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Why Baseball is Not Physics

Hu's on first.

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The concept of state explained in baseball.

The Concept of State

To expound the concept of state, let us consider the state of a baseball game.

I mean the state of the game as a baseball game, not the physics of the matter and energy of the stadium and everything in it. Expressing the state of the game requires much information, but no so much information as the location of every blade of grass or the aerodynamics of a knuckle ball. I really know next to nothing about baseball, so I cannot expect to get things entirely right or even to know the right questions. Not the first time.

These are some possible facts in a game which may or may not be useful:

• The score. It takes two numbers to express the score, and each of them has to be associated with a team. The names of the teams do not matter, but the numbers need to be associated with each of the Visitor and Home teams because this is very pertinent to whether the game is over after eight and a half innings.
• The inning. This is a positive integer. In theory (in some leagues) there is no upper limit on this number, but in Major League Baseball there has never been a game that went much beyond twenty-something innings.
• The half of the inning. Two values traditionally known as Top and Bottom.
• Runners. Each of the bases 1st, 2nd, and 3rd can have a runner or not. This can be represent by one octal digit from 0 to 7. (Or can it?)
• The number of outs. This can be 0, 1, or 2.
• The number of strikes. This can be 0, 1, or 2.
• The number of balls. This can be 0, 1, 2, or 3.
• The batter's number. This could be a one- or two- digit number.
• The numbers of the batters in the rest of the batting order in order. For now I neglect the possibility of pinch hitters.

These are just a few of the basics, what you want to know if you turn on the TV in the middle of game. It is the situation, which by and large is just another way of saying this is the state of the game. The color commentator can provide plenty of other information, much of which is entirely irrelevant to the state of the game. He might tell you the pitcher's eyes are a beautiful shade of blue. That is not something you need to know to understand the state of the game. He also can provide much statistical information, and you may be interested in that it has something to do with the probability of various outcomes, but that is getting ahead of ourselves. If you listen closely you will realize there are more aspects to the state than we have mentioned.

• Where the game is being played. We do not really care what major corporation paid to have its name attached to the stadium, but we might need to know the physical dimensions of the outfield. Generally, of course, this is not going to change during the game, but there is a slight possibility of a game being suspended in one city and resumed in another.
• Which way the wind blows. This is another factor of the many affecting the probabilities of various things happening. For now I am neglecting them because I am only trying to describe the system, not predict it. It is pointless to think about probabilities until we have some grasp of the possibilities.
• The identities of the defensive players. This is beyond the scope of the present example, but obviously there are statistics which affect to probabilities of the possible outcomes.

State Vector Abbott

So, on a Saturday afternoon, in your den, you turn on the TV for a baseball game. But this time you are not coming in at the middle. You are right on time. The National Anthem has been sung. The ceremonial first pitch has been thrown out. The umpired has yelled, "Play Ball!" The Home team has taken the field, and the first visiting batter has entered the batter's box. I am not entirely sure the order of these preliminary events, but as physicists are wont to say, the game has been prepared.

Now, let us construct a state vector for the game at this point. we will call this vector

{$$\LARGE \left| Abbott \right\rangle$$}

The point here is: The stuff in this symbol (which is a Dirac ket) is just a label. It can be anything, including numbers, letters, symbols, emojis and any combination of the above. In cases where numbers or letters are used it is important to understand that the label is not involved in any operation or calculation.

Although I have seen texts in which authors have gone the other way, the ket represents a column vector. What is more, kets are generally used for vectors in a vector space over the complex numbers. Now, a baseball state, at least so far as I am going to define one, can be represented entirely in non-negative integers. By a tremendous stroke of luck, the non-negative integers are a subset of the complex numbers. So maybe we will be okay.

{$$\LARGE |Abbott \rangle = \begin{pmatrix} a_1 \\ a_2 \\ \dots \\ a_n \end{pmatrix}$$}

The {$a_1 , a_2, \text{and } a_n$} are elements of the vector, and the dots represent any number of elements that have been omitted. As I have mention, the non-negative integers are sufficient for our purposes, but in general the elements could be complex numbers. Also, I'll be renaming the elements so they are slightly easier to remember.

Let us start with the score. The score is two numbers, one for the visitors and one for the home team. For convenience we name these variables {$s_v \text{and } s_h$}. At the beginning of the game, which is where we are, both these numbers are 0. Here I also introduce the general baseball vector, just to help us keep track of things.

