The Strange Unsorting Machine for Nefarious Purposes

Question

Good Evening Golfers!

Your challenge is to completely unsort a series of numbers.

Input

Exactly 100 integers will be fed to your program. Your program may accept the input either as a file, or via stdin. Each integer will be separated by a newline character.

Those 100 integers will range from the minimal to the maximal values of a signed integer in your chosen language.

There will be no duplicate values. The values may be ordered, unordered or partially ordered - your program should be able to handle each case.

Output

The output must be each of the 100 integers, completely unsorted, each separated by a newline character. The output may be via stdout, or to a file.

Completely Unsorted means that no value is adjacent to any value which it would be adjacent to if the list were completely sorted in an ordered sequence.

Score

Pure code-golf. 1 point per character, and lowest score wins. There is a bonus of -100 for any solution using no built in or library sorting functions. There is a bonus of -20 for any solutions using no built in random number functions.

I have tried to define this question as completely as possible. If you have any questions, please ask. If you have any comments on how I could do better next time, please let me know.

Fore!

Answer 1

GolfScript (score 27 - 120 = -93)

~].,{{.2$<{\}*}*]}*.(;+2%n*

Note: that $ is referencing an element on the stack. There is sorting, but it's done with an manually coded bubble sort.

Thanks to Howard, for -90 => -92; and Ilmari, who inspired -92 => -93.

Answer 2

Python -26

(94-120): New, crude approach. Keep popping lowest elements into new list to get the elements sorted, then iterate:

t=l=[]
i=N=100
exec't=t+[input()];'*N+'l+=[t.pop(t.index(min(t)))];'*N+'print l[i%N];i+=3;'*N

Python -13

(107-120): First approach: Removes four lowest elements at a time, then print these four in another order:

exec'l=[]'+'+[input()]'*100
while l:
 a,b,c,d=eval('l.pop(l.index(min(l))),'*4)
 for e in[b,d,a,c]:print e

Answer 3

Mathematica -56 44 4 (95-120) = -25

Edit:

This version relies on neither built-in functions for sorting lists, nor randomization functions.

Riffle[RotateLeft[#[[All, 2]], 2], #[[All, 1]]] &
[Partition[l //. {x___, a_, b_, y___} /; b < a :> {x, b, a, y}, 2]]

Answer 4

C: 11 (131 - 120)

The programm reads from stdin and does a simple insert sort, after that it prints the nth together with th n+50th number, like many of the other solutions.

main(){int*i,a[101],*j=a;while(scanf("%d",a)>0)for(i=++j;i-->a;)i[1]=*i>=*a?*i:*(i=a);while(a<(i=j-50))printf("%d\n%d\n",*i,*j--);}

Answer 5

J, -63 (57-120) characters

Since everyone else is going down the self-written sort route...

,50(}.,.{.)($:@([-.m),~m=.]#~]=<./)^:(0<#),".[;._2[1!:1[3

Doesn't use any random number function, nor any built-in sort.

Uses a simple recursive selection sort to sort the input.

Answer 6

Python 2: 90 char

n=100
s=sorted(int(raw_input())for i in range(n))
for i in range(n):print s[(4*i+4*i/n)%n]

lazy attempt but just for starters

Answer 7

Python 48 = (148 - 100)

from random import*
x=[input()for i in range(100)]
while any(abs(x[i]-x[i+1])>1 for i in range(99)):n=randint(1,99);x=x[n:]+x[:n]
for j in x:print j

Haven't tested this because it isn't guaranteed (or likely) to run in any reasonable amount of time, but should work in theory given infinite time.

Answer 8

Python 27 (147 - 100 - 20)

R=range
def S(L):
 for i in R(len(L)-1):
    if L[i]>L[i+1]:L[i:i+2]=[L[i+1],L[i]];S(L)
a=map(input,['']*100);S(a)
for i in R(100):print a[i/2+i%2*50]

Note: the spaces before if L[i]>... should be a tab but apparently show up as spaces in a code block.