Difference of the square of the sum

Question

Find the difference between the square of the sums and sum of the squares.

This is the mathematical representation:

\$\left(\sum n\right)^2-\sum n^2\$

Your program/method should take two inputs, these are your lower and upper limits of the range, and are inclusive. Limits will be whole integers above 0.

Your program/method should return the answer.

You may use whichever base you would like to, but please state in your answer which base you have used.

Test case (Base 10)

5,9      970
91,123   12087152
1,10     2640

This is usual code-golf, so the shorter the answer the better.

Answer 1

Python 2, 43 bytes

f=lambda a,b,s=0:b/a and 2*a*s+f(a+1,b,s+a)

Test it on Ideone.

How it works

Call the function defined in the specification \$g(a, b)\$. We have that

$$ \newcommand{\sumab}[2]{\sum_{a \le #1 \le b} #2 \:} \begin{align} g(a,b) & = \left( \sumab n n \right)^2 - \sumab n {n^2} \\ & = \sumab {i,j} {ij} - \sumab n {n^2} \\ & = \sumab {i<j} {ij} + \sumab {i=j} {ij} + \sumab {j<i} {ij} - \sumab n {n^2} \\ & = 2 \sumab {j<i} {ij} \\ & = 2 \sumab {i} {\sum_{a \le i < j} ij} \\ & = \sumab {i} {\left( 2i \sum_{a \le j < i} j \right)} \end{align} $$

Define the function \$f(x, y, s)\$ recursively as follows.

$$ f(x, y, s) = \begin{cases} 2xs + f(x+1, y, s+x) & \text{ if } x \le y \\ 0 & \text{ if} x > y \end{cases} $$

By applying the recurrence relation of \$f(a, b, 0)\$ a total of \$b - a\$ times, we can show that:

$$ \begin{align} f(a, b, 0) & = 2\cdot0 + f(a+1, b, a) \\ & = 2\cdot0 + 2\cdot(a+1)\cdot a + f(a+2, b, a+(a+1)) \\ & = 2\cdot0 + 2\cdot(a+1)\cdot a + 2\cdot(a+2)\cdot(a+(a+1)) + f(a+3, b, a+(a+1)+(a+2)) \\ & \vdots \\ & = \sumab {i} {\left( 2i \sum_{a \le j < i} j \right)} + f\left(b+1, b, \sumab j j\right) \\ & = \sumab {i} {\left( 2i \sum_{a \le j < i} j \right)} \\ & = g(a, b) \end{align} $$

This is the function f of the implementation. While b/a returns a non-zero integer, the code following and is executed, thus implementing the recursive definition of f.

Once b/a reaches 0, we have that b > a and the lambda returns False = 0, thus implementing the base case of the definition of f.

Answer 2

MATL, 9 bytes

&:&*XRssE

Try it online!

Explanation

&:   % Inclusive range between the two implicit inputs
&*   % Matrix of all pair-wise products
XR   % Upper triangular part of matrix, without the diagonal
ss   % Sum of all elements of the matrix
E    % Multiply by 2. Implicit display

Example

These are the partial results of each line for inputs 5 and 9:

1. &:

   5 6 7 8 9
   

2. &:&*

   25 30 35 40 45
   30 36 42 48 54
   35 42 49 56 63
   40 48 56 64 72
   45 54 63 72 81
   

3. &:&*XR

   0 30 35 40 45
   0  0 42 48 54
   0  0  0 56 63
   0  0  0  0 72
   0  0  0  0  0
   

4. &:&*XRss

   485
   

5. &:&*XRssE

   970
   

Answer 3

Jelly, 9 8 bytes

rµS²_²S$

Try it online!

r         inclusive range from first input to second input
 µ        pass the range to a new monadic chain
  S       the sum
   ²      squared
    _     minus...
     ²S$  the squares summed

Thanks to FryAmTheEggman for a byte!

Answer 4

Python 2, 45 bytes

lambda a,b:(a+~b)*(a-b)*(3*(a+b)**2+a-b-2)/12

Closed form solution - not the shortest, but I thought it'd be worth posting anyway.

Explanation

Let p(n) be the nth square pyramidal number, and t(n) be the nth triangular number. Then, for n over the range a, ..., b:

• ∑n = t(b)-t(a-1), and
• ∑n² = p(b) - p(a-1)
• So (∑n)²-∑n² = (t(b)-t(a-1))² - (p(b) - p(a-1)).

