## The four fours problem

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

51: -~(44+sqrt(4)+4)

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

52: 4! + 4! + √(4) + √(4)
53: 4! + 4! + 4 + φ(√(4))
54: 4! + 4! + 4 + √(4)
55: 4! + 4! + Γ(4) + φ(√(4))
56: 4! + 4! + 4 + 4
57: 4! + 4! + 4/(.4...)
... meaning repeating of 4 (that is 4/0.444... = 4/(4/9) = 9)
58: 4! + 4! + Γ(4) + 4
59: (Γ(4)! / 4!) * √(4) - φ(φ(4))
60: 44 + 4^(√(4))

EDIT: unicode ftw! and changed 58: 4! + 4! + 4/.4 to 4! + 4! + Γ(4) + 4
Last edited by poizan42 on Wed May 02, 2007 6:44 pm UTC, edited 1 time in total.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

I thought φ was only defined for integer inputs? I think this is illegal. You could do floor(sqrt(4)), but that's really ugly. Elsewhere you have φ(φ(4)), which is better.

61=(gamma(4)! / 4!) * sqrt(4) + φ(φ(4))
62=(gamma(4)! / 4!) * sqrt(4) + sqrt(4)
63=(4^4)/4-φ(φ(4))
64=(4^4)/(sqrt(4)*sqrt(4))
65=(4^4)/4+φ(φ(4))
66=(4^4)/4+sqrt(4)
67=<4/4,4/.4>
68=(4^4)/4+4
69=<4-φ(φ(4)),4+4>
70=<4,4/.44...>
Last edited by skeptical scientist on Wed May 02, 2007 12:46 pm UTC, edited 1 time in total.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

skeptical scientist
closed-minded spiritualist

Posts: 6135
Joined: Tue Nov 28, 2006 6:09 am UTC
Location: San Francisco

By the by, way back at 3, isn't (4+4+4)/4 much neater than messing around with sqrts? And (4*4 + 4)/4 for 5. Also for 7, 44/4 - 4. How far can you get only using the basic +-*/ functions? I can get to 12.

Edit: Also, I know I'm being very annoying here, but 60 = 44 + 4^(sqrt(4))? Surely 44 + 4*4?
PaulT

Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

71: 4!*~-4-4/4
72: 4!*(4-4/4)
73: 4!*~-4+4/4
74: 4!+4!+4!+sqrt(4)
75: 4!+4!+4!+~-4
76: 4!+4!+4!+4
77: 4!+4!+4!-~4
78: 4!*~-4+sqrt(4)+4
79: 4!*~-4+4+~-4
80: 4!*4-4*4
81: (~-4)**(~-4)+4-4

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

skeptical scientist wrote:I thought φ was only defined for integer inputs? I think this is illegal. You could do floor(sqrt(4)), but that's really ugly. Elsewhere you have φ(φ(4)), which is better.

What?? Since when wasn't sqrt(4) = 2 an integer??

PaulT wrote:Edit: Also, I know I'm being very annoying here, but 60 = 44 + 4^(sqrt(4))? Surely 44 + 4*4?

sqrt(4) = 2
4*4 = 4^(sqrt(4)) = 4^2 = 16
44 + 16 = 60
where's the problem?

EDIT: merged posts

EDIT2:
hotaru: what does that ~ mean? and ** is that exponentiation?
Last edited by poizan42 on Wed May 02, 2007 1:50 pm UTC, edited 1 time in total.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

I did this upto 40 or 44 or so In year 9 for some reward at school or something.

I only used Factorial/Multiply/Divide/Add/Subtract/Power and I might have used forth root, but I can't remember.

The teacher had to cheat and use 44 or .4 for some of them (at least, I thought that was cheating).
"I knew you had it in you all along."
"Greatness?"
"No. Better than that. Goodness."