{$$\LARGE |baseball \rangle = \begin{pmatrix} s_v \\ s_h \\ \dots \\ a_n \end{pmatrix}$$}

{$$\LARGE |Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ \dots \\ a_n \end{pmatrix}$$}

Next we need to keep track of the inning and half inning. For convenience, we will call the top of the inning 0 and the bottom 1.

{$$\LARGE |baseball \rangle = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ \dots \\ a_n \end{pmatrix}$$}

{$$\LARGE |Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \dots \\ a_n \end{pmatrix}$$}

The only thing here is that there is no inning zero. Then, in inning time, the next significant thing is the number of outs. At the beginning of the game there are no outs. I will use the little letter o and hope not to confuse myself by mistaking it for zero.

{$$\LARGE |baseball \rangle = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ \dots \\ a_n \end{pmatrix}$$}

{$$\LARGE |Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ \dots \\ a_n \end{pmatrix}$$}

Next we consider the base runners. I said they could be represented by one octal digit, which is true. But it is not very convenent. So let us take it as three binary digits. We will call them {$r_1 , r_2, \text{and } r_3$} for runner on first, runner on second, and runner on third. I want to say one will represent a runner on the base, and zero will represent no runner on the base. At the beginning of the game, there are no runners on base, so they are all zero. But this is just wrong.

No base can have more than one runner on it. So it is right that there is only zero or one runners on a base. But the runners are not all the same. Suppose there is a guy on second and a guy on third. The guy on second cannot score or advance unless the guy on third scores or is put out. So if the next thing is that there is just a guy on third, that could happen if the guy on third scored or was put out and the guy on second advanced. Alternatively, the guy on second might have been caught off base and put out. So if the outs increase there is only a guy on third, we don't know which one of them is put out. So somewhere down the line, it will pay to keep track of which player is on which base.

If a base is empty, it will still have a value of zero. But if someone is on base, that base will have that player's number. Each base can either be unoccupied (zero) or have a non-zero number (the player's uniform number). Shortly there will more to say about uniform numbers.

{$$\Large |baseball \rangle = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ \dots \\ a_n \end{pmatrix}$$}

{$$\Large |Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \\ 0 \\ 0 \\ \dots \\ a_n \end{pmatrix}$$}

Next is the count. I will use k for strikes and b for balls. K is the scorecard code for struck out. Generally check boxes are used for strikes and balls, but we will not try to imitate that. Also, scorecard scoring allows for extended strikes, which occur when foul fly balls are not fielded after there are already two strikes on the batter. In theory there could be many of these. I do not know what the record is, but you do not have to watch very much baseball to see cases of three or four extended strikes. I'm just going to pretend they do not happen for the time being. so strikes can be an integer from 0 to 2, and balls can be an integer from 0 to 3.

Why are strikes not a number between 0 and 3? Why did I say strikes are a number between 0 and 2? Because on the third strike, the at-bat is over, and either the game is over or there is a different batter at bat who has 0 strikes.

{$$\large |baseball \rangle = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ b \\ k \\ \dots \\ a_n \end{pmatrix}$$}

{$$\large |Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \dots \\ a_n \end{pmatrix}$$}

Now the final thing that I will add to this simplified model of a baseball state is the batting order. Every player on the roster has a unique number which is a one- or two-digit integer. There have been various schemes to make ranges of numbers correspond to positions either on the field or in batting order, but the history of this is sketchy, and the official rules only require that a player have a number and that it be unique on the roster.

Now the uniform numbers are a peculiar kind of number. They are in the form of non-negative integers, but you cannot sensibly do things with them that you could do with non-negative integers. You can, for example, add them or subtract them, but it is nonsense to do so. You cannot (now) add one to a player's number and get the number of the next player in the batting order. But for present purposes we can define these operations on a player's number:

1. A player's number is non-zero (to save the bases thing)
2. A player's number plus 0 is just the same as the player's number.
3. A player's number times 1 is the player's number.
4. A player's number times 0 is 0.

I am ignoring substitutions and designated hitters and such for now, since I would have to learn much more baseball than I want to and the state vector, which is already a healthy size, might get enormous.

