Making any integer with four 2s

I feel like the second you allow functions you've thrown the spirit of the game.

Ex, the gamma function is (n-1)! So now you're making 7 with four twos and a one. You've broken the spirit.

If I can hide numbers in a function call... It's trivially easy to always succeed.

> use any mathematical operations

Okay, then this is easy, just use the successor function.

  S(n) = n+1

  6 = 2*2*2-2
  7 = S(2*2*2-2)
  8 = S(S(2*2*2-2))

Etc.

See also: "Representing numbers using only one 4" written by a 26-year-old Donald Knuth in 1964 (https://www.jstor.org/stable/2689238 reprinted as Chapter 10 of his Selected Papers on Fun and Games) — it uses the single digit 4, and the three operations √x (square root), ⌊x⌋ (floor, i.e. greater integer not greater than), and x! (factorial), and ends with a (still unsolved) conjecture about whether every integer can be represented in this way.

The appendix (written for the book in 2011) points out an earlier (1962) 1.5-page paper π in Four 4's by J. H. Conway and M. J. T. Guy, written when they were students at Cambridge, that has a similar idea: https://archive.org/details/eureka-25/page/18/mode/1up?view=...

For example,

    5 = ⌊√√√√√(4!)!⌋

because 24! lies between 5^{32} and 6^{32}.

Maybe it's just me, but writing sqrt(2+2) instead of sqrt(2*2) or sqrt(2^2) was such an odd choice. It obfuscates the reason why 2=sqrt(2+2), and unnecessarily so.

I think my preference is more towards conciseness.

I made a stack machine with single character instructions and needed to solve a variation of this problem. I had just the digits 0 through 9. The characters '23' would be push 2 followed by push 3. To actually represent the number 23 you would use

    45*3+
 or something similar.

That left me with the problem of how to encode each integer in the fewest characters.

Tools at hand.

        The digits 0 through 9
        'P':  Pi
        '*':  (a * b),
        '/':  (a / b),
        '-':  (a - b),
        '+':  (a + b),
        's':  sin(a),
        'c':  cos(a),
        'q':  sqrt(a),
        'l':  log(a),
        '~':  abs(a),
        '#':  round(a),
        '$':  Math.floor(a),
        'C':  clamp(a),
      
        '<': min(a, b),
        '>': max(a, b),
        '^':  pow(a, b),
        'a': atan2(a, b),
        '%': positiveMod(a, b),
        '!': (1 - a),
        '?': (a <= 0 ? 0 : 1)
        'o':   a xor b scaled by c;   ((a*c) xor (b*c))/c

        'd':  duplicate the top stack entry
        ':':  swap the top two stack entries
        ';':  swap the top and third stack entries

I have wondered about revisiting the stack machine with a complex number stack to see what I can come up with.

(Next time I post something like this I am not going to use my phone)

> There's just one small wrinkle: it uses three instances of the digit 2, not four.

One small wrinkle, if you ignore the fact that the root notation conceals exponentiation by 1/2, by making that common value a default.

That's a lot of hidden 2's!

Related: there as a reverse engineering/CTF challenge (which shall remain nameless to prevent you from cheating) where my solution involved injecting shellcode that adds specific number to the stack pointer. But your shellcode -- including the number(s) you add -- can only involve bytes from the ascii alphanumeric set.

So I used a SAT solver to find a combination of numbers, not using prohibited bytes, that add up to the number I really want.

https://docs.google.com/presentation/d/19K7SK1L49reoFgjEPKCF...

This reminds me of this mobile game Tchisla[0] where you have to build all numbers up to 1000 (10000?) using only a given digit and a couple of operators (including sqrt and !)

It was a lot of fun, you tend to develop strategies and the game has a simple, efficient UX. Fair warning, it is very time consuming.

[0] https://apps.apple.com/fr/app/tchisla-number-puzzle/id110062...

There's the classic “four fours”, which I learned as a child in the book “The Man Who Counted”.

https://en.wikipedia.org/wiki/Four_fours

https://en.wikipedia.org/wiki/The_Man_Who_Counted

This is amazing, but there are a lot of 2's hiding in those sqrt symbols

Here are some values that are (understandably) not listed on the blog. They happen only due to the limited precision of floating point formats.

