Minimum Scalar Product

Question

Minimum Scalar Product

The inspiration for this code golf problem is from Google's code jam competition. The premise behind the problem is, given the input of two vectors of varying lengths, find the minimum possible scalar. A scalar can be found using the following formula:

x1 * y1 + x2 * y2 + ... + xn * yn

The problem, however, is that multiple values for the scalar can be found depending on the order of the numerals in the input case (seen below). Your goal is to determine the minimum possible scalar integer solution by plugging in the input case numbers into the equation and solving for it. You may use every number in the input only once, and must use all of the numbers.

Allow me to provide an example with the following vectors.

Input

3
1 3 -5
-2 4 1

Output

-25

The first integer on the line represents the number of numbers, n, in each vector. In this case, we have three numbers in each vector.

The number n may vary with each test case, but there will always be two vectors.

In the example input, the minimum scalar product would be -25.

(-5 * 4) + (1 * 1) + (3 * -2) = 25

Rules

• You may only use each integer in both vectors once.
• You must use all integers in the vectors.
• Your output must only include the final product
• I'll select the solution with the least amount of code, which follows all of the specifications listed above, in any language!

Hint: You don't need to brute force this problem, unless it makes your code shorter. There is a specific method involved in finding the minimum spanning scalar :).

Answer 1

Jelly, 6 bytes

ṢṚ×Ṣ}S

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Using brute force is equally short:

Œ!×S€Ṃ

How it works

ṢṚ×Ṣ}S  Main link. Arguments: u (vector), v (vector)

Ṣ       Sort the components of u.
 Ṛ      Reverse.
   Ṣ}   Sort the components of v.
  ×     Multiply the results, element by element.
     S  Compute the sum of the products.

Answer 2

Seriously, 6 bytes

,SR,S*

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Explanation:

,SR,S*
,SR     input first vector, sort, reverse
   ,S   input second vector, sort
     *  dot product

Answer 3

APL, 15 bytes

{+/⍺[⍒⍺]×⍵[⍋⍵]}

This is a dyadic function that accepts arrays on the left and right and returns an integer. It uses the same approach as my Julia answer: dot product of the sorted arrays, one descending and one ascending.

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Answer 4

MATL, 6 bytes

Code:

SiSP*s

My first MATL answer :)

Explanation:

S       # Sort the first array
 iS     # Take the second array and sort it
   P    # Flip the array
    *   # Multiply both arrays with each other
     s  # Sum of the result

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Answer 5

Mathematica, 30 17 bytes

-13 bytes by murphy

Sort@#.-Sort@-#2&

Function, input is vector1(list),vector2(list) Several revisions:

Plus@@(Sort@#*Reverse@Sort@#2)&(*me*)
Total[Sort@#*Reverse@Sort@#2]& 
Sort@#.Reverse@Sort@#2&        (*alephalpha*)
Sort@#.Sort[#2,#>#2&]&         (*murphy*)
Sort@#.SortBy[#2,-#&]          (*me*)
Sort@#.-Sort@-#2&              (*murphy*)

Answer 6

Julia, 32 25 bytes

x->y->-sort(-x)⋅sort(y)

This is an anonymous function that accepts two arrays and returns an integer. To call it, assign it to a variable and do f(x)(y).

For inputs x and y, we simply compute the dot product of x sorted in reverse order with y sorted. We get x in reverse sorted order by negating all values, sorting, then negating again.

Saved 7 bytes thanks to Dennis!

Answer 7

Pyth - 14 8 bytes

I think I figured out the trick.

s*VSQ_SE

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Answer 8

Javascript ES6, 69 bytes

a=>b=>a.sort((x,y)=>x-y).map((x,y)=>i+=b.sort((x,y)=>y-x)[y]*x,i=0)|i

Wow, this is way too long.

Answer 9

Perl 6, 33 30 bytes

{sum @^a.sort Z*@^b.sort.reverse}

Answer 10

CJam, 11 Bytes

q~$\$W%.*:+

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Answer 11

C++, 124 bytes

#include<algorithm>
int m(int*a,int*b,int n){std::sort(a,a+n);std::sort(b,b+n);int r=0;while(--n>=0)r+=a[n]**b++;return r;}

ungolfed:

#include<algorithm>
int m(int*a,int*b,int n){
 std::sort(a,a+n);
 std::sort(b,b+n);
 int r=0;
 while(--n>=0)
  r+=a[n]*(*b++);
return r;
}

At first i used std::greater<int>() for the sort in b but just reversing the order in the summation is easier.

Answer 12

Haskell, 59 bytes

import Data.List
v?u=sum$zipWith(*)(sort v)$reverse$sort u

Answer 13

Python 3, 136 bytes

def mdp(n, a, b):
    a = list(reversed(sorted(a)))
    b = sorted(b)
    res = sum([a[i] * b[i] for i in range(len(a))])
    return res

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Answer 14

RETURN, 29 bytes

[{␆␃}\{␆}␄␅[¤¥][×␌]#}␁[¤][+]#]

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Replace any ␆␃␄␇ with their unprintable counterparts.

Anonymous lambda that leaves result on stack2. Usage:

""{1 3 0 5-}""{0 2- 4 1}[{␆␃}\{␆}␄␅[¤¥][×␌]#}␁[¤][+]#]!

Explanation

[                                 ]  lambda
 {␆␃}                              sort and reverse first stack
       \{␆}                         sort second stack
            ␄␅                     transpose and flatten
               [  ][  ]#             while loop
                ¤¥                     check if 2 items exist in stack
                    ×                  if so, multiply top 2 items
                     ␌                 and push to stack2
                        }␁          switch to stack2
                           [¤][+]#   sum stack2

Answer 15

J, 14 bytes

+/@(*|.)&(/:~)

Uses the same principle as the others.

Explanation

+/@(*|.)&(/:~)  Input: x on LHS and y on RHS
        &(/:~)  Sort both x and y
     |.         Reverse the sorted y
    *           Multiply the sorted x and reversed sorted y elementwise
+/@             Reduce the products using addition and return

Answer 16

Ruby, 44 bytes

->a,b{a.sort.reverse.zip(b.sort).sum{_1*_2}}

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Answer 17

Japt -x, 6 bytes

Íí*VÍÔ

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Answer 18

Uiua, 9 bytes

/+×⇌∩⊏∩⊸⍏

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/+×⇌∩⊏∩⊸⍏
     ⊏ ⊸⍏ # sort
    ∩ ∩   # both
   ⇌      # reverse one
  ×       # multiply
/+        # sum

Answer 19

JavaScript (Node.js), 62 bytes

a=>a.map(x=>x.sort((x,y)=>x-y))[i=0].map(x=>i+=a[1].pop()*x)|i

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Based on Mama Fun Roll's

-4B by Shaggy by switching Input format

Answer 20

Scala 3, 76 bytes

(x,y)=>x.sorted(Ordering[Int].reverse).zip(y.sorted).map{case(a,b)=>a*b}.sum

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