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YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

Tomasz Grysztar wrote: 
I used a simple expression generator (on top of a digit permutation) and fasmg's "eval" to find the above ones.


See if you can find an expression for each of these numbers with your program:
42722830  using seven digits, from 1 to 7
I believe this one is doable because I have already worked out a very close value by hand:
(5^6 + 1) x 2734 = 42721484
But I need the exact value.
The pi approximation will be correct to 7 decimal places!
888582403  using six digits, either {2, 4, 5, 7, 8, 9} or {3, 4, 5, 6, 7, 8}
This one may not be doable. Still, give it a try.
The pi approximation will be correct to 10 decimal places!
I am happy to share the honors with you!
Please bring me some good news on New Year Day!

31 Dec 2016, 12:23 

bitRAKE
Joined: 21 Jul 2003
Posts: 2624
Location: dank orb

3(5167)/(24+89) another in the set of 6 correct places, lol.
There are a lot of possible patterns.

02 Jan 2017, 03:37 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

So, the best answers as of now are still correct to 6 decimal places:
T.G.: (2+6*9/743)*8^(1/5) = 3.14159288348 ...
bitRAKE: 3(5167)/(24+89) = 3.14159292035 ...
Come on! Something even better is out there!

02 Jan 2017, 04:06 

bitRAKE
Joined: 21 Jul 2003
Posts: 2624
Location: dank orb

The power searches take a long time...
Three) 3+4^6^1/(7^2*589)
I can search any pattern you think might be fruitful.
_________________ The generation of random numbers is too important to be left to chance  Robert R Coveyou

02 Jan 2017, 05:05 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

bitRAKE wrote: 
3+4^6^1/(7^2*589)


Just correct to 3 decimal places. Not good enough by the 2017 standard!
bitRAKE wrote: 
I can search any pattern you think might be fruitful.


Then try this one:
42722830  using seven digits, from 1 to 7
I need the exact value.
The pi approximation will be correct to 7 decimal places!
Thanks!

02 Jan 2017, 05:22 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

Just found one, which has the same value as 355/113:
3^1 + (4^2)/((9+5)*8 + 7  6) = 3.14159292035 ...
Still correct to 6 decimal places.

02 Jan 2017, 07:41 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

So, the best answers as of now are still correct to 6 decimal places:
T.G.: (2+6*9/743)*8^(1/5) = 3.14159288348 ...
bitRAKE: 3(5167)/(24+89) = 3.14159292035 ...
YONG: 3^1 + (4^2)/((9+5)*8 + 7  6) = 3.14159292035 ...
Come on! Something even better is out there!

02 Jan 2017, 07:43 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6602
Location: Kraków, Poland

My simple "eval" script was able to search only through specific types of expressions, so to give it a try I wrote a proofofconcept script in fasmg that does a brute force search through all possible expressions in RPN notation:
Code: 
aim = 3.141592653589793238462643383279502884197169399375105820974
delta = 1 ; start showing results when they are within this distance from the aim
digits db '1234'
digits_count = $digits
operators db '+*/^'
operators_count = $operators
ln2 = 0.693147180559945309417232121458176568075500134360255254120
while 1
repeat 1 shl (digits_count1), separators:0
; Create numbers from digits:
position = 0
repeat digits_count, index:0
load digit:byte from digits+position
position = position + 1
number#index = digit'0'
while separators shr (position1) and 1
load digit:byte from digits+position
position = position + 1
number#index = number#index*10 + digit'0'
end while
if position >= digits_count
numbers_count = %
break
end if
end repeat
variations_count = 1
repeat numbers_count1
variations_count = variations_count*operators_count
end repeat
repeat variations_count, variation:0
; Prepare distribution of operators:
operators_count0 = 0
counter = numbers_count1
while 1
repeat counter
operators_count#% = 1
end repeat
; Evaluate:
variation_cursor = variation
repeat numbers_count, index:0
stack =: number#index
repeat operators_count#index
tmp = stack
restore stack
operator_index = variation_cursor mod operators_count
variation_cursor = variation_cursor / operators_count
load operator:1 from operators+operator_index
if operator = '+'
stack = stack + tmp
else if operator = ''
stack = stack  tmp
else if operator = '*'
stack = stack * tmp
else if operator = '/'
stack = float stack / tmp
else if operator = '^'
if tmp < 20 & tmp > 20 & tmp = trunc tmp
tmp = trunc tmp
inv = 0
if tmp < 0
inv = 1
tmp = tmp
end if
sq = stack
stack = 1
while tmp
if tmp and 1
stack = stack * sq
end if
sq = sq * sq
tmp = tmp shr 1
end while
if inv
stack = float 1/stack
end if
else
x = float stack
if x > 0
ln = 2*(x1)/(x+1)
repeat 4
k = trunc (ln/ln2)
r = ln  k*ln2
term = 1
exp = term
repeat 12
term = term*r/(float %)
exp = exp + term
end repeat
exp = exp shl k
ln = ln + 2*(xexp)/(x+exp)
end repeat
x = ln*tmp
k = trunc (x/ln2)
r = x  k*ln2
term = 1
stack = term
repeat 12
term = term*r/(float %)
stack = stack + term
end repeat
stack = stack shl k
else
stack = 0
end if
end if
else
err 'unknown operator'
end if
end repeat
end repeat
result = stack
restore stack
d = resultaim
if (d > 0 & d < delta)  (d < 0 & d > delta)
if d > 0
delta = d
else
delta = d
end if
; Display generated expression:
variation_cursor = variation
repeat numbers_count, index:0
if % > 1
display ' '
end if
repeat 1, x: number#index
display `x
end repeat
repeat operators_count#index
operator_index = variation_cursor mod operators_count
variation_cursor = variation_cursor / operators_count
load operator:1 from operators+operator_index
display ' ',operator
end repeat
end repeat
out showfloat result
display ' = ',out,13,10
end if
; Next distribution of operators:
counter = 1
repeat numbers_count1, index:0
if operators_count#index > 0
operators_count#% = operators_count#% + 1
counter = operators_count#index  1
operators_count#index = 0
break
end if
end repeat
if counter < 0
break
end if
end while
end repeat
end repeat
; Next permutation of digits:
i = digits_count1
while i >= 0
load b:byte from digits+i
i = i  1
if i >= 0
load a:byte from digits+i
if a < b
break
end if
end if
end while
if i < 0
break
end if
j = i + 1
while j < digits_count1
load c:byte from digits+j+1
if c > a
b = c
j = j + 1
else
break
end if
end while
store b:byte at digits+i
store a:byte at digits+j
i = i + 1
j = digits_count1
while i < j
load x:byte from digits+i
load y:byte from digits+j
store x:byte at digits+j
store y:byte at digits+i
i = i + 1
j = j  1
end while
end while


