[Guest]

Here's the game: You are manipulating three different characters, $, A, 1. You will be putting these characters into different sequences or groups, such as: $A1, $A$A11, etc.

Here are the rules:

If you have a group with "A" at the end, you may add "1" to it. So if you have $A$11A you can make it $A$11A1, if you have $AA, you can make it $AA1, etc.

If you have a group that is $x where x is any set of $,A or 1, you may add x. So if you have $A1, you can make it $A1A1, if you have $aa11 you can make it $aa11aa11, ect.

If you have "AAA" anywhere in your group, you can change that to "1". So if you have $AAA$ it becomes $1$. If you have $11AAAA it can become $111A or $11A1. etc.

Edit (because I'm silly and didn't copy and paste correctly first time):

Rule 4: If you ever get "11" anywhere in a group, you can drop it. So $A11 can become $A, $111 can become $1, and $A11A can become $AA. (Sorry about the initial omission - see later post)

Your goal is to get from $A to $1. Here's your start:

$A

(now you can do $A1, or $AA - the choice is yours)

If you solve it, show your progression. Shortest route to $1 wins!

[Guest]

Here's the game: You are manipulating three different characters, $, A, 1. You will be putting these characters into different sequences or groups, such as: $A1, $A$A11, etc.

Here are the rules:

If you have a group with "A" at the end, you may add "1" to it. So if you have $A$11A you can make it $A$11A1, if you have $AA, you can make it $AA1, etc.

If you have a group that is $x where x is any set of $,A or 1, you may add x. So if you have $A1, you can make it $A1A1, if you have $aa11 you can make it $aa11aa11, ect.

If you have "AAA" anywhere in your group, you can change that to "1". So if you have $AAA$ it becomes $1$. If you have $11AAAA it can become $111A or $11A1. etc.

Your goal is to get from $A to $1. Here's your start:

$A

(now you can do $A1, or $AA - the choice is yours)

If you solve it, show your progression. Shortest route to $1 wins!

Would the same work the oposite way? like if you had three ones could it turn to an a?

[Guest]

Here's the game: You are manipulating three different characters, $, A, 1. You will be putting these characters into different sequences or groups, such as: $A1, $A$A11, etc.

Here are the rules:

If you have a group with "A" at the end, you may add "1" to it. So if you have $A$11A you can make it $A$11A1, if you have $AA, you can make it $AA1, etc.

If you have a group that is $x where x is any set of $,A or 1, you may add x. So if you have $A1, you can make it $A1A1, if you have $aa11 you can make it $aa11aa11, ect.

If you have "AAA" anywhere in your group, you can change that to "1". So if you have $AAA$ it becomes $1$. If you have $11AAAA it can become $111A or $11A1. etc.

Your goal is to get from $A to $1. Here's your start:

$A

(now you can do $A1, or $AA - the choice is yours)

If you solve it, show your progression. Shortest route to $1 wins!

Perhaps I don't understand something about the rules, but I don't see how you can accomplish this goal. I say this because the only way listed to remove extraneous letters is to turn three A's into a 1. There is no way listed to remove extra 1's. On a side note, there is no way to create or destroy $'s, based on the starting conditions. Therefore, by definition, once you have a "1" plus any number of "A's", you have failed because you cannot remove the "A" without creating another "1" and you cannot remove a "1."

If you are attempting to convert "$A" to "$1", you must create a single "1" while simultaneously removing all "A's"

Based on the above analysis, there are four possible permutations, all leading to a failure state:

1. $A -> $A1 (Failure)

2. $A -> $AA -> $AAAA -> $A1 (Failure)

3. $A -> $AA -> $AAAA -> $1A (Failure)

4. $A -> $AA -> $AA1 (Failure)

I very much like the idea behind this puzzle, so I hope I am misunderstanding something. Help me out?

[Guest]

Perhaps I don't understand something about the rules, but I don't see how you can accomplish this goal. I say this because the only way listed to remove extraneous letters is to turn three A's into a 1. There is no way listed to remove extra 1's. On a side note, there is no way to create or destroy $'s, based on the starting conditions. Therefore, by definition, once you have a "1" plus any number of "A's", you have failed because you cannot remove the "A" without creating another "1" and you cannot remove a "1."

If you are attempting to convert "$A" to "$1", you must create a single "1" while simultaneously removing all "A's"

Based on the above analysis, there are four possible permutations, all leading to a failure state:

1. $A -> $A1 (Failure)

2. $A -> $AA -> $AAAA -> $A1 (Failure)

3. $A -> $AA -> $AAAA -> $1A (Failure)

4. $A -> $AA -> $AA1 (Failure)

I very much like the idea behind this puzzle, so I hope I am misunderstanding something. Help me out?

I came to the same conclusion. I also wanted to know, how(in the OP), do you get a sequence with a "$" in the middle of some of your example sequences? I don't see anywhere in the rules about being able to replicate a $ign. Also in rule 3, do the "AAA" have to be next to each other, or for every 3 As you have, you can replace it with a 1?

