/2 hour the strings will NOT necessarily be half burned length-wise, only time-wise) How do you use these strings to time exactly 45 minutes without using any other time-keeping device?
Light one string at one end and the other string at both ends at 0:00. When the second string burns out it will be 0:30; light the first string from the other end. When that one's completely gone too, it's 0:45, you Amish pyro. Burning precious string-clocks? Shame on you!
:D hah! that's right!Quote:
Originally posted by solarflare
Light one string at one end and the other string at both ends at 0:00. When the second string burns out it will be 0:30; light the first string from the other end. When that one's completely gone too, it's 0:45, you Amish pyro. Burning precious string-clocks? Shame on you!
So do you have a questoin?
Do I? (Hint: there are only two possible answers and 'yes' is wrong)Quote:
Originally posted by SeventhStar
:D hah! that's right!
So do you have a questoin?
I'll try to revive this! Again ;)
A warden of a prison comes back to the prison from a party very drunk. In his state, he accidentally turns the key to open all 100 of the jail cells.He realizes what he's done, so he goes back and turns the key on every other cell (thus, now cell 1 is open, cell 2 is closed, cell 3 is open, etc.). He realizes that he hasn't closed all of them, so he goes back and turns thekey on every third cell (thus, now cell 3 is closed, and cell 6, which heclosed when he turned every other key, is now open). Still drunk, he goesback and turns the key on every 4th, then every 5th, 6th, 7th, etc. until heturns the key on every 100th cell (thus, only on cell #100). After all this,which cells remain open?
The first one.
nope! That's not correct :) sorryQuote:
Originally posted by Elrond
The first one.
Every cell that whose # happens to be a perfect sqare. In other words:
The 1st cell, 4th cell, 9th cell, 16th cell, 25th cell, 36th cell, 49th cell, 64th cell, 81st cell, and 100th cell are all open.
- Kevin
P.S. The prisoners must have been twice as drunk as the prison guard if they didn't wake up from all the noise and escape when they had the chance. :D
Uhm.. a keik:) :p
Just a sec for a joke...:)Sorry
yep That's right :DQuote:
Originally posted by KevinHall
Every cell that whose # happens to be a perfect sqare. In other words:
The 1st cell, 4th cell, 9th cell, 16th cell, 25th cell, 36th cell, 49th cell, 64th cell, 81st cell, and 100th cell are all open.
- Kevin
Heh :D That's right tooQuote:
Originally posted by KevinHall
P.S. The prisoners must have been twice as drunk as the prison guard if they didn't wake up from all the noise and escape when they had the chance. :D
I forfeit my turn at authoring a question. Anyone got a new question?
I am sorry I wasso wrong...:)
Ok I hawve a new one :D
A monk needed to go from his monastery to another, located much higher in the mountains. So off he went. Sometimes he went fast, sometimes slow, but finally he got there. He finished his job there and started going back to his monastery. He reached it a lot faster because he was descending instead of climbing. As a whole he went really uneven in both directions. So question is: If the monk's monastery is called A and the other one is B is there a point (C) that is positioned in time so that AC (in time) is equal BC (in time) in the monk's walk.
Prove your suggestions.
There is.
On his way back, obviously at his start, the lenght of time between B and his position (P) is zeor, that is smaller than the time between A and P: t(BP) < t(AP).
The more he goes down, the higher t(BP) is (as time pass), and the smaller t(AP) is (as he is going closer from his start from A).
When he reaches A, t(AP) is 0, while t(BP) his higher than that: t(BP) > t(AP).
There must be a point somewhere where t(BP) goes from less than to more than t(AP): t(BP) == t(AP). This point is C.
can you tell how to find that pointQuote:
Originally posted by Elrond
There is.
On his way back, obviously at his start, the lenght of time between B and his position (P) is zeor, that is smaller than the time between A and P: t(BP) < t(AP).
The more he goes down, the higher t(BP) is (as time pass), and the smaller t(AP) is (as he is going closer from his start from A).
When he reaches A, t(AP) is 0, while t(BP) his higher than that: t(BP) > t(AP).
There must be a point somewhere where t(BP) goes from less than to more than t(AP): t(BP) == t(AP). This point is C.
400Km
Do you mean this problem is taken from some book and 400 km s the answer when time and distance are given ?
If so, why does he ask this question again ?
I am sorry I just think because this thread title is Questions and Answers, there must be someone who couldnot do his/her homework joinin to make questions and waiting for help from some others...
No, the answer is 55,... or not!
I am not sure whether point C is supposed to be a space coordinate, a time coordinate, or a space-time coordinate.
C as a space coordinate:
* The problem states that "As a whole he went really uneven in both directions". I assume this means that his paths were random. C cannot exist a space coordinate with this interpretation.
** This means that his two paths do not have to cross (when viewed as space paths).
** Since his two paths do not have to cross, then the only points in common at the end-points of his travel, the two monasteries (points A and B) and there cannot in general be a midpoint C common to both paths.
* If "As a whole he went really uneven in both directions" means that the monk followed the same path, but just traveled it at different speeds, (and we consider space to be a static Euclidean space) then the answer would be yes, point C can exist as a space coordinate because time is now irrelevant (we are only concerned with space coordinates.)
