Oh ****!
(thats what i can say now :D ) i just joined the thread and i read last few mails not the whole lot So soory :)
i'll soon post a new one ;)
Aamir!
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Oh ****!
(thats what i can say now :D ) i just joined the thread and i read last few mails not the whole lot So soory :)
i'll soon post a new one ;)
Aamir!
you know i have a coool riddle.
here it is:
three poker players (A, B and C)played three games.In the first game A lost and B and C doubled their money, In the second game B lost and A and C doubled their money, In the third the first game C lost and A and B doubled their money. In the end every one had $24. How much did they have in the beggining.
by the way here is the up-to-date score:
1. Gabriel 6pts
2. me 5pts
3. Saturno, Simon 4pts
4. Elrond 3pts
5. solarflare, John - 2pts
6. dimm_coder, Xplorer - 1pt
ps solarflare please ask questions that can only have one logical answer... and if an answer doesn't deni the condition it must be considered correct
A had $39, B had $21 and C had $12.Quote:
Originally posted by SeventhStar
you know i have a coool riddle.
here it is:
three poker players (A, B and C)played three games.In the first game A lost and B and C doubled their money, In the second game B lost and A and C doubled their money, In the third the first game C lost and A and B doubled their money. In the end every one had $24. How much did they have in the beggining.
If we say x, y, and z is what they had at the beginning, and a, b, and c is what each of them lost in each game, we can write the correct equations (and feed them in Maple :) )
Code:restart;
> eq1:=x+y+z=72;
> eq2:=x-a+2*y+2*z = 72;
> eq3:=2*(x-a)+2*y-b+4*z=72;
> eq4:=4*(x-a)=24;
> eq5:=2*(2*y-b)=24;
> eq6:=4*z-c=24;
eq1 := x + y + z = 72
eq2 := x - a + 2 y + 2 z = 72
eq3 := 2 x - 2 a + 2 y - b + 4 z = 72
eq4 := 4 x - 4 a = 24
eq5 := 4 y - 2 b = 24
eq6 := 4 z - c = 24
> solve({eq1, eq2, eq3, eq4, eq5, eq6}, {x, y, z, a, b, c});
{y = 21, a = 33, z = 12, b = 30, c = 24, x = 39}
You're one again correct Gabriel... ;) and win a point
Ask a question now
Cool! This is indeed a good problem to solve! The tricky part is that they all have the same distance travelled = 120! But the problem is the total distance travelled by the FLY..The graphical journey of the fly might beQuote:
So... first question:
Two camels are galloping towards each other through a desert with a speed of 60 km/h (fast isn't it). A fly keeps flying from the nose of one camel to the nose of the other with a speed of 200 km/h (even faster). Two hours after starting the gallop the camels meet. Each one has passed 120 km. But what distance has the fly flown?
trick question
<- -
>- - - - -> - ->
<- - - - - - - - - - -< - - - - - <
- - - - -> - - - - - -> - - - - - - - - - - -> - - - - - - - - ->
--------------------------------------------------------------------------
desert
Ooohh nnooo!! sorry guyz I was lost!!
my!Quote:
Originally posted by Thread1
Cool! This is indeed a good problem to solve! The tricky part is that they all have the same distance travelled = 120! But the problem is the total distance travelled by the FLY..The graphical journey of the fly might be
<- -
>- - - - -> - ->
<- - - - - - - - - - -< - - - - - <
- - - - -> - - - - - -> - - - - - - - - - - -> - - - - - - - - ->
--------------------------------------------------------------------------
desert
That was asked pretty long ago, man! :D
Gabriel, will you ask a question or do you leave the the place of the asker open...
Go on, aks one if you have a good one.
I'll ask one but wait 5-10 mins because I've got to translate it first.
You are given names of numbers in an unknown language.
Names of the base numbers do not have same roots and do not have morphological alterations when combined with others.
