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Title: For the Honor of Hufflepuff Post by Icarus on Jun 21st, 2003, 6:44pm This was too much! As a prefect of Hufflepuff you had to speak out! Things at Hogwarts have been going downhill for years now - ever since that kid first showed up. You set back and let it pass the first year - after all, when the school purposely gathers all the bad ones into one house, you know that they're being singled out for nasty treatment. So it was no surprise when Dumbledore, in truly sadistic fashion, let Slytherin believe they had won the house cup only to snatch it from their hands with a carefully contrived "last-minute awarding of points". But you didn't realize then that it would become a new tradition!. Year after year, Dumbledore has awarded a few extra points to Gryffindor - for ever more flimsy reasons. Always just enough to give them the victory. This time it is truly ridiculous. Hufflepuff came into the final ceremony with a 310 point lead over the lazy Gryffindors, only to hear Dumbledore announce he is awarding 315 points to Harry Potter "for parting his hair in a most intriguing fashion"! So you have stood in protest, joined not only by the rest of proud Hufflepuff, but also by Ravenclaw and Slytherin as well. Even some of Gryffindors appear to be chagrined by the blatant favoritism. Dumbledore cannot ignore this show of unity. "Let there then be a challenge.", he announces, "To the winner will go the house cup!" The challenge agreed upon involves Peano's balls. There are one each of these balls for every natural number. The number of each ball is written on it in a special magical script. The balls are stored in bottomless urns. The urns have a dispenser where a few of the lowest numbered balls in the urn are displayed in a line. Three urns are brought, with all of the balls in one of them. Dumbledore removes ball #2 from the dispenser, and you see balls #3 through #6 roll down against #1, and ball #7 appear. But when he drops ball #2 back into the top, you see that balls #3 through #7 suddenly roll back up, and ball #2 comes rolling through them until it returns to the second position. "The rules of the engagement are thus", Dumbledore announces. "You have each been given an empty urn. You, young Hufflepuff, will take balls two at a time from the full urn and place them in your own. Each time you do this Mr. Potter will take a single ball from your urn to be placed in his. I will cast Zeno's Infinite Acceleration upon you both. This spell will allow you to do each action twice as fast as the one before. The spell is only broken when actions would become instantaneous. You have, I hope, advanced far enough in your study of Arithmancy to understand that this will bring the task of moving all the balls into the realm of the possible? The contest will end after all the balls have been removed from the original urn." "Be aware that cheating will not be tolerated! The only balls that will be counted are the Peano balls. Any other balls found in your urn (and we can tell whether or not such exist - magic is quite useful for that), will result in a most horrendous punishment. The only means by which you may move a ball is with your hands as described in the rules. Magicking a ball from one urn to another will be punished in the same fashion. Violators will be required to attend Professor McGonagall's singing recital!" Harry is looking worried, you notice. "But sir, if he puts two balls in his urn, and I take one of them to put in mine, we shall always have the same number of balls. How can either of us win?". "Oh no, Mr. Potter, The winner is not determined by the number of balls in the two urns! If the Hufflepuff urn contains any balls at all when the contest ends, then Hufflepuff will be declared the winner. Surely they cannot accuse me of favoritism under conditions such as these!" Harry is now on the verge of panic, but Dumbledore bends over and whispers something to him quietly. As Dumbledore begins the spell, you see a smile slip across Harry's face. Clearly the fix is in. How does Dumbledore expect Harry to win? And what can you do to overcome it? Comment 1: The first question is answered by the Impish Pixie puzzle. The second is not. Comment 2: Apologies to J.K. Rowling for the twisting of her creation. But not for pointing out the unfairness of Dumbledore's actions! ;) Comment 3: Torg Potter and the Sorcerer's Nuts (http://www.sluggy.com/daily.php?date=020902&mode=weekly) Edited to bring the story a little more in line with Rowling books. (I really don't think Dumbledore would have ever used the word "doable", for instance!) Another change was to replace "proctor" with "prefect" - an office actually occuring in the stories. I have also made changes to address the issue that Wowbagger expresses in the third post. |
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Title: Re: For the Honor of Hufflepuff Post by BNC on Jun 21st, 2003, 11:26pm Hmm, let's see if I learned anything from your explanations to the "impish pixie"... [hide] Since the urns will display the six lowest-number balls, you don't start from 1. First pick 5,6. Harry will take 5. You now take 7,8; Harry will take 6. And so forth. Continue until the urn is empty except for balls 1-4. Take bals 3-4. Harry will take 3. Now take balls 1-2. The original urn is empty. Your urn should contain balls 1,2,4. [/hide] Now, as I'm unliterated in things magical and infinitical, that's probably wrong. Can't wait to learn why :) |
