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## 35 posts in this topic

Posted (edited) · Report post

Hmmm .... I only found a far-fetched explanation.

@Vineetrika, is it making sense?

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
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Posted (edited) · Report post

Ok, 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 - 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
.

Else
alpha
=
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
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Posted · Report post

Ok, 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 - 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
.

Else
alpha
=
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?

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.

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Posted (edited) · Report post

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!

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

```ALABAMA - A3739768

ARKANSAS - A4842191

CONNECTICUT - N6444583

DELAWARE - E4975557

FLORIDA - O6319251

GEORGIA - R6435841

HAWAII - A5175716

KANSAS - A5287938

LOUISIANA - I7545683

MAINE - I6358396

MARYLAND - A5286668

MINNESOTA - N6556793

MISSOURI - S6433358

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
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Posted · Report post

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

```ALABAMA - A3739768

ARKANSAS - A4842191

CONNECTICUT - N6444583

DELAWARE - E4975557

FLORIDA - O6319251

GEORGIA - R6435841

HAWAII - A5175716

KANSAS - A5287938

LOUISIANA - I7545683

MAINE - I6358396

MARYLAND - A5286668

MINNESOTA - N6556793

MISSOURI - S6433358

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!
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Posted · Report post

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

at least these two:

KANSAS - K3852136

MARYLAND - M3852999

Thank you for checking

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Posted · Report post

Ok, 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 - 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

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

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
.

Else
alpha
=
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.

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

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

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Posted · Report post

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

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 ...

IOWA - W5727359

Number of distinct vowels n=3.

First character of password is W.

Writing the command starting from 3rd position, repeating the command until I get a 8-letter word:

WAIOWAIO

Idea 1

-------

Now trying the simple recurrence: password = password[i-1] + i + x

Next character is A <=M so x[2]=0.

password[2] = password[1] + 2 + x[2] = 3 + 2 + 0 = 5

Next character is I<=M so x[3]=0.

password[3] = password[2] + 3 + x[3] = 5 + 3 + 0 = 8 !!!! instead of 7

Next character is O>M so x[4]=1.

password[4] = password[3] + 4 + x[4] = 7 (corrected) + 4 + 1 = 12.

Next character is W>M so x[5]=1.

password[5] = password[4] + 5 + x[5] = 2 + 5 + 1 = 8 !!! again difference

Idea 2

-------

So to patch this up, I came with the idea of adding n

This actually worked for other words except that it gave somewhat shifted results when you get a password>10.

E.g. it would give 7,8,9 where it should, but it would give 10 instead of 1, 11 instead of 2, etc.

Idea 3

-------

So to patch this up, I came with a weird modulo function that would shift the output as needed (since Mod(a)=a%9 gives 0..8 output)

Weirdmod(a) = (a-1)%9+1. It gives what I needed: WeirdMod (x)=x for x=2..9 and WeirdMod(10)=1, WeirdMod(11)=2 and so on.

Again it worked for the first 2 characters for every command, except IOWA

See below.

Next character is A <=M so x[2]=0.

password[2] = password[1] + n + 2 + x[2] = 0 + 3 + 2 + 0 = 5

Next character is I<=M so x[3]=0.

password[3] = password[2] + n + 3 + x[3] = 5 + 3 + 3 + 0 = 11. WeirdMod(11)= 1 instead of 7.

Next character is O>M so x[4]=1.

password[4] = password[3] + n+ 4 + x[4] = 7 (corrected) + 3+4 + 1 = 15. WeirdMod(15)=6 instead of 2

Idea 4

-------

So to patch idea 4 up, I came with the idea of alpha. After you pass the beginning of the commands, you substract the nr of letters in command.

E.g. for IOWA - W5727359

The running command (starting from 3rd position, repeating the command)

WAIOWAIO

and alpha is

004444

It becomes 4 (nr of letters in command) when it reaches the beginning of the word (first I in case of IOWA).

Idea 4 password = WeirdMod(password[i-1] + n + i + x - alpha)

Next character is A <=M so x[2]=0.

alpha[2]=0

password[2] = WeirdMod(password[1] + n + 2 + x[2] -alpha[2])= WeirdMod(0 + 3 + 2 + 0 - 0) = WeirdMod(5)=5

Next character is I<=M so x[3]=0.

alpha[3]=4 since we started the word again.

password[3] = WeirdMod(password[2] + n + 3 + x[3] -alpha[3])= WeirdMod(5 + 3 + 3 + 0 - 4) = WeirdMod(7)=7 (correct this time).

Next character is O>M so x[4]=1.

alpha[4]=4 since we started the word again.

password[4] = WeirdMod(password[3] + n + 4 + x[4] -alpha[4])= WeirdMod(7 + 3 + 4 + 1 - 4) = WeirdMod(11)=2 (correct this time).

Next character is W>A so x[5]=1.

alpha[5]=4 since we started the word again.

password[5] = WeirdMod(password[4] + n + 5 + x[5] -alpha[5])= WeirdMod(2 + 3 + 5 + 1 - 4) = WeirdMod(7)=7 (correct this time).

And so on...

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 · Report post

Sorry, my logic was flawed too.

Correction 1: x

x = 1 if the letter under observation is <= M

x = 2 if the letter under observation is > M

Correction 2: formula

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

Correction 3:

if the password 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 ...

IOWA - W5727359

Number of distinct vowels n=3.

First character of password is W.

Writing the command starting from 3rd position, repeating the command until I get a 8-letter word:

WAIOWAIO

password[2] = 0 (prev.) + 4 (position of A in IOWA) + 1 (<= M) = 5

password[3] = 5 (prev.) + 1 (position of I in IOWA) + 1 (<= M) = 7

password[4] = 7 (prev.) + 2 (position of O in IOWA) + 2 (> M) = 11 = 1 + 1 = 2

password[5] = 2 (prev.) + 3 (position of W in IOWA) + 2 (> M) = 7

password[6] = 7 (prev.) + 4 (position of A in IOWA) + 1 (<= M) = 12 = 1 + 2 = 3

password[7] = 3 (prev.) + 1 (position of I in IOWA) + 1 (<= M) = 5

password[8] = 5 (prev.) + 2 (position of O in IOWA) + 2 (> M) = 9

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Posted (edited) · Report post

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.

Edited by araver
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