You know that complicated numbers are formed thusly:
1. if the part 'i' is between two numbers the new one is their sum ('2'i'5' is 7)
2. if two numbers are written right after one another this means multiplication ('2''5' = 10)
you know these numbers:
14 = farokobarwonhul
29 = omaryhutarhul
37 = babumbarwonhul
39 = ganalibarwonhul
48 = babusgutarhul
52 = farokadonzesgutarhul
59 = babutfarokhul
64 = gutaragutarhul
71 = urukigutarhul
89 = barwonybabuhul
106 = omarifarokhul
112 = gutarobabuhul
128 = omarturukhul
149 = ganalsurukhul
199 = barwoniganalhul
208 = barwonadonzemganalhul
233 = gutarmomarhul
242 = donzemomarhul
none of the numbers is one-letter. there are some one-letter particles apart from 'i' that indicate some kind of order. The whole system's idea is the formation of numbers in numerical intervals.
questions:
write in this language:
178, 189, 351, 372
what is:
'barwonadonzesbarwonhul'
and
'omarydonzehul'
:D GOOOD LUCK!!!!
(it took me 2 days to solve it let me see you)
I don't understand what you mean by:
Quote:
if the part 'i' is between two numbers the new one is their sum ('2'i'5' is 7)
for exapmle if
blabla - is a base number and it's 76
and
slon - is a base number and it's 30
then
blablaislon - is a formed number and it's 106 (76+30)
(sloniblabla is the same)
slonblabla is 2280 (76x30 or 76*30 if you like)
(blablaslon is the same)
It's probably quite easy if you know how to work with Maple or MATLAB or so. I don't have the time to write it out.
nope! the system is quite different. I'm afraid they won't help a bit but it's quite a complicated task. Maybe solarflare will have something for tonight :DQuote:
Originally posted by Simon666
It's probably quite easy if you know how to work with Maple or MATLAB or so. I don't have the time to write it out.
Is this pointed towards me? :DQuote:
Originally posted by Simon666
It's probably quite easy if you know how to work with Maple or MATLAB or so. I don't have the time to write it out.
Yep, I think it is a (huge) set of equations again, in spite of what SeventhStar says. However, you need to try out and feed in several integers to see what integers give good results for the different substrings (hul,momar,gutar)Quote:
Originally posted by Gabriel Fleseriu
Is this pointed towards me? :D
if you say so...Quote:
Originally posted by Simon666
Yep, I think it is a (huge) set of equations again, in spite of what SeventhStar says. However, you need to try out and feed in several integers to see what integers give good results for the different substrings (hul,momar,gutar)
;) :p
but dont forget this
none of the numbers is one-letter. there are some one-letter particles apart from 'i' that indicate some kind of order. The whole system's idea is the formation of numbers in numerical intervals.
Like "a" an "y". :cool:Quote:
Originally posted by SeventhStar
there are some one-letter particles apart from 'i' that indicate some kind of order
maybe
:D ;) :p :) :cool:
That's a toughy, I'll have to think for a while on this one ;).
Question about the numbers problem: how is the order of operations handled? Is it same as in arabic mathematics, or just left to right, or some other system?
Also, are the base numbers all positive integers?
Well the system is pretty bizzare...Quote:
Originally posted by solarflare
Question about the numbers problem: how is the order of operations handled? Is it same as in arabic mathematics, or just left to right, or some other system?
Also, are the base numbers all positive integers?
And yes all base numbers are positive integers
I found a mistake in my question... :rolleyes:
well it's not a real mistake.
i wrote (at the end):
what is:
'barwonadonzesbarwonhul'
but i wanted to write
what is:
'barwonadonzesbarwonhulhul'
sorry though I doubt it's a big problem for anybody
I think that if today noone answers the problem, tommorow I'll tell the answer and write myself a point ;)
Maybe the next one must be easier :rolleyes:
Are you sure there aen't multiple mistakes in what you wrote down? Shouldn't it be
barwonyoroganalhul
or some other letter instead of
barwoniganalhul
?
Definitively forget this please!!Quote:
Originally posted by Doctor Luz
I think is a good idea to let few days to answer and make more difficult questions.
Please go back to the easy questions. Do it for my mental health, I don't want to get mad, I don't want to have more nightmares about hairy numbers and letters with pointed teeth eating me, I want to be able to sleep again.
Please...
;) :p
LOL :D:D:DQuote:
Originally posted by Doctor Luz
Definitively forget this please!!
Please go back to the easy questions. Do it for my mental health, I don't want to get mad, I don't want to have more nightmares about hairy numbers and letters with pointed teeth eating me, I want to be able to sleep again.
Please...
