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Archive-name: puzzles/archive/pickover/part3
Last-modified: 17 Aug 1993 Version: 4 See reader questions & answers on this topic! - Help others by sharing your knowledge ==> pickover/pickover.12.p <== Title: Cliff Puzzle 12: Slides in Hell From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider a metallic slide with 10 large holes in it equally spaced from top to bottom. If you attempt to slide down the slide you have a 50% probability of sliding through each hole in the slide into an oleaginous substance beneath the slide during each encounter with a hole. 1. If you were a gambling person, which hole would you bet a person would fall through? 2. If you were a gambling person, how many attempts would it require for a person to slide from the top of the slide to the bottom without falling through a single hole. 3. If all the people on earth lined up to go down the slide, and they slid down a more horrifying slide with 100 holes at a rate of 1 person per second, when would you expect the first person to arrive at the bottom of the slide without falling through. An hour? A day? A decade? ... Received: from uoft02.utoledo.edu by watson.ibm.com (IBM VM SMTP V2R2) with TCP; Title: Cliff Puzzle 12: Slides in Hell >Consider a metallic slide with 10 large holes in it equally spaced from >top to bottom. If you attempt to slide down the slide you have a 50% >probability of sliding through each hole in the slide into an >oleaginous substance beneath the slide during each encounter with a >hole. > >1. If you were a gambling person, which hole would you bet a person >would fall through? None. The best chance is the first hole but I got a 50-50 chance. Why bother? (2nd hole is 1/4, 3rd 2**-3, ...) >2. If you were a gambling person, how many attempts would it require >for a person to slide from the top of the slide to the bottom without >falling through a single hole. No gurantee. Each slide is an independent event. Now, if you are talking mere probability, on the average, one in 1024 slides may make it through all 10 holes. >3. If all the people on earth lined up to go down the slide, and they >slid down a more horrifying slide with 100 holes at a rate of 1 person >per second, when would you expect the first person to arrive at the >bottom of the slide without falling through. An hour? A day? A decade? Again, can't tell. It could be the first one, it could be none. Probablity can not foretell actual events. But if you have infinite number of people sliding down till eternity, on the average, you may see 1 person slide over all holes every (2**100)/(365*24*69*6) years. This number is many times bigger than the world population for now. ==> pickover/pickover.12.s <== ------------------------- In article <1992Oct23.160130.166012@watson.ibm.com> you write: : Consider a metallic slide with 10 large holes in it equally spaced from : top to bottom. If you attempt to slide down the slide you have a 50% : probability of sliding through each hole in the slide into an oleaginous : substance beneath the slide during each encounter with a hole. : : 1. If you were a gambling person, which hole would you bet a person : would fall through? The chance of falling thru the first hole is 50%. For the second hole, it is (.5)(.5) = 25%, the thrid is (.5)^3 = .125. The chance by the tenth hole is about .0097 %. Obviously, since I am limited to one hole, I would place my money on hole #1 (best chance). : 2. If you were a gambling person, how many attempts would it require : for a person to slide from the top of the slide to the bottom without : falling through a single hole. The sum of the prob for falling thru a hole is .5 + .5^2 + .5^3 +...+.5^10. This is about 99.902% = .99902. So about 98 times out of 100000, someone will make it through without falling. This is about 1 time out of 1020. So give or take about 1020 tries.... : : 3. If all the people on earth lined up to go down the slide, and they : slid down a more horrifying slide with 100 holes at a rate of 1 person : per second, when would you expect the first person to arrive at the : bottom of the slide without falling through. : An hour? A day? A decade? ... The prob for falling thru the last hole is .5^100 = 7.88x10^-31. There must be some chance less than this that one WILL make it thru the slide. The MIN number of tries that it must take is 1/.5^100 = 1.26x10^30. At the given rate this is about 9.647 x 10^23 years, much older than the universe if I remeber correctly. Also, the chance of making it must be GREATER than .5^101. or with all the math, the MAX amount of time is 1.929x10^24 years. So give or take about 1.5x10^24 years.... -- Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD! // | Senior, Chemical Engineering, Univ. of Toledo \\ // Only the | Summer Intern, NASA Lewis Research Center \ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu / --------+ How do YOU spell 'potato'? How 'bout 'lousy'? +---------- "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L ------------------------- In rec.puzzles you write: >Title: Cliff Puzzle 12: Slides in Hell >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please send me your name, >address, affiliation, e-mail address, so I can properly credit you if Jeff Rogers Rensselaer Polytechnic institute rogerj@rpi.edu >Consider a metallic slide with 10 large holes in it equally spaced from >top