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Title: Re: For the Honor of Hufflepuff Post by wowbagger on Jun 23rd, 2003, 4:40am I would agree with your general idea, BNC. Hoewever, I feel a bit uneasy about the step asking us to continue "[hide]until the urn is empty except for balls 1-4[/hide]". Maybe Dumbledore should elaborate on the whole procedure anyway: 1. Are the wizards-to-be able to read the special magical script on Peano's balls? And can they read it before taking a ball from an urn? 3. Are the six lowest-numbered Peano's balls of an urn ordered in any way, such that one can infer which is the least one, for example, from its position? (Only necessary if we can't read the script, of course.) |
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Title: Re: For the Honor of Hufflepuff Post by Icarus on Jun 23rd, 2003, 3:38pm Are we over-analyzing the puzzle? You may assume that the balls not removed do not change places (or perhaps the balls are always organized by increasing number in the same way). So if you remove the highest numbered balls, the others will remain where you left them. The "magic script" is only there because a real script for every natural number cannot be fit on a fixed-sized ball. |
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Title: Re: For the Honor of Hufflepuff Post by James Fingas on Jun 24th, 2003, 9:07am I don't see how anything you, as the proctor of Hufflepuff, could do to combat the simple strategy Harry Potter could employ of "take the lowest-numbered ball". To put some bounds on things: 1) You can put ball 2N+4 into your urn on turn N at the earliest (starting from turn 1) 2) Once you put ball 2N+4 into the urn, you will have at most (2N+3)-N turns before it is removed. (there are already N balls with smaller numbers gone, and there are only 2N+3 smaller balls total). 3) Once Harry takes a ball, it can never be put back. So if Harry can always pick the smallest-numbered ball (by the original marking scheme), you're stuck. Maybe you can fiddle with the markings so that every turn you put in ball 2N and ball "1". |
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Title: Re: For the Honor of Hufflepuff Post by Icarus on Jun 26th, 2003, 7:24pm James - There is nothing you can do to leave balls in your urn at the "omega-point", when an infinite number of transactions have first occured. But what about after? |
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Title: Re: For the Honor of Hufflepuff Post by Jamie on Jun 27th, 2003, 1:21am Does this work? [hide] On each of your turns you take the middle two of the six balls. This is equivalent to the Impish Pixie game, but with the balls numbered from 3 rather than 1. We know that that game ends with your urn empty, so the final turn of this game will start with your urn empty, and Harry's containing every ball except for 1 and 2, which are in the original urn. You put those two balls into your urn, Harry takes one, leaving you victorious. [/hide] Two things bother me about this. Firstly, [hide]I'm not sure why you picked six, rather than four[/hide]. More importantly, [hide]the game ends when the actions would have become instantaneous, not when the original urn is empty. Surely this happens whenever there are a finite number of balls left. In fact, given this, I'm not sure how your comment about the omega-point can be valid, Isn't the game over at that point?[/hide]. Nice puzzle. I particularly like the fact that you went to the effort of constructing a fun story to set the scene. |
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Title: Re: For the Honor of Hufflepuff Post by visitor on Jun 27th, 2003, 6:33am These solutions that involve leaving the lowest balls in the urn until the end kind of show the ridiculousness of the whole riddle. Any riddle that has you do an infinite number of steps, one at a time, and ever finish makes no sense, even if you move infinitely fast. If you put zeno's spell on Harry and Hufflepuff, from our point of view the time may come when they're finished. But from their point of view they'll never finish, yet Hufflepuff's supposed to put ball numbered infinity into his urn followed by ball 3 and 4, or maybe just 4, depending on whether infinity is even or odd. If the process ever comes to an end, then we should be able to film it with an infinitely fast movie camera and watch the film backwards. Here it looks like Harry takes his turn by taking a ball out of his own urn and putting it into Hufflepuff's. Then Hufflepuff takes two balls out of his own urn and puts them into the original urn. An impressive trick, considering the fact that his urn was empty to start with. I understand the notion that there exists a one-to-one correspondence between the members of the two different infinite sets that's supposed to allow the impish pixie riddle to work, but it only works as a logical concept, not as a real world, one-at-a-time process. You might as well have Dumbledore tell Harry to take an infinite number of balls and divide them evenly among 0 urns, and then when you're done combine the balls in all the 0 urns into 1 urn to get infinity back. |
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Title: Re: For the Honor of Hufflepuff Post by James Fingas on Jun 27th, 2003, 1:34pm Icarus, My question is: if you put in balls after you get to the omega point, how long does it take? |