Gelsamel
Lame and emo

Posts: 8083
Joined: Thu Oct 05, 2006 10:49 am UTC
Location: Melbourne, Victoria, Australia

hmm no one said the successor function succ(x) (or S(x)) wasn't allowed :/
But well... that's not very fun

82 = sqrt(4)^(gamma(4))+4!-gamma(4)
Last edited by poizan42 on Wed May 02, 2007 2:33 pm UTC, edited 1 time in total.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

82 = 4^(5/2) / .4 + sqrt(4)

mattmacf

Posts: 90
Joined: Sat Sep 02, 2006 11:19 pm UTC

mattmacf wrote:82 = 4^(5/2) / .4 + sqrt(4)

1. too slow!
2. you aren't allowed to do 4^(5/2)
3. you lose!
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

poizan42 wrote:
PaulT wrote:Edit: Also, I know I'm being very annoying here, but 60 = 44 + 4^(sqrt(4))? Surely 44 + 4*4?

sqrt(4) = 2
4*4 = 4^(sqrt(4)) = 4^2 = 16
44 + 16 = 60
where's the problem?

Well, there's no problem per se. But where's the elegance?
PaulT

Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

poizan42 wrote:
mattmacf wrote:82 = 4^(5/2) / .4 + sqrt(4)

2. you aren't allowed to do 4^(5/2)

4^(5/2) is, in LaTeX notation, \sqrt[.4]{4]; in other words, the 0.4th root of 4.
EvanED

Posts: 3929
Joined: Mon Aug 07, 2006 6:28 am UTC

poizan42 wrote:hotaru: what does that ~ mean? and ** is that exponentiation?

~
and yes, ** is exponentiation.

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

poizan42 wrote:
mattmacf wrote:82 = 4^(5/2) / .4 + sqrt(4)

1. too slow!
2. you aren't allowed to do 4^(5/2)
3. you lose!

Haha, my bad. I explained the .4th root thing a couple posts ago (I think it's on the first page). And yes, I was very late posting that. I tend to multitask when posting and get sidetracked terribly easily

I hope I can make it up to you with
83 = (4/.4...)^sqrt(4) + sqrt(4)

...where the .4... = 0.44444444... repeating infinitely. Usually you'd use the bar on top of it to denote that, but I don't believe you can do that with ASCII

mattmacf

Posts: 90
Joined: Sat Sep 02, 2006 11:19 pm UTC

hotaru wrote:
poizan42 wrote:hotaru: what does that ~ mean? and ** is that exponentiation?

~
and yes, ** is exponentiation.

a FORTRANer?

edit: nevermind spotted the link now
in ur beanz makin u eveel

evilbeanfiend

Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

84 = 44 * sqrt(4) - 4

(I'm pretty sure I got that one right this time )
EvanED

Posts: 3929
Joined: Mon Aug 07, 2006 6:28 am UTC

mattmacf wrote:Haha, my bad. I explained the .4th root thing a couple posts ago (I think it's on the first page).

oic. By why not use 4^(1/.4) ? I think it's more clear that you are only using fours that way.

85 = floor((4^4*√(4))/Γ(4))
86 = Γ(4)!/(4+4) - 4
87 = √(4)^Γ(4) + 4! - φ(φ(4))
88 = Γ(4)!/(4+4) - √(4)
EDIT: got beaten on that one.
Last edited by poizan42 on Wed May 02, 2007 6:23 pm UTC, edited 3 times in total.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

88=44+44

Hooray!
PaulT

Posts: 80
Joined: Wed Feb 21, 2007 1:39 pm UTC
Location: Manchester, UK

89 = Γ(4)!/(4+4) - φ(φ(4))
90 = √(4)^Γ(4) + 4! + √(4)
91 = Γ(4)!/(4+4) + φ(φ(4))
92 = Γ(4)!/(4+4) + √(4)
93 = 4!*4 - √(4) - φ(φ(4))
94 = 4!*4 - 4 + √(4)
95 = 4!*4 - √(4) + φ(φ(4))
96 = 4!*4 + 4 - 4
97 = 4!*4 + √(4) - φ(φ(4))
98 = 4!*4 + 4 - √(4)
99 = 4!*4 + 4 - φ(φ(4))
100 = 4!*4 + Γ(4) - √(4)
101 = 4!*4 + 4 + φ(φ(4))
102 = 4!*4 + 4 + √(4)
103 = 4!*4 + Γ(4) + φ(φ(4))
104 = 4!*4 + Γ(4) + √(4)
105 = 4!*4 - φ(φ(4)) + 4 + Γ(4)
106 = 4!*4 + 4 + Γ(4)

