7 ms·
Who Can Name the Bigger Number?
- rookie 16y agoa more interesting challenge would be the biggest number using each of the ten digits only once. ready, go.
- sesqu 16y ago9876543210 If I'm also allowed an operator, 3↑⁹⁸⁷⁶⁵⁴¹⁰2
- rookie 16y agoha, yes allowed operators, as many as youd like.
- sesqu 16y agoThen the best I can come up with on no notice is 2 ↑^(4 ↑^(6 ↑^(8↑^(9↑10)) 7) 5) 3. That would be trivial to increase substantially by using some other notation for the hyperoperation, but I'm going with the presumably most familiar one. I'm sure there are bigger numbers - specifically, Wikipedia mentions Conway chains, where a→b→c = a ↑^c b, and so 2→3→4→5→6→7→8→9→10 should be quite large indeed.
- baddox 16y agoExcellent! I hadn't thought about it, but at a glance I'm pretty sure your ordering and grouping does generate the maximal result. However, the parent comment said as much punctuation as you like, which allows for unbounded expressions with unary operators like the factorial!
- deleted 16y ago[deleted]
- pmb 16y ago9876543210!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
- lakeeffect 16y ago10!^2!^3!^4!^5!^6!^7!^8!^9!
- Retric 16y agoHA! 2^3!^4!^5!^6!^7!^8!^9!^10! ;-) (2! = 2) and (2^x) > (x ^2) where x > 3. Then again 2^3^4^5^6^7^8^9^(10!!!!!!!!!) is larger. Also if it's it's simply a question of symbols 9!! is larger than 9^9 so 9 followed by n ! is probably your best bet without going into higher level math like BB(BB(...(9))...)
- kenjackson 16y agoOr what if you restricted the number of symbols to be no more than 20 symbols total.
- die_sekte 16y ago10→9→8→7→6→5→4→3→2 Somehow I think that Conway's chained arrow notation is cheating.
- spoonboy 16y agoJust for fun, here's my shot: The number of steps required to compute the solution to the travelling salesman problem for each quark of every subset of every permutation of quarks in the universe in space-time.
- InclinedPlane 16y agoI think that breaks the rules. Here's something simple I came up with: P(1) = 10 P(n+1) = 10^P(n) Q(n) = P(n)^P(n) Z(1) = Q(1) Z(n+1) = Q(Q(n)) And then: Z(10^100)
- btilly 16y agoIn a somewhat restricted version of this contest some time ago the best entry was my: B(b, c, 0) = 0 i < b => B(b, c, i + j * b) = i + B(b, c, j) * c T(b, 0) = b * b T(b, n+1) = T(T(b, n), B(b, T(b, n), n)) Z = T(2, T(2, T(2, 9))) It is much, much larger than the number you just threw out. See http://bentilly.blogspot.com/2010/03/large-numbers.html http://bentilly.blogspot.com/2010/03/large-numbers.html for an explanation.
- shalmanese 16y agoThat's just 2^10^83
- vecter 16y agoI'm not sure if quarks have "positions" in the exact sense of the word, so I don't think there's a metric you could apply here to get distances for which to solve TSP. Suppose there was a metric between quarks. To be clear, are you saying: take all quarks in the universe. Take their powerset P. For each set S in P, create the complete graph K_{|S|} where edge lengths are distances. Let TSP(x, K_{|S|}) be the number of steps required to solve TSP. Sum TSP(K_{|S|}) for all S in P. Let n be the number of quarks in the universe. The size of the powerset is 2^n. For each S in P, since TSP is solvable in exponential time, we'll upper bound the number of steps by 2^n. In other words, for each powerset, you're cost for solving TSP is at most 2^n. Therefore, the number of steps is at most 2^n * 2^n = 2^(2n), which is far smaller than say Ackermann(6, 6).
