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The Mathematics of Quatrainment

Finding the Solution Loran Briggs and Daniel McBride

College of the Redwoods Eureka, CA 95501, USA

December 17, 2010

Abstract

The game of Quatrainment is a simple game where you flip over tiles to make a 4 by 4 gird look like another target 4 by 4 grid. This is an example of a finite-state machine. In this presentation, we will be discussing our method of finding a unique solution to any quatrainment game using the fewest number

• f moves possible. We will go through the rules of the game, figuring out

whether or not there is a solution to every game, how to find a solution, as well as our program which finds the quickest solution to the game.

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Overview

Setup and Goal Rules Base Two Systems Is a Solution Always Possible? The Solution Algorithm

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Setup and Goal

Starting Grid X X X X X X X Target Grid X X X X X X X

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Rules

Corner Cell Selected. Inner Cell Selected Edge Cell Selected

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Base Two System

Four possibilities when adding in base two: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 1 1 1 1 1 1 1 + 1 1 1 1 1 1 = 1 1 1 1 1 1 1 1 1

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Is a Solution Always Possible?

First we need to define our moves in the form of 4 × 4 matrices. M1 = 1

1 1 1 1 1

• M2 =

1

1 1

• M3 =

1 1 1

• M4 =

1 1 1 1 1 1

• M5 =

1

1 1

• M6 =

1 1 1 1 1

• M7 =

1 1 1 1 1

• M8 =

1 1 1

• M9 =

1 1 1

• M10 =

1 1 1 1 1

• M11 =

1 1 1 1 1

• M12 =

1 1 1

• M13 =

1 1 1 1 1 1

• M14 =

1 1 1

• M15 =

1 1 1

• M16 =

1 1 1 1 1 1

• Loran Briggs and Daniel McBride (CR)

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Are the Inputs Linearly Independent?

Is every target array reachable from any starting array? Are the input moves independent of each other? α1M1 + α2M2 + α3M3 + α4M4 + α5M5 + α6M6 + α7M7 + α8M8 + α9M9+ α10M10 + α11M11 + α12M12 + α13M13 + α14M14 + α15M15 + α16M16 = 0 Is the only solution the trival solution? α0 = α1 = α2 = α3 = α4 = α5 = α6 = α7 = α8 = α9 = α10 = α11 = α12 = α13 = α14 = α15 = α16 = 0

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To do this we expand our Mi matrices and multiply them with our α′s which gives     α1 α1 α1 α1 α1 α1     +     α2 α2 α2     +     α3 α3 α3     +     α4 α4 α4 α4 α4 α4     +     α5 α5 α5     +     α6 α6 α6 α6 α6     + · · · +     α16 α16 α16 α16 α16 α16     =        

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Each cell indiviually must equal zero, form the zero matrix α1 + α2 + α5 = 0 α1 + α3 + α4 + α6 = 0 α1 + α2 + α4 + α7 = 0 α3 + α4 + α8 = 0 α1 + α6 + α9 + α13 = 0 α2 + α5 + α6 + α7 + α10 = 0 α3 + α4 + α6 + α7 + α8 + α11 = 0 α4 + α7 + α12 + α16 = 0 α1 + α5 + α10 + α13 = 0 α6 + α9 + α10 + α11 + α13 + α14 = 0 α7 + α10 + α11 + α12 + α15 + α16 = 0 α4 + α8 + α11 + α16 = 0 α9 + α13 + α14 = 0 α10 + α13 + α15 + α16 = 0 α11 + α13 + α14 + α16 = 0 α12 + α15 + α16 = 0

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The previous sixteen equations form this matrix here. Rα = 0 :                             1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1                                                         α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 α12 α13 α14 α15 α16                             =                                                        

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In reduced row echelon form, matrix R yeilds:                             1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1                             Since there is a pivot in every row, R is non-singular and therefore our input moves are linearly independent.

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The Solution Algorithm

If we replace the zero vector from before with a target vector we call p. We can now take our equation: Rα = p Solving for α gives R−1Rα =R−1p α =R−1p Now to find R−1.

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Using a MATLAB program named bin_solve, we have our R−1 R−1 =                             1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1                            

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So now we take our initial matrix and our final matrix add them together to get

• ur matrix P, we convert our P matrix into a vector p. We then find our α by

α = R−1p For example: X X X X X X X Initial Grid X X X X X X X Target Grid

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We wrote a program in MATLAB which will help us. First we convert our grids to base 2 matrices. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Add them together to get: 1 1 1 1 This is our P matrix

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We then convert our P matrix into the vector p on the left. p =                             1 1 1 1                             α = R−1p =                             1 1 1 1 1 1 1 1                             Now, using MATLAB we multiply R−1p to get our α on the right.

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We then convert our α back into the following matrix:     1 1 1 1 1 1 1 1    

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QUESTIONS?

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