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Published at Oct 28 2019
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Instructions

Test suite

Solution

Find the difference between the square of the sum and the sum of the squares of the first N natural numbers.

The square of the sum of the first ten natural numbers is (1 + 2 + ... + 10)Â² = 55Â² = 3025.

The sum of the squares of the first ten natural numbers is 1Â² + 2Â² + ... + 10Â² = 385.

Hence the difference between the square of the sum of the first ten natural numbers and the sum of the squares of the first ten natural numbers is 3025 - 385 = 2640.

You are not expected to discover an efficient solution to this yourself from first principles; research is allowed, indeed, encouraged. Finding the best algorithm for the problem is a key skill in software engineering.

Suppose you're solving **halting-problem**:

- Start a REPL, either in your favorite editor or from the command line.
- Type
`(load "halting-problem.scm")`

at the prompt. - Test your code by calling
`(test)`

from the REPL. At first this should result in failed tests. - Develop your solution in "halting-problem.scm" and reload the file to run the tests again.

You can see more information about failing test cases by passing
arguments to the procedure `test`

.
To see the failing input call `(test 'input)`

and to see the input and output together call `(test 'input 'output)`

.

Problem 6 at Project Euler http://projecteuler.net/problem=6

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
(define test-fields '(input output who))
(define (test-run-solution solution input)
(if (procedure? solution) (apply solution input) solution))
(define (test-success description success-predicate
procedure input output)
(call/cc
(lambda (k)
(with-exception-handler
(lambda (e)
(k `(fail
(description . ,description)
(input . ,input)
(output . ,output)
(who . ,procedure))))
(lambda ()
(let ([result (test-run-solution procedure input)])
(unless (success-predicate result output)
(error 'exercism-test
"test fails"
description
input
result
output)))
`(pass . ,description))))))
(define (test-error description procedure input)
(call/cc
(lambda (k)
(with-exception-handler
(lambda (e) (k `(pass . ,description)))
(lambda ()
(test-run-solution procedure input)
`(fail
(description . ,description)
(input . ,input)
(output . error)
(who . ,procedure)))))))
(define (run-test-suite tests . query)
(for-each
(lambda (field)
(unless (and (symbol? field) (memq field test-fields))
(error 'run-test-suite
(format "~a not in ~a" field test-fields))))
query)
(let-values ([(passes failures)
(partition
(lambda (result) (eq? 'pass (car result)))
(map (lambda (test) (test)) tests))])
(cond
[(null? failures) (format #t "~%Well done!~%~%") 'success]
[else
(format
#t
"~%Passed ~a/~a tests.~%~%The following test cases failed:~%~%"
(length passes)
(length tests))
(for-each
(lambda (failure)
(format
#t
"* ~a~%"
(cond
[(assoc 'description (cdr failure)) => cdr]
[else (cdr failure)]))
(for-each
(lambda (field)
(let ([info (assoc field (cdr failure))])
(format #t " - ~a: ~a~%" (car info) (cdr info))))
query))
failures)
(newline)
'failure])))
(define (test . args)
(apply
run-test-suite
(list
(lambda ()
(test-success "square of sum 1" = square-of-sum '(1) 1))
(lambda ()
(test-success "square of sum 5" = square-of-sum '(5) 225))
(lambda ()
(test-success "square of sum 100" = square-of-sum '(100)
25502500))
(lambda ()
(test-success "sum of squares 1" = sum-of-squares '(1) 1))
(lambda ()
(test-success "sum of squares 5" = sum-of-squares '(5) 55))
(lambda ()
(test-success "sum of squares 100" = sum-of-squares '(100)
338350))
(lambda ()
(test-success "difference of squares 1" =
difference-of-squares '(1) 0))
(lambda ()
(test-success "difference of squares 5" =
difference-of-squares '(5) 170))
(lambda ()
(test-success "difference of squares 100" =
difference-of-squares '(100) 25164150)))
args))
```

```
;;;; difference-of-squares.scm
;;;; an exercism exercise
(import (rnrs (6)))
(load "test.scm")
(define (square-of-sum n)
"return (1 + ... + n)^2"
(expt (/ (* n (+ n 1)) 2) 2))
(define (sum-of-squares n)
"return (1^2 + ... + n^2)"
(* (/ n 3) (+ n 1) (+ n 1/2)))
(define (difference-of-squares n)
"return square-of-sum n - sum-of-squares n"
(- (square-of-sum n) (sum-of-squares n)))
```

A huge amount can be learned from reading other peopleâ€™s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

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