- deleted 16y ago[deleted]
- robinhouston 16y ago“Who can name the bigger number? Whoever has the deeper paradigm.” In 2001, David Moews ran a BIGNUM BAKEOFF competition on the comp.lang.c newsgroup, where each entry had to be a C program (512 non-whitespace characters or fewer) returning a number from main(). The winner is the program that returns the largest number, assuming an arbitrarily-large int type. http://groups.google.com/group/comp.lang.c/browse_thread/thread/1511694ab8570ccc/540218fdfa4b2ab8 http://groups.google.com/group/comp.lang.c/browse_thread/thr... After reading Scott Aaronson's wonderful essay, you won't be surprised to hear that the contest was won by a theoretical computer scientist, Ralph Loader. (Second place was taken by Heiner Marxen, who is mentioned in Scott's article in connection with the Busy Beaver problem.) Ralph Loader's remarkable program is at http://homepages.ihug.co.nz/~suckfish/busy/reduced.c http://homepages.ihug.co.nz/~suckfish/busy/reduced.c, and the documented and unmangled version in http://homepages.ihug.co.nz/~suckfish/busy/busy.tar.gz http://homepages.ihug.co.nz/~suckfish/busy/busy.tar.gz David Moews' discussion of the entries is at http://djm.cc/bignum-results.txt http://djm.cc/bignum-results.txt
- tel 16y agoI find it hilarious that three people thought to return the highest possible number by underflowing unsigned ints (that are being treated as arbitrarily large ints). It's a funny hack even if it doesn't teach you anything like Loader's does.
- amalcon 16y agoMy first thought was along those lines: (int)((unsigned int)-1 >> 1)==INT_MAX
- mseebach 16y agoI don't see how that is relevant when the problem assumes an arbitrarily large int-type. It's trivial to return the largest possible value (INT_MAX) for any fixed-size int-type, but a such does not exist. Basically, many posters are missing the point that it's a mathematics challenge, not a C-challenge, which is excusable since it's posted in a C-group.
- arethuza 16y agoGraham's Number? http://en.wikipedia.org/wiki/Graham%27s_number http://en.wikipedia.org/wiki/Graham%27s_number
- baddox 16y agoLet g = Graham's number. Then use Knuth's up arrow notation. Then (and this is perhaps cheating) use a non-computable function like the busy beaver function, which grows faster than ANY computable function. That last step may in fact not be cheating, because it might be possible to prove a lower bound and therefore prove that it's higher than the opponent's number. Pretty notation here: http://mathbin.net/50035 http://mathbin.net/50035
- jmatt 16y agoThis is what I was thinking. Knuth's up arrow on Graham's number. I like the addition of the busy beaver function. A creative improvement for sure. So far I think you win :) for the lazy amongst us: http://en.wikipedia.org/wiki/Knuths_up-arrow_notation http://en.wikipedia.org/wiki/Knuths_up-arrow_notation
- Timothee 16y agoIf you use Busy Beavers, shouldn't you go all out and do something like that: BB(BB(BB(BB(BB(…g↑↑↑↑g))))) or a mix thereof? edit: I'd actually try BB(BB(BB(BB(BB(…g↑↑↑↑g)))))+1, just in case.
- baddox 16y agoThere's no end to that. You could define a function BZ(n) that equals BB(BB(BB...(g↑↑↑...g)...)) with n nested BB's and n up arrows. Then you could define a function RBB(n) that equals BZ(BZ(BZ...(g↑↑↑...g)...)) with n nested BZ's and n up arrows.
- joe_the_user 16y agoIf the busy beaver function runs on a computer, it is computable, unless I missed something pretty interesting.
- lakeeffect 16y agoPut the Factorial, exclamation mark after each exponent.
- diziet 16y agoSomething like (9||||9)! Using knuth's up arrow notation http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation And a factorial for good measure
- baddox 16y agoI certainly demolished your number in this post: http://news.ycombinator.com/item?id=1539866 http://news.ycombinator.com/item?id=1539866 Also, why use | when you can use ↑? ;)
- diziet 16y agoYou're right, you've got me certainly beat by unimaginable orders of magnitude. How does g↑(sub_g)g compare with BB(g)?
- baddox 16y agoI have no idea how they compare. The former is incomprehensible and the latter is both incomprehensible and non-computable. My intuition tells me the latter is much larger, because a computer could "easily" calculate the former.
- buzzblog 16y agoMy daughter at bedtime says she's giving me infinity times infinity time infinity kisses. That's the biggest number, I don't care what any of you math whizzes have to say about the matter. ... OK, let's call it the best not the biggest.
- Angostura 16y agoIt's going to be a long night.
- Avshalom 16y ago2 things 1) http://qntm.org/planar http://qntm.org/planar obviously not the largest number but pretty big. 2) http://www.ephraimkishon.de/en/my_favorite_stories.htm http://www.ephraimkishon.de/en/my_favorite_stories.htm The story Jewish Poker is at least one take on the old joke mentioned in the intro, and a fun read.
