Create an application that determines if a sequence of numbers is sorted

Question

Another batch of the weekly game with code bowling as the previous incarnation is older than a week and is pretty well studied. As an update:

Code-Bowling is a challenge for writing the most obscure, non-optimized, terrible and bastardized code possible. Basically, the exact opposite is Code-Golf.

Challenge :

Create a program that accepts a sequence of numbers and determines if they are in ascending order.

An example :

$ ./myprogram 1 2 7 10 14
true

$ ./myprogram 7 2 0 1
false

rules

In fact, they are not. It can be a console application, it can be a web page, it can be anything. It just needs to be a standalone program that takes numbers and returns numbers. The format and methods are 100% up to you.

, , , !

Answer 1

, " ". , 1, . , , , , . , .

function askMom($num1, $num2) {
    $chance = mt_rand(0,2);
    if ($chance>1) {
        return askDad($num1, $num2);
    } else {
        return $num1 <= $num2;
    }
}
function askDad($num1, $num2) {
    $chance = mt_rand(0,4);
    if ($chance>1) {
        return askMom($num1, $num2);
    } else {
        return $num1 <= $num2;
    }
}

function parentSort(array $numbers) {
    for ($i = 0; $i < count($numbers)-1; $i++) {
        $chance = mt_rand(0,1);
        if ($chance) {
            if (askMom($numbers[$i], $numbers[$i+1])) {

            } else {
                return false;
            }
        } else {
            if (askDad($numbers[$i], $numbers[$i+1])) {

            } else {
                return false;
            }
        }
    }

    return true;
}

Answer 2

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char** argv){
    int a, b;
    if (argc > 2){
        sscanf(argv[1], "%d", &a);
        sscanf(argv[2], "%d", &b);
        if (a<=b) 
            return main(argc-1, argv+1);
        printf("false");
        exit(0);
    };
    printf("true");
    return 0;
};

Answer 3

O (n!) , , (. "" ), ( ).

def value(list):
    sum = 0
    for i in range(len(list)-1):
        sum = sum + (list[i]-list[i+1])**2.0
    return sum

def drop(lst, i):
     if i + 1 >= len(lst):
         return lst[:i]
     else:
         return lst[:i] + lst[i+1:]

class list_permute:
    def __init__(self, lst):
        self.lst = lst
        self.i = -1
        self.subiter = None
    def __iter__(self):
        return self
    def next(self):
        if len(self.lst) == 1:
            if self.i == -1:
                self.i = self.i + 1
                return self.lst
            else:
                raise StopIteration()

        if self.subiter != None:
            try:
                return [self.lst[self.i]] + self.subiter.next()
            except StopIteration:
                self.subiter = None

        if self.subiter == None:
            self.i = self.i + 1
            if self.i >= len(self.lst):
                raise StopIteration()
            else:
                self.subiter = list_permute(drop(self.lst, self.i))
            return self.next()

def test(list):
    given = value(list)
    for i in list_permute(list):
        if value(i) < given:
            return False

    # Test for false positive
    if list[0] > list[len(list)-1]:
        return False
    return True

list = []
print "Feed me your numbers (end with ^C)"
try:
    while True:
        try:
            list.append(int(raw_input()))
        except ValueError:
            print "NaN"
except (KeyboardInterrupt, EOFError):
    pass

print test(list)

Answer 4

. , , . 0 255...

array_shift($argv);
$str = str_repeat(chr(0), 256);
foreach ($argv as $key => $element) {
    $str[(int) $element] = chr($key + 1);
}
$str = str_replace(chr(0), '', $str);
$hex = unpack('H*', $str);
for ($i = 1; $i < strlen($str); $i++) {
    if (substr($hex[1], $i * 2 - 2, 2) != dechex($a)) {
        echo "False\n";
        die();
    }
}
echo "True\n";

, (1 2 5 4 1 2 0 4 3, , , . , 1 1.

( , ):

array_shift($argv);
$vals = array_flip($argv);
ksort($vals);
echo array_values($vals) == range(0, count($vals) - 1) ? "True\n" : "False\n";

Answer 5

, , ...

/* Either #define macros FIRST, SECOND, THIRD, etc. here, or do so on the
 * command line when "compiling" i.e.
 * $ gcc -D FIRST=1 -D SECOND=5 -D THIRD=42
 */

#define min(X, Y) ((X) < (Y) ? (X) : (Y))

#define pairsorted(X, Y) (min((X), (Y)) == (X) ? 1 : 0)

#if defined (FIRST) && defined (SECOND) && pairsorted(FIRST, SECOND)
#if defined (THIRD) && pairsorted(SECOND, THIRD)
#if defined (FOURTH) && pairsorted (THIRD, FOURTH)
#if defined (FIFTH) && pairsorted (FOURTH, FIFTH)
#error "Sorted!"
#elif !defined (FIFTH)
#error "Sorted!"
#else /* FIFTH is defined and < FOURTH */
#error "Not sorted!"
#endif /* FIFTH */

#elif !defined (FOURTH)
#error "Sorted!"
#else /* FOURTH is defined and < THIRD */
#error "Not sorted!"
#endif /* FOURTH */

#elif !defined (THIRD)
#error "Sorted!"
#else /* THIRD is defined and < SECOND */
#error "Not sorted!"
#endif /* THIRD */

#elif !defined (SECOND)
#error "Sorted!"
#else /* SECOND is defined and < FIRST */
#error "Not sorted!"
#endif /* SECOND */

#ifndef SECOND
#error "I need at least two values to compare"
#endif

(ab) C script ( sortedcpp.c):

#!/bin/bash

ORDINALS=(ZEROTH FIRST SECOND THIRD FOURTH FIFTH)
VALUES=(0 $@)

for i in 1 2 3 4 5; do
  if [ $i -le $# ]
    then
      flags="$flags -D ${ORDINALS[$i]}=${VALUES[$i]}"
  fi
done

output=`gcc $flags sortedcpp.c 2>&1`

echo $output | sed -e 's/sortedcpp.c:[0-9]*: error: #error \"\(.*\)\"/\1/'

Answer 6

Yay, Python!

def IsSorted(lst):
    return sorted(lst) == lst

Answer 7

For the first time I used dynamic programming to make things worse. It has the complexity of time and space O (n²)

    #include <stdio.h>
    int main (int argc, char **argv)
    {
      int is_ordered[1000][1000];
      int list[1000];
      int i,j;

      for(i = 1; i < argc; i++)
        sscanf(argv[i],"%d", &list[i-1]);

      for (i = 0; i < argc -2; i++)
      {
        if (list[i] < list[i+1])
          is_ordered[i][i+1] = 1;
        else
          is_ordered[i][i+1] = 0;
      }

      for (i = 2; i < argc -1; i++)
        for (j = 0; j < (argc - 1 - i); j++)
        {
          if (is_ordered[j+1][i+j] && is_ordered[j][i+j-1])
            is_ordered[j][j+i] = 1;
          else
            is_ordered[j][j+i] = 0;
        }

      if(is_ordered[0][argc-2])
        printf("True\n");
      else
        printf("False\n");
      return 0;
    }