The four fours problem

[hotaru]

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

[poizan42]

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

[skeptical scientist]

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...>

[PaulT]

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]

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]

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??

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?

[Gelsamel]

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).

[poizan42]

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)

[mattmacf]

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

[poizan42]

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

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

[PaulT]

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]

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]

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

~
and yes, ** is exponentiation.

[mattmacf]

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]

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

~
and yes, ** is exponentiation.

a FORTRANer?

edit: nevermind spotted the link now

[EvanED]

84 = 44 * sqrt(4) - 4

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

[poizan42]

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.

[PaulT]

88=44+44

Hooray!

[poizan42]

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!

[poizan42]

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

[poizan42]

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]

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]

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)

[Torn Apart By Dingos]

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]

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

[evilbeanfiend]

yes clearly this is a puzzle which calls for elegant solutions rather than scalable solutions.

[cmacis]

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

[kyrmse]

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

3 = (4+4+4)/4

[hotaru]

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

[blob]

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)

[fortyseventeen]

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?

[ZeroSum]

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]

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)

[hotaru]

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...

[Patashu]

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...

[gmalivuk]

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

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

[Alpha Omicron]

I spent much of this school day playing this game. It gets rather easy when you allow yourself enough obscure functions.

[hotaru]

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]

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.