Okey... think I should stop now

EDIT: unicode ftw!
Last edited by poizan42 on Thu May 03, 2007 8:59 am UTC, edited 1 time in total.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

### Getting small numbers

I just thought it was fun to find different ways to get to the numbers between 1 and 10 using a different number of fours (and very useful for the four fours problem )

1 with 1 four: φ(φ(4))
1 with 2 fours: 4/4
1 with 3 fours: 4 - √(4) - φ(φ(4)) or just φ(φ(4)) + 4 - 4
1 with 4 fours: 4/4 * 4/4

2 with 1 four: √(4)
2 with 2 fours: 4 - √(4)
2 with 3 fours: √(4) + 4 - 4
2 with 4 fours: 4 - 4 + 4 - √(4) or 4/4 + 4/4

3 with 1 four: π(Γ(4)) where π is the prime counting function
was: floor(ln(4!))
could also be: ~-4 (but I don't like it - it's like the predecessor function - you can get anything by applying it repeatedly to some big value)
3 with 2 fours: 4 - φ(φ(4))
3 with 3 fours: √(4) + 4/4
3 with 4 fours: 4 - √(4) + 4/4

4 with 1 four: 4
4 with 2 fours: √(4) + √(4)
4 with 3 fours: 4 + 4 - 4
4 with 4 fours: 4 + (4-4)*4

5 with 1 four: π(π(φ(Γ(Γ(4)))))
was: round(log((4!)!))
5 with 2 fours: 4 + φ(φ(4))
5 with 3 fours: 4 + 4/4
5 with 4 fours: 4 + √(4) - 4/4

6 with 1 four: Γ(4)
6 with 2 fours: 4 + √(4)
6 with 3 fours: 4 + 4 - √(4)
6 with 4 fours: 4 - 4 + 4 + √(4)

7 with 1 four: floor(exp(2))
or: ~-φ(4!)
7 with 2 fours: Γ(4) + φ(φ(4))
7 with 3 fours: Γ(4) + 4/4
7 with 4 fours: Γ(4) + √(4) - 4/4

8 with 1 four: φ(4!)
was: floor(Γ(log(√(√(√(√(exp(4))))))))
8 with 2 fours: 4 + 4
8 with 3 fours: 4 + √(4) + √(4)
8 with 4 fours: 4 - 4 + 4 + 4

9 with 1 four: π(4!)
was: cototient(round(√(Γ(4)!)))
and: floor(√(exp(exp(log(√(√(√(√((4!)!)))))))))
9 with 2 fours: φ(4!) + φ(φ(4))
9 with 3 fours: φ(4!) + 4/4
9 with 4 fours: φ(4!) + √(4) - 4/4

10 with 1 four: π(π(Γ(Γ(4))))
was: floor(log(exp(4!)))
10 with 2 fours: Γ(4) + 4
10 with 3 fours: Γ(4) + √(4) + √(4)
10 with 4 fours: Γ(4) + 4 + 4 - 4

Someone who can find something better for the ones using floor and round?

