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

Thank you for checking

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Guest Message by DevFuse

Started by Vineetrika, Nov 11 2010 03:29 PM

34 replies to this topic

Posted 22 November 2010 - 01:31 PM

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

Spoiler for Right

Thank you for checking

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Posted 22 November 2010 - 02:07 PM

You got me up to this point and then I lost youOk, it is less far-fetched once you can actually see

... 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 - denotedpassword [ 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<=Mthenx[i]=0elsex[i]=1

I am not sure I understand the alpha and weirdword concept but my logic is ...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

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

Thenpassword[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.

Define previous digit of password (

password[i] = password[i-1] + i + x[i]

Please let me know if this is not making sense. I hope you enjoyed solving this one.

Posted 22 November 2010 - 03:36 PM

You got me up to this point and then I lost you

I am not sure I understand the alpha and weirdword concept but my logic is ...

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

password[i] = password[i-1] + i + x[i]

Please let me know if this is not making sense. I hope you enjoyed solving this one.

I enjoyed it very much, thanks for the challenge

Well, I can explain how I arrived at my rather complicated formula because of IOWA. This state systematically broke every relation I was able to come up ...

Spoiler for

I understand that it's not the function either of you use and it just *coincidentally* gives the same results for 8 letter passwords, but I still don't see a simple function

Can you express it in terms of a recurrence and show it step by step for IOWA?

Thank you.

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Posted 22 November 2010 - 05:09 PM

Sorry, my logic was flawed too.

Correction 1: x[i]

x[i] = 1 if the letter under observation is <= M

x[i] = 2 if the letter under observation is > M

Correction 2: formula

password[i] = password[i-1] + n + x[i] => where n is the position of the ith letter in the command

Correction 3:

if the password[i] value is greater than 10, add the digits and if the sum of the digits is again greater than 10, sum the digits again .. this is recursive until you get a single digit.

I am not sure if this formula is accurate either ... so ...

Correction 1: x[i]

x[i] = 1 if the letter under observation is <= M

x[i] = 2 if the letter under observation is > M

Correction 2: formula

password[i] = password[i-1] + n + x[i] => where n is the position of the ith letter in the command

Correction 3:

if the password[i] value is greater than 10, add the digits and if the sum of the digits is again greater than 10, sum the digits again .. this is recursive until you get a single digit.

I am not sure if this formula is accurate either ... so ...

Spoiler for Step by Step for IOWA

Posted 22 November 2010 - 10:08 PM

Yes, your formula is indeed correct.

Just compiled them both in a spreadsheet.

They give the same results for all commands including ABORT.

Surprisingly, the methods may yield different results for 2 and 3 letter words, but are equal for longer-than-5 letter-words and 4 letter-words, except those with 4 distinct vowels.

It seems that my WeirdMod function + alpha + constant difference in definition of x (mine's is 0..1, yours is 1..2) just gives in a more complicated way the same array you do naturally with "position of the letter in command". E.g. for IOWA 41234123 is obtained by 2 variables and one function in my case. Not sure how exactly your digit adding function (if >10) is simulated with my variables, but it seems to work.

That's why it always seemed unnatural to me. I went on on a wrong path and constructed artificial variables just to fit the data and bring me back to it. Guess this is kind of a physicist nightmare

Thank you again for the challenge ... twice

I really enjoyed first discovering my*alternate-oh-so-complicated-but-still-fits-the-data-so-it-must-be-true-right* theory and then your *simple-oh-so-simple-how-did-i-miss-that-wait-they-both-cannot-get-the-same-results-can-they-oh-my-god-they-can-how-the-hell-is-that-possible* theory.

Just compiled them both in a spreadsheet.

They give the same results for all commands including ABORT.

Surprisingly, the methods may yield different results for 2 and 3 letter words, but are equal for longer-than-5 letter-words and 4 letter-words, except those with 4 distinct vowels.

It seems that my WeirdMod function + alpha + constant difference in definition of x (mine's is 0..1, yours is 1..2) just gives in a more complicated way the same array you do naturally with "position of the letter in command". E.g. for IOWA 41234123 is obtained by 2 variables and one function in my case. Not sure how exactly your digit adding function (if >10) is simulated with my variables, but it seems to work.

That's why it always seemed unnatural to me. I went on on a wrong path and constructed artificial variables just to fit the data and bring me back to it. Guess this is kind of a physicist nightmare

Thank you again for the challenge ... twice

I really enjoyed first discovering my

**Edited by araver, 22 November 2010 - 10:08 PM.**

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