- ars 16y agoDup: http://news.ycombinator.com/item?id=47408 http://news.ycombinator.com/item?id=47408 http://news.ycombinator.com/item?id=951095 http://news.ycombinator.com/item?id=951095 http://news.ycombinator.com/item?id=492615 http://news.ycombinator.com/item?id=492615 http://news.ycombinator.com/item?id=514955 http://news.ycombinator.com/item?id=514955
- codexon 16y agoVery interesting that it was submitted so many times yet only now has it gotten this many votes.
- ars 16y agoA comment with this link did get some interest though: http://news.ycombinator.com/item?id=378776 http://news.ycombinator.com/item?id=378776
- unwantedLetters 16y agoWhat surprises me is that hacker news still let these dupes get posted. I was under the impression that a URL once submitted cannot be resubmitted. I've submitted duplicate articles before (unknowingly) and it simply acts as an upvote for an article that has already been submitted. I was thinking perhaps it lets you re-post after a certain predefined amount of time, but between two of the postings, there were only 18 days.
- astine 16y agoIf you add URL parameters, you can bypass the uniqueness requirements. Theoretically, you could submit the same URL an infinite number of times, with different combinations of parameters and never trigger the uniqueness check.
- noonespecial 16y ago/Eagles music, singing/ welcome to the hotel David Hilbert...
- Rhapso 16y ago"The traveling salesman problem asks for the shortest route connecting a set of cities, given the distances between each pair of cities. The rub is that the number of possible routes grows exponentially with the number of cities." Is it just me or does the traveling salesman problem have (cities-1)! solutions? thus it is not exponential? and is in fact far more annoying?
- Eliezer 16y agoWhat I'd want to say here is "the largest number nameable using 1000 symbols in second-order logic" (not original, I think someone else won a contest that way). But the fact that someone can ask "Which formulation of second-order logic?" probably defeats this unless you can also, in those fifteen seconds, cite a standard formulation of second-order logic and a standard axiomatization of number in that formulation, otherwise it's an ambiguous specification. HOL is a standard and formal formulation of higher-order logic that I can think of in fifteen seconds, but I wouldn't be able to think of a standard formulation of Peano Arithmetic that was already in HOL. Actually, now that I think of it, "the largest number nameable using 1000 symbols in HOL plus a successor symbol" might work, since you can define addition, multiplication, etc. in higher-order logic using only a successor symbol. Interesting fact: This is a fundamental jump up from first-order logic, which is a fundamental jump up from Busy Beaver, and then once you're there, it's operating at the highest level of abstraction that I know about or have ever heard of. There are things you can do to make that number bigger - picking a different ordinal for the order of logic, recursing over how you get the number of symbols - but they all amount to saying "plus one", and not jumping to a qualitatively different level of abstraction. Once you abstract over second-order logic you are, so far as I know, done.
- deleted 16y ago[deleted]
- andrewljohnson 16y agoRead, then comment.
- Eliezer 16y agoI beg your pardon? What did I miss? (In case it happens to be you and not I who missed something: the said number is formally specified though uncomputable, just as much as saying BusyBeaver(100); it is not paradoxical like using English to talk about English; and it is also vastly larger than anything you can express compactly by talking about Busy Beavers and oracle machines, which was the top level of the original article.)
- 16y ago
- joe_the_user 16y agoOne of the fundamental points of Kolmagorov's algorithmic information theory is that the function for the largest number algorithmically representable by n bits of information is incomputable. http://en.wikipedia.org/wiki/Kolmogorov_complexity http://en.wikipedia.org/wiki/Kolmogorov_complexity
- tman 16y agoa1 = 10^10 a2 = a1^a1 ... a_n = a_(n-1)^a_(n-1) My number is a_(a_(10^10))
- Jun8 16y agoSurely not every day, but frequently enough, I read a submission and related comments on HN and say to myself, "Ahhh, this is why I continue to come to HN!". This is definitely one of those.
- Timothee 16y agoI have seen it hovering here and there on the front page, and finally opened it. No regrets, it was probably the best submission I've read here.
- rw 16y agoDefinitely check out his other essays too!
- dennisgorelik 16y agoIt looks like many commentators did not bother to read the full article. Neither did I, but I remember not to use "infinity" and "number of atoms in universe".
- exit 16y ago> Be precise enough for any reasonable modern mathematician to determine exactly what number you’ve named, by consulting only your card and, if necessary, the published literature. but there isn't enough entropy left before cosmic heat death to evaluate 9^(9^9).
- richieb 16y agoAleph Null?