The four fours problem

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

poizan42
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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.

skeptical scientist
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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.
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PaulT
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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?

hotaru
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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

poizan42
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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.

Gelsamel
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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).
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poizan42
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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.

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

poizan42
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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!

PaulT
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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?

EvanED
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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.

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

~
and yes, ** is exponentiation.

mattmacf
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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 evilbeanfiend
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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
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EvanED
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84 = 44 * sqrt(4) - 4

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

poizan42
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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.

PaulT
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88=44+44

Hooray!

poizan42
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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
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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
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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

Cosmologicon
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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.

umbrae
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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)

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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!

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

evilbeanfiend
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yes clearly this is a puzzle which calls for elegant solutions rather than scalable solutions.
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This was far more fun and useful than doing particle dynamics.
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kyrmse
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If I'm not mistaken, the simplest (IMHO) solution for 3 has not yet been posted:

3 = (4+4+4)/4

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

blob
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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)
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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.
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ZeroSum
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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.

blob
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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)
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hotaru
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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...

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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...

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

That feels more like cheating than phi(4), definitely.
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I spent much of this school day playing this game. It gets rather easy when you allow yourself enough obscure functions.
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hotaru
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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)?

poizan42
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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.