What is the solution to the MI string puzzle?

[croxbearer]

I don't know if you have encountered this problem in Logic and Mathematics. Anyway, you may try this:

You are given with an initial string of MI.

Here are the rules:

i) If you have a string that ends with I you can add a U
ii) If you have an Mx, (i.e., an a string that starts with an M, with other characters succeeding it) you can double the characters succeeding the M, i.e., you may lengthen your string by writing Mxx
ii) If you have an III part on your string, you may replace it with a U
iv) If you have a UU in your string, you can drop this altogether, i.e., starting with an MUUU, you can drop the UU, to get MU.

Now, the question is: Could you produce a string containing the characters, MU? (without of course trespassing the bounds of the rules)

 

[Rogerio]

YES!
initial string "MI"
(rule 2) -> "MII"
(rule 2) -> "MIIII"
(rule 3) -> "MUI" : this string contains the characters MU !

But it's impossible to get the string "MU" as final result.

 

[croxbearer]

I am sorry. May I rephrase the question with:

Could you produce a string containing exactly the characters, MU?

 

[NateTG]

I am sorry. May I rephrase the question with:

Could you produce a string containing exactly the characters, MU?

Obviously not, since you can never get a string that does not contain at least one I.

Proof:
Consdier how the various operations alter the number of I's mod 3:
1,3, and 4 do not change it at all, and 2 multiplies the total number of I's by two.
Thus you cannot get from a string that has a number of I's that is not zero mod 3 to one that is. Specificall you cannot get from a string that contains one I to one that contains zero.

 

[Xargoth]

You can double the x where $$M_x$$

3I is equal to U

$$\frac{2^n}{3}$$

You are doomed to end up with 1 "I" remaining..

 

[gnpatterson]

It is possible to get MIIIII (5 I's) so why shouldn't it be possible to get more numbers?

mi->mii->miiii->miiiiiiii->miiiiiiiu->miiiiiuu->miiiii

 

[gnpatterson]

OK I think i see it now, It seems to be impossible to get to 3 from 1 by doubling, i can get any number mod 3 = 1 or 2.

 

[Xargoth]

Oh i was trying to say minimum # of I can only be 1, i didn't mean to rule out "2"