to bottom. If you attempt to slide down the slide you have a 50% >probability of sliding through each hole in the slide into an oleaginous >substance beneath the slide during each encounter with a hole. > >1. If you were a gambling person, which hole would you bet a person >would fall through? The first one. There's only a 50% chance of them getting past it, and a small chance of them falling into each succeeding hole. hole # percent chance of reaching and falling into 1 50 2 25 3 12.5 4 6.25 5 3.125 6 1.5625 7 0.78125 8 0.390625 9 0.1953125 10 0.09765625 > >2. If you were a gambling person, how many attempts would it require >for a person to slide from the top of the slide to the bottom without >falling through a single hole. The chances for reaching each succeeding hole are the same as reaching and falling into the previous one. Therefore, the chances of passing all the holes are the same as reaching and falling into the last hole (see previous answer for stats), which makes the probability .0009765625, so statistically, 1024 slides would be required to guarantee reaching the bottom. If I was a gambling person, I'd probably bet about half this, because the actual events can happen in any order, and on average, I'd guess that he'd get down in about 512 slides. > >3. If all the people on earth lined up to go down the slide, and they >slid down a more horrifying slide with 100 holes at a rate of 1 person >per second, when would you expect the first person to arrive at the >bottom of the slide without falling through. >An hour? A day? A decade? ... This is solved similarly; it is represented by powers of 2. To successfully get past the last hole, it would require (statistically, at least) 2^100 or (by my trusty pocket calculator) 1.2676506 *10^30 slides. More significant figures? dc! Which gives 1267650600228229401496703205376. In similar logic as the last problem, I'd expect about half that, or 633825300114114700748351602688 slides. How much time would this be? Excluding leap years, I calculate 20098468420665737593491 years. That's 20 sextillion years, significantly more than the age of the universe, by about 11 orders of magnitude. So I'd guess that no one will ever reach the bottom, they'll all try and fail (assuming everyone only gets to go once), or die waiting in line. Diversion -- "I can see 'em | "Want me to create a diversion?" I can see 'em | Diversion Someone wake me when it's over" | rogerj@rpi.edu ------------------------- In article <1992Oct23.160130.166012@watson.ibm.com> you write: Title: Cliff Puzzle 12: Slides in Hell >Consider a metallic slide with 10 large holes in it equally spaced from >top to bottom. If you attempt to slide down the slide you have a 50% >probability of sliding through each hole in the slide into an >oleaginous substance beneath the slide during each encounter with a >hole. > >1. If you were a gambling person, which hole would you bet a person >would fall through? None. The best chance is the first hole but I got a 50-50 chance. Why bother? (2nd hole is 1/4, 3rd 2**-3, ...) >2. If you were a gambling person, how many attempts would it require >for a person to slide from the top of the slide to the bottom without >falling through a single hole. No gurantee. Each slide is an independent event. Now, if you are talking mere probability, on the average, one in 1024 slides may make it through all 10 holes. >3. If all the people on earth lined up to go down the slide, and they >slid down a more horrifying slide with 100 holes at a rate of 1 person >per second, when would you expect the first person to arrive at the >bottom of the slide without falling through. An hour? A day? A decade? Again, can't tell. It could be the first one, it could be none. Probablity can not foretell actual events. But if you have infinite number of people sliding down till eternity, on the average, you may see 1 person slide over all holes every (2**100)/(365*24*69*6) years. This number is many times bigger than the world population for now. ------------------------- Some answers to your questions: 1. As the puzzle states there is a 50% chance of falling into each hole, I would bet a person would fall into the first hole -- in a large enough sample, 1/2 of the people will fall through the first hole, 1/4 through the second, 1/8 through the third, etc. 2. In a large sample, 1/(2^10) people would make it all the way down the slide without falling through any of the holes (1/1024). This means that 1023 out of 1024 people would fall through a hole. Using the formula (1023/1024)^x=1/2, we can determine out of the first x people to go down the slide, there is a 50% chance that one person will make it down without falling through a hole. The answer to this equation is x=709.4 Thus I would bet that a person would make it all the way down on one of the first 710 attempts. 