Basic sort, with annoying bug

Question

Your input is a list/sequence/vector/array of 5-255 positive integers, not necessarily unique. You may assume whatever input format is most suitable, and that each integer (as well as the quantity of integers) is chosen uniformly at random from the range 5-255.

The goal is to output the same list, in the same (or equivalent) format, but sorted into increasing (nondecreasing) order. A common early exercise in learning a language. Submissions to include:

1. An answer which works correctly and achieves the goal; and

2. A second answer which contains an annoying bug. Between 1% and 10% of the time, the output needs to be a list in the correct format and containing the correct elements, but in the wrong order (any order except correctly sorted). The rest of the time, the program must work correctly and achieve the goal.

The two answers must have Levenshtein distance one; that is, we can get one from the other by deleting one byte, or adding one byte, or changing one byte.

Scoring as usual in code-golf (based on the shorter of your two answers), with usual loopholes forbidden.

10% bonus (decrease to score) if the annoying bug is input-independent, i.e. using the same input again doesn't reproduce the bug (except between 1% and 10% of the time).

Answer 1

Python 3, 36 bytes

Bug-free version, 37 bytes

lambda l:sorted(l,reverse=l[-9:]==[])

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Annoying version, 36 bytes

lambda l:sorted(l,reverse=l[9:]==[])

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This depends on the input and therefore does not qualify for the bonus.
It has a probability of around 2% to fail. It fails when the input length is lower then 10.

Combined with LyricLy's answer this gets 34 bytes:

lambda l:sorted(l)[::l[9:]>[]or 1]
lambda l:sorted(l)[::l[9:]>[]or-1]

Answer 2

05AB1E, 5*0.9 = 4.5 bytes

Working solution

{TLΩi

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Explanation

{      # sort input
 TL    # push the range [1 ... 10]
   Ω   # pick a random number in the range
    i  # if true (equal to 1), do nothing

Solution containing the bug

Gives the wrong solution 10% of the time (independent on input).

{TLΩiR

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Explanation

Same as the working solution, except it reverses the list if the number picked is true.

Answer 3

Jelly, 7 * (100% - 10%) = 6.3 bytes

Ṣ¹⁵X:¤¡

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Buggy version:

ṢẊ⁵X:¤¡

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In both links, there's a test harness that will run the code 100 times, each time with the list you give as the argument, and then return the results.

The probability of each input length is:

0.1 - 0.1/(length!)

So for length 1 there's a 0% probability, for length 2 5%, for length 3 8.83̅%, for length 4 9.583̅% etc. until length ∞ which has a 10% probability.

Answer 4

Python 3, score 58 57 - 10% = 51.3

Saved a byte thanks to ovs.

Bug-free version, 57 bytes

lambda m:sorted(m)[::random()>.1or 1]
from random import*

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Bugged version, 57 bytes

lambda m:sorted(m)[::random()>.1or-1]
from random import*

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I decided to try a solution that uses the bonus. It doesn't beat the other Python answer, but I had fun thinking it up.

Answer 5

C, 71*0.9 = 63.9 bytes

Bug-free:

c(int*a,int*b){return*a-*b;}f(a,n)int*a;{if(rand()%1<9)qsort(a,n,4,c);}

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Buggy:

c(int*a,int*b){return*a-*b;}f(a,n)int*a;{if(rand()%10<9)qsort(a,n,4,c);}

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Answer 6

Groovy, 31 bytes

Bugged solution:

{a->a.sort()[a[9]?0..-1:-1..0]}

Working solution:

{a->a.sort()[a[0]?0..-1:-1..0]}

The Groovy subscript operator (the getAt method) returns null for lists if the index is bigger than the size. So if there's a ninth element it'll stay the same as the sorted list, but if not (1.99203187% chance) it'll be reversed. However there will always be a first element because the list's size is always bigger than or equal to 5. So the 0 in a[0] could be swapped with 1, 2, 3 or 4.

Answer 7

Wolfram Language (Mathematica), 29 bytes

This is 26.1 bytes with the bonus, but I'm not entirely sure I earn the bonus; on already-sorted input, both versions always produce sorted output.