This expression reduces to that in the code.

Answer 5

05AB1E, 8 bytes

ŸDOnsnO-

Explained

ŸD       # range from a to b, duplicate
  On     # sum and square first range
    s    # swap top 2 elements
     nO  # square and sum 2nd range
       - # take difference

Try it online

Answer 6

Mathematica, 21 bytes

Tr[x=Range@##]^2-x.x&

An unnamed function taking two arguments and returning the difference. Usage:

Tr[x=Range@##]^2-x.x&[91, 123]
(* 12087152 *)

There's three small (and fairly standard) golfing tricks here:

• ## represents both arguments at once, so that we can use prefix notation for Range. Range@## is shorthand for Range[##] which expands to Range[a, b] and gives us an inclusive range as required.
• Tr is for trace but using it on a vector simply sums that vector, saving three bytes over Total.
• x.x is a dot product, saving four bytes over Tr[x^2].

Answer 7

Labyrinth, 28 24 bytes

?:?:}+=-:(:(#{:**+**#2/!

Try it online!

Explanation

Since loops tend to be expensive in Labyrinth, I figured the explicit formula should be shortest, as it can be expressed as linear code.

Cmd Explanation                 Stacks [ Main | Aux ]
?   Read M.                     [ M | ]
:   Duplicate.                  [ M M | ]
?   Read N.                     [ M M N | ]
:   Duplicate.                  [ M M N N | ]
}   Move copy to aux.           [ M M N | N ]
+   Add.                        [ M (M+N) | N ]
=   Swap tops of stacks.        [ M N | (M+N) ]
-   Subtract.                   [ (M-N) | (M+N) ]
:   Duplicate.                  [ (M-N) (M-N) | (M+N) ]
(   Decrement.                  [ (M-N) (M-N-1) | (M+N) ]
:   Duplicate.                  [ (M-N) (M-N-1) (M-N-1) | (M+N) ]
(   Decrement.                  [ (M-N) (M-N-1) (M-N-2) | (M+N) ]
#   Push stack depth.           [ (M-N) (M-N-1) (M-N-2) 3 | (M+N) ]
{   Pull (M+N) over from aux.   [ (M-N) (M-N-1) (M-N-2) 3 (M+N) | ]
:   Duplicate.                  [ (M-N) (M-N-1) (M-N-2) 3 (M+N) (M+N) | ]
*   Multiply.                   [ (M-N) (M-N-1) (M-N-2) 3 ((M+N)^2) | ]
*   Multiply.                   [ (M-N) (M-N-1) (M-N-2) (3*(M+N)^2) | ]
+   Add.                        [ (M-N) (M-N-1) (3*(M+N)^2 + M - N - 2) | ]
*   Multiply.                   [ (M-N) ((M-N-1)*(3*(M+N)^2 + M - N - 2)) | ]
*   Multiply.                   [ ((M-N)*(M-N-1)*(3*(M+N)^2 + M - N - 2)) | ]
#   Push stack depth.           [ ((M-N)*(M-N-1)*(3*(M+N)^2 + M - N - 2)) 1 | ]
2   Multiply by 10, add 2.      [ ((M-N)*(M-N-1)*(3*(M+N)^2 + M - N - 2)) 12 | ]
/   Divide.                     [ ((M-N)*(M-N-1)*(3*(M+N)^2 + M - N - 2)/12) | ]
!   Print.                      [ | ]

The instruction pointer then hits a dead end and has to turn around. When it now encounters / it attempts a division by zero (since the bottom of the stack is implicitly filled with zeros), which terminates the program.

Answer 8

Haskell, 34 bytes

a#b=sum[a..b]^2-sum(map(^2)[a..b])

Usage example: 91 # 123 -> 12087152.

Nothing to explain.

Answer 9

Brachylog, 24 bytes

:efL:{:2^.}a+S,L+:2^:S-.

Expects the 2 numbers in Input as a list, e.g. [91:123].