3. As 2^100=1.2676*10^30 (roughly), and (including leaps years under the Gregorian calendar) there are 31556952 seconds in the average year, then statistically one person should make it down the slide every 4.017*10^22 YEARS. However, and this is a very rough estimate, I figure the log of (1-1/(1.2676*10^30)) to be about -5.5*10^(-29). [I'm doing the calculations on a scientific calculator which only has 10 places.] Thus, using the formula xlog(1-1/2^100)=log(1/2), I get x=5.5*10^27. Thus, there's about a 50% chance that after 5.5*10^27 seconds, someone will have made it down the slide. To be on the safe side, I'd bet only if I were given at least 6*10^27 seconds, a value which equals 1.901*10^20 YEARS. I hope this answers the questions. Ted Schuerzinger email: J.Theodore.Schuerzinger@Dartmouth.EDU snailmail: HB 3819 Dartmouth College Hanover, NH 03755 USA In case you're wondering, I'm just a junior at Dartmouth who's interested in puzzles like these. I'm not even a math major -- I'm a double major in government and Russian. ------------------------- In article <1992Oct23.160130.166012@watson.ibm.com> you write: >Title: Cliff Puzzle 12: Slides in Hell >From: cliff@watson.ibm.com >Consider a metallic slide with 10 large holes in it equally spaced from >top to bottom. If you attempt to slide down the slide you have a 50% >probability of sliding through each hole in the slide into an oleaginous >substance beneath the slide during each encounter with a hole. > >1. If you were a gambling person, which hole would you bet a person >would fall through? There's a 50% chance of falling through the first hole, 25% the second, 2^-n the n'th. If the odds offered were the same, I'd go for the first hole. >2. If you were a gambling person, how many attempts would it require >for a person to slide from the top of the slide to the bottom without >falling through a single hole. You expect to make it 1 out of 1024 times; after 710 tries, the chance of someone succeeding exceeds 1/2. (Log base (1023/1024) of 1/2 is 709.4). >3. If all the people on earth lined up to go down the slide, and they >slid down a more horrifying slide with 100 holes at a rate of 1 person >per second, when would you expect the first person to arrive at the >bottom of the slide without falling through. >An hour? A day? A decade? ... Never. OK, 1/2^100 will make it. There being under 2^33 people on the planet, ... After 4.2e22 years, the expected number of people who succeeded is 1; after about 2.9e22 years, the chance of someone having succeeded is about 1/2. Like I said, never. Seth sethb@fid.morgan.com ------------------------- In rec.puzzles you write: >1. If you were a gambling person, which hole would you bet a person >would fall through? If the pay-back odds were the same regardless of the hole, then obviously, I'd bet on the first hole! There's a 1:2 chance the person falls through the first hole, a 1:4 combined chance of the person falling though the second hole, etc... >2. If you were a gambling person, how many attempts would it require >for a person to slide from the top of the slide to the bottom without >falling through a single hole. 1024 is the median value for this case... There's a 1:2**n chance of a person falling through the nth hole, having missed all of the holes before n. Since the probability of falling through = the probability passing over the hole safely (vs not ever getting there), the probability that a person makes it to the end is also 1:1024. >3. If all the people on earth lined up to go down the slide, and they >slid down a more horrifying slide with 100 holes at a rate of 1 person >per second, when would you expect the first person to arrive at the >bottom of the slide without falling through. >An hour? A day? A decade? ... There is a 1:2**(100-Log2(5 billion people)) chance that somebody makes it through... Given a finite # of people on the planet (approx 5 bil.) I think we'll run out first... --Joseph Zbiciak im14u2c@camelot.bradley.edu ------------------------- Subject: Re: Cliff Puzzle 12: Slides in Hell (SPOILER) Newsgroups: rec.puzzles References: <1992Oct23.160130.166012@watson.ibm.com> In article <1992Oct23.160130.166012@watson.ibm.com>, Cliff Pickover writes: > Consider a metallic slide with 10 large holes in it equally spaced from > top to bottom. If you attempt to slide down the slide you have a 50% > probability of sliding through each hole in the slide into an oleaginous > substance beneath the slide during each encounter with a hole. > 1. If you were a gambling person, which hole would you bet a person > would fall through? The probability of falling into hole i is (1/2)^i, so your best bet would be hole 1. > 2. If you were a gambling person, how many attempts would it require > for a person to slide from the top of the slide to the bottom without > falling through a single hole. The probability of success is p = (1/2)^10, and as each trial is independant the expected number of trials before success is 1/p or 2^10. > 3. If all the people on earth lined up to go down the slide, and they > slid down a more horrifying slide with 100 holes at a rate of 1 person > per second, when would you expect the first person to arrive at the > bottom of the slide without falling through. In this case the number of expected trials is 2^100, which is much larger than the total number of people. > An hour? A day? A decade? ... Try about 10^24 years. As another problem, assuming a large enough supply of sliders estimate when the slide will wear through from friction. ------------------------- In article <1992Oct23.160130.166012@watson.ibm.com> you write: >Title: Cliff Puzzle 12: Slides in Hell >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please send me your name, >address, affiliation, e-mail address, so I can properly credit you if >you provide unique information. PLEASE ALSO directly mail me a copy of >your response in addition to any responding you do in the newsgroup. I >will assume it is OK to