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Title: Re: For the Honor of Hufflepuff Post by Icarus on Jun 27th, 2003, 5:49pm James, as you know, the doubling effect means that it will take as long to reach the [omega]-point from the end of the first step, as the first step itself took. At this point, the spell breaks, and all remaining motion takes the same amount of time as it normally would. So to toss in the two or four remaining balls would not be a problem. Jamie - The solution that BNC and you have found does work. This is the trick that rmsgrey points out in the Impish Pixie puzzle. The "empty urn" result assumes that we are dealing with a set ordered exactly like the natural numbers - the ordering expressed by the infinite ordinal omega ([omega]). By holding back a couple balls to toss in after the [omega]-point, you are switching the ordering to the ordinal [omega]+2. There was no particular reason for exactly 6 balls. It just had to be greater than or equal to 4 - so that you could always leave two behind. I chose a larger number to try and avoid being too obvious. I think now that I should have just said "several balls", or some such. That would have sufficiently implied "more than 3". on 06/27/03 at 06:33:23, visitor wrote:
Under the conditions stated in the puzzle, you would finish in a finite amount of time. In fact you would finish an infinite number of steps in exactly the same amount of time it takes you to complete 2 steps while unspelled. And you never, ever "move infinitely fast", which is impossible for logical reasons, not just physical ones. Quote:
Here you are making assumptions about the nature of this spell. One can make other assumptions by which this is not a problem. Perhaps they must only fix a strategy in their minds, and the spell moves their hands according to the strategy, while they themselves experience time just as everyone else in the room does. And please don't bring the physics up. By definition magic violates the laws of physics. Since the only thing about the spell that is needed for the puzzle is its doubling effect, I did not, and still do not, see the need to completely describe it. Quote:
"FOUL!", cries Dumbledore. "Really, young sir, did you think that after I specifically warned you that only Peano's Balls would be allowed, that I was bluffing? Apparently also, you have been lax in your study of Arithmancy, to believe that we would mistake "infinity" for a natural number! The House cup is Gryffindors! As for you, young sir, the Recital begins at 8!" Quote:
No-can-do, not even with magic. "Infinitely fast" is a logical impossibility. You could use the same spell on the camera, but it would still require an infinite amount of film. How could you watch it from the end? It has none. A point that you miss with your next statement: Quote:
This assumes that there is a "last" move among those leading up to the "omega point" where the spell collapses, which will be the first move as described on the reversed film. But there is no such last move. Watching your infinite film backwards at normal speeds would require starting it "infinitely far" back in time. Quote:
Are you under the impression that Hogwarts and magic exist in the real world? This is not a real world problem. That does not make it a "ridiculous" problem. Most puzzles are not truly real world problems, in case you haven't noticed. What is required of a puzzle such as this is internal consistency and a logically deducible answer. Both exist in the puzzle, but not in your concluding statement. Quote:
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Title: Re: For the Honor of Hufflepuff Post by jade_conundrum on Jul 8th, 2003, 9:37pm um...sorry about this, but could you guys and gals be so kind as to explain to me what the "omega point" is and how it correlates to this problem? The way I see it is that Hufflepuff will always have more balls than harry does huff: takes 3,4 has 2 balls harry: takes 3 has 1 ball huff takes 5,6 has 3 balls harry: takes 4 has 2 balls huff: takes 7,8 has 4 balls harry: takes 5 has 3 balls huff: takes 9,10 has 5 balls harry: takes 6 has 4 balls etc... also, it should not matter if you leave balls 1 and 2 at the beginning, because if you do then even if hufflepuff reaches the ball marked infinity then, with two numbers left over, the actual count would be infinity-2, not the actual infinity. And it all really depends upon weather or not infinity is reached with two or one balls left (even or odd) which would decide if harry could take the last ball or not. of course, what really matters is who reaches infinite speed first, for that would be when the game is over. That would further lead to either hufflepuff having 3 balls left (1 + 2 from pick up) or 2 (harry reaches infinity and the game ends with harry not able to compleate his move, allowing hufflepuff the win anyway. given that the game ends as soon as one can move at infinity, not when a turn ends with a person who moved at infinity) blah blah blah, please feel free to tear apart my post and point out every little thing that is wrong with my reasoning. |
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Title: Re: For the Honor of Hufflepuff Post by BNC on Jul 9th, 2003, 12:19am on 07/08/03 at 21:37:28, jade_conundrum wrote:
No need to do either of these. All you need is a peek at the impish pixie (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1049309591;start=0) riddle |
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Title: Re: For the Honor of Hufflepuff Post by Jamie on Jul 9th, 2003, 1:09am on 07/08/03 at 21:37:28, jade_conundrum wrote:
The puzzle only states that the spell is broken when the moves would have become instaneous. It doesn't actually tell us the conditions for the game to end, but I assume it's when the urn is empty. |