Explanation

:efL                     Find the list L of all integers in the range given in Input
    :{:2^.}a             Apply squaring to each element of that list
            +S,          Unify S with the sum of the elements of that list
               L+:2^     Sum the elements of L, then square the result
                    :S-. Unify the Output with that number minus S

Answer 10

Matlab, 30 29 28 bytes

Using Suever's idea of norm gives us 2 bytes less

@(x,y)sum(x:y)^2-norm(x:y)^2

Old (simple) version:

@(x,y)sum(x:y)^2-sum((x:y).^2)

Answer 11

Octave, 27 23 bytes

@(x,y)sum(z=x:y)^2-z*z'

Creates an anonymous function named ans which accepts two inputs: ans(lower, upper)

Online Demo

Explanation

Creates a row vector from x to y (inclusive) and stores it in z. We then sum all the elements using sum and square it (^2). To compute the sum of the squares, we perform matrix multplication between the row-vector and it's transpose. This will effectively square each element and sum up the result. We then subtract the two.

Answer 12

Java, 84 77 characters, 84 77 bytes

7 bytes smaller due to Martin Ender and FryAmTheEggMan, thank you.

public int a(int b,int c){int e=0,f=0;for(;b<=c;e+=b,f+=b*b++);return e*e-f;}

Using the three test cases in the original post: http://ideone.com/q9MZSZ

Ungolfed:

public int g(int b, int c) {
    int e = 0, f = 0;
    for (; b <= c; e += b, f += b * b++);
    return e*e-f;
}

Process is fairly self-explanatory. I declared two variables to represent the square of the sums and the sum of the squares and repeatedly incremented them appropiately. Finally, I return the computed difference.

Answer 13

JavaScript (ES6), 46 bytes

f=(x,y,s=0,p=0)=>x<=y?f(x+1,y,s+x,p+x*x):s*s-p

Answer 14

JavaScript (ES6), 50 37 bytes

f=(n,m,s=0)=>n>m?0:2*n*s+f(n+1,m,n+s)

Now a port of @Dennis♦'s Python solution.

Answer 15

Factor, 48 bytes

[ [a,b] [ [ sq ] map sum ] [ sum sq ] bi - abs ]

An anonymous function.

[ 
  [a,b] ! a range from a to b 
  [ 
    [ sq ] map sum ! anonymous function: map sq over the range and sum the result 
  ] 
  [ sum sq ] ! the same thing, in reverse order
  bi - abs   ! apply both anon funcs to the range, subtract them and abs the result
]

Answer 16

Haskell, 36 bytes

m#n=sum[2*i*j|i<-[m..n],j<-[i+1..n]]

---

λ> m # n = sum [ 2*i*j | i <- [m..n], j <- [i+1..n] ]
λ> 5 # 9
970
λ> 91 # 123
12087152
λ> 1 # 10
2640

Note that

$$\left( \sum_{k=m}^n k \right)^2 - \sum_{k=m}^n k^2 = \cdots = \sum_{k_1=m}^n \sum_{k_2=m\\ k_2 \neq k_1}^n k_1 k_2 = \sum_{k_1=m}^n \sum_{k_2=k_1+1}^n 2 \,k_1 k_2$$

Answer 17

Perl 6,  36 32  31 bytes

{([+] $_=@_[0]..@_[1])²-[+] $_»²}
{([+] $_=$^a..$^b)²-[+] $_»²}
{[+]($_=$^a..$^b)²-[+] $_»²}

Test it

Explanation:

{ # bare block with placeholder parameters $a and $b

  [+](# reduce with &infix:<+>
      # create a range, and store it in $_
      $_ = $^a .. $^b
  )²
  -
  [+] # reduce with &infix:<+>
    # square each element of $_ ( possibly in parallel )
    $_»²
}

Test:

#! /usr/bin/env perl6
use v6.c;
use Test;

my @tests = (
  (5,9) => 970,
  (91,123) => 12087152,
  (1,10) => 2640,
);

plan +@tests;

my &diff-sq-of-sum = {[+]($_=$^a..$^b)²-[+] $_»²}

for @tests -> $_ ( :key(@input), :value($expected) ) {
  is diff-sq-of-sum(|@input), $expected, .gist
}

1..3
ok 1 - (5 9) => 970
ok 2 - (91 123) => 12087152
ok 3 - (1 10) => 2640

Answer 18

Aheui (esotope), 108 bytes(36 chars)

방빠방빠쌍다쌍상싸사타빠밤타빠받다파반다받상싸사빠따따다따따받밤따나망히

Try it online!

---

Takes two inputs from user(separated by line feed or whitespace) and print result of following function;

\$f(a,b)=((a-b)\cdot(a-b-1)\cdot(3\cdot(a+b)^2+a-b-2))/12\$

Answer 19

APL, 23 20 bytes

-/+/¨2*⍨{(+/⍵)⍵}⎕..⎕

Works in NARS2000.