describe your answer in any article or >publication I may write in the future, with attribution to you, unless >you state otherwise. Thanks, Cliff Pickover > > * * * > >Consider a metallic slide with 10 large holes in it equally spaced from >top to bottom. If you attempt to slide down the slide you have a 50% >probability of sliding through each hole in the slide into an oleaginous >substance beneath the slide during each encounter with a hole. > >1. If you were a gambling person, which hole would you bet a person >would fall through? I'd bet that they fell through the first hole. The probability of that happening is 50%. The probability of them falling through the second hole is: P(didn't fall through the first)*P(fell through the second) = 50%*50% = 25% In general, P(falls through hole n)= P(no fall through 1)*P(no fall through 2)*...*P(no fall through n-1) *P(fell through hole n). For this problem, P(falls through hole n) is (50%)^n, where n is the hole # from the top. >2. If you were a gambling person, how many attempts would it require >for a person to slide from the top of the slide to the bottom without >falling through a single hole. (Hey, after the first failed attempt, they're screwed, no?) P(success)=P(no fail)=P(no fall 1)P(no fall 2)...P(no fall 10) =50%^10 =1/1024 They should make it at least one time in 1024. >3. If all the people on earth lined up to go down the slide, and they >slid down a more horrifying slide with 100 holes at a rate of 1 person >per second, when would you expect the first person to arrive at the >bottom of the slide without falling through. >An hour? A day? A decade? ... Oh, one in about 4.02*10^22 years... I wouldn't hold my breath. -Richard ------------------------- 1. I would bet on the first hole, as there is a 0.5 probability of a person's falling into it, which is the highest such probability. 2. The probability of reaching the end of the slide on a particular try is 1/2^10 = 1/1024. In 709 tries, there is an approximately 0.5 probability of 3. Beats me - the even money bet is for a number of tries (approximately) equal ((2^100 - 1)/(2^100)) calculate it. -- _______________________________________________________________________ Dan Blum Institute for the Learning Sciences Room 327 blum@ils.nwu.edu 1890 Maple Ave., Evanston, IL 60201 708-467-2306 "Let it be granted that a controversy may be raised about any question, and at any distance from that question." Lewis Carroll _______________________________________________________________________ ==> pickover/pickover.13.p <== Title: Cliff Puzzle 13: Ladders to Heaven From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider the following scenario. A standard ladder stretches from each country on the earth upward a distance equal to the distance from the earth to the moon. Assume: 1. the ladder is made out of a strong metal such as titanium, which will not break. 2. the ladder is inclined at a very steep angle, 70 degrees, for each country. 3. there is a breathable atmosphere. 4. the people (or teams of people) are allowed to use standard mountain climbing and camping gear, e.g. ropes, backpacks, etc. but not sophisticated electrical mechanisms, engines, etc. 5. a reward is given to whomever reaches the top of the ladder first: 1 million dollars to that person. In addition the country's national debt is wiped out. Questions: 1. Approximate how long it would take a person (or team of people) to reach the top of the ladder. Days? Weeks? Years? 2. Which country would be the first? 3. Is there any novel method you would suggest to achieve this goal? 4. Is this task impossible to carry out. ==> pickover/pickover.13.s <== ------------------------- Interesting puzzle... Just one question though: Is there a moon, i.e. is it possible to use the gravitational field of the moon to your advantage by "falling upwards" once you have reached the point where the moon's gravity is bigger than the erath's (and do we also assume that the the climber(s) must survive the fall?? :-) or shall we assume that the earth is alone in the universe? Spyros Potamianos potamian@hpl.hp.com ------------------------- Newsgroups: rec.puzzles Subject: Re: Cliff Puzzle 13: Ladders to Heaven References: <1992Oct23.193252.108077@watson.ibm.com> Organization: The Chrome Plated Megaphone of Destiny >1. Approximate how long it would take a person (or team of people) to >reach the top of the ladder. Days? Weeks? Years? Note that after you're 22,300 miles from the earth's axis, you get to "fall" the rest of the way, as long as you don't lose contact with the ladder. >2. Which country would be the first? It has already been pointed out that countries on the equator have an advantage. I suppose you could consider that countries with a large national debt have extra motivation. :-) >3. Is there any novel method you would suggest to achieve this goal? I would suggest a bicycle-like vehicle clamped to the ladder. By pulling a light but strong rope on a pulley (perhaps obtained form the same source as this fantastic ladder material), riders could be changed fairly quickly, thanks to a crew of brawny pulley-pullers with a variable-geared linkage to the rope. For the rider to pull this ever-longer rope seems impossible, but I think shorter segments could be lifted and linked. Or the ground crew could help the rider by pulling down rope from a hub of lesser diameter than the wheels of the