  128              = √(2 / √√(√2 - (2 / √2)))
  8192             = √√(2 / ((√2 * √2) - 2))
  16384            = (2 / √√(√2 - (2 / √2)))
  67108864         = √(2 / ((√2 * √2) - 2))
  134217728        = (2 / √(√2 - (2 / √2)))
  4503599627370496 = (2 / ((√2 * √2) - 2))
  9007199254740992 = (2 / (√2 - (2 / √2)))
  6369051672525773 = (√2 / (√2 - (2 / √2)))

I found these by accident a long time ago but kept them because they do "work". Try to input one expression in the lil box in https://www.wolframalpha.com/?source=nav and they will quickly evaluate to these values; the charade goes away after you press Enter and get the (mathematically) correct answer.

My old solvers from what feels like a previous life: https://madflame991.blogspot.com/2013/02/four-fours.html https://madflame991.blogspot.com/2013/02/return-of-four-four...

That was fun

Very clever, but using an arbitrary number of square roots seems almost cheating since it’s practically another symbol for a “2” (exponent of 1/2)

> I've read about this story in Graham Farmelo's book The Strangest Man: The Hidden Life of Paul Dirac, Quantum Genius.

"The Strangest Man": https://en.wikipedia.org/wiki/The_Strangest_Man

Four Fours: https://en.wikipedia.org/wiki/Four_fours :

> Four fours is a mathematical puzzle, the goal of which is to find the simplest mathematical expression for every whole number from 0 to some maximum, using only common mathematical symbols and the digit four. No other digit is allowed. Most versions of the puzzle require that each expression have exactly four fours, but some variations require that each expression have some minimum number of fours.

I think I saw this one on an ancient arab math problems book. But it may be apocryphal, not sure how many tools they would have had, factorial symbols?

At any rate they invented algebra so maybe there's something to it

Edit: It was the man who counted, definitely apocryphal, as it was written in the 20th century

Related numberphile video which goes into a different variation of using all digits in ascending and descending order: https://www.youtube.com/watch?v=-ruC5A9EzzE

but in this case there is a unsolved gap!

So, the formula is really about making any integer with three 2s, but historical precedent calls the game with four 2s more interesting, so the (stronger) result is monkey-patched by replacing a 2 with sqrt(2+2).

Wow, it’s been ages since I stumbled upon a math problem that’s not only a blast to solve but also tells a tale! This one’s got it all: easy to grasp, deliciously complex, and comes with a side of drama where some genius totally threw a wrench in the works for everyone else. Bravo!

I like these games, and I would say more fun when using a larger number that has more factors, for example 120. 120 is among the superior highly composite numbers:

https://en.wikipedia.org/wiki/Superior_highly_composite_numb...

I came across a similar video a couple weeks ago about how many ways you can turn a 2 into a 4... https://youtu.be/VEQOv61Gveg

I kinda feel that's cheating and each square root requires a two to use it.

Is using a primorial permitted?

    7 = (2+2)#+2/2

https://en.wikipedia.org/wiki/Primorial

You don't need the gamma function to get to 7. You can stay at an Algebra 1 level.

I solved the puzzle for 1-10 before looking at the answers, and this was my solution for 7:

⌊√222⌋/2

or more readably:

floor(sqrt(222)) / 2

I used to play this game when a kid but with four number 4 instead. Just operators (+,!,/,-,*, ^, etc)

Am I missing something or 7 is simply 2 + 2 + 2 + 2/2?

All those are allowed, so what's the problem?

  Getting to 7 is notoriously difficult

What am I missing?

2 * 2 * 2 - 2/2

This is just Peano arithmetic with extra steps

i thought the famous puzzle was 4, 4s

9 = 22 / 2 - 2

Is 7 really notoriously difficult to define?

7 = 2/2 + 2 + 2 + 2

2/2+2/2…

Then you just add it multiple times

And if 0 is an integer.

2/2-2/2