## Minimum Scalar Product

### August 10, 2012

We begin with a function to compute the sum of the pair-wise products of two vectors, which we represent as lists:

`(define (scalar xs ys) (sum (map * xs ys)))`

Then the minimum scalar product is found when one vector is sorted in increasing order and the other vector is sorted in decreasing order; there is no need to generate and test all possible permutations:

`(define (msp xs ys) (scalar (sort < xs) (sort > ys)))`

Here are the two sample problems:

```> (msp '(1 3 -5) '(-2 4 1)) -25 > (msp '(1 2 3 4 5) '(1 0 1 0 1)) 6```

We used `sum` and `sort` from the Standard Prelude. You can run the program at http://programmingpraxis.codepad.org/ucF1geyn.

Pages: 1 2

### 14 Responses to “Minimum Scalar Product”

```import Data.List

minScalar :: (Ord a, Num a) => [a] -> [a] -> a
minScalar a = sum . zipWith (*) (sort a) . reverse . sort
```
2. Jussi Piitulainen said

Is there a simple proof of the assertion that this always works? I looked briefly but failed to find one on the web and I need to attend to other things. I did find some other pages that simply assert that this pair of permutations produces the minimum. I found a purported counter-example that failed to be a counter-example. And I wrote a test program, below, that seems to confirm the assertion by always returning an empty set of permutations (of the second vector) that produce an even smaller result. So I believe it but if there is a memorable proof, I’d like to know.

```from itertools import permutations
from operator import mul
from random import sample

def subminima(first, second):
first, second = sorted(first), sorted(second, reverse = True)
allegedminimumscalarproduct = sum(map(mul, first, second))
return { p for p in permutations(second)
if sum(map(mul, first, p)) < allegedminimumscalarproduct }

pop = tuple(range(-10, 11)) + tuple(range(-4, 5)) + tuple(range(-3, 4))

def test():
for _ in range(10):
print(subminima(sample(pop, 5), sample(pop, 5)))
```
3. programmingpraxis said

It’s not a proof, but a simple observation is that the maximum scalar product is produced by multiplying the largest item from each vector, then the second largest, and so on; then the minimum scalar product is the opposite. I’ll ask at a couple of web sites I know.

4. swuecho said

perl6 version

```
sub min_scalar(@xs,@ys) {
[+] @xs.sort <<*>> @ys.sort.reverse;
}

# test
say min_scalar([1..5],[1,0,1,0,1]); # 6
say min_scalar([1,3,-5],[-2,4,1]);  # -25
```
5. Mike said

``` Here's an attempt at a proof:  Consider two vectors:      A = <a_1, a_2, ..., a_n> sorted highest to lowest, so a_i >= a_j if i < j; and     B = <b_1, b_2, ..., b_n> sorted lowest to highest, so b_i <= b_j if i < j.  The scalar product (A dot B) of these two vectors is:      a_1*b_1 + a_2*b_2 + ... + a_i*b_i + ...  Assume there exist a B' in which the positions of b_i and b_j have been exchanged so that:                                  A dot B' < A dot B      ... + a_i*b_j + ... + a_j*b_i + ... < ... + a_i*b_i + ... + a_j*b_j + ...  all the terms in the '...' are the same on both sides, so this reduces to:                        a_i*b_j + a_j*b_i < a_i*b_i + a_j*b_j  which can be rearranged:                           a_i*(b_j - b_i) < a_j*(b_j - b_i)  By definition of B, b_i <= b_j for i < j.      If b_i == b_j, then B' == B, and (A dot B') == (A dot B).          If b_i < b_j, then (b_j - b_i) > 0, so:                                     a_i < a_j  This contradicts that A is sorted so that a_i >= a_j for i < j. Therefore, the assumption must be false. ```

6. […] today’s Programming Praxis exercise, our goal is to calculate the minimum scalar product of two vectors. […]

7. infgeoax said

My Python solution:

def min_scalar_product(v1, v2):
return sum([x * y for x, y in zip(sorted(v1), sorted(v2, reverse=True))])

8. programmingpraxis said

I asked at http://www.reddit.com/r/math/comments/xzsi8/minimum_scalar_product/ and got this answer from Luonos. I think that’s pretty much the same as Mike’s proof.

Requires -std=c++0x option when compiled with g++.

```#include <iostream>
#include <vector>
#include <algorithm>

template< class IT1, class IT2,
class T = decltype ( typename std::iterator_traits<IT1>::value_type()
* typename std::iterator_traits<IT2>::value_type() ) >
T minimumScalarProduct( IT1 a, IT1 aend, IT2 b, IT2 bend, T init = 0 )
{
typedef typename std::iterator_traits<IT1>::value_type IT1_ValueType;
typedef typename std::iterator_traits<IT2>::value_type IT2_ValueType;
std::vector< IT1_ValueType > va( a, aend );
std::vector< IT2_ValueType > vb( b, bend );
std::sort( std::begin(va), std::end(va) );
std::sort( std::begin(vb), std::end(vb), std::greater< IT2_ValueType >() );
return std::inner_product( std::begin(va), std::end(va), std::begin(vb), init );
}

int main( int argc, char *argv[] )
{
std::vector<int> a { 5, 4, 3, 2, 1 };
int b[] = { 1, 0, 1, 0, 1 };
std::cout
<< minimumScalarProduct( std::begin(a), std::end(a), std::begin(b), std::end(b) )
<< std::endl;
return 0;
}
```
10. A Go solution:

```
package main

import (
"fmt"
"sort"
)

type Reverse struct {
sort.Interface
}

func (r Reverse) Less(i, j int) bool {
return r.Interface.Less(j, i)
}

func ScalarProduct(v1, v2 []int) int {
p := 0
for i := 0; i < min(len(v1), len(v2)); i++ {
p += v1[i] * v2[i]
}
return p
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

func MinScalarProduct(v1, v2 []int) int {
sort.Ints(v1)
sort.Sort(Reverse{sort.IntSlice(v2)})
return ScalarProduct(v1, v2)
}

func main() {
fmt.Printf("%d\n", MinScalarProduct([]int{1, 3, -5}, []int{-2, 4, 1}))
fmt.Printf("%d\n", MinScalarProduct([]int{1, 2, 3, 4, 5}, []int{1, 0, 1, 0, 1}))
}

```
11. […] Another post from Programming Praxis, this time we’re to figure out what is the minimum scalar product of two vectors. Basically, you want to rearrange two given lists a1, a2, …, an and b1, b2, …, bn such that a1b1 + a2b2 + … + anbn is minimized. […]