34 replies to this topic

[araver]

Spoiler for And I think that

my logic works better

Spoiler for since it is less complex:

[octopuppy]

Anyone else cracked this yet?

The reason I ask is because I was thinking of doing another CtC, but wondering where to pitch the difficulty level.
In all previous CtC's I've done, I have given some thought to how people are going to solve it, and tried to leave a "way in" or at least give the hackers a fair chance. Vineetrika's offering here seems to have given less consideration to how you're going to solve it. The 1st letter is a bit of a start but not much of one really. As a result, I actually found this to be a far more satisfying challenge, I really enjoyed bashing away at it. In the end it was a great puzzle, so thanks for that Vineetrika. I've also noticed that in my algorithms, people have sometimes picked up on clues I did not intend to leave. So maybe I should post one with no intentional weaknesses. However, if nobody else can crack this, I'd be wasting my time if I were to up the ante from here. Come on guys, this is solvable, show me you can do it!

[araver]

Anyone else cracked this yet?

Come on guys, this is solvable, show me you can do it!

I've cracked the second character a little after you posted the ABORT command, but since there was no longer any pressure, I admit I kind of left it alone for a while

Spoiler for Second character:

Select next letter in command (position n+1, where n = number of distinct vowels).
If this letter is <= M then x=0, else x=1.
Write n + 2 + x in password.

[Vineetrika]

araver, when you can solve the second character, what is stopping you from solving the rest ?

[araver]

araver, when you can solve the second character, what is stopping you from solving the rest ?

As I said, since there was no pressure I gave the problem some rest as I was blocked going around in circles and trying the same things over and over again. I will give it another try with a fresh perspective

[araver]

Hmmm .... I only found a far-fetched explanation.
@Vineetrika, is it making sense?

Spoiler for So far

First character
--------------------
n = number of distinct vowels
The n-th letter in command is the first character of the password.

Second character

--------------------
If (n+1)-th letter in command is <=M then x=0 else x=1
n+2+x is the second character in the password



Third character

--------------------
If (n+2)-th letter in command is <=M then y=0 else y=1
If (n+2)-th letter in command is the first letter of the command (considering that command is read continuously from left to right) then subtract number of letters) (4 in case of IOWA) i.e. alpha= - nrletters else alpha =0.
(n+2+x)+(n+3+y) + alpha gives a number. If it is equal or greater than 10, subtract 9 from it. (sort of a weird modulo 9?)

Fourth character
----------------------
If (n+3)-th letter in command is <=M then z=0 else z=1
If (n+3)-th letter in command is the first letter of the command (considering that command is read continuously from left to right) then subtract number of letters (5 in case of IDAHO, 4 for UTAH and OHIO) i.e. beta= - nrletters else beta =0.
(n+2+x)+(n+3+y) + (n+4+z)+beta gives a number. If it is equal or greater than 10, subtract 9 repeatedly from it until you arrive in [1..9]. (sort of a weird modulo 9?)
EDIT: Also, if you've passed the beginning of the command add 1 before subtracting 9's (case of IOWA).

Edited by araver, 20 November 2010 - 12:41 PM.

[araver]

Ok, it is less far-fetched once you can actually see

Spoiler for

... a recurrence.

First character
--------------------
n = number of distinct vowels
The n-th letter in command is the first character of the password.

Second to eight character of password
-----------------------------------------------------
For the i-th letter of the password - i=2..8 - denoted password [ i ].

Find the (n+i-1) letter of command (considering that command is read continuously from left to right)
If the letter you found is <=M then x[i]=0 else x[i]=1

If you finish the word once before finding this letter alpha[i] = nr of letters of command.

Else If you finish the word the second time before finding this letter alpha[i] = -1.

Else alpha[i] = 0.

Define previous digit of password (password[i-1]) as 0 for i=2 and actual previous digit of password for i=3..8

Then

password[i] = WeirdMod(password[i-1]+n+i+x[i]-alpha[i])

where WeirdMod gives shifted 1..9 output instead of 0..8 normal modulo operation.

WeirdMod(a)=(a-1)%9+1

Still pondering at a more natural way to describe this.

So, if it is correct and we move on to the next CtC challenge, I had a question: I also have developed an algorithm, I'm not sure of the difficulty though. So if it at least sounds less difficult than the one octopuppy suggests ... should we do it next or after?

Edited by araver, 21 November 2010 - 12:36 PM.

[octopuppy]

Ok, it is less far-fetched once you can actually see

Spoiler for

... a recurrence.

First character
--------------------
n = number of distinct vowels
The n-th letter in command is the first character of the password.

Second to eight character of password
-----------------------------------------------------
For the i-th letter of the password - i=2..8 - denoted password [ i ].

Find the (n+i-1) letter of command (considering that command is read continuously from left to right)
If the letter you found is <=M then x=0 else x=1

If you finish the word once before finding this letter alpha = nr of letters of command.

Else If you finish the word the second time before finding this letter alpha = -1.

[i]Else alpha[i] = 0.

[i]Define previous digit of password (password[i-1]) as 0 for i=2 and actual previous digit of password for i=3..8

Then

password[i] = WeirdMod(password[i-1]+n+i+x[i]-alpha[i])

where WeirdMod gives shifted 1..9 output instead of 0..8 normal modulo operation.

WeirdMod(a)=(a-1)%9+1

Still pondering at a more natural way to describe this.

So, if it is correct and we move on to the next CtC challenge, I had a question: I also have developed an algorithm, I'm not sure of the difficulty though. So if it at least sounds less difficult than the one octopuppy suggests ... should we do it next or after?

I don't think that's quite it, though I am confused. Let's see what Vineetrika says. An easier way to check if you're on the money might be to just encode a few random words!
I think it's best if you host the next game, I'm a bit busy at the moment.

[araver]

I don't think that's quite it, though I am confused. Let's see what Vineetrika says. An easier way to check if you're on the money might be to just encode a few random words!

Spoiler for

Better than random, these are the last states according to the algorithm I stated before:

ALABAMA - A3739768
ARKANSAS - A4842191
COLORADO - O4186568
CONNECTICUT - N6444583
DELAWARE - E4975557
FLORIDA - O6319251
GEORGIA - R6435841
HAWAII - A5175716
KANSAS - A5287938
LOUISIANA - I7545683
MAINE - I6358396
MARYLAND - A5286668
MINNESOTA - N6556793
MISSOURI - S6433358
NEBRASKA - E4176558
OKLAHOMA - K4965447
RHODEISLAND - D6444582
VERMONT - E5187714
WASHINGTON - S5219136
WYOMING - Y5176583

I think it's best if you host the next game, I'm a bit busy at the moment.

Then I will open a next CtC challenge soon, later today, while waiting for Vineetrika.

Edited by araver, 21 November 2010 - 06:14 PM.

[octopuppy]

Spoiler for

Better than random, these are the last states according to the algorithm I stated before:

ALABAMA - A3739768
ARKANSAS - A4842191
COLORADO - O4186568
CONNECTICUT - N6444583
DELAWARE - E4975557
FLORIDA - O6319251
GEORGIA - R6435841
HAWAII - A5175716
KANSAS - A5287938
LOUISIANA - I7545683
MAINE - I6358396
MARYLAND - A5286668
MINNESOTA - N6556793
MISSOURI - S6433358
NEBRASKA - E4176558
OKLAHOMA - K4965447
RHODEISLAND - D6444582
VERMONT - E5187714
WASHINGTON - S5219136
WYOMING - Y5176583

I'd agree with that except you miscounted the vowels on a couple. Looks like you got it!