I agree this one is probably (:D) a bit difficult and if I am correct and it contains some mistakes it is impossible.
nope no mistakes!Quote:
Originally posted by Simon666
Are you sure there aen't multiple mistakes in what you wrote down? Shouldn't it be
barwonyoroganalhul
or some other letter instead of
barwoniganalhul
?
I cheked it 5-6 times but there are no typos. The condition is correct.
It IS possible. I solved it. All by my little self.Quote:
Originally posted by Simon666
LOL :D:D:D
I agree this one is probably (:D) a bit difficult and if I am correct and it contains some mistakes it is impossible.
The problem with that one isnot just that it's difficult, it's that it requires a lot of time to think about it, and I don't have enough time to spend on it :(
If anyone else wants to work on it but can't find a place to start, I have proven that 'barwon'=1, 'ganal'=33, and 'hul'=6.
Man solarflare! Aren't you sleeping! It's probably 7 in the morning in Philadelphia.
:D
:D
:D
Well alright. I'll look once tomorrow so SeventhStar, please wait just a little while longer.
If you like I'll post the solve tomorow at 12:00 (CET)
Nope, 9:30.... say, as long as you're here, I have a question about the operators in your problem (which you probably won't answer, and that's okay).
What is the complexity of the operators... are they limited to just one or two functions? Because for some of them that are used just twice or so in the whole list you gave, you could make up some complicated thing they do that would fit in with the numbers. I am currently guessing that each operator does the following:
blahoblahh (blah o blahh) = (some constant)*(blah)+(some constant)*(blahh)+(some constant), where the operator used (in this case, 'o') defines the three constants.
noooo! then it'd be impossible to solve!Quote:
Originally posted by solarflare
Nope, 9:30.... say, as long as you're here, I have a question about the operators in your problem (which you probably won't answer, and that's okay).
What is the complexity of the operators... are they limited to just one or two functions? Because for some of them that are used just twice or so in the whole list you gave, you could make up some complicated thing they do that would fit in with the numbers. I am currently guessing that each operator does the following:
blahoblahh (blah o blahh) = (some constant)*(blah)+(some constant)*(blahh)+(some constant), where the operator used (in this case, 'o') defines the three constants.
operators are just operations and some of them are sort of indexes like it's between x and y. but that's a big clue and I wont tell you any more...
I know this is not the way you intended it, but I just realized you can write:
178 = ganaliganaliganaliganaliganalihulihulibarwon
189 = ganaliganaliganaliganaliganalihulihulihulihul
351 = ganalhuliganaliganaliganaliganalihulihulihulibarwonibarwonibarwon
372 = ganalhuliganaliganaliganaliganaliganalihulibarwonibarwonibarwon
Although that still doesn't help at all for the last 2 questions.
I can't seem to figure out what this means.Quote:
Originally posted by SeventhStar
operators are just operations and some of them are sort of indexes like it's between x and y. but that's a big clue and I wont tell you any more...
The 'operators' either
- perform one standard operation each (by standard, I mean can be expressed with one symbol, such as * / + - ^, etc.)
- act as an index, in which case the numbers still get multiplied together (because it's like there's no operator between them)
Well then, if this is the case, there either must be a lot of "index operators" or else some of the operations are repeated.
Probably everything I just said is wrong... now I'm really confused.
I was thinking for example that y in p1yp2p3 could mean:
p1 * ( p2 + p3)
Thus imply some sort of order, while
( p1 * p2 ) + p3
would be a different result ofcourse. Perhaps it is something like that.
I don't think that's what 'y' does, because it is involved in too many prime numbers. I think it's probably simple subtraction.
Wow! I've looked at this thread from time to time, but you guys are crazy!!! The current problem is great! I just wish I had time to work on it...
I'm just going to grab the problem off and work on it at home this weekend, since I don't want the surprise spoiled (and it looks like solarflare is almost there and I don't want to see his answer!)
If you are correct I would like to make a few additions. It can help, though I'm still stuck.Quote:
Originally posted by solarflare
If anyone else wants to work on it but can't find a place to start, I have proven that 'barwon'=1, 'ganal'=33, and 'hul'=6.