EDIT: found something better for 8 and 9 - 9 is still ugly though
EDIT2: better ones for 3, 5, 9 and 10 using the prime counting function
Last edited by poizan42 on Fri May 04, 2007 1:13 pm UTC, edited 4 times in total.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

Because i can... and i don't have anything better to do

107 = Γ(Γ(4)) - 4!/√(4) - φ(φ(4))
108 = 4!*4 + 4!/sqrt(4)
109 = Γ(Γ(4)) - 4!/√(4) + φ(φ(4))
110 = Γ(Γ(4)) - Γ(4) - √(4) - √(4)
111 = Γ(Γ(4)) - 4 - 4 - φ(φ(4))
112 = 4!*4 + 4*4
113 = Γ(Γ(4)) - 4 - √(4) - φ(φ(4))
114 = Γ(Γ(4)) - √(4) - √(4) - √(4)
115 = Γ(Γ(4)) - √(4) - √(4) - φ(φ(4))
116 = Γ(Γ(4)) - 4 + 4 - 4
117 = Γ(Γ(4)) - 4 + 4/4
118 = Γ(Γ(4)) - √(4) + 4 - 4
119 = Γ(Γ(4)) - φ(φ(4)) + 4 - 4
120 = Γ(Γ(4)) + (4 - 4)/4
121 = Γ(Γ(4)) + φ(φ(4)) + 4 - 4
122 = Γ(Γ(4)) + √(4) + 4 - 4
123 = Γ(Γ(4)) + √(4) + 4/4
124 = Γ(Γ(4)) + 4 + 4 - 4
125 = Γ(Γ(4)) + 4 + 4/4
126 = Γ(Γ(4)) + √(4) + √(4) + √(4)
127 = Γ(Γ(4)) + 4 + √(4) + φ(φ(4))
128 = Γ(Γ(4)) + 4 + √(4) + √(4)
129 = Γ(Γ(4)) + 4 + 4 + φ(φ(4))
130 = Γ(Γ(4)) + 4 + 4 + √(4)
131 = Γ(Γ(4)) + Γ(4) + 4 + φ(φ(4))
132 = Γ(Γ(4)) + Γ(4) + 4 + √(4)
133 = Γ(Γ(4)) + Γ(4) + Γ(4) + φ(φ(4))
134 = Γ(Γ(4)) + Γ(4) + Γ(4) + √(4)
135 = Γ(Γ(4)) + 4*4 - φ(φ(4))
136 = Γ(Γ(4)) + Γ(4) + Γ(4) + 4
137 = Γ(Γ(4)) + 4*4 + φ(φ(4))
138 = Γ(Γ(4)) + 4*4 + √(4)
139 = Γ(Γ(4)) + 4! - 4 - φ(φ(4))
140 = Γ(Γ(4)) + 4*4 + 4
141 = Γ(Γ(4)) + 4! - 4 + φ(φ(4))
142 = Γ(Γ(4)) + 4! - 4 + √(4)
143 = Γ(Γ(4)) + 4! - 4/4
144 = Γ(Γ(4)) + 4! * 4/4
145 = Γ(Γ(4)) + 4! + 4/4
146 = Γ(Γ(4)) + 4! + Γ(4) - 4
147 = Γ(Γ(4)) + 4! + √(4) + φ(φ(4))
148 = Γ(Γ(4)) + 4! + √(4) + √(4)
149 = Γ(Γ(4)) + 4! + 4 + φ(φ(4))
150 = Γ(Γ(4)) + 4! + 4 + √(4)
151 = Γ(Γ(4)) + 4! + Γ(4) + φ(φ(4))
152 = Γ(Γ(4)) + 4! + 4 + 4
153 = Γ(Γ(4)) + 4! + φ(4!) + φ(φ(4))
154 = Γ(Γ(4)) + 4! + φ(4!) + √(4)
155 = Γ(Γ(4)) + φ(Γ(Γ(4))) + 2 + φ(φ(4))
156 = Γ(Γ(4)) + 4! + φ(4!) + 4
157 = Γ(Γ(4)) + φ(Γ(Γ(4))) + 4 + φ(φ(4))
158 = Γ(Γ(4)) + 4! + φ(4!) + Γ(4)
159 = Γ(Γ(4)) + φ(Γ(Γ(4))) + Γ(4) + φ(φ(4))
160 = Γ(Γ(4)) + 4! + 4*4