As you can see, I set up the sample above so that it searches through expressions composed from digits 14 only. It gets very slow quickly for larger sets of digits (to monitor the script I made a tweaked version of fasmg that shows the messages generated with DISPLAY immediately instead of buffering them*). The next step should be to rewrite this prototype script in assembly, though I do not have time for this now. With a native implementation exhaustive search through expressions generated from a sets of 6 or even 7 digits may be doable. But 8 and more is probably going to be unattainable with this method.
(The script uses the "showfloat" macro I once shared in the other thread.)
Some interesting results from my test searches with small sets of digits: with digits 14 the best possible approximation of pi is 3+2/14 = 3.1428571429..., interestingly digits 15 yield no better result. For a set of digits 16 some better results start coming in, for example 1*(354)^(2/6)=3.141380652... but I have not yet finished this search.
YONG wrote: 
See if you can find an expression for each of these numbers with your program: (...)


With this script I can now try to search through expressions generated with 6 digits. With a native implementation perhaps the exhaustive search with 7 digits would also be possible.
___
* This is now in the official fasmg releases as a hidden "v2" option.
Last edited by Tomasz Grysztar on 08 Sep 2017, 17:04; edited 9 times in total

02 Jan 2017, 09:38 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

Am I dreaming?
How come the above script  written by T.G.  actually has some comments?

02 Jan 2017, 11:49 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6602
Location: Kraków, Poland

And another bug in fasmg found thanks to these scripts...

02 Jan 2017, 15:26 

bitRAKE
Joined: 21 Jul 2003
Posts: 2624
Location: dank orb

Code: 
5)
296^5/(814)^7+3
(278^4+1)/59^6+3
(6/8)^5 * (3/9)^1 * (7/4)^2
[6^(43)+7^(92)]/8^(1+5)


...some lesser solutions.

02 Jan 2017, 18:53 

bitRAKE
Joined: 21 Jul 2003
Posts: 2624
Location: dank orb

Tomasz Grysztar wrote: 
And another bug in fasmg found thanks to these scripts...


Does it also cover multiple operators? Specifically, A*B comes to mind. I was thinking "/" could be replaced by a monotonic negation  actually, I think it's called something else. Where each value can be subtracted from zero by itself. This probably increases the complexity rather than the intent.
I like how you choose to iterate between the operators  very concise.
_________________ The generation of random numbers is too important to be left to chance  Robert R Coveyou

02 Jan 2017, 19:58 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6602
Location: Kraków, Poland

bitRAKE wrote: 
Tomasz Grysztar wrote: 
And another bug in fasmg found thanks to these scripts...