Code:
Solved:
39 = ganalibarwonhul
199 = barwoniganalhul
Unsolved:
14 = farokobarwonhul
29 = omaryhutarhul
37 = babumbarwonhul
48 = babusgutarhul
52 = farokadonzesgutarhul
59 = babutfarokhul
64 = gutaragutarhul
71 = urukigutarhul
89 = barwonybabuhul
106 = omarifarokhul
112 = gutarobabuhul
128 = omarturukhul
149 = ganalsurukhul
208 = barwonadonzemganalhul
233 = gutarmomarhul
242 = donzemomarhul
Starting from urukigutarhul = 71
gutar=11 uruk = 5
gutar=10 uruk = 11
gutar= 9 uruk = 17
gutar= 7 uruk = 29
gutar= 5 uruk = 41
gutar= 4 uruk = 47
gutar= 3 uruk = 53
Starting from omarifarokhul = 106
farok = 17 omar = 4
farok = 16 omar = 10
farok = 15 omar = 16
farok = 13 omar = 28
farok = 12 omar = 34
farok = 11 omar = 40
farok = 10 omar = 46
farok = 8 omar = 58
farok = 5 omar = 76
farok = 4 omar = 82
farok = 3 omar = 88
farok = 2 omar = 94
Impossible:
farok = 1 omar =100 because farok = 1 = barwon
farok = 6 omar = 58 because farok = 6 = hul
farok = 7 omar = 64 because omar = 64 = gutaragutarhul
farok = 9 omar = 52 because omar = 52 = farokadonzesgutarhul
farok = 14 omar = 22 because farok = 14 = farokobarwonhul
gutar = 1 uruk = 65 because gutar = 1 = barwon
gutar = 2 uruk = 59 because uruk = 59 = babutfarokhul
gutar = 6 uruk = 35 because gutar = 6 = hul
gutar = 8 uruk = 23 because gutarhul = 48 = babusgutarhul
This is a tough problem, and my two days have not netted me any conclusions. This is a huge set of parallel operator equations and the nature of the operators is not very well specified, and it is hard to make a lot of progress into them. But I see everyone is sharing their work so we can all figure it out (which I think would be more fun than getting the answer...), so I wanted to share what I have found out so far. Much of it seems to have already been found out by solarflare, but I wanted to show the manipulations involved, because I think they are illustrative.
First, I noticed that the numbers requested could be built from the rules we already knew, so that it was not necessary to actually solve any of the operator equations prior. This is because 178 = 29 + 149 is an element of {S + S} and 189 = 48 + 52 + 89, 351 = 48 + 71 + 242, and 372 = 52 + 112 + 208 are elements of {S + S + S}. So we get the names
- omaryhutarhuliganalsurukhul (178)
- babusgutarhulifarokadonzesgutarhulibarwonybabuhul (189)
- babusgutarhuliurukigutarhulidonzemomarhul (351)
- farokadonzesgutarhuligutarobabuhulibarwonadonzemganalhul (372)
Now I found the first two by playing around with the numbers and the last two I had to be more systematic and enumerated the 3-partitions of the last digits in the partition summing to 0, 10, or 20 + the last digit of the target sum. I think there is some good algorithms out ther for automating this type of thing if it pops up later...
Now, I normally approach these type of problem by brute force at first until certain patterns make themselves more familiar to me. So I decided to systematize the names with abstract labels. My first chomskian parse of the words has led me to:
- farok = A
- barwon = B
- hul = C
- omar = D
- hutar = E (?)
- babu = F
- ganal = G
- gutar = H (?)
