introduzione

In questa sfida dovresti dividere un numero intero in due pezzi. Dal momento che a nessuno piace ottenere il pezzo più piccolo di torta, il tuo obiettivo è quello di essere il più equo possibile. Ad esempio, se si desidera dividere il numero intero 7129in due parti, ci sono 3 modi possibili per farlo.

7,129, 71,29e 712,9sono tutte possibilità, ma71,29 è il modo più giusto di dividerlo in due pezzi perché riduce al minimo la differenza tra i due:

7 129 -> |7-129| = 122
71 29 -> |71-29| = 42
712 9 -> |712-9| = 703

Sfida

Dato un numero intero determinare il miglior modo possibile di partizionarlo come descritto sopra e riportare la differenza risultante.

Regole

• La divisione ha senso solo per numeri interi di lunghezza almeno due, l'ingresso sarà sempre ≥ 10
• L'input può essere un numero intero, un elenco di cifre o una stringa
• Non è necessario gestire input non validi

Casi test

Devi solo segnalare la differenza risultante, il partizionamento è qui solo a scopo illustrativo:

10 -> 1,0 -> 1
11 -> 1,1 -> 0
12 -> 1,2 -> 1
13 -> 1,3 -> 2
101 -> 1,01 -> 0
128 -> 12,8 -> 4
313 -> 3,13 -> 10
1003 -> 1,003 -> 2
7129 -> 71,29 -> 42
81128 -> 81,128 -> 47
999999 -> 999,999 -> 0
9999999 -> 999,9999 or 9999,999 -> 9000

Brachylog , 12 11 byte

La mia prima risposta su Brachylog

Prendi input come stringa

{~cĊịᵐ-ȧ}ᶠ⌋

Provalo online!

Spiegazione:

ᶠ will find all possible outputs for the predicate in {…} and store them in a list. ~c says that the output is a list that, when concatenated, is equal to the input. Next Ċ asserts that the the output of ~c has length 2.

ịᵐ converts both elements of the output into integers (this gets rid of leading 0s), -ȧ takes the absolute difference of the two elements.

Once we have our list of all possible outputs, we get the minimum element with ⌋

Haskell, 48 bytes

f n=minimum[abs$n`div`10^k-n`mod`10^k|k<-[1..n]]

[1..n] makes this too slow for the larger test cases.

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05AB1E, 9 bytes

Code:

ā¤âε£ÆÄ}W

Uses the 05AB1E encoding. Try it online!

Explanation

ā            # Get the array [1, 2, .., len(input)]
 ¤â          # Cartesian product with the last element, (e.g. for input 12345:
               [[1, 5], [2, 5], [3, 5], [4, 5], [5, 5]])
   ε   }     # For each element:
    £        #   Get the substrings (12345 [3, 5] £ --> [123, 45])
     Æ       #   Reduce by subtraction
      Ä      #   Get the absolute value
        W    # Take the minimum of all results

Python 2, 64 bytes

lambda n:min(abs(int(n[i:])-int(n[:i]))for i in range(1,len(n)))

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Perl 6, 40 bytes

{min map {abs [-] @$_},m:ex/^(.+)(.+)$/}

Test it

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣

  min
    map
      {
        abs
          [-]    # reduce with &infix:«-»
            @$_  # the input of this inner block as a Positional
      },

      # split ｢$_｣ into 2 in every possible way
      m
      :exhaustive
      /^ (.+) (.+) $/
}

C, 94 bytes

c,r,d,a;f(n){for(c=1,r=0,d=n<11?1:n;n;r+=n%10*c,c*=10,n/=10)a=abs(r-n),d=r&&a<d?a:d;return d;}

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Python 2, 51 bytes

f=lambda n,k=1:min(abs(n/k-n%k),k/n*n or f(n,10*k))

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Prolog (SWI), 195 189 154 117 112 bytes

35 bytes saved thanks to Eminga

A*H:-findall(X,(between(0,A,I),r(A,I,X)),L),sort(L,[H|_]),!.
r(A,B,C):-Z is 10**B,divmod(A,Z,X,Y),C is abs(X-Y).

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This is my first try at prolog golfing so it may be a bit horrendous. Here is how it works.

At the highest level we have *. * takes A and H, and determines if H is the smallest way to split A.

    A*H:-
      findall(X,(between(0,A,I),call(r,A,I,X)),L),
      sort(L,[H|_]),
      !.

The first line here uses a technique from this SO post, to essentially perform a map of the predicate r(A) over the integers from 0 to A. Since r confirms the values of each partitions this will give us the values of all possible partitions, plus a whole load of extra junk. All these partitions will be stored in L in no particular order. Once that is done we sort the list to find the smallest element. We then use a cut to prevent backtracing.