** Choose a point C that is equidistant from A and B for the first trip so that distance(AC) = distance(CB).
** Since the space is static and Euclidean, distance(XY) = distance(YX) for all time.
** So we may rewrite the equation distance(AC) = distance(CB) as distance(CA) = distance(BC).
** Using the math identity (I forgot the name for this) that if X=Y then Y=X, we can rewrite distance(CA) = distance(BC) as distance(BC) = distance(CA).
** But the equation distance(BC) = distance(CA) describes the return trip. So yes, point C exists as the midpoint in space coordinates assuming (a) the monk followed the same space-path and (b) we are assuming a static Euclidean space.
* I suppose there exists the possibility that we are dealing with static non-Euclidean space. I am not sure that in general the previous "proof" (and I use that term lightly) would be true. Then we could also be talking about a non-static space. But really, this means that the supposed path itself changes and my first "proof" would be sufficient here as well to state that point C cannot exist,
C as a time coordinate:
* Assuming that the monk is not infinately fast at traveling and working, the second travel happened after his first travel. C cannot exist as a time coordinate with this setup.
** This means we have t(A1) < t(B1) < t(B2) < t(A2).
** There is no solution where a fixed t(C) exist so that t(A1) < t(C) < t(B1) < t(B2) < t(C) < t(A2).
** So t(C) cannot exist as a mid-point for both time paths.
* If the monk is infinately fast at traveling and working, then t(A1) = t(C) = t(B1) = t(B2) = t(C) = t(A2), and in this case C does exist as a time coordinate because all time coordinates are the same.
C cannot exist as a space-time coordinate in general:
* C can only exist as a space-time coordinate if the monk is infinately fast at traveling and working and travels the same path both ways.
** That is the only solution that satisfies C as time coordinate.
** If everything happend as the same time, then we only need to consider the space at one point in time. This removes all concern about whether the space is static or not.
** Then the only solution possible from a space-viewpoint is that the monk follow the same path (see first agument for C as a space coordinate).
** But if the monk traveled infinately fast and traveled the same space path, then he really did not travel "uneven in both directions."
** So we have generated a contridiction and proved that C cannot exist as a space-time coordinate in general.
Kevin, I assumed that when SevenStar says:
"point (C) that is positioned in time so that AC (in time) is equal BC (in time) in the monk's walk."
he talks about a time of walk relative to the beginning of each travel. It means that when he leaves A, the time is set to zero, for his first trip, and when he leaves B on his way back, the time is set to zero for his second trip.
In this case, if he takes the same path on both ways, there will be a point where t(AC) = t(BC).
Now, as the monk waks unevenly, I don't think it is possible to know exactly where that point is. The best idea would says that it is somewhere in the section of his first trip between his start from A and the point he has reached after walking the same length of time that he has walked on his way back.
Well then is C a point in space, or a point in time (where time is reset to zero)? If C is a point in space, there is nothing to say that the same space-point C will equally divide the monk's path in time.Quote:
Originally posted by Elrond
Kevin, I assumed that when SevenStar says:
"point (C) that is positioned in time so that AC (in time) is equal BC (in time) in the monk's walk."
he talks about a time of walk relative to the beginning of each travel. It means that when he leaves A, the time is set to zero, for his first trip, and when he leaves B on his way back, the time is set to zero for his second trip.
In any case I made assumptions, you made assumptions. If people are making too many assumptions, then the question needs clarification. Seventh Star, is C supposed to be a space point or a time point (or a space-time point)? Is time reset to zero at the beginning of each trip? What exactly do you mean by As a whole he went really uneven in both directions"? Did we walk unevenly in speed, path, or both?
Right, the answer needs clarification. And what is the expected answer? The exact location of the point C? Or the best possible estimation?
Rolle's theorem (a version of the mean value theorem)
How does Rolle's theorem apply here?
What is Rolle's theorem?
If f(x) is continuous on [a,b] and differntiable on (a,b) and f(a)=f(b), then there is some point c that is part of the interval (a,b) where f'(c)=0.
Well there cannot be any possible estimation becaouse the monk was going very unsteady...Quote:
Originally posted by Elrond
Right, the answer needs clarification. And what is the expected answer? The exact location of the point C? Or the best possible estimation?
Sorry for the second questioen it wasn't correctly put :rolleyes:
Anyway the expected answer was:
"If you have two monks that are the same and they both start going simultaniously the first from A to B and the second from B to A, they will meet in point C"
I see now. The question now makes more sense now that I know the answer. Interesting question. A few assumptions did need to be made (like that the monk followed the same path in both of his walks). What's thenext one?
Who knows? :D I don't!Quote:
Originally posted by KevinHall
What's thenext one?
If you have a question, go ahead!
Rolle's theorem is equivalent to the intermediate value theorem (equivalent in the mathematical sense that if one assumes the validity of one, all valid conclusions of the axiom set are conclusions to if one assumes the other). They both codify the topological / analytic result that if a continuous function (like the assumption that the monk cannot teleport) takes on values f(a) and f(b) (with f(a) < f(b)), then for any value f(a) < c < f(b) there exist a value d between a and b such that f(d) = c. It is the mathematical theorem required to prove SeventhStar's "If you have two monks that are the same and they both start going simultaniously the first from A to B and the second from B to A, they will meet in point C" (which in my opinion is only a nonrigorous restatement of the problem ;)).
Just being cryptic to increase (decrease?) my coolness factor...
Galathaea.... Your avatar is a beauty :)