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YONG
revolution wrote: Let's see if others here are happy to accept this notation for BBG. For example, let's see if we can come up with a zeroless pandigital approximation for pi. The rules are:  Every digit from 1 to 9 must be used exactly once.  No decimal point.  Only basic arithmetic operations are allowed: +, , *, /, (), power. Here is my submission: 23519/7486 = 3.14173123163 ... It is correct to three decimal places! Refer to: http://mathworld.wolfram.com/PandigitalNumber.html 

28 Dec 2016, 03:52 

YONG
revolution wrote:


28 Dec 2016, 10:02 

revolution
YONG wrote: Accept whatever notation he would like to use. And then just let him win. 

28 Dec 2016, 14:11 

Tomasz Grysztar
YONG wrote: I don't think many forum members are interested in this old game any more. Maybe we can start a new thread with a new game. 397*65/8214 

28 Dec 2016, 16:03 

revolution
Tomasz Grysztar wrote: 397*65/8214 

28 Dec 2016, 16:33 

YONG
revolution wrote:


29 Dec 2016, 01:50 

YONG
My new submission:
(876*34 + 25)^(1/9) = (29784 + 25)^(1/9) = 29809^(1/9) = 3.14159149039 ... Correct to five decimal places! 

29 Dec 2016, 07:00 

sleepsleep
is this trial and error or you guys got group of wizards behind?


29 Dec 2016, 07:26 

YONG
sleepsleep wrote: is this try and error or you guys got group of wizards behind? 

29 Dec 2016, 07:36 

Tomasz Grysztar
YONG wrote: My new submission: (sqrt(2)1)*4983/657 = 3.141592361195... I also found (1627/389)^(4/5) = 3.14159618798... but it is correct only to five places and not as good as yours. sleepsleep wrote: is this try and error or you guys got group of wizards behind? 

29 Dec 2016, 20:58 

YONG
Tomasz Grysztar wrote: your list of basic operations did not include roots ... So, the best submissions as of now are correct to 5 decimal places. See if any smart forum members out there can come up with something even better! YONG: (876*34 + 25)^(1/9) = 3.14159149039 ... T.G.: (1627/389)^(4/5) = 3.14159618798 ... 

30 Dec 2016, 02:01 

Tomasz Grysztar
Got a new one, obtained through two distinct pandigital expressions:
6745/(2138+9) = (3594)/(12768) = 3.1415929203... 

30 Dec 2016, 18:59 

revolution
Tomasz Grysztar wrote: 3.1415929203... 

30 Dec 2016, 19:28 

YONG
The 355/113 fraction is a wellknown approximation for pi. That smells a bit like plagiarism to me ...
But the 6745/(2138+9) expression is fine. Good job! So, the best answers as of now are correct to 6 decimal places. T.G.: 6745/(2138+9) = ( 359  4 ) / ( 127  6  8 ) = 3.1415929203 ... Come on! Let's see if some smart forum members can come up with something even better! 

31 Dec 2016, 02:45 

revolution
There are only 9! (362,880) ways to arrange nine digits. Writing a program to enumerate all possible expressions would be challenging but doable. And then add in the evaluator and comparator and this could be solved by brute force.


31 Dec 2016, 02:56 

YONG
revolution wrote: There are only 9! (362,880) ways to arrange nine digits. 

31 Dec 2016, 04:28 

revolution
YONG wrote:


31 Dec 2016, 05:11 

Tomasz Grysztar
revolution wrote: There are only 9! (362,880) ways to arrange nine digits. Writing a program to enumerate all possible expressions would be challenging but doable. And then add in the evaluator and comparator and this could be solved by brute force. To generate all possible expressions I think it would be better to use RPN to deal with the operation ordering without brackets. So then it would go like this: take a permutation of digits, partition it into N (between 1 and 9) numbers and then distribute N1 operators anywhere between or after these numbers, remembering that to generate a correct RPN expression you can put at most M2 operators in the part of expression before the Mth number. 

31 Dec 2016, 10:12 

Tomasz Grysztar
My "eval" searcher found me another one:
(2+6*9/743)*8^(1/5) = 3.14159288348... It is still correct only to 6 places, but it is slightly closer to pi 

31 Dec 2016, 12:14 

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