Next we have the definition of r. First r calculates the two results of the split naming them X and Y.

r(A,B,C):-
  Z is 10**B,
  divmod(A,Z,X,Y),
  C is abs(X-Y).

Then we assert that C is the difference of them and is positive.

  C is abs(X-Y).

Haskell, 68 65 bytes

f x=minimum[abs$read(take i x)-read(drop i x)|i<-[1..length x-1]]

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Explanation

minimum              -- Minimum of ...
 [abs$               -- The absolute value of ...
  read(take i x)     -- The first i characters of x
  -                  -- Minus ...
   read(drop i x)    -- The last i characters of x
 |i<-[1..length x-1] -- From i=1 to i=length x - 1
 ]

Charcoal, 14 bytes

Ｉ⌊Ｅθ↔⁻Ｉ…θκＩ✂θκ

Try it online! Link is to verbose version of code. Conveniently I get to use the 2-arg variant of Slice. Explanation:

   θ            Input string
  Ｅ             Map over characters
        θ   θ   Input string
         κ   κ  Current map index
       …        Mold to length (i.e. head)
           ✂    Slice (i.e. tail)
      Ｉ   Ｉ     Cast to integer
     ⁻          Subtract
    ↔           Absolute value
 ⌊              Minimum
Ｉ               Cast to string
                Implicitly print

Jelly, 9 8 bytes

ḌÐƤḊạḌƤṂ

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-1 byte thanks to Dennis. Input is a list of digits.

Explanation

ḌÐƤḊạḌƤṂ
ḌÐƤ          Convert to integer from decimal for all Ƥostfixes. [1,2,3]->[123,23,3]
   Ḋ         Remove the first element ->[23,3]
     ḌƤ      Convert to integer from decimal for all Ƥrefixes [1,2,3]->[1,12,123]
    ạ        Absolute difference. [23,3]ạ[1,12,123]->[22,9,123]
       Ṃ     Minimum

Funky, 159 134 99 bytes

s=>{S={}fori=1i<#s i++{S[#S]=(((v=s::sublist)(0,i)::reduce@..-v(i)::reduce@..)^2)^.5};math.min...S}

Actually fitting the spec is shorter it seems.

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Retina, 36 bytes

\B
,$'¶$`
\d+
$*
(1*),\1

Om`^.*
\G1

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Explanation

\B
,$'¶$`

This generates all possible partitions on separate lines, as well as a trailing line with the original input.

\d+
$*

Convert each number in each partition to unary.

(1*),\1

Remove a maximal and equal amount of 1s from both parts of each partition (i.e. remove the minimum, and subtract it from the maximum, which gives the absolute difference).

Om`^.*

Sort the lines.

\G1

Count the 1s on the first line, which gives the minimal absolute difference.

J, 32, 27 23 bytes

-5 bytes thanks to FrownyFrog! -4 bytes if input is a string.

[:<./}:@(".\)|@-1}.".\.

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Original: Takes a number as input

(".\(}:@[([:<./|@-)}.@])".\.)@":

How it works:

                             @": - convert the number to list of chars and
(".\                    ".\.)    - form all prefixes/suffixes and convert them to numbers
    (}:@[          }.@])         - drop the last prefix / first suffix
         (     |@-)              - find the absolute differences
          [:<./                  - find the minimum

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JavaScript (ES6), 64 bytes

Takes input as a string.

f=([c,...s],l=0)=>c?Math.min(Math.abs((l+=c)-s.join``),f(s,l)):l

Test cases

f=([c,...s],l=0)=>c?Math.min(Math.abs((l+=c)-s.join``),f(s,l)):l

console.log(f("10"     )) // -> 1
console.log(f("11"     )) // -> 0
console.log(f("12"     )) // -> 1
console.log(f("13"     )) // -> 2
console.log(f("101"    )) // -> 0
console.log(f("128"    )) // -> 4
console.log(f("313"    )) // -> 10
console.log(f("1003"   )) // -> 2
console.log(f("7129"   )) // -> 42
console.log(f("81128"  )) // -> 47
console.log(f("999999" )) // -> 0
console.log(f("9999999")) // -> 9000

Commented

f = ([c, ...s],           // c = current character, s = array of remaining characters
                l = 0) => // l = left part of the integer, initialized to 0 (see footnote)
  c ?                     // if c is defined:
    Math.min(             //   return the minimum of:
      Math.abs(           //     1) the absolute value of:
        (l += c) -        //       the updated left part
        s.join``          //       minus the right part
      ),                  //     end of Math.abs()
      f(s, l)             //     2) the result of a recursive call
    )                     //   end of Math.min()
  :                       // else:
    l                     //   stop the recursion by returning l (now equal to the input)

Non-recursive (ES7), 65 bytes

Takes input as a string.

