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HellBound Hackers | Challenges | Timed Challenges

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Timed 4


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Posted on 11-09-08 00:41
Honestly I have no idea what they mean with their rules. Do they mean swap the characters, add to the ascii values, or do they mean make a new string selecting the odd characters plus the second?

Thanks,
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RE: Timed 4

ynori7
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Posted on 11-09-08 02:09
They gave you pseudo code. If you can't understand that then you aren't ready for this challenge.


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RE: Timed 4


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Posted on 13-09-08 20:19
final += final[i]?

I understand pseudo code, but that isn't correct, unless they mean append or something, and I don't want to waste my time.
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RE: Timed 4

ynori7
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Posted on 13-09-08 20:41
Think of final as an empty string, and you just add characters to it in the right spot as you get to them while you trace through your scrambled string.


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RE: Timed 4


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Posted on 13-09-08 21:22
The_Gman wrote:
final += final[i]?

I understand pseudo code, but that isn't correct, unless they mean append or something, and I don't want to waste my time.


You code python, thats valid code for appending to a string.


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RE: Timed 4

Mouzi
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Posted on 14-09-08 11:57
The_Gman wrote:
final += final[i]?

I understand pseudo code, but that isn't correct, unless they mean append or something, and I don't want to waste my time.


IMO you're right. There is actually an error in the pseudo code.
Code
FOR i = 0; i < str.length; i++
IF isEvenOrZero(i)
final += final[i]
ELSE
final += final[i+2]
END FOR
final += final[1]
RETURN final



In the line where it says final += final[i] the final[i] would be null, because the final is a nonexistent variable at this point. It should actually say final += str[i], because str is the variable referred to on the first line. I believe str in this case refers to the string given in the challenge. Same goes for the next ones too (final[i+2] and final[1]). On the last line the final variable would still be null.


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Edited by Mouzi on 14-09-08 12:11
You would try to hack it anyways.
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RE: Timed 4


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Posted on 15-09-08 02:47
That makes a lot more sense.. I'll post in ten minutes with [hopefully] my success Smile
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RE: Timed 4


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Posted on 15-09-08 03:24
wait wait, what about the last odd character? str[i+2] clearly won't work.

And we're trying to REVERSE that algorithm, correct?

something like this? (although this won't run)
Code

code = 'x'*len(word)
code[1] = word[len(word)-1]
for i in range (len(word)):
    if (i%2 == 0):
        code[i] = word[i]
    else:
        code[i+2] = word[i]