Classic problem: Four 4's

[Ken_g6]

I found a version of the four 4's problem in an old book awhile back. Four those who haven't seen it befour, the challenge is to create fourmulas, using no more than four 4's, that evaluate to other integers.

This version is more stringent on the operations allowed than most. The operations allowed are listed below. The book claims that all positive integers up to 119 can be created this way. I have been unable to create three of those integers, so I'd like to see what you can come up with. If the book is correct, though, all integers from 1 through at least 130 can be created.

Allowed unary operations: (each example in parentheses uses one 4)
Square (4^2) - this is the only place where 2's are allowed. Also note below that exponentiation other than this is not allowed.
Factorial (4!)
Decimal (.4)

Allowed binary operations: (each example in parentheses uses two 4's)
+ (4+4)
- (4-4)
* (4*4)
/ (4/4)
Concatenate (44); also before (4.4) or after (.44) a Decimal.
Any number of parentheses are also allowed. Standard order of operations applies when parentheses are missing.

Since it's usually easy to fill out an expression with more 4's (e.g. 8 = 4+4 or 4^2/4+4^2/4), I'll determine best solutions first by fewest 4's, and second by shortest total length.

Wacky, would you mind setting up a scoreboard from 0 to 160 or so? I'll start populating it:

N: 4's,len expression (Author, date)
8:   2, 3: 4+4 (Ken_g6, 9/28)

But I won't do any more unless you guys get desperate.

[cheesehead]

Originally Posted by Ken_g6

This version is more stringent on the operations allowed than most.

I've never seen another version that allowed squaring as a unary operator. OTOH versions that I've seen allowed floor, ceiling, square root and exponentiation, so this'll be more difficult in some cases.

Notation: I'll use 4^sq to denote the unary squaring, so as not to introduce any "2".

Here's some easy or obvious ones, not necessarily of minimal length, just to initiate a few table entries:

4 = 4

6 = 4! / 4

10 = 4 / .4
11 = 44 / 4

16 = 4^sq

24 = 4!

32 = 4^sq + 4^sq

36 = 4!^sq / 4^sq

44 = 44

48 = 4! + 4!

64 = 4^sq * 4

72 = 4! * 4 - 4!

80 = 4^sq * 4 + 4^sq

96 = 4! * 4

100 = ( 4 / .4 )^sq

110 = 44 / .4
111 = 44.4 / .4

128 = 4^sq * 4 + 4^sq * 4

144 = 4!^sq / 4

256 = 4^sq * 4^sq

[cheesehead]

Originally Posted by cheesehead

256 = 4^sq * 4^sq

256 = 4^sqsq

[Wacky]

I'll set up a scoreboard soon.

Obviously:
0 = 4-4
1 = 4/4
2 = (4+4)/4
3= (4+4+4)/4
5 = 4 + 4/4
7 = 4 + 4 - 4/4
8 = 4 + 4
9 = 4 + 4 + 4/4
12 = 4 + 4 + 4

[grandpascorpion]

13 = 4sq -4 + 4/4
15 = 4sq - 4/4

[Numbers]

Originally Posted by ken_g6

The book claims that all positive integers up to 119 can be created this way

Originally Posted by ken_g6

If the book is correct, though, all integers from 1 through at least 130 can be created.

Originally Posted by ken_g6

Wacky, would you mind setting up a scoreboard from 0 to 160 or so?

Originally Posted by cheesehead

256 = 4^sqsq

Did I miss something ?

[grandpascorpion]

17 = 4sq + 4/4
18 = 4/.4 + 4 + 4
19 = 4! - 4 - 4/4
20 = 4! - 4
21 = 4! -4 + 4/4
23 = 4! - 4/4

25 = 4! + 4/4
26 = 4sq + 4/.4
27 = 4! + 4 - 4/4
28 = 4sq + 4sq - 4
29 = 4! + 4 + 4/4
30 = (4+4/4)!/4
31 = 4sq + 4sq - 4/4
32 = 4sq + 4sq
33 = 4sq + 4sq + 4/4
34 = 4! + 4/.4
35 = 4! + 44/4
37 = 4!sq / 4sq + 4/4
39 = 4! + 4sq - 4/4
40 = 4! + 4sq
41 = 4! + 4sq + 4/4

95 = (4! * 4sq-4)/4
96 = 4! * 4sq/4
97 = (4! * 4sq+4)/4
109 = (44 -.4)/.4
120 = (4+4/4)!
121 = (44/4)sq

[akruppa]

7 = (4!+4)/4
60 = 4!/.4

[grandpascorpion]

14 = 4sq - (4+4)/4
22 = 4! - (4+4)/4
43 = 44 - 4/4
45 = 44 + 4/4
47 = 4! + 4! - 4/4
49 = 4! + 4! + 4/4
52 = 4! + 4! + 4

56 = 4*4sq-4-4
60 = 4*4sq-4
63 = 4*4sq-4/4
65= 4*4sq+4/4
81 = ((4-4/4)sq)sq

[Orgasmic Troll]

hmm. The way I remember this was that you had to use exactly 4 4's

I don't think squaring should be included, I can't logically fathom why that's allowed. Perhaps they mean square roots?

Also, is (.4...) or .4 with an overbar allowed? (4/.4..) is a nice way to get 9 :)

Has anyone ever looked at a systematic way of enumerating all possible answers?

[Ken_g6]

Originally Posted by Numbers

Did I miss something ?

You are missing something, and that is the list of numbers between 1 and 160 for which I could and couldn't find solutions. But I want you to try to discover those for yourself, as I may have missed solutions for some of them.

Originally Posted by TravisT

hmm. The way I remember this was that you had to use exactly 4 4's

That's the way it was in the book, but I explained earlier that since it's easy to add 4's, I'll go for the smallest number of 4's possible.

Originally Posted by TravisT

I don't think squaring should be included, I can't logically fathom why that's allowed. Perhaps they mean square roots?

Also, is (.4...) or .4 with an overbar allowed? (4/.4..) is a nice way to get 9 :)

Has anyone ever looked at a systematic way of enumerating all possible answers?

It's this way because it's from a book for doing math on a calculator that has an x^2 button, and because it's more challenging this way. I found several web pages using overbars, square roots, and of course the general method of square roots and logs mentioned in the "Generating 2005" puzzle. I like the description from this page:

In summary, this puzzle depends entirely on what rules you choose. It has more to do with manipulating the symbolism in clever ways than on any mathematical truths, and if mathematical notation had evolved differently, the outcome of the puzzle would be quite different too.

This particular set of rules seems to allow a wide range of solutions, but makes them harder to find than some other rule sets.

P.S. There's a better solution for 12 (and thus 3).

Edit: I guessed 160 as a rough upper limit on the contiguous integers from 1 that have four 4's solutions. We can go higher if anyone wants.