Answer 20

MATL, 11 bytes

&:ts2^w2^s-

Try it online!

Explanation:

&:           #Create a range from the input
  t          #Duplicate it
   s2^       #Sum it and square it
      w      #swap the two ranges
       2^s   #Square it and sum it
          -  #Take the difference

Answer 21

PowerShell v2+, 47 bytes

Two variations

param($n,$m)$n..$m|%{$o+=$_;$p+=$_*$_};$o*$o-$p

$args-join'..'|iex|%{$o+=$_;$p+=$_*$_};$o*$o-$p

In both cases we're generating a range with the .. operator, piping that to a loop |%{...}. Each iteration, we're accumulating $o and $p as either the sum or the sum-of-squares. We then calculate the square-of-sums with $o*$o and subtract $p. Output is left on the pipeline and printing is implicit.

Answer 22

Pyth, 11 bytes

s*M-F#^}FQ2

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s*M-F#^}FQ2
       }FQ    Compute the range
      ^   2   Generate all pairs
   -F#        Remove those pairs who have identical elements
 *M           Product of all pairs
s             Sum.

Answer 23

Japt, 10 bytes

õV
x²aUx ²

Try it

Answer 24

Jelly, 6 bytes

rµÄḋḊḤ

Recent improvements to the Jelly language allow a compact implementation of the \$g\$ function from my Python answer.

Try it online!

How it works

rµÄḋḊḤ  Main link. Arguments: a, b (integers)


r       Range; yield R := [a, ..., b].
 µ      Begin a monadic chain with argument R.
  Ä     Accumulate; take the cumulative sum of R.
    Ḋ   Deque; yield [a+1, ..., b].
   ḋ    Take the dot product, ignoring the last term of the cumulative sum.
     Ḥ  Unhalve; double the result.

Answer 25

Vyxal, 8 bytes

ṡ:²∑$∑²ε

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How?

ṡ:²∑$∑²ε
ṡ        # Inclusive range between (implicit) second input and (implicit) first input
 :       # Duplicate this range
  ²      # Square each of the duplicate
   ∑     # Summate
    $    # Swap so the other range is at the top
     ∑   # Summate
      ²  # Then square
       ε # Absolute difference between the two

Answer 26

Thunno 2, 8 bytes

IDS²s²S-

Attempt This Online!

Explanation

IDS²s²S-  # Implicit input
ID        # Inclusive range; duplicate
  S²      # Sum the list; square the sum
    s     # Swap so the range is back on top
     ²S   # Square each item; sum the list
       -  # Take the difference
          # Implicit output

Answer 27

CJam, 17 bytes

q~),>_:+2#\2f#:+-

Test it here.

Explanation

q~       e# Read and evaluate input, dumping M and N on the stack.
),       e# Increment, create range [0 1 ... N].
>        e# Discard first M elements, yielding [M M+1 ... N].
_        e# Duplicate.
:+2#     e# Sum and square.
\2f#:+   e# Swap with other copy. Square and sum.
-        e# Subtract.

Alternatively, one can just sum the products of all distinct pairs (basically multiplying out the square of the sum, and removing squares), but that's a byte longer:

q~),>2m*{)-},::*:+

Answer 28

JavaScript (ES6), 67 bytes

a=>b=>([s=q=0,...Array(b-a)].map((_,i)=>q+=(s+=(n=i+a),n*n)),s*s-q)

Test Suite

f=a=>b=>([s=q=0,...Array(b-a)].map((_,i)=>q+=(s+=(n=i+a),n*n)),s*s-q)
e=s=>`${s} => ${eval(s[0])}` // template tag format for tests
console.log(e`f(5)(9)`)
console.log(e`f(91)(123)`)
console.log(e`f(1)(10)`)

Answer 29

J, 29 bytes

Port of Doorknob's Jelly answer.

[:(+/@(^&2)-~2^~+/)[}.[:i.1+]

Usage

>> f = [:(+/@(^&2)-~2^~+/)[}.[:i.1+]
>> 91 f 123x
<< 12087152

Where >> is STDIN, << is STDOUT, and x is for extended precision.

Answer 30

Pyke, 11 bytes

h1:Ds]MXXs-

Try it here!

h1:         - inclusive_range(input)
   Ds]      -     [^, sum(^)]
      MX    -    deep_map(^, <--**2)
         s  -   ^[1] = sum(^[1])
          - -  ^[0]-^[1]