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Title: Re: For the Honor of Hufflepuff Post by Icarus on Jul 9th, 2003, 8:05pm on 07/08/03 at 21:37:28, jade_conundrum wrote:
While BNC is correct that your questions are answered in the impish pixie thread, I will try to address them more directly. The "omega point" is that point in time when an infinite number of moves have been made. By the conditions of the Zeno spell, if the first move takes exactly one minute to perform, the "omega point" is reached exactly 2 minutes after the start of the contest, and it is at this point that the spell breaks. It is called the "omega point", because a small omega is commonly used to represent the least infinite ordinal number. Quote:
At the end of every FINITE step, Hufflepuff and Gryffindor are tied. However, one of the anti-intuitive properties of infinite processes is that this does not hold at the omega point. To see this, consider the location of the balls at the omega point. If you do not leave any behind, but always take the lowest numbered balls, you will move ball #n to your urn on the [ (n-1)/2 ]th turn. Harry moves it from your urn to his on the nth turn. These are the only two moves it makes, so it's position at the omega point is well-defined: it is in the Gryffindor urn. But this is true for every natural number n. Thus all the balls are in the Gryffindor urn, and yours must be empty. This does not make any sense intuitively, because our intuition is based entirely upon finite processes. But logically, it is the only possibility. (In the impish pixie puzzle, some argue for other possible models for the situation - but all of these models are even more opposed to "common sense" imo. I have tried to narrowly define the problem here so as to preclude any alternatives.) Quote:
"Is NO ONE paying attention to their Arithmancy lessons," Dumbledore grumbles. "Yes, I know you have - put your hand down Hermione. If you would be so kind as to recall my explanation of the Peano balls, surely you will notice that I said each ball has on it a natural number, and for every natural number, there is a ball. Surely you don't believe 'infinity' is a natural number, now do you?" Part of the nature of infinity is that at the omega point, there is no "previous step". Between any particular previous step and the omega point are an infinite number of intervening steps. So - there is no ball "infinity", and there is no final step at the omega point. Next: what is "infinity-2"? If you have an infinite number of something and you take two away, how many are left? Still an infinite number! If you come across some of my other posts on inifinity (such as the ones in the 0.999... thread), I have explained that there are three different types of infinite numbers in widespread use (there are others as well - it's just a matter of definition). All of them agree that for any infinity, infinity - 2 = infinity. In particular note that the function f(n)=n+2 matches up one-to-one the set of all natural numbers to the set of all natural numbers >=3. Even though the latter set has two fewer members than the former, the two sets must be the same size. Quote:
I wouldn't depend on the weather if I were you! It's been particularly nasty here tonight. I would have replied much earlier, but I was stuck away from home listening to the storm trackers talk about tornados passing through my neighborhood. False alarm - when I did get home, the trash cans out on curb hadn't even been knocked over. (Sorry - I couldn't resist, whether or not it was wise.) Again you are being mislead by treating the infinite as if it were finite. At step k, you take balls 2k-1 and 2k (again, if you ignore the strategy). Harry takes ball k. At the omega point, you have had a step for every natural number k. Since every natural number is of the form 2k or 2k-1, you have moved all the balls in pairs, with none left over. Unfortunately, Harry has also moved all the balls, with none left over. If you don't follow the strategy, you are left looking forlornly at your empty urn, wondering where all the balls went! Quote:
Since your moves are syncronized by the rules, you get one turn and Harry gets one turn, you both reach the omega point at the same time. However, the game ends when all the balls have been moved - not when the omega point is reached. It only ends at the omega point if you failed to leave any balls behind. (I thought I had said this, but I must have removed it when I modified the puzzle before. I have edited the problem to say so explicitly again.) By leaving the two balls behind, you have effectively given yourself two extra balls to be played after the omega point has been reached. You could look at it this way: before the contest begins, you place stickers on all the balls: For each ball >= 3, the sticker is 2 less than the original number. Balls 1 and 2, you relabel A and B. Effectively you now have all the Peano balls, and 2 extras that you can use without violating the rules. You proceed as before, starting with the balls stickered 1 & 2, and continuing. At the omega point, you are left with an empty urn and all the balls in Harry's urn. But, you have 2 more balls to play. You place them in your urn. Harry takes 1, and the game ends with 1 ball in your urn, and infinitely many in Harry's. But 1 ball was all you needed to win. Quote:
My purpose in replying is not to tear apart your reasoning, but to show you the things you have not yet considered, so that you can expand your understanding. I enjoy infinities, and like to point out their odd behavior with others. |
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