## Jumping Jack

### March 22, 2013

Here is our function to compute the minimum number of steps:

`(define (jack n)`

(define (diff-parity? t n)

(not (= (modulo t 2) (modulo n 2))))

(let loop ((k 0) (t 0))

(if (or (< t (abs n)) (diff-parity? t (abs n)))

(loop (+ k 1) (+ t k 1))

k)))

The basic idea is to make jumps until you equal or exceed the absolute value of the target, then continue making jumps until the parity of the target and the parity of the counter *t* are equal, since you can only subtract an *even* number.

The function that computes the jumps themselves is similar. Once you have passed the target, *d* is the number you have to subtract.

`(define (jacks n)`

(define (diff-parity? t n)

(not (= (modulo t 2) (modulo n 2))))

(if (negative? n)

(map (lambda (x) (- x)) (jacks (- n)))

(let loop ((k 0) (t 0))

(if (or (< t n) (diff-parity? t n))

(loop (+ k 1) (+ t k 1))

(let loop ((d (/ (- t n) 2)) (xs (range k 0)) (zs (list)))

(if (null? xs) zs

(if (<= (car xs) d)

(loop (- d (car xs)) (cdr xs) (cons (- (car xs)) zs))

(loop d (cdr xs) (cons (car xs) zs)))))))))

It’s easy to check the output by checking that the sum of `(jacks n)`

is equal to *n*, though I don’t know any easy way to check that the result is minimal. Here’s an example:

`> (jack 12)`

7

> (jacks 12)

(-1 2 3 4 5 6 -7)

The sum of (-1 2 3 4 5 6 -7) is 12; the signs indicate the direction of the jumps: left, right, right, right, right, right, left.

We used `range`

from the Standard Prelude. You can run the program at http://programmingpraxis.codepad.org/NVxW2bGK.

Pages: 1 2

[…] today’s Programming Praxis exercise, our goal is to determine the smallest amount of sequential numbers […]

My Haskell solution (see http://bonsaicode.wordpress.com/2013/03/22/programming-praxis-jumping-jack/ for a version with comments):

Assume the destiny number is k, the step is n

we are looking for a minimal value of n that n(n+1) = 2k + 4x where x is the sum of a sub set of [1…n] so x can be any positive integer(no larger than n(n+1)/2)

import math

k = 12

n = int(math.sqrt(2 * k)) + 1

while (n * (n + 1) – 2 * k) % 4 != 0:

n += 1

print n

Python – do a breadth first search of possibilities and return the first working solution

Here is my Ruby solution (note that I’m currently learning Ruby, so feel free to suggest stylistic improvements):

module JumpingJack

# Solves the puzzle; to be used like

# JumpingJack.jump_to(122)

# => [1, 2, 3, 4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16]

def JumpingJack.jump_to(number)

number >= 0 ? jump_to_positive_number(number) : reverse_sign(jump_to_positive_number(-number))

end

private

# The algorithm works in two steps:

# – In the first step, we create the shortest array of the form [1, 2, 3, 4, 5, …] so that

# the sum s of the contained numbers satisfies ‘s – number % 2 == 0′. If ‘s – number == 0′

# we are done. In any case, the resulting array has exactly the length we are looking for.

# – In the second step, we adjust the signs of the numbers from the first array so that their

# sum equals the given number.

#

def JumpingJack.jump_to_positive_number(number)

n = 0

sum = 0

ret = []

loop do

diff = sum – number

return ret if diff == 0

break if diff > 0 && diff % 2 == 0

n += 1

sum += n

ret <= number

ret[-i] = -val

break if diff == number

sum = diff

end

end

ret

end

def JumpingJack.reverse_sign(numbers)

numbers.map { |i| -i }

end

end

[\sourcecode]

Here is my Ruby solution (this time correctly formatted – please remove the other comment):