Amber Truhanovitch October 31, 2007
POW 3 Group-C
MU
Problem Statement:
Given the rules of the MIU-system, can MI be converted into MU is the main problem.
It is known that (rule 1) If you possess a string whose last letter is I, you can add on a U at the end.
(rule 2 ) Suppose you have Mx. Then you ay add Mxx to the collection.
(rule 3) If III occurs in one of the strings in your collection, you may make a new string with U in place of III.
And finally rule 4, If UU occurs inside one of your strings, you can drop it.
So, with these rules, we must understand what the MU-puzzle is and how it works. And after doing some practice problems, try to figure out if MU is a theorem of the MIU-system and can be derived from the axiom MI.
Process:
The first step in solving this problem is to understand the problem. With this, we must first do a practice problem to understand how the MIU-system even works.
When we solve one of the problems, we must find it the string is a theorem. To find this we can only use the four rules, which are stated above. The catch is, you can only do them forwards not backwards. Meaning III can become U but U can not become III.
Practice 1. MI-MUIU
Yes MUIU is a theorem of the system because MI can be converted into MUIU.
Rule 2- MI- MII Rule 2 MII- MIIII Rule 3 MIIII- MUI Rule 1 MUI- MUIU
Practice 2. MI-MUII
Yes MUII is a theorem of the system because MI can be converted into MUII.
MI-MII-MIIII-MIIIU-MUIU-MUIUUIU-MUIIU-MUIIUUIIU-MUIIIIU-MUIUU-MUIUUUIUU-MUII
Practice 3. MI-MIUU
MI-MII-MIIII-MIIIIIIII-
MIUUI-MIUUIIUUI-MIUUIII-MIUUIIIIUUIII-MIUUUIUUIII-MIUIIII-MIUUI-MIUUIIUUI-MIUUIII-MIUUU-MIU
MIIIIIIIIIIIIIIII-MIUUUUU-MIUUU-MIU-
Hahaha I don’t know.
So now that we some what understand what to do to try and solve strings with the axiom MI, we must try and find out if we can derive MU from MI.
MI- MII-MIIII-MUI-MUIU-MUIUUIU-MUIIU-MUIIUUIIU-MIIIIU-MIUU
-MI-MIU-MIUIU-MIUIUIU
-MI-MII-MIIU-MIIUIIU
-MI-MII-MIIII-MIUU-MIUUIUU-MII
-MI-MII-MIIII-MIUU-MIUUIUU-MIUUIUUIUUIUU-MIIII-MUI
OR
-MIUUIUU-MIUUI-MII
-MIUUIUU-MIUUIIUUI-MIIIUUI-MIIIUUIIIIUUI-MUUUIIIIUUI-MUIIIIUUI-MIUUUI- MIUI-MIUIIUI-MIUIIUIIUIIU
So with all this work out, I have come to a conclusion that this “tree” system for solving that I have going is impractical because I either end up in the same spot, and don’t realize it or there are thousands of different combinations. Especially within a series of I’s (like IIIIIIIII ) So the tree system is not working for me.
Now I must look for a different way to go about the problem.
I have done a lot of different combinations, and I think the best thing right now would be to look and try to distinguish patterns.
The I count is the part of the problem that sticks out the most to me. By examining the problems, I have found that the I count should be in multiples of threes. Because it begins with an I. Also, we know that two of the rules do not affect the I’s at all. And the other two rules change the count but never a non-multiple of three into a multiple of three. And this means that we can never get I to equal to zero. And therefore, MU is not a theorem of the MIU-system.
Solution and generalization:
In this problem we are using the MIU-system to determine if MU is a theorem of the system. We are using the four rules and the axiom MI. When using the rules, we can only work forwards not backwards.
Here are the rules as follows:
rule 1- If you possess a string whose last letter is I, you can add on a U at the end.
rule 2 - Suppose you have Mx. Then you ay add Mxx to the collection.
rule 3 - If III occurs in one of the strings in your collection, you may make a new string with U in place of III.
rule 4 - If UU occurs inside one of your strings, you can drop it.
The solution of the problem is that MU is not a theorem of the MIU-system. Although there are lots of strings that are, MU is not one of them.
After doing out all of the strings that I could think of, I realized it was easier to look for patterns and see if anything could be generalized.
The I count (the I’s in a string ex. IIII) is the most important part that I missed.
Rules one and four do not affect the I count at all. Rule three reduces it by three (which is a common number in the system). So with that said, now we know that rules one, three and four can not make the I-count to three. Now rule two. This rule, doubles the I count. And it can not create a multiple of three from a non multiple of three to start with. ( if 3/2n is must divide n also because it does not divide 2.)
The I count starts with one, MI. Now, no one of the rules can affect the I-count to create a multiple of three which is needed to make a final I-count 0. And without the ability to make it zero, we can not possible change MI to MU. And our final answer is that MU is not a theorem of the MIU-system.
Self-assessment:
I do not think I did too badly on this POW. I’m sure I could have done better thought, if I spent more time looking deeper into more algebraic issues to derive the proof, but I am not very good at it so, I couldn’t. I heard a lot of people talking about it, but patterns work well for me, so I wanted to stick with that. I also, as POW’s go on, I see how guess and check is not the best way to solve problems, especially ones like this POW. But for the most part, I am happy with my POW and I learned a lot! J
Oh and by the way, oh thing I did realize is,
MU is a system—
Not a cow –(but I wish it was, I would like it a lot more, hahaha)
MOOOO