If someone doesn't have something that can calculate φ(Γ(Γ(4))) it's = φ(120) = 32
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

poizan42 wrote:I just thought it was fun to find different ways to get to the numbers between 1 and 10 using a different number of fours (and very useful for the four fours problem )

1 with 1 four: φ(φ(4))
2 with 1 four: √(4)
3 with 1 four: floor(ln(4!))
4 with 1 four: 4
5 with 1 four: round(log((4!)!))
6 with 1 four: Γ(4)
7 with 1 four: floor(exp(√(4)))
8 with 1 four: floor(Γ(log(√(√(√(√(exp(4))))))))
9 with 1 four: cototient(round(√(Γ(4)!)))

Man, this could make the greatest Sudoku ever!

Incidentally, one way of making any number with three 4's is:

N = -ln[ln(√√ ... √√4)/ln(4)]/ln(√4)

Where the number of square roots in the innermost parentheses is equal to N. This is pretty cheap, but I think it sets a lower limit.

Cosmologicon

Posts: 1806
Joined: Sat Nov 25, 2006 9:47 am UTC
Location: Cambridge MA USA

Nifty.

Then you can just do it such that you receive N+4, and then subtract 4 for your solution, using four fours.

(Math hax)

umbrae

Posts: 336
Joined: Sat Sep 09, 2006 2:44 pm UTC

Or, if we allow Peano's successor function S, we can write every N as N=S(S(S(...S(4/4)...))). Or for more breaking-the-rules-and-ruining-the-fun smart-assery, we can write every N as {{},{{}},{{{}}},...}, requiring no fours at all!

Torn Apart By Dingos

Posts: 817
Joined: Thu Aug 03, 2006 2:27 am UTC

you can also do -~-~-~...-~(4/4)

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

yes clearly this is a puzzle which calls for elegant solutions rather than scalable solutions.
in ur beanz makin u eveel

evilbeanfiend

Posts: 2650
Joined: Tue Mar 13, 2007 7:05 am UTC
Location: the old world

This was far more fun and useful than doing particle dynamics.
li te'o te'a vei pai pi'i ka'o ve'o su'i pa du li no
Mathematician is a function mapping tea onto theorems. Sadly this function is irreversible.
QED is Latin for small empty box.
Ceci nâ€™est pas une [s]pipe[/s] signature.

cmacis

Posts: 754
Joined: Wed Dec 13, 2006 5:22 pm UTC

If I'm not mistaken, the simplest (IMHO) solution for 3 has not yet been posted:

3 = (4+4+4)/4

kyrmse

Posts: 1
Joined: Thu May 10, 2007 2:46 pm UTC
Location: Brasil

i: (-4/4)^(sqrt(4)/4)

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

There's a similar game where you use the digits of the current year...

1 = 2-0!+0*7
2 = 2+0*0*7
3 = -2-0!-0!+7

(they don't have to be in order ... that's just a bonus)
Avatar yoinked from Inverloch.

"Unless ... unless they kill us, then animate our corpses as dead zombies to fight for them. Then I suppose they've taken our lives, AND our freedom."
- Elan, OOTS 421.

blob

Posts: 328
Joined: Thu Apr 05, 2007 8:19 pm UTC

hotaru wrote:you can also do -~-~-~...-~(4/4)

Um, yeah...that's pretty much a major hack. Bits don't exist in the analytical world of integers, nor to binary representations of negative numbers.

EDIT: As such, I'm gonna try a redo on 51:

51 = 4! + 4! + phi(phi(4+4)), or 4/phi(4), or phi(sqrt(4)+sqrt(4))...

hmm...is phi() too much of a cheat, too?
Last edited by fortyseventeen on Thu May 10, 2007 8:54 pm UTC, edited 1 time in total.
Quick, what's schfifty-five minus schfourteen-teen?

fortyseventeen

Posts: 88
Joined: Fri Mar 02, 2007 3:41 am UTC
Location: SLC, UT, USA

Cosmologicon wrote:
poizan42 wrote:I just thought it was fun to find different ways to get to the numbers between 1 and 10 using a different number of fours (and very useful for the four fours problem )

1 with 1 four: φ(φ(4))
2 with 1 four: √(4)
3 with 1 four: floor(ln(4!))
4 with 1 four: 4
5 with 1 four: round(log((4!)!))
6 with 1 four: Γ(4)
7 with 1 four: floor(exp(√(4)))
8 with 1 four: floor(Γ(log(√(√(√(√(exp(4))))))))
9 with 1 four: cototient(round(√(Γ(4)!)))