vehicle. >4. Is this task impossible to carry out. No. I thought it might be impossible to halt at the far end of the ladder and return, due to centrifugal acceleration, but that acceleration turns out to be only about 5 cm/s^2. __________________________________________________________ Matt Crawford matt@severian.chi.il.us Java Man ------------------------- > How do we get food to the people? I would have the riders change so often that they'd only need some high-carbohydrate snacks and a couple quarts of fluid. I think the brawny ground crew could pull up the next rider, with his supplies and another pulley and segment of rope, at an acceleration of about 0.5 g or better. That would be under 90 minutes for each shift- change up to the synchronous orbit level. I haven't figured out yet how to link each new piece of rope that's pulled up with a rider to the pulley that's at the high point reached by the previous rider. Linking is easy, but it would be nice to find a way that lets the next pulled-up rider go from one segment to the other without interruption. Well, since the sky-buckets at Disneyland do this trick at each end, I know it can be done. I didn't know you'd written any books, but it was clear you're working on one now. Sure, send a list, but I have access to some on-line catalogs, so maybe I can find them anyway. Matt Crawford ------------------------- > Consider the following scenario. A standard ladder stretches from each > country on the earth upward a distance equal to the distance from the > earth to the moon. > > Assume: > 1. the ladder is made out of a strong metal such as > titanium, which will not break. > 2. the ladder is inclined at a very steep angle, 70 degrees, for > each country. > 3. there is a breathable atmosphere. > 4. the people (or teams of people) are allowed to use standard > mountain climbing and camping gear, e.g. ropes, backpacks, etc. but not > sophisticated electrical mechanisms, engines, etc. > 5. a reward is given to whomever reaches the top of the ladder > first: 1 million dollars to that person. In addition the country's > national debt is wiped out. I would imagine that one would be able to fashion a hot air balloon given condition 4. Also, given condition 3, the hot air balloon would be able to cover the entire distance. One would then only need to attach a sliding hookup between the ladder and the balloon and wait. ===M.Graf==graf@island.com================================================== ==> pickover/pickover.14.p <== Title: Cliff Puzzle 14: Geography Genuflection From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * 1. How would the world be different today, geopolitically speaking, if the ancient land masses had never drifted apart and, therefore, today's world consisted of a single supercontintent? 2. What would today's world be like if the land mass which formed the Greek peninsula never existed? 3. What would today's world be like if the land bridge which joined Alaska to Asia never existed? 4. Why do all the major peninsulas on earth point south? See for example: Italy, Greece, Florida, and Baja, and the tips of Africa, South America, India, Norway, Sweden, Greenland, and many other landmasses. ==> pickover/pickover.14.s <== ------------------------- In rec.puzzles you write: >If you respond to this puzzle, if possible please send me your name, >address, affiliation, e-mail address, so I can properly credit you if >you provide unique information. > Mike Neergaard University of Wisconsin neergaar@math.wisc.edu I'm not a professional at this sort of thing, so I just summarized my conclusions. I'm sure they would be ripped to shreds by any competent whatsit-type-individual-who-knows-all-about-this-kind-of-stuff. >1. How would the world be different today, geopolitically speaking, if >the ancient land masses had never drifted apart and, therefore, >today's world consisted of a single supercontintent? We would all speak German. >2. What would today's world be like if the land mass which formed the >Greek peninsula never existed? > We would know a low more about fluid dynamics. >3. What would today's world be like if the land bridge which joined >Alaska to Asia never existed? Christopher Columbus would be a national hero, instead of being vulnerable to counter-claims of genocide. America would have been settled several decades later, due to a dearth of demonstrable natural resources. >4. Why do all the major peninsulas on earth point south? See for >example: Italy, Greece, Florida, and Baja, and the tips of Africa, >South America, India, Norway, Sweden, Greenland, and many other >landmasses. I just work here . . . -- I really don't make any claim at all to know what I'm talking about. Actually, I make no claim to know what YOU'RE talking about, either. In fact, now I've forgotten what we were talking about . . . ------------------------- In article <1992Oct26.140330.142282@watson.ibm.com> you write: >Title: Cliff Puzzle 14: Geography Genuflection >From: cliff@watson.ibm.com > >If you respond to this puzzle, if possible please send me your name, >address, affiliation, e-mail address, so I can properly credit you if >you provide unique information. PLEASE ALSO directly mail me a copy of >your response in addition to any responding you do in the newsgroup. I >will assume it is OK to describe your answer in any article or >publication I may write in the future, with attribution to