- donze = I
- uruk = J
for the object labels and
- o = (a)
- y = (b)
- m = (c)
- s = (d)
- a = (e)
- t = (f)
for the operators. I am concerned about the hutar / gutar parsing because the original question seems to say that two different "words" would not have the same root. The three possible explanations I have are: the "h" in hutar is a mistype, I have misunderstood the "root" rule, or "g" and "h" are operators on "utar". The latter would significantly change my analysis so far, but I have done some analysis in the other 2. Translating the known decompositions under the above abstract labels and ordering by operation, I get:
From this I saw the same thing solarflare must have seen, which is the two (+) pairs that involve B, C, and G. Since (+) is a known operator (unlike the (a) - (f) operators), this seems a good system to start with. This system solves toCode:(+) pairs
G(+)B(*)C = 39
J(+)H(*)C = 71
D(+)A(*)C = 106
B(+)G(*)C = 199
(a) pairs
A(a)B(*)C = 14
H(a)F(*)C = 112
(b) pairs
D(b)E(*)C = 29
B(b)F(*)C = 89
(c) pairs
F(c)B(*)C = 37
H(c)D(*)C = 233
I(c)D(*)C = 242
(d) pairs
F(d)H(*)C = 48
G(D)J(*)C = 149
(e) pair
H(e)H(*)C = 64
(f) pairs
F(f)A(*)C = 59
D(f)J(*)C = 128
triples
A(e)I(d)H(*)C = 52
B(e)I(c)G(*)C = 208
Now I brute force tested the endpoints and came up with two possible solutions: B = 1, G = 33, C = 6 or B = 1, G = 6, C = 33. I have some heuristics for choosing the first one (as solarflare has), but I still don't have a definitive proof for either one.Code:G = 39 - BC
C = (39 +/- sqrt(39^2 - 4B(199-B))) / (2B)
Looking at the two remaining unsolved questions, I see I have to find out the objects B, D, and I as well as the operators (b), (d), and (e). (e) has a very interesting member relationship with H to itself that seems a possibility to attack. H also has relations to D and I through the (c) operator. If E and H are the same (possibility of mistype) then there is an additional D relationship to H through the (b) operator. Thus there seems sufficient information to solve for the latter name question if we can figure out the operators.
This is where I think we all might be having problems. Its hard to figure out enough operators to test out and a brute force enumeration. I'll give a list of all the types I've been trying out. I think it was very important the mention of intervals as the center of the language's reasoning, and SeventhStar said something very cryptic about it being something like between X and Y. I came up with
- distance -- eg. 20(dist.)25 = 5
- forward reflection -- eg. 30(fr.)33 = 36
- backward reflection -- eg. 17(br.)22 = 12
- midpoint -- eg. 44(mp.)56 = 50
- addition -- eg. 65(+)31 = 96
- lattice count, multiplication -- eg. 12(*)20 = 240
- modulo circle mapping -- eg. 32(%)15 = 2
But I would love to know what possibilities you all are thinking about. This is currently where I am at. I'm trying out various operators on the 3 equations I mentioned earlier to see if I can crack the second name 1st.
Where is every one at? Let's all share and try to figure it out, cause its huge!
The reason I use distance in the above is because it is guaranteed to be nonnegative, which I believe SeventhStar mentioned as the domain of analysis for the operators. However, what is your reasoning solarflare on your conclusion about 'y'? This is extremely important for solving D(b)H(*)C = 29 if H=E.
I have also checked operators that are linear operators like the placement operators of normal number systems
so each operator may be distinguished by a direction and a multiplier. But then the given equations are not complete enough to specify the result (I believe, from dimensional analysis). So I haven't had much success in this direction.Code:A(forward lin.)B = AC + B
A(backward lin.)B = A + CB
Also, I have tried index operators into Fibonacci codes. Something like
but now I think that might be kind of silly. The reason I was think of this was because the sum and multiplication of the size of 1 runs have some special properties in Fibonacci code, which could have been a reason for the number system (and can basically said to be a number system of intervals).Code:A(fib. #n)B = 11...A 1s.. 100...n 0s...011...B 1s...1
I have been trying to figure out (e) for some time because of the H relationship I mentioned earlier, and I could see where it could be several operators (giving me different H). This (e) is important because it is found in both triples given and the one triple asked for (name #1). It seems to have some special operation that can be used (possibly unambiguously? -- order of operations) with another operation. If anyone has made progress on this I would love to know their reasoning.
SeventhStar, if you could comment on hutar/gutar, it would help me alot. Is the 'h' a mistype or are h and g separate operators? Or is the analysis where they are different correct?
S Galathaea! You are really great !!!
Oh my god!!! I'm sososososo sorry!!!!!! I've missed this. there is no 'hutar' it's 'gutar'.... :rolleyes:Quote:
Originally posted by galathaea
...
SeventhStar, if you could comment on hutar/gutar, it would help me alot. Is the 'h' a mistype or are h and g separate operators? Or is the analysis where they are different correct?
I'm really sorry guys... My I'm a stupid ********. Sorry.
I already requested if there were spelling errors. :rolleyes: Are you sure there aren't any further? Like "i" instead of "y" or vice versa?Quote:
Originally posted by SeventhStar
Oh my god!!! I'm sososososo sorry!!!!!! I've missed this. there is no 'hutar' it's 'gutar'.... :rolleyes:
I'm really sorry guys... My I'm a stupid ********. Sorry.