The New Sudoku Players' Forum

[Serg]

Hi!

8 used 7 times

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8! / [ 8 + 8 + sqrt (8+8) ]

- sqrt (sqrt (8+8) )



Well, you introduced very productive new idea: 2014 = 8!/20 - 2;
It will be exploited multiple times to construct new formulas (999_Springs showed such examples). Expression sqrt(8+8) = 4 looks like simple, but I wasn't managed to find it. Congratulations for this formula!

Serg

[Serg]

Hi!

9 used 8 times

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9! / [ 9 x ( 9+9 ) + 9 + 9 ]

- ( sqrt ( 9 ) ) !
  / sqrt ( 9 )



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( 999 + 9 - 9/9 )

* ( sqrt ( 9 ) ) !
  / sqrt ( 9 )



3rd version comes from my above 6s,
with 6 rewritten as
( sqrt ( 9 ) ) !

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  ( sqrt ( 9 ) ) !!   x sqrt ( 9 )

- ( sqrt ( 9 ) ) !!   /   ( sqrt ( 9 ) ) !
- ( sqrt ( 9 ) ) !!   / ( ( sqrt ( 9 ) ) ! x ( sqrt ( 9 ) ) ! )

- ( sqrt ( 9 ) ) !


Expression "sqrt(9)!" looks rather unusal from mathematical point of view. (It is unusal to see an expression before factorial sign.) First example - variation about "2014 = 8!/20 - 2", but with your expression "3!/3 = 2". The second example is new idea "2014 = (999+9-1) x 2". Well done!

Serg

[Serg]

Hi, 999_Springs!
I am glad to see new competitor. Some your ideas are brilliant.

(2*2^2)!/(22-2)-2

Nice implementation of Pat's idea "2014 = 8!/20 - 2". I was slightly corrected your formula according to Pat's post - substituted "2*2^2" by "2*2*2", because the last expression doesn't need brackets to show true sequence of operations.

(4+4)!/(4!-4)-sqrt4

Next nice implementation of Pat's idea "2014 = 8!/20 - 2". Well done!

5^5-5555/5

Brilliant formula! (My favorite among published ones.) Original idea, very short representation, containing minimum of operation signs and having no brackets. Nice work!

(9!*sqrt(9)!*sqrt(9)!)/(9*sqrt(9)!!)-sqrt(9)!/sqrt9

Next (but not trivial) nice implementation of Pat's idea "2014 = 8!/20 - 2". Well done!
Congratulation on setting 4 new records!

Serg

[Serg]

Hi!
I've found new formula having 12 zeroes (yet another implementation of Pat's idea "2014 = 8!/20 - 2):
2014 = (0!+0!) x ((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0! - 0!

I am updating the table with best results here.

I propose to stop this game on February, 1 (24:00), and determine the champion by the following rule. The champion must have the most number of best results (counting the best formulas) at the end of game, each best formula increment player's score by 1 point. If some digit representation has more than one best formulas, each formula costructor should get proper fraction of 1 point. For example, player A has 2 best N-digit formula for digit "M", player B has 1 best N-digit formula for the same digit (totally there are 3 best presentation with "M" digits). I this case player A will get 2/3 points, player B will get 1/3 points. At the moment Pat has 4 points, 999_Springs has 4 points, Serg has 2 points.

Serg

[Smythe Dakota]

What if somebody submits multiple N-count solutions for the digit "M", in order to break ties favorably to himself, but those solutions aren't significantly different from each other?

For example, 1, 1!, and sqrt(1) are all interchangeable. So, if somebody has a solution with 6 1's, he could actually submit 729 solutions, thus grabbing 729/730 of a tie-break point versus his opponent's 1/730, rather than 1/2 versus 1/2 which really seems more fair.