you, unless >you state otherwise. Thanks, Cliff Pickover > > * * * > Okay, administrative trivia first. My name is Martin Eiger, you don't need my address (home or business?), I don't want you citing my affiliation if you quote me, and my e-mail address is mie@thumper.bellcore.com. >1. How would the world be different today, geopolitically speaking, if >the ancient land masses had never drifted apart and, therefore, >today's world consisted of a single supercontintent? My theory is that mankind would never have evolved. The dominant species would still be some sort of mammal, but not us. This renders a large number of geopolitical questions irrelevant. For example, elephant-like creatures are unlikely to care whether there is one or two Germanys. >2. What would today's world be like if the land mass which formed the >Greek peninsula never existed? A tough one, since I'm not up on my Greek influences in the evolution of civilization. My guess is that civilization would have evolved anyway, probably not too differently than it did. It might not have evolved as fast, i.e., we might now be where we were a thousand years ago or so, but over the long haul, human history would follow a similar course. >3. What would today's world be like if the land bridge which joined >Alaska to Asia never existed? Pretty much the same, I bet. People would have colonized North America anyway. After all, they got to Hawaii, so somebody could probably have gotten to North America. And whether or not people colonized North America from across the Pacific, people from Europe would have paved the place over just the same. >4. Why do all the major peninsulas on earth point south? See for >example: Italy, Greece, Florida, and Baja, and the tips of Africa, >South America, India, Norway, Sweden, Greenland, and many other >landmasses. First of all, you have to define what's a major peninsula. Secondly, I don't like your list. Norway and Sweden are on the same peninsula, and Greenland is an island, not a peninsula. And third, there are plenty of perfectly fine peninsulas that don't point south: Alaska, Siberia, Michigan (two peninsulas for the price of one), Yucatan, Arabia (points kind of southeast), and Iberia, for instance. And fourth, you missed a few good southern-pointing ones, such as Korea, Crimea, the Sinai, and the one that kind of points from eastern Siberia toward Japan that I'm sure has a name but I don't know it. So while there are lots of peninsulas pointing lots of directions, a majority of them do seem to point south, and I have no idea why. ==> pickover/pickover.15.p <== Title: Cliff Puzzle 15: Cherries in Wine Glasses From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider a 9x9 grid of beautiful crystal wineglasses. Throw 32 cherries at the grid. A glass is considered occupied if it contains at least one cherry. (With each throw a cherry goes into one of the glasses.) How many different patterns of occupied glasses can you make? (A glass with more than one cherry is considered the same as a glass with one cherry in the pattern). 2. Same as above except that you place 8 cherries in glasses (x,y) and then determine the other positions by placing cherries at (x,-y), (-x,y), (-x,-y) leading to 32 cherries in the grid. Consider the array of glasses centered at the origin. How many different patterns of occupied glasses can you make? (A glass with more than one cherry is considered the same as a glass with one cherry in the pattern). 3. Can your results be extrapolated to an NxN grid with M cherries thrown at it for both problems? ==> pickover/pickover.15.s <== In article <1992Oct30.173903.108937@watson.ibm.com> you write: : Consider a 9x9 grid of beautiful crystal wineglasses. Throw 32 cherries : at the grid. A glass is considered occupied if it contains at least one : cherry. (With each throw a cherry goes into one of the glasses.) How : many different patterns of occupied glasses can you make? (A glass with : more than one cherry is considered the same as a glass with one cherry : in the pattern). Assuming that rotated patterns are allowed, then it is (simply) sum( 81!/(81-n)! , n=1->32) . Since, if a total of n different classes are filled, then the number of combinations is 81!/(81-n)!. Since there can be from 1 to 32 glasses filled, the total # is just the sum of these... : : 2. Same as above except that you place 8 cherries in glasses (x,y) and : then determine the other positions by placing cherries at (x,-y), : (-x,y), (-x,-y) leading to 32 cherries in the grid. Consider the array : of glasses centered at the origin. How many different patterns of : occupied glasses can you make? (A glass with more than one cherry is : considered the same as a glass with one cherry in the pattern). This limitation basically reduces the number of available spots, from 9x9 to 5x5. Also, I only have to worry about 8 occupied spaces. Soo... #of comb. = sum( (25!/(25-n)!, n=1->8) : : 3. Can your results be extrapolated to an NxN grid with M cherries : thrown at it for both problems? With a odd N, and M = 4k (evenly divs by 4), then for 1.... #of comb = sum( (N^2)!/(N^2-n)! , n=1->M) for 2.... #of comb = sum( (((N+1)/2)^2)!