Bug-free version (29 bytes)

If[RandomReal[]<0.,#,Sort@#]&

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Annoying version (30 bytes)

If[RandomReal[]<0.1,#,Sort@#]&

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Answer 8

PHP, 70 bytes

Bug-free version, 70 bytes

<?unset($argv[0]);((rand(1,9)?"":r).sort)($argv);echo join(" ",$argv);

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Bugged version, 70 bytes

<?unset($argv[0]);((rand(0,9)?"":r).sort)($argv);echo join(" ",$argv);

Try it online!

The bugged version sorts in reverse order 10% of the time (based on a random number generator).

Answer 9

Python 2, 26 bytes

Buggy:

lambda l:l[9:]and l.sort()

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Outputs by modifying the input list. Sorts the list only if its length is at least 10. The non-buggy version replaces the 9 with a 0 to always sort.

Working:

lambda l:l[0:]and l.sort()

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We can modify the function to return the list at the cost of 4 bytes, for 30 bytes total:

lambda l:l[9:]and l.sort()or l

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---

25 bytes with some stretches of the rules:

[list,sorted][id(0)%17>0]

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Outputs a function literal that either sorts or is the identity, using id(0) as a random source. Change > to >= to fix, or 0 to ~0.

Answer 10

Husk, 6 bytes

Buggy version:

?OIV¦9

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Correct version:

?OIVK9

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Explanation

These programs are completely deterministic. In fact, Husk has currently no support at all for random numbers.

?  V    If any element
    ¦9  is divisible by 9 (in buggy version),
    K9  is truthy when replaced by 9 (in correct version),
 O      sort the list,
  I     otherwise return it unchanged.

I claim that the output of the buggy program is not sorted with a probability between 1% and 2%. Denote by N = 251 the number of possible values of the elements. The probability that a random list of length L contains no multiples of 9 is ((N-K)/N)^L, where K is the number of values divisible by 9 (in our case K = 28). The total probability is the average of this for 5 ≤ L ≤ 255, which is about 1.98%. Some of these lists are false positives, since they are already sorted. The probability of a random list of length L to be sorted is at most ((N+N*(N-1)/2)/N^2)^⌊L/2⌋: if we break the list into chunks of length 2, each of the ⌊L/2⌋ chunks must be sorted. The total probability of a list being sorted is bounded by the average of the above for 5 ≤ L ≤ 255, which is about 0.30%. Thus the probability of the function returning a non-sorted list is between 1.67% and 1.98%.

Answer 11

Bash, 26 bytes

Correct version

s=n
sort -${s:RANDOM%20<0}

Try it online! or check the probabilities.

Bugged version

s=n
sort -${s:RANDOM%20<1}

Try it online! or check the probabilities.

Takes input as newline-separated numbers. Uses the builtin variable RANDOM, which always returns a (pseudo)random number in the range 0 - 32767. Using %20 results in about a 5% failure rate (thanks @Titus for clarifying issues with %10).

This randomness means the failure rate is independent of the input, but this does require that the input array includes at least one number with more than one digit, because the failure output is sorted lexicographically.

Alternate version, 27 bytes

((RANDOM+20))||cat&&sort -n

Bugged version replaces the + with a %. Try it online or try it bugged.

Answer 12

Pyth, score 8 * 0.9 = 7.2

First snippet (correct one):

h.uSNT.S

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Second snippet (bugged one):

O.uSNT.S

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Saved two bytes (and 1.8 score) thanks to isaacg!

Answer 13

JavaScript (ES6), 24 bytes

Bug-free version (at least for integers in the range 0-2147483647, so anything in the given range):

a=>a.sort((a,b)=>a-b>>0)

Buggy version:

a=>a.sort((a,b)=>a-b>>1)

Depends on a) the engine's sorting algorithm and b) the input list containing two values in the wrong order that differ by 1. (If the probability of that turns out to be too low then the 1 can be increased, but by the time you get to 8 it simply won't sort anything in the range 5-255.)