Bill Smythe

[999_Springs]

idea: 2014=6^4+6!-2

can construct

3!*3!^3+3!!-3!/3

and similarly

6*sqrt(6^6)+6!-(6+6)/6

and can rewrite 3 as sqrt(9)

sqrt(9)!*sqrt(9)!^sqrt9+sqrt(9)!!-sqrt(9)!/sqrt9

[Serg]

Hi, 999_Springs!
Really original formula construction idea! Well done!

idea: 2014=6^4+6!-2

can construct

3!*3!^3+3!!-3!/3

and similarly

6*sqrt(6^6)+6!-(6+6)/6

and can rewrite 3 as sqrt(9)

sqrt(9)!*sqrt(9)!^sqrt9+sqrt(9)!!-sqrt(9)!/sqrt9

Did you do it manually or maybe use some computer assistance?

I updated my results table. You are leader at the moment. You have 6 points (6 formulas), Pat and me have 2 points each.
Good luck!

Serg

[Serg]

Hi, Bill!

What if somebody submits multiple N-count solutions for the digit "M", in order to break ties favorably to himself, but those solutions aren't significantly different from each other?

For example, 1, 1!, and sqrt(1) are all interchangeable. So, if somebody has a solution with 6 1's, he could actually submit 729 solutions, thus grabbing 729/730 of a tie-break point versus his opponent's 1/730, rather than 1/2 versus 1/2 which really seems more fair.

I think you treat this problem too serious. I am sure all known at the moment competitors will not produce "new" formulas by substitution "1" by "1!", etc.

Serg

[999_Springs]

Hi!
I've found new formula having 12 zeroes (yet another implementation of Pat's idea "2014 = 8!/20 - 2):
2014 = (0!+0!) x ((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0! - 0!

nice formula, but there is an easy simplification that removes a zero

2014 = (0!+0!) x ( ((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0! )

[Serg]

Hi, 999_Springs!

Hi!
I've found new formula having 12 zeroes (yet another implementation of Pat's idea "2014 = 8!/20 - 2):
2014 = (0!+0!) x ((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0! - 0!

nice formula, but there is an easy simplification that removes a zero

2014 = (0!+0!) x ( ((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0! )

Good observation! I updated the list of best formulas by your formula for "0".

Well, today it is the last day of our competition. I spent a lot of time attacking "7" digit's formula. 9 digits seemed to be too high for "7" digit formula. But I could find just another 9-digits formulas:

2014 = (7!+7!)/7 + 7 x (77+7) - 7 - 7;
2014 = (7!+7!)/7 + 7 x sqrt(7!+7/7) + 77;

Serg

[Serg]

Hi, people!
The game is over now. 999_Springs is the champion of the game. Congratulations to him! He is really smart.
12 formulas use 6 different construction ideas. Pat is the author of the most fruitful idea: 2014 = 8!/20 - 2. (It is used in 4 formulas.)
Here is final version of the formulas for all digits.

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2014 = (0!+0!) x (((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0!);              [999_Springs; 11 digits]

2014 = (1+1)^11 - (1+1+1) x 11 - 1;                                      [Serg;        10 digits]

2014 = (2x2x2)!/(22-2) - 2;                                              [999_Springs;  7 digits]

2014 = 3! x (3!^3) + (3!)! - 3!/3;                                       [999_Springs;  6 digits]

2014 = (4+4)!/(4!-4) - sqrt(4);                                          [999_Springs;  5 digits]

2014 = 5^5 - 5555/5;                                                     [999_Springs;  7 digits]

2014 = 6 x sqrt(6^6) + 6! - (6+6)/6;                                     [999_Springs;  7 digits]

2014 = ((7!/7-7x7) x (7+7+7) + 7)/7;                                     [Pat;          9 digits]
2014 = (7!+7!)/7 + 7 x (77+7) - 7 - 7;                                   [Serg;         9 digits]
2014 = (7!+7!)/7 + 7 x sqrt(7!+7/7) + 77;                                [Serg;         9 digits]

2014 = 8!/(8+8+sqrt(8+8)) - sqrt(sqrt(8+8));                             [Pat;          7 digits]

2014 = sqrt(9)! x (sqrt(9)!^sqrt(9)) + (sqrt(9)!)! - sqrt(9)!/sqrt(9);   [999_Springs;  6 digits]


Serg

[Smythe Dakota]

How about the following as the next contest:

Find the shortest representation of 2014 using two digits 0 through 9. That way, instead of having just 10 separate contests, you'd have 45. (I almost put an exclam after that 45, but then I realized you might think I meant 45 factorial.)