/(((N+1)/2)^2-n)! , n=1->M/4) -- Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD! // | Senior, Chemical Engineering, Univ. of Toledo \\ // Only the | Summer Intern, NASA Lewis Research Center \ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu / --------+ How do YOU spell 'potato'? How 'bout 'lousy'? +---------- "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L ==> pickover/pickover.16.p <== Title: Cliff Puzzle 16: Undulating Squares From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * A square number is of the form y=x**2. For example, 25 is a square number. Undulating numbers are of the form: ababababab... For example, the following are undulating numbers: 1717171, 282828, etc. 1. Are there any undulating square numbers? 2. Are there any undulating cube numbers? ==> pickover/pickover.16.s <== ------------------------- In article <1992Oct30.175102.142177@watson.ibm.com> you write: : 1. Are there any undulating square numbers? 11^2 = 121 : 2. Are there any undulating cube numbers? 7^3 = 343 (yes, I know they're short, but they qualify!) -- Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD! // | Senior, Chemical Engineering, Univ. of Toledo \\ // Only the | Summer Intern, NASA Lewis Research Center \ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu / --------+ How do YOU spell 'potato'? How 'bout 'lousy'? +---------- "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L ------------------------- In article <1992Oct30.204134.97881@watson.ibm.com> you write: >Hi, I was interested in non-trivial cases. Those with greater >than 3 digits. Award goes to the person who finds the largest >undulating square or cube number. Thanks, Cliff 343 and 676 aren't trivial (unlike 121 and 484 it doesn't come from obvious algebraic identities). The chance that a "random" number around x should be a perfect square is about 1/sqrt(x); more generally, x^(-1+1/d) for a perfect d-th power. Since there are for each k only 90 k-digit undulants you expect to find only finitely many of these that are perfect powers, and none that are very large. But provably listing all cases is probably only barely, if at all, possible by present-day methods for treating exponential Diophantine equations, unless (as was shown in a rec.puzzles posting re your puzzles on arith. prog. of squares with common difference 10^k) there is some ad-hoc trick available. At any rate the largest undulating power is probably 69696=264^2, though 211^3=9393931 comes remarkably close. --Noam D. Elkies ------------------------- In article <1992Oct30.175102.142177@watson.ibm.com>, you write... >1. Are there any undulating square numbers? > Other than the obvious 11**2, 22**2, and 26**2, there is 264**2 which equals 69696. >2. Are there any undulating cube numbers? > Just 7**3 as far as I can tell, though I'm limited to IEEE computational reals. PauL M SchwartZ (-Z-) | Follow men's eyes as they look to the skies v206gb6c@ubvms.BitNet | the shifting shafts of shining pms@geog.buffalo.edu | weave the fabric of their dreams pms@acsu.buffalo.edu | - RUSH - ==> pickover/pickover.17.p <== Title: Cliff Puzzle 17: Weird Recursive Sequence From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider the simple yet weird recursive formula a(n) = a(a(n-1)) + a(n-a(n-1)) The sequences starts with a(1) = 1, and a(2) = 1. The "future" values at higher values of n depend on past values in intricate recursive ways. Can you determined the third member of the sequence? At first, this may seem a little complicated to evaluate, but you can being slowly, by inserting values for n, as in the following: a(3) = a(a(2)) + a(3-a(2)) a(3) = a(1) + a(3-1) = a(3) = 1+1 = 2 Therefore, the 3rd value of the sequence a(3) is 2. The sequence a(n) seems simple enough: 1, 1, 2, 2, 3, 4, 4, 4, 5, ... Try computing a few additional numbers. Can you find any interesting patterns? The prolific mathematician John H Conway presented this recursive sequence at a recent talk entitled "Some Crazy Sequences." He noticed that the value a(n)/n approaches 1/2 as the sequence grows and n becomes larger. Can you find a value, N, above which the sequence the value of a(n)/n is always within 0.05 of the value 1/2? (In other words, .eq vbar a(n)/n -1/2 vbar lt 0.05. The bars indicate the absolute value.) A difficult problem? you ask. John Conway offered $10,000 to the person to find the s-m-a-l-l-e-s-t such N. A month after Conway made the offer, Colin Mallows of AT&T solved the $10,000 question: N = 3,173,375,556. Manfred Shroeder has noted that the sequence is "replete with appealing self-similarities that contain the clue to the problem's solution." Can you find any self-similarities? As I write this, no-one on the planet has found a value for the smallest N such that a(n)/n is always within 0.01 of the value 1/2. .eq (vbar a(n)/n -1/2 vbar lt 0.01. ) ==> pickover/pickover.17.s <== ------------------------- In article <1992Nov06.160358.101157@watson.ibm.com> you write: : Title: Cliff Puzzle 17: Weird Recursive Sequence : Consider the simple yet weird recursive formula : a(n) = a(a(n-1)) + a(n-a(n-1)) The first 32 terms, and the ratio a(n)/n for each is as follows... n a(n) a(n)/n 1 1 1.0 2 1 1.0 3 2 .666 4 2 .5 5 3 .6 6 4 .666 7 4 .5714 8 4 .5 9 5 .5555 10 6 .6 11 7 .6363 12 7 .5833 13 8 .6153 14 8 .5714 15 8 .5333 16 8 .5 17 9 .5294 18 10 .5555 19 11 .5789 20 12 .6 21 12 .5714 22 13 .5909 23 14 .6086 24 14 .5833 25 15 .6 26 15 .5769 27 15 .5555 28 16 .5714 29 16 .5517 30 16 .5333 31 16 .5161 32 16 .5 33 17 .... and so and.... off the top, we can see that on the 2^k (k a pos. int) terms, the ratio goes to .5 between each of these, the ratio goes