s=>Math.min(...[...s].map(c=>((l+=c)-s.slice(++i))**2,i=l=0))**.5

Test cases

let f =

s=>Math.min(...[...s].map(c=>((l+=c)-s.slice(++i))**2,i=l=0))**.5

console.log(f("10"     )) // -> 1
console.log(f("11"     )) // -> 0
console.log(f("12"     )) // -> 1
console.log(f("13"     )) // -> 2
console.log(f("101"    )) // -> 0
console.log(f("128"    )) // -> 4
console.log(f("313"    )) // -> 10
console.log(f("1003"   )) // -> 2
console.log(f("7129"   )) // -> 42
console.log(f("81128"  )) // -> 47
console.log(f("999999" )) // -> 0
console.log(f("9999999")) // -> 9000

Commented

s =>                            // given s
  Math.min(...                  // get the minimum value in the result of this map():
    [...s].map(c =>             //   for each character c in s:
      ((l += c)                 //     append c to l (the left part)
                - s.slice(++i)) //     and subtract the right part from it
      ** 2,                     //     square the result
      i =                       //     start with i = 0 (split position)
      l = 0                     //     and l = 0 (left part, see footnote)
    )                           //   end of map()
  )                             // end of Math.min()
  ** .5                         // return the square root of the smallest square

Note: In both versions, l is coerced to a string on the first iteration. Normally, we should be careful about leading zeros in a numeric literal: 0123 - 10 === 73 because 0123 is parsed as an octal value (this is now deprecated, but still valid in non-strict mode). But '0123' - '10' === 113, the leading zero being this time ignored. So, it's sound to do so.

From the specification of the abstract operation ToNumber applied to a string:

A StringNumericLiteral that is decimal may have any number of leading 0 digits

APL (Dyalog), 27 bytes

{⌊/|-/⍎¨↑⊂∘⍵¨↓1,∘.=⍨⍳¯1+≢⍵}

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How?

¯1+≢⍵ - length of n minus 1

∘.=⍨⍳ - identity matrix

      1,∘.=⍨⍳3
1 1 0 0
1 0 1 0
1 0 0 1

1, - prepend 1 for each row

↓ - split by rows

⊂∘⍵¨ - for each, partition the string by it

      1 0 1 0 ⊂ '7129'
┌──┬──┐
│71│29│
└──┴──┘

↑ - flatten

-/ - reduce each pair with subtraction

| - take absolute values

⌊/ - minimum

APL (Dyalog), 35 bytes

{⌊/|-/⍎¨(⊂∘⍵⍤1)1,∘.=⍨⍳¯1+≢⍵}

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Jelly, 11 bytes

ŒṖṖLÐṂḌạ/€Ṃ

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-3 bytes thanks to dylnan

How it works

ŒṖṖLÐṂḌạ/€Ṃ - Main link. Argument: n (integer)    e.g    7129
ŒṖ          - Partitions of n's digits;                  [[7, 1, 2, 9], [7, 1, [2, 9]], [7, [1, 2], 9], [7, [1, 2, 9]], [[7, 1], 2, 9], [[7, 1], [2, 9]], [[7, 1, 2], 9], [7, 1, 2, 9]]
  Ṗ         - Remove the final element                   [[7, 1, 2, 9], [7, 1, [2, 9]], [7, [1, 2], 9], [7, [1, 2, 9]], [[7, 1], 2, 9], [[7, 1], [2, 9]], [[7, 1, 2], 9]]
    ÐṂ      - Keep the lists with the minimum...         [[7, [1, 2, 9]], [[7, 1], [2, 9]], [[7, 1, 2], 9]]
   L        -   length
      Ḍ     - From digits                                [[7, 129], [71, 29], [712, 9]]
        /   - Reduce...
         €  - ...each...
       ạ    - ...by absolute difference                  [122, 42, 703]
          Ṃ - Take the minimum                           42

Python 2, 58 bytes

lambda n:min(abs(n%10**i-n/10**i)for i in range(len(`n`)))

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Pyth, 15 bytes

hSmaFvMdfqlT2./

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MATL, 15 bytes

"GX@:&)UwU-|vX<

Input is a string representing the integer.

Try it online! Or verify all test cases.

Explanation

"         % Implicit input. Do the following as many times as input length
  G       %   Push input
  X@      %   Push iteration index (1-based), k
  :       %   Range: gives [1 2 ... k]
  &)      %   Two-ouput reference indexing: gives a substring with the first
          %   k characters in the input and then a substring with the rest
  U       %   Convert to number
  wU      %   Swap, convert to number
  -|      %   Absolute difference
  v       %   Vertically concatenate stack. This concatenates the obtained
          %   absolute difference with the minimum so far; does nothing in 
          %   the first iteration
  X<      %   Minimum of array
          % Implicit end. Implicit display

Wolfram Language (Mathematica), 66 bytes

Min@Abs[#-#2&@@FromDigits/@#&/@ReplaceList[#,{a__,b__}:>{a}|{b}]]&

Takes a list of digits.