Bill Smythe

[999_Springs]

so i wrote a program to do this stuff for me (i hope computer use is allowed here?)

here are the two-digit results that i got; i only included those that were strictly better than the results for each of the constituent digits

stuff: Show

smaller digit 0:
3!!+(3!)^(3+0!)-3+0!
6*sqrt(6^6)+6!-0!-0!
(7!/(7-0!-0!)-0!)*(0!+0!)
sqrt(9)!!+(sqrt(9)!)^(sqrt9+0!)-sqrt9+0!

smaller digit 1:
5^5-1111
6*sqrt(6^6)+6!-1-1
7!-17*(177+1)

smaller digit 2:
3!!+3!^(2+2)-2
((2+5)!-5)/5*2
6!+6^(6-2)-2
(7!/(7-2))*2-2
8!/((8+2)*2)-2
sqrt(9)!!+sqrt(9)!^(2+2)-2

smaller digit 3:
3!!+3!^4-sqrt4
53*(35+3)
3!!+sqrt(3!^8)-8+3!

smaller digit 4:
6!+6^4-sqrt4
8!/(4!-4)-sqrt4
sqrt(9)!!+sqrt(9)!^4-sqrt4

smaller digit 5:
6!+65*5!/6-6
(7!+7!)/5-7+5
(8!/5-8)/sqrt(8+8)

smaller digit 6:
6!+6!+7*(6+76)
6!+sqrt(6^8)-8+6

smaller digit 7:
(sqrt(9)!^sqrt9+7)*9+7

smaller digit 8:
sqrt(9)!+sqrt(sqrt(9)!^8)-8+sqrt(9)!

almost all of these are variants of 8!/20-2 or 6!+6^4-2

slightly surprisingly, it found nothing new for the single-digit cases

[Serg]

Hi, 999_Springs!

so i wrote a program to do this stuff for me (i hope computer use is allowed here?)

here are the two-digit results that i got; i only included those that were strictly better than the results for each of the constituent digits

stuff: Show

smaller digit 0:
3!!+(3!)^(3+0!)-3+0!
6*sqrt(6^6)+6!-0!-0!
(7!/(7-0!-0!)-0!)*(0!+0!)
sqrt(9)!!+(sqrt(9)!)^(sqrt9+0!)-sqrt9+0!

smaller digit 1:
5^5-1111
6*sqrt(6^6)+6!-1-1
7!-17*(177+1)

smaller digit 2:
3!!+3!^(2+2)-2
((2+5)!-5)/5*2
6!+6^(6-2)-2
(7!/(7-2))*2-2
8!/((8+2)*2)-2
sqrt(9)!!+sqrt(9)!^(2+2)-2

smaller digit 3:
3!!+3!^4-sqrt4
53*(35+3)
3!!+sqrt(3!^8)-8+3!

smaller digit 4:
6!+6^4-sqrt4
8!/(4!-4)-sqrt4
sqrt(9)!!+sqrt(9)!^4-sqrt4

smaller digit 5:
6!+65*5!/6-6
(7!+7!)/5-7+5
(8!/5-8)/sqrt(8+8)

smaller digit 6:
6!+6!+7*(6+76)
6!+sqrt(6^8)-8+6

smaller digit 7:
(sqrt(9)!^sqrt9+7)*9+7

smaller digit 8:
sqrt(9)!+sqrt(sqrt(9)!^8)-8+sqrt(9)!

almost all of these are variants of 8!/20-2 or 6!+6^4-2

slightly surprisingly, it found nothing new for the single-digit cases

Impressive results! What is your method?

I should say possible operations were not strictly defined. Besides operations being used some mathematical functions can be used too ("log", "sin", etc.). This consideration is more important for 2-digit formulas. For example, expression "sin(45 deg.)" is valid to my mind.

Serg

[Smythe Dakota]

Could you use "log 32 base 2" when trying to create 5 from 2 and 3?

Bill Smythe