Recover the mutated source code (for robbers)

Question

This is the companion thread to the main Recover the mutated source code (link) challenge. If you have successfully cracked a cop's answer, post your solution as an answer to this question.

As a reminder, here are the robber rules from the main challenge again:

The robber will attempt to change the cop's program (which completes Task #1) into a program that completes Task #2 (not necessarily the original program written by the cop) in the edit distance specified by the cop.

An already-cracked submission cannot be cracked again (only the first robber who cracks a submission gets credit).

After cracking a submission, please do the following:

• Post an answer to this challenge's accompanying question (link), providing the language, your solution, and a link to the original answer.
• Leave a comment with the text "Cracked" that links to your posted answer.
• Edit the cop's answer if you have edit privileges (if you do not, either wait until someone else with the required privileges does so for you or suggest an edit).

And scoring:

If a robber successfully cracks a cop's submission, the robber's score goes up by the edit distance of that submission. For example, a robber that cracks a submission with an edit distance of 3 and one with a distance of 5 earns 8 points. The robber with the highest score wins. In the event of a tie, the robber who earned the score first wins.

Leaderboard

There are no cracked submissions yet.

Answer 1

Python 2, FryAmTheEggman

x=n=1;j=input();
while j>2:
    x,n=n,x+n;j-=1;
    ##while~-all(n%i for i in range(2,n)):n+=1;
print n

Used 12 edits. Put in an extra # to make it 13.

Answer 2

Python 2, Sp3000

from fractions import*
n=input()
k,=P=[1]
while n>len(P):k+=1;z=reduce(lambda x,y:x+y,P[~1:]);P+=[z]#*(gcd(z,k)<2)
print P[-1]

Answer 3

J, grc

   f=:3 :'{.+/(!|.)i.y'

   f 45
1134903170

Answer 4

Python 3, Sp3000

x=n= int(input())       # 3
P = [1,1]               #+2 = 5
k = 2
while n >=len(P):       #+1 = 6
 k += 1
 for x in P:
  if k%x ==~0: break    #+1 = 7
 else: P += [P[-2]+x]   #+7 = 14
print(x)

Answer 5

Python 3, matsjoyce

a,c,n=1,1,int(input())
while n-1:
 #c+=1
 ##########list(map(c.__mod__,range(2,46))).count(0):
 a,c=a+c,a
 n-=1
print(c)

The Fibonacci program was strangely already in there...only needed 5 edits to get it.

Answer 6

CJam by Martin Büttner

T1l~({_2$+}*p];

Takes input n from STDIN

Answer 7

Python 2, Pietu1998

f=lambda p,i:p if p[45:]else f(p+[i]#f all(i%q for q in p[1:])else
,p[-1]+i)
print f([1,1,1],2)[input('!')]

I used 9 edits to get a Fibonacci program.

Answer 8

JAGL, globby

T~2]d]2C{cSdc+c]}wSP

It might not be the most efficient approach, it it is almost definitely not the Cop's code, but it works, and its 12 away.

Answer 9

Ruby, histocrat

p [x=1,y=1,*(1..200).map{|i|z=y;y+=(y*x**(i-1)+x%2).divmod(i)[2-1]?x:1;x=z;y}].-([x-1])[gets.to_i-1]