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Clean, 106 83 bytes

import StdEnv
@n#f=toInt o(%)n
=hd(sort[abs(f(0,i)-f(i+1,size n))\\i<-[0..size n]])

Defines the function @, taking a string.

Mostly self-evident, the only tricky bit is f=toInt o(%)n: This takes the toInt class of functions, and composes it (o) with the curried slice operator class (%) already supplied with the first argument (n). Since there is only one type (String, equivalent to {#Char}) which has overloads for both % and toInt the line actually compiles, whereas normally it's hard to compose functions when golfing due to the lack of contextual information given to the compiler.

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Jelly, 12 bytes

JṬ€œṗ€⁸Ḍạ/€Ṃ

A monadic link taking a list of digits and returning the integer.

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How?

JṬ€œṗ€⁸Ḍạ/€Ṃ - Link: list of digits     e.g. [7,1,2,9]
J            - range of length               [1,2,3,4]
 Ṭ€          - untruth €ach                  [[1],[0,1],[0,0,1],[0,0,0,1]]
      ⁸      - chain's left argument         [7,1,2,9]
   œṗ€       - partition at truthy for €ach  [[[],[7,1,2,9]],[7,[1,2,9]],[[7,1],[2,9]],[[7,1,2],9]]
       Ḍ     - undecimal (vectorises)        [[0,7129],[7,129],[71,29],[712,9]]
         /€  - reduce €ach by:
        ạ    - absolute difference           [7129,122,42,703]
           Ṃ - minimum                       42

Pyth, 10 bytes

hSaMv<./Ql

Test suite

Takes input as a string.

This uses one of Pyth's more recent features, which is that applying a function to a list defaults to mapping the function over the list, if no other behavior is defined. This means that v applied to a list of list of strings evaluates all of the strings.

hSaMv<./Ql
hSaMv<./QlQ    Implicit variable
      ./Q      Form all partitions of the input string.
               Split it in all possible ways, maintaining the order.
               Partitions are ordered from shortest to longest.
     <   lQ    Take the prefix as long as the input string.
               This keeps just the splits into one and two pieces.
    v          Evaluate. All strings are converted to numbers.
  aM           Map the absolute difference function.
hS             Minimum

Note that the list of splits allows the split into 1 piece, but the value of this will always be larger than the minimum, so it is safely ignored.

Tcl, 116 bytes

foreach d [split [set b [set R $argv]] {}] {append L $d
regexp .(.+) $R - R
set b [expr min($b,abs($L-$R))]}
puts $b

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Explanation

b ← R ← input number
for each digit (d) in the input number:
  L += d
  strip first digit off of R using a regular expression
  b ← min( b, distance between L and R )
print b

It works by using a regex trick allowing a degenerate final case that will always compute to greater than the minimum difference. For “12345” the values are:

1 2345 → 2344
12 345 → 333
123 45 → 78
1234 5 → 1229
12345 5 → 12340 (degenerate case)

Ruby, 44 bytes

->a{((w=1)..a).map{((a%w*=10)-a/w).abs}.min}

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APL+WIN, 31 bytes

⌊/|(⍎¨m↓¨⊂n)-⍎¨(m←⍳¯1+⍴n)↑¨⊂n←⎕

Prompts for screen input of integer as a string.

Explanation:

m←⍳¯1+⍴n Create a list of numbers from 1 to length of string - 1

↑¨⊂n←⎕ Using m create a nested vector taking successively characters from the front of the string defined by m

⍎¨ Convert from character to integer

(⍎¨m↓¨⊂n) Using m create a nested vector dropping successively characters from the front of the string defined by m 

⌊/| take the minimum absolute value after subtracting the two vectors of integers

Perl 5, 51 41 + 1 (-p) = 42 bytes

$\=$_;$\=$\>($"=abs$'-$`)?$":$\while//g}{

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inspired by @Nahuel-Fouilleul's comment

C# (.NET Core), 112 107 + 18 = 125 bytes

n=>Enumerable.Range(1,n.Length-1).Min(i=>System.Math.Abs(int.Parse(n.Remove(i))-int.Parse(n.Substring(i))))

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The count includes the 18 bytes in using System.Linq;. Takes input as a string.

• 5 bytes saved by Caius Jard!

Common Lisp, 131 bytes

My first time participating in code golf and I decided to use Lisp, since I like it.

Here is my solution:

(defun f (s) (loop for i from 1 below (length s) minimizing (abs (- (parse-integer (subseq s 0 i)) (parse-integer (subseq s i))))))

Input needs to be a string, not integer or list.