up and then drops back to .5 (ignoring the variances due to integer arithmatic) the value of n at the maximum in each jump is halfway between the two 2^k points. The value of a(n) at those points seems to be 2^(k-1) - f(k), where f(k) is some function that I cannot determine without more computing power.... *sniff* Therefore, we must find a value of x such that... (2^(x-1)-f(x))/2^x - .5 <.05 (or whatever) or f(x)/2^x < .05 and then N would be .5*(2^x-2^(x-1)) if I could see the next terms up to 128, I might be able to calculate it... -- Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD! // | Senior, Chemical Engineering, Univ. of Toledo \\ // Only the | Summer Intern, NASA Lewis Research Center \ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu / --------+ How do YOU spell 'potato'? How 'bout 'lousy'? +---------- "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L ------------------------- In article <1992Nov06.160358.101157@watson.ibm.com> you write: >John Conway offered $10,000 to the person to find the s-m-a-l-l-e-s-t >such N. A month after Conway made the offer, Colin Mallows of AT&T >solved the $10,000 question: N = 3,173,375,556. As I pointed out in my posting, this is incorrect, and differs from Mallows' correct answer published in his article. But a bit of investigation shows that the above N is hardly a random guess, either. Conway's sequence is best understood by analyzing it on "levels", where the k'th level is the set of integers between 2^k and 2^(k+1). It turns out that Mallows' correct answer, 6083008742, lies on level 32, and the largest candidate answer on level 31 is N=3173375556, the number quoted above. Where did you see the above value of N given as the answer to Conway's question? -tal kubo@math.harvard.edu p.s. As I found out when I edited my posted response to your message, you either use lines longer than 80 characters in your postings, or else your editor appends extra linefeeds to each line. Since both conditions could be problematic for a lot of people who read your messages on rec.puzzles, you might want to correct this condition. ==> pickover/pickover.18.p <== Title: Cliff Puzzle 18: Difficult Nested Roots From: cliff@watson.ibm.com If you respond to this puzzle, if possible please send me your name, address, affiliation, e-mail address, so I can properly credit you if you provide unique information. PLEASE ALSO directly mail me a copy of your response in addition to any responding you do in the newsgroup. I will assume it is OK to describe your answer in any article or publication I may write in the future, with attribution to you, unless you state otherwise. Thanks, Cliff Pickover * * * Consider the following nested set of square roots. .eq ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>> Here, G indicates "Googol" or 10**100. The "<" and ">" symbols indicate where the beginning and ends of the the nested roots. 1. What is the value for in this infinite set of nested roots. 2. What is the next term under the root? Hint: In 1911, the famous mathematical prodigy Srinivasa Ramanujan posed the following question (#298) in a new mathematical journal called the :Journal of the Indian Mathematical Society. .eq ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>> ==> pickover/pickover.18.s <== ------------------------- In article <1992Nov11.221749.129578@watson.ibm.com> you write: : Title: Cliff Puzzle 18: Difficult Nested Roots : From: cliff@watson.ibm.com : Consider the following nested set of square roots. : : ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>> : : Here, G indicates "Googol" or 10**100. : The "<" and ">" symbols indicate where the beginning and ends of the : the nested roots. : : 1. What is the value for in this infinite set of nested roots. : 2. What is the next term under the root? : Hint: : In 1911, a twenty-three-year-old Indian clerk named Srinivasa Ramanujan : posed the following question (#298) in a new mathematical journal called : the Journal of the Indian Mathematical Society. : : ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>> : Doing a n-depth thing-a-ding on this..... n=1 v=1 2 1.732 3 2.236 4 2.5598 5 2.7551 6 2.867 .... 20 2.99999376 .... so I expect that the sum is actually 3. Or in the general case when the 2 (or the G from above) is replaced by m, then the evaluation of the series is m+1. This CAN be shown as follows.... m+1 = sqrt(1+m sqrt(1+(m+1)*sqrt(....)) m^2 + 2m +1 = 1 + m *sqrt(1 + (m+1)*sqrt(...)) m^2 + 2m = m*sqrt(1+(m+1)*sqrt(...)) m+2 = sqrt(1+(m+1)*sqrt(1+(m+2)*sqrt(...)) Thus if m+1 is then sum when the series is based off m, then m+2 is then sum when the series is based off m+1. Since this works for m=2 (as shown above), then it must work for all whole numbers (mathematical induction is such a wonderful thing...) Therefore, the sum with m=G is G+1. The next term, as show above, is (1+(m+2)*sqrt(1+....)) -- Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD! // | Senior, Chemical Engineering, Univ. of Toledo \\ // Only the | Summer Intern, NASA Lewis Research Center \ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu / --------+ How do YOU spell 'potato'? How 'bout 'lousy'? +---------- "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L User Contributions: |
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rec.puzzles Archive (pickover), part 30 of 35
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