Man, this could make the greatest Sudoku ever!

Incidentally, one way of making any number with three 4's is:

N = -ln[ln(√√ ... √√4)/ln(4)]/ln(√4)

Where the number of square roots in the innermost parentheses is equal to N. This is pretty cheap, but I think it sets a lower limit.

You can make that a four 4s solution (excepting 1, I guess) by replacing with one of the √√ pairs with a 4th root.
ZeroSum
Cooler than Jeff

Posts: 2903
Joined: Tue May 08, 2007 10:10 pm UTC

ZeroSum wrote:
Cosmologicon wrote:Incidentally, one way of making any number with three 4's is:

N = -ln[ln(√√ ... √√4)/ln(4)]/ln(√4)

Where the number of square roots in the innermost parentheses is equal to N. This is pretty cheap, but I think it sets a lower limit.

You can make that a four 4s solution (excepting 1, I guess) by replacing with one of the √√ pairs with a 4th root.

Or just replace one of the 4s with √(4*4)
Avatar yoinked from Inverloch.

"Unless ... unless they kill us, then animate our corpses as dead zombies to fight for them. Then I suppose they've taken our lives, AND our freedom."
- Elan, OOTS 421.

blob

Posts: 328
Joined: Thu Apr 05, 2007 8:19 pm UTC

fortyseventeen wrote:EDIT: As such, I'm gonna try a redo on 51:

51 = 4! + 4! + phi(phi(4+4)), or 4/phi(4), or phi(sqrt(4)+sqrt(4))...

hmm...is phi() too much of a cheat, too?

i think (4!-4+.4)/.4 is better...

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

cmacis wrote:This was far more fun and useful than doing particle dynamics.

Useful?

Maybe if a terrorist ever points a gun to your head and demands you solve the four fours problem...

Patashu

Posts: 377
Joined: Mon Mar 12, 2007 8:54 am UTC

hotaru wrote:i think (4!-4+.4)/.4 is better...

That feels more like cheating than phi(4), definitely.
In the future, there will be a global network of billions of adding machines.... One of the primary uses of this network will be to transport moving pictures of lesbian sex by pretending they are made out of numbers.
Spoiler:
gmss1 gmss2

gmalivuk
Archduke Vendredi of Skellington the Third, Esquire

Posts: 20293
Joined: Wed Feb 28, 2007 6:02 pm UTC
Location: Here and There

I spent much of this school day playing this game. It gets rather easy when you allow yourself enough obscure functions.
Here is a link to a page which leverages aggregation of my tweetbook social blogomedia.

Alpha Omicron

Posts: 2765
Joined: Thu May 10, 2007 1:07 pm UTC

gmalivuk wrote:
hotaru wrote:i think (4!-4+.4)/.4 is better...

That feels more like cheating than phi(4), definitely.

which part of it feels more like cheating than phi(4)?

hotaru

Posts: 943
Joined: Fri Apr 13, 2007 6:54 pm UTC

hotaru wrote:
gmalivuk wrote:
hotaru wrote:i think (4!-4+.4)/.4 is better...

That feels more like cheating than phi(4), definitely.

which part of it feels more like cheating than phi(4)?

The part about the hidden zero? The part about doing things relying on base 10? phi(4) is not relying on any specific representation of the number - nor is it relying on hiding something.
poizan42

Posts: 20
Joined: Sun Mar 04, 2007 10:32 am UTC
Location: Denmark

Previous