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Given a non-empty list of digits 0 though 9, output the smallest number that can be produced by an expression formed by reordering these digits and introducing exponentiation signs ^, with adjacent digits getting concatenated as multi-digit numbers. Exponentiation is evaluated as right-associative.

For example, [4, 2, 3, 3] could become 2^34^3 which evaluates as 2^(34^3), but simply writing 2334 is smallest here. You can assume that any array containing a 0 gives a minimum of zero (a special case), or that it's achievable by an expression like 0^rest, or like 00 for just zeroes.

The input list can be taken as any ordered sequence, but not as an unordered (multi)set.

Test cases:

[2,3] => 8 (2^3) [3,3] => 27 (3^3) [4,3] => 34 (34) [1,2,3] => 1 (1^2^3) [2,2,2] => 16 (2^2^2) [2,3,2] => 81 (2^3^2) [3,0] => 0 (Special, or maybe 0^3) [0,1,0] => 0 (Special, or maybe 0^10)

Shortest code in each language wins. I'll accept an answer just to give the +15 rep without any specific accepting standard.

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    \$\begingroup\$ "Acception only act as a +15." Are you saying that when/if you accept an answer, it will only count as 15 reputation? \$\endgroup\$ Commented May 11, 2020 at 12:05
  • \$\begingroup\$ I like the challenge concept, but your text is hard to read and it took me a few tries to understand it. \$\endgroup\$ Commented May 11, 2020 at 12:05
  • \$\begingroup\$ @ouflak That means if someone is accept, it only mean he get 15 reputation, I don't claim any accepting standard \$\endgroup\$ Commented May 11, 2020 at 12:07
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    \$\begingroup\$ For golfers who might want to write a non-brute-force solution, it looks like the minimal result is the digits concatenated in sorted order, except for smaller values of 0 for any list with a 0, 1 for any list with a 1, and these six values produced from power expressions: 2^2=4, 2^3=8, 2^4=16, 3^3=27, 2^2^2=16, 3^2^2=81. \$\endgroup\$ Commented May 11, 2020 at 12:23
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    \$\begingroup\$ Probably worth including [2,3,2] in the examples. \$\endgroup\$ Commented May 11, 2020 at 17:16

6 Answers 6

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05AB1E, 12 11 bytes

œε.œJ€.«m}ß

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Explanation:

œ # Get all permutations of the (implicit) input-list ε # Map each permutation to: .œ # Get all its partitions J # Join each inner-most list together € # For each inner list: .« # Right-reduce it by: m # Taking the power }ß # After the map: pop and push the flattened minimum # (after which it is output implicitly as result)
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Jelly, 10 bytes

Œ!*@/€;ḌƊṂ

A monadic link accepting a list of digits yields an integer.

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How?

Checks digital and power-wise evaluations of all permutations even though we only need to check forward-sorted-digital, forward-sorted-power-wise, and reverse-sorted-power-wise (and only for [3,2,2]), because it's far terser.

Note that there is no need to check any mixture of digital and power-wise evaluations (they can never be strictly smaller than one of the three previously mentioned evaluations)

Œ!*@/€;ḌƊṂ - Link: list, L Œ! - all permutations (of L) -> P Ɗ - last three links as a monad: € - for each (p in P): / - reduce with: @ - using swapped arguments: * - exponentiation Ḍ - un-decimal (vectorises across P) ; - concatenate (these two lists of numbers) Ṃ - minimum
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  • \$\begingroup\$ Notice that you happen to use 0^1^0^0 on [0,0,0,1] and 0000 on [0,0,0,0]. Unexpected work :) \$\endgroup\$ Commented May 11, 2020 at 19:34
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Japt -g, 12 bytes

á Ër!p mDìÃn

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  • \$\begingroup\$ Fails for [2,2,3], which results in 64 instead of 81. \$\endgroup\$ Commented May 11, 2020 at 16:30
  • \$\begingroup\$ You don't write \$2{^2} {^3}\$, but \$3^{2^2}\$ is fine \$\endgroup\$ Commented May 11, 2020 at 16:33
  • \$\begingroup\$ @KevinCruijssen, 2**3**2=64 \$\endgroup\$ Commented May 11, 2020 at 17:04
  • \$\begingroup\$ "and power must go from right to left" -> 2**(3**2) = 512 \$\endgroup\$ Commented May 11, 2020 at 17:09
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    \$\begingroup\$ It means that the precedence is to be treated as right-most first, i.e. 2**3**2 is to be \$2^{(3^2)}=2^9=512\$. \$\endgroup\$ Commented May 11, 2020 at 17:14
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Io, 77 bytes

Based on xnor's observation in the comment section. -12 bytes thanks to Arnauld's idea.

method(x,y :=x sort;if(y==list(2,2,3),81,y join()asNumber min(y reduce(**))))

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Explanation

method(x, // Anonymous method with param x y := x sort // Assign y to sorted x if(y == list(2,2,3), // Edge case treatment 81, // (It's a kinda weird case that goes right to left) ,y join()asNumber// Join the sorted digit-list into a single number. min( // Minimum with: y reduce(**)))) // the sorted list reduced by the power function.
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JavaScript (ES7),  65  62 bytes

Fixed thanks to @xnor and @l4m2
Saved 3 bytes thanks to @l4m2

a=>Math.min(x=a.sort().join``,a[0]&&eval(x-223?a.join`**`:81))

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How?

The general case is to look for the minimum of the concatenation of the digits in ascending order and an exponent chain, also in ascending order.

But as noticed by @xnor, there are a few special cases:

  • if the list contains a \$0\$, the result is always \$0\$
  • for [3,2,2], the minimum value is \$3^{2^2}=81\$
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    \$\begingroup\$ Be careful with the [3,2,2] input \$\endgroup\$ Commented May 11, 2020 at 13:31
  • \$\begingroup\$ @xnor Thank you for the notification. Hopefully fixed now. \$\endgroup\$ Commented May 11, 2020 at 13:46
  • \$\begingroup\$ Fail on [0,0,1] \$\endgroup\$ Commented May 11, 2020 at 15:40
  • \$\begingroup\$ @l4m2 Thanks. Updated. \$\endgroup\$ Commented May 11, 2020 at 16:02
  • \$\begingroup\$ Fail on [3,2,2,0,0]. I consider x*a[0] in numbers for min instead \$\endgroup\$ Commented May 11, 2020 at 16:08
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perl -alp, 83 bytes

%_=(22,4,23,8,24,16,33,27,222,16,223,81);$_=join"",sort@F;$_=/0/||/1/?$&:$_{$_}||$_

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Reads a list of white space separated numbers from STDIN, writes the answer to STDOUT. If the input contains a 0, output 0, else, if the input contains a 1, output a 1. Else, if it's one of 6 special cases, output the corresponding value. Otherwise, output the sorted input.

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