Answer 14

PHP, 62 bytes

inspired by Jo´s solution (and I just noticed: it´s a port of Justin Mariner´s):

working (sorts ascending):

unset($argv[0]);(r[rand()+20].sort)($argv);echo join(_,$argv);

buggy (approx. 5% chance of descending sort):

unset($argv[0]);(r[rand()%20].sort)($argv);echo join(_,$argv);

Run with -nr

Answer 15

Pushy, 9 bytes - 10% = 8.1

Bugged Solution:

g0TUn?};_

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Working Solution:

g1TUn?};_

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The bugged code does the following:

g0TUn?};_

g          \ Sort the stack correctly
 0TU       \ Generate random(0, 10)
    n? ;   \ If this is zero:
      }    \    Cyclically shift the stack right once
        _  \ Print the result

The fixed code simply changes 0 to 1. As random(1, 10) will never be 0, the if statement will never be executed.

Answer 16

MATL, 7*0.9 = 6.3 6*0.9 = 5.4 bytes

Buggy version:

Gr.9<?S

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Explanation:

G        % Grab input
 r       % Push a random number between 0 and 1
  .9     % Push 0.9
    <    % Check if the random number is smaller than 0.9
     ?   % If true
      S  % Sort input
         % Implicit output

Bug-free version:

Gr9<?S

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Explanation:

G       % Grab input
 r      % Push a random number between 0 and 1
  9     % Push 9
   <    % Check if the random number is smaller than 9 (always true)
    ?   % If true
     S  % Sort the input
        % Implicit output     

Answer 17

Jq 1.5, 42 bytes

Buggy

sort|if length%13<=0then reverse else. end

Working (delete the =)

sort|if length%13<0then reverse else. end

Assuming line lengths are uniform in range [5,255] about 7% will trigger the bug

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Answer 18

J, 22.5 bytes (25 bytes - 10%)

with bug:

[:]`(1&A.)@.(0.1>?@0:)/:~

without bug:

[:]`(0&A.)@.(0.1>?@0:)/:~

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Answer 19

R, 30*.9 = 27 bytes

(buggy)

function(l)sort(l,runif(1)<.1)

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(not buggy)

function(l)sort(l,runif(1)<.0)

The buggy version sorts in decreasing=T 10% of the time, sampling from a uniform(0,1) distribution. The un-buggy version is always decreasing=F

Answer 20

Röda, 42 bytes - 10% = 37.8

Bug-free:

{sort key={|i|currentTime|[_%20//19*0+i]}}

Buggy:

{sort key={|i|currentTime|[_%20//19*i+i]}}

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This uses the currentTime function to create random numbers. It seems that their distribution varies a little between machines. The ratio 20//19 can be adjusted to get different results for no byte penalty (unless it is smaller than 99//98).

Answer 21

Octave, 36*0.9 = 32.4 bytes

Buggy version:

@(x)sort(x)(shift(1:end,+(rand<.1)))

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Bug-free version:

@(x)sort(x)(shift(1:end,+(rand<.0)))

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This sorts the vector, then shifts all numbers one to the right if a random number is less than 0.1.

Answer 22

Java 8, 45 34.2 (50 38 - 10%) bytes

Normal version:

a->{if(Math.random()>0.)a.sort(null);}

Explanation:

Try it here.

a->{                    // Method with ArrayList<Integer> parameter and no return-type
  if(Math.random()>0.)  //  If the random 0-1 double is larger than 0:
    a.sort(null);       //   Sort the input-List
}                       // End of method

---

Bugged version (51 39 bytes):

a->{if(Math.random()>0.1)a.sort(null);}

LD of 1: 1 added.

Explanation:

Try it here.

a->{                     // Method with ArrayList<Integer> parameter and no return-type
  if(Math.random()>0.1)  //  If the random 0-1 double is larger than 0.1:
    a.sort(null);        //   Sort the input-List
}                        // End of method

Answer 23

JavaScript, 25*0.9=22.5

new Date%25?x:x.sort()

input x