Does it also cover multiple operators? Specifically, A*B comes to mind. I was thinking "/" could be replaced by a monotonic negation  actually, I think it's called something else. Where each value can be subtracted from zero by itself. This probably increases the complexity rather than the intent.
I like how you choose to iterate between the operators  very concise.


My implementation does not have unary operators. You could insert any number of unary operators in any place in a RPN expression. For unary minus it does not make sense to put more than one in a row anywhere, but for example if we had allowed square root (which is also an unary operator), the number of potential expressions would become infinite.
Whether the formulation of problem allows to unary minus is another issue. This probably should be clarified.

02 Jan 2017, 20:06 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

Tomasz Grysztar wrote: 
but for example if we had allowed square root (which is also an unary operator) ...
Whether the formulation of problem allows to unary minus is another issue. This probably should be clarified.


Allowing square root means freeing up at least two digits:
^(1/2) becomes √
Allowing the radical sign of the higher roots means freeing up at least one digit:
^(1/n) becomes ^n √
Allowing unary minus in the power means freeing up at least one digit:
(1/x)^n becomes x^(n)
A wise forum member once said, "Something something keeping up the standards something something."
So, I would say we better stick to those restrictive rules and maintain the difficulty of the problem.

03 Jan 2017, 03:48 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

bitRAKE wrote: 
Code: 
5)
296^5/(814)^7+3
(278^4+1)/59^6+3
(6/8)^5 * (3/9)^1 * (7/4)^2
[6^(43)+7^(92)]/8^(1+5)


...some lesser solutions.


In (278^4+1)/59^6+3, there is no need to write "+1". Just write "1".
The expression ( 6 / 8 )^5 * (3/9)^1 * (7/4)^2 gives 0.24224853515. Something is wrong with your program.

03 Jan 2017, 03:57 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6602
Location: Kraków, Poland

YONG wrote: 
888582403  using six digits, either {2, 4, 5, 7, 8, 9} or {3, 4, 5, 6, 7, 8}
This one may not be doable. Still, give it a try.


I've been running my fasmg script in the background searching with these two sets of digits, but as you suspected no expression yielded such result. The closest number that it found is:
Code: 
[RPN] 5 2 * 9 4 78 /  ^ = (5*2)^(9(4/78)) = 888623816.274...


If we raise this to the power 1/(3*6), which is what I suspect you wanted to do, we get an approximation of pi, but it is much worse than the ones we already have:
Code: 
((5*2)^(9(4/78)))^(1/(3*6)) = 3.1416007876788536...


The search for 42722830 that you also asked for is practically out of reach with fasmg script, because the number of expressions for seven digits is much larger. But I we write a native implementation of the searcher, then such search may become possible. But brute force scans with 8 and 9 digits are going to be out of reach anyway.

05 Jan 2017, 09:12 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

@T.G.: Thank you very much.
So, six digits is basically the limit for your script. Then give the following a try:
924269  using six digits:
{2, 5, 6, 7, 8, 9},
{3, 4, 5, 7, 8, 9},
{2, 4, 5, 6, 7, 8},
{2, 3, 5, 6, 7, 9}, or
{2, 3, 4, 6, 8, 9}.
(924269)^(1/(3*4))
= (924269)^(1/(2*6))
= (924269)^(1/(3+9))
= (924269)^(1/( 4 + 8 ))
= (924269)^(1/(5+7))
= 3.14159260217 ... (correct to 7 decimal places)
If I can use one more digit (4), here is the solution:
(6^3 * 4279 + 5)^(1/(8+4)) = 924269^(1/12) = 3.14159260217 ...
Please give me some good news this time!

05 Jan 2017, 10:20 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6602
Location: Kraków, Poland

I have continued to look for 888582403 with other sets of digits, because there are also other ways to make 1/18 power. And I found this one:
Code: 
[RPN]41 5 ^ 9 7 ^  8 * = (41^59^7)*8 = 888585856


Still not perfect, but a bit better, and the pi approximation it gives is:
Code: 
((41^59^7)*8)^(2/36)=3.14159333180...



05 Jan 2017, 13:56 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

Tomasz Grysztar wrote: 
there are also other ways to make 1/18 power


Good point!
1/12 = 2/24 = 2/( 3 * 8 ) = 2/(4*6) = 4/48 = 4/( 6 * 8 ) = 8/96
So, please also try:
{1, 4, 5, 6, 7, 9},
{1, 3, 5, 7, 8, 9},
{1, 2, 3, 5, 7, 9}, and
{1, 2, 3, 4, 5, 7}
for 924269.
Thanks!

06 Jan 2017, 01:33 

YONG
Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N  114° 10' E

Any number n within the range
888582131 =< n =< 888582639
can give a pi approximation correct to 7 (or more) decimal places.
Come on!

06 Jan 2017, 04:57 



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