Oh and yea txmom2 question as well.

And the above, how do you eliminate 1s?

[Guest]

What if you had:

$x

$xx

Can you just put another single x? because it doesn't say you have to worry about the other x

Why not just do

$x

$xx

$xxx

where all x's are A's

$A

$AA

$AAA

$1

[Guest]

My thought is in reverse, but still presents the same problems:

$1

To arrive at the solution, you must use rule 3. This means the second to last step must be:

$AAA

Rule 1 must add 1's, and no rule gives a provision for removing them. As 1's aren't present here, rule 1 can't be used to arrive here. Rule 2 can only double the symbols following a $ sign. There are an odd number of symbols after $ sign, suggesting that rule 2 cannot produce this value. Rule 3 provides a system for creating 1's from A's. As there are no 1's present, rule 3 can also not produce this value. Either racingnat is correct in his assumption, or we're missing information.

[Guest]

Can I just take $1 out of my wallet?

[Guest]

What if you had:

$x

$xx

Can you just put another single x? because it doesn't say you have to worry about the other x

Why not just do

$x

$xx

$xxx

where all x's are A's

$A

$AA

$AAA

$1

I thought of that too, but then if you look at the rule, it basically says when you apply that rule you are copying your sequence and pasting it at the end of it, or basically doubling your previous sequence. That's what I gather out of it anyway.

so $x, then $xx, then $xxxx, then $xxxxxxxx.....so on. It makes it so you cannot have 3 As in just 3 moves. That would be too obvious I'd think.

or $AA11A1AA11A1 rule two applied=$AA11A1AA11A1AA11A1AA11A1

Who knows tho.

[Guest]

"If you have a group that is $x where x is any set of $,A or 1, you may add x."

Nothing says my group must end where my caracters do.

So, I may group the caracters any way I wish to.

The shortest way of doing that (and the only one I can see, if you don't consider a group could be made of no elements) is the one racingnat1230 suggested:

$(A) -> $AA

$(A)A -> $AAA

$(AAA) -> $1

[Guest]

James, the examples may suggest you that, but it isn't specified by the rules. Anyway, if you think otherwise there's no way of solving it, since the last step must be $AAA and there would be no way to get AAA with a function that doubles your string, unless you subtract them by turning them into 1's, what would ruin the possibility of a solution, since you can't dispose of 1's.

[itachi-san]

is there supposed to be a rule about manipulating the '$'? or am I just missing something? for instance, how do you ever get: $A$A11 like in the first sentence of the OP?

[Guest]

James, the examples may suggest you that, but it isn't specified by the rules.

To me thats how I understood the rule. If $x=your current group and you apply rule 2 you add x. Sounds specific to me. I don't know. Or are we assuming that the part that says you may add x, x can be anything you want? If so then take your $A to start, apply rule 2, call x=AA, add it for $AAA, and there you go $1.

I'd be a little disappointed if you could solve this with using one rule(I suppose rule 3 also to change AAA to 1) in 2 or 3 moves. I thought of that way back before any answers were given, but thought that would be too simple. Unless that was the intention, and throw us off by adding in rules 1 and 3. edit: I geuss rule 3 is needed to convert "AAA" to a 1

I also am not following your example either..you said that your sequence does not end with your last character? Where does it end then? If you start with $, anything following that, ends with the last character you put down in that sequence. Or am I missing something there?

Anyway, if you think otherwise there's no way of solving it, since the last step must be $AAA and there would be no way to get AAA with a function that doubles your string, unless you subtract them by turning them into 1's, what would ruin the possibility of a solution, since you can't dispose of 1's.

Thats why my post #4 and Rattiars post #3 we were asking to make sure the OP was correct.

Edited June 18, 2009 by James8421

[Guest]

How about this...

$A

$AA

10($AA)+$A = $AAA

$AAA = $1

[Guest]

Ok, I admit it was kind of a joke. It's a classic puzzle from a book called GEB (Godel, Escher, Bach), called the MU puzzle. It's unsolvable as presented. I won't give away how to solve it, because I know that one of you at least will go find GEB and read it and I wouldn't want to spoil the fun.

[Guest]

I also am not following your example either..you said that your sequence does not end with your last character? Where does it end then? If you start with $, anything following that, ends with the last character you put down in that sequence. Or am I missing something there?

It was not my example you followed, that would be impossible since I posted after you. By the "last character" thing I meant you needn't choose the last character of the whole string as the last character of a group that you define, that just means you can group characters any way you wish.

Your understanding of the rule is acceptable, too, but that would lead the problem not to have an answer, that is why I explored the text more deeply, to another understanding. Well, but now we know it was only a joke.

Writersblock, you could present us the original puzzle, then.

Edited June 18, 2009 by Lehanius

[Guest]

Ok, I admit it was kind of a joke. It's a classic puzzle from a book called GEB (Godel, Escher, Bach), called the MU puzzle. It's unsolvable as presented. I won't give away how to solve it, because I know that one of you at least will go find GEB and read it and I wouldn't want to spoil the fun.

