Talk:Four Fours/197/Solutions/1
I feel like we need to limit what kinds of aritmetical operations are allowed for this puzzle.
I think the floor and ceiling functions should be disallowed, as repeated applications of factorials, Gamma, sqrt and ceil/floor on $4$ can produce a lot of different integers.
(In Four Fours/Historical Note:
- He finishes with a reference to an article by Donald Ervin Knuth in which it is proved that all positive integers up to $208$ can be expressed with nothing but one $4$, instances of the square root sign, the factorial sign, and parentheses.
was this Knuth's method?)
Similarly:
- $\map \Gamma 4 !! \times 4 + \dfrac {\sqrt 4} {.4} = 48 \times 4 + 5 = 197$
could be allowed, but we cannot allow multifactorials more than the double, e.g.:
- $\paren {4!} \underbrace{!!\dots!}_{16} + \dfrac {4! - 4} 4 = 24 \times \paren {24 - 16} + 5 = 197$
would be too much.
We have currently used addition, subtraction, multiplication, division, factorial, Gamma function, square root, (multiple roots), exponentiation;
and for $4$'s, concatenation, decimal point and repeated decimals notations.
I am still strugging to find a representation of $197$ using those only. --RandomUndergrad (talk) 05:31, 15 February 2022 (UTC)
- What about the successor function $\map s 4 = 5$?
- I'm for being inclusive. Those who are reading round the subject will be able to evaluate the result for themselves. "Oh dear, you can't do $197$ without using the floor function."
- Knuth's solution is worth jstoring for. I'll post it up in due course as it adds an interesting slant to the question.
- As has been determined long back, you can't do $113$ without using Gamma. Some suggest you shouldn't include the square root sign because it implicitly implies a $2$. Some don't allow $44$ or $.4$ and so on because they are shorthand for an expression which requires a $10$ to join the party. So it goes. If we impose a gatekeeping policy we'll be limiting our experience.
- Just my NSHO. --prime mover (talk) 06:24, 15 February 2022 (UTC)