Given that there is obviously a trick to it, I know of one that will work.

If you write your sequence on a sheet of paper then you can create a sequence that ends in 1 and fold the paper so that the $ and the 1 are next to each other. It's a cheesy trick, but the only way to solve a few "draw this shape without lifting your pen" problems that I know.

[Guest]

Here's the book if you are intrigued: Gödel, Escher, Bach: an Eternal Golden Braid by Douglas R. Hofstadter

I won't present the original puzzle here to avoid potential issues with copyright. It is basically the same puzzle, except the author calls it the MU puzzle and uses the characters M for my $, I for my A, and U for my 1: so MIU. You are to go from MI to MU following the same rules.

Again, I don't want to spoil the fun for those who read the book. However, I will say that the solution is not a "trick,"such as folding paper.

Hint: try "mapping" the original characters to integers and translate the original "rules" to arithmetical "rules." Then solve.

[Guest]

OY! The chagrin! I just now went over all the responses and realized that the confusion arises because in my original post I accidentally omitted rule 4! Shame on me.

Rule 4: If you ever get "11" anywhere in a group, you can drop it. So $A11 can become $A, $111 can become $1, and $A11A can become $AA.

[Guest]

DeeGee *might* be on the right track, but violated rules by adding "$A" which is not allowed.

[Guest]

OY! The chagrin! I just now went over all the responses and realized that the confusion arises because in my original post I accidentally omitted rule 4! Shame on me.

Rule 4: If you ever get "11" anywhere in a group, you can drop it. So $A11 can become $A, $111 can become $1, and $A11A can become $AA.

well then!

if you start of with $A and change that to $AA then to $AAAA and so on till you reach 132 A's, it would then be a multiple of three. you could then convert it to 42 ones and 6 A's. the 42 ones could be removed, because they would just be a bunch of 11's. That would leave you with 6 A's. Because there's an A at the end, you could add a one. That would give you 6 A's and a one. change the 6 A's to 2 ones, then you have three ones. omit two ones, and you have $1!

I hope I did my math right...

[Guest]

well then!

if you start of with $A and change that to $AA then to $AAAA and so on till you reach 132 A's, it would then be a multiple of three. you could then convert it to 42 ones and 6 A's. the 42 ones could be removed, because they would just be a bunch of 11's. That would leave you with 6 A's. Because there's an A at the end, you could add a one. That would give you 6 A's and a one. change the 6 A's to 2 ones, then you have three ones. omit two ones, and you have $1!

I hope I did my math right...

Hi texmom2..

132? rule 2 I believe is just a power of 2s sequence...

1,2,4,8,16,32,64,128,256. I think anyways.

I'm not sure the new rule even helps. You still need to reduce the As down to 3 of them, then convert them to a 1. I don't know.

I saw the original...

unsolveable

[Glycereine]

well then!

if you start of with $A and change that to $AA then to $AAAA and so on till you reach 132 A's, it would then be a multiple of three. you could then convert it to 42 ones and 6 A's. the 42 ones could be removed, because they would just be a bunch of 11's. That would leave you with 6 A's. Because there's an A at the end, you could add a one. That would give you 6 A's and a one. change the 6 A's to 2 ones, then you have three ones. omit two ones, and you have $1!

I hope I did my math right...

You can never reach a multiple of 3 my multiplying by anything other than a multiple of 3 (3,6,9, etc). Multiplying the number of A's by 2 will only get you larger factors of 2.... 2,4,8,16,32,64,128... (not 132) which will never be divisible by 3 and thus never reduce down to 0 A's.

Alternatively, by adding a 1 at the end of the string, all you do is make it so you can cancel out another 1 (useful if you had a multiple of 3 A's, which isn't possible).

Unsolvable.

[Guest]

Hi texmom2..

132? rule 2 I believe is just a power of 2s sequence...

1,2,4,8,16,32,64,128,256. I think anyways.

I'm not sure the new rule even helps. You still need to reduce the As down to 3 of them, then convert them to a 1. I don't know.

I saw the original...

unsolveable

I multiplied wrong, that's how!

[Glycereine]

I multiplied wrong, that's how!

doh!

[Guest]

well then!

if you start of with $A and change that to $AA then to $AAAA and so on till you reach 132 A's, it would then be a multiple of three. you could then convert it to 42 ones and 6 A's. the 42 ones could be removed, because they would just be a bunch of 11's. That would leave you with 6 A's. Because there's an A at the end, you could add a one. That would give you 6 A's and a one. change the 6 A's to 2 ones, then you have three ones. omit two ones, and you have $1!

I hope I did my math right...

not quite, if you keep going from $x to $xx they way you do, you will get a sequence that is 2^n with n>=0, 2^7 is 128, you will never get 132.

[Guest]

So, after reading the posts, i think the question can be translated as -- Is there an integer that is a multiple of 3 in the series 2,4,8,16,32,... (or 2^n series).

Solving this may provides the actual solution... Any comments?