The player at bat is Hu, and his uniform number is 60. I will call the number of the player at bat {$P_B$}. The number of the player in the on-deck circle will be called {$P_{OD}$}. And the numbers of the rest of the batting order, in order, will be {$P_3, P_4, \dots P_9$}

{$$|baseball \rangle = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ b \\ k \\ P_B \\ P_{OD} \\ P_3 \\ P_4 \\ P_5 \\ P_6 \\ P_7 \\ P_8 \\ P_9 \end{pmatrix}$$}

{$$|Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 60 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \end{pmatrix}$$}

Whew!

The dots have gone away because I am done. I have defined as much of the state vector for baseball as I care to and {$\left| Abbott \right\rangle$} is complete.

State Vector Costello

The pitcher is winding up when tragedy strikes. You realize you have not got a beer. So you go to the kitchen, you stumble over the cat, you discover there is only one cold beer left in the refrigerator, so you get a six-pack out of the pantry and put it in the freezer to chill, you take the cold beer. It is a bottle, and you do yourself damage trying to twist off the cap, give up, and find the bottle opener. You open your beer and return to the den. Now you find the game in a different state:

{$$\left|Costello \right\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 60 \\ 0 \\ 0 \\ 0 \\ 0 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \\ 60 \end{pmatrix}$$}

What does it mean?

The number of the player at bat is the number of the player who was in the on-deck circle. There are no balls or strikes, so you deduced he has not been pitched to yet. There is a runner on first and his uniform number is 60. The only conceivable interpretation is: Hu's on first!

Now, I am going to spoil both jokes.

First I spoil the pun on Abbott and Costello's comedy routine by pointing out that they were discussing the identity of the first baseman, not the identity of a runner on first base.

Second, I have to spoil the punchline of my article by saying now why baseball is not physics. From {$\left|Costello \right\rangle$} you cannot tell what has happened.

I have much more to say about how to manipulate the baseball state vector as I have defined it, and I will say it, but it clearly is fatally defective. Even though I know a previous state in this game, namely {$\left|Abbott \right\rangle,$} I have no idea how we got to {$\left|Costello \right\rangle$}.

I have seen lists of 23 ways a player can get on first base. Some of them seem to me to be degenerate, by which I mean something that happens under exactly the same rule being called by two different names. Some of these things, such as a fielder's choice cannot happen unless there is already a runner on base. But you do not have to be a lawyer to think of several ways of getting on first base. A player can make a base hit, can be hit by a ball, or can get a walk (base on balls). {$\left|Costello \right\rangle$} could have come about several ways, and we do not know which one. Information has been lost. This is a physical no-no. In physical systems information cannot be lost.

So when you reach a state, if you cannot trace back how you got to that state. In this case Hu could have been hit by a ball. He could have got a base hit. Maybe you were in the kitchen long enough for Hu to get a base on balls. Maybe you were there for a longer time than you thought. It is not especially likely, but it is possible for the pitcher to have intentionally walked the first batter, or for a nervous or cold pitcher to unintentionally throw four balls in his first four pitches, and so forth. Likewise, for the dropped third strike; you could have been gone long enough for that to happen.

We could patch up our state vector, adding such things as hits, errors, pitches thrown, and so forth and that would eliminate much of the ambiguity. We already have some ambiguity because although we are ready to keep track of the strikes and balls, we have not kept tract of the order in which they occurred.

Now, in the baseball state, we do not have to have an actual history. We have to know how within the rules of the game, we got to where we are. Now, in the state Costello, Hu is on first and the pitcher is about to deliver the first pitch to the second batter. And to the rules of baseball, it does not matter at all how Hu got to first. He is on first base and that is it.

We do not have to be so fine grained as to account for every pitch. Eventually, we want to get there, but we do not have to do that yet. And, really, have you tried reading the Major League Baseball rules?

The Base Hit Equivalent Operator

Given the Costello state, we know the Abbott state had to have occurred and occurred before the Costello state. We are not consider pinch runners or any kinds of substitutions, so if Hu is on first, what happened before is that he must have been at bat. He was not born on first base, and the only way he could get there within the rules is that he was at bat.

It is true we strongly suspect (we know) several things could have happened to get him to first base. And eventually we want to know about all those various intermediate states, and so forth, if for no other reason than to gather statistics in order to estimate the probability that he gets to first base in other at-bats. But for now, we (I) just do not know enough to figure out the possible intermediate states. So I am going to invent a hack.

Actually, this is looking more and more like physics after all.

I am going to call this hack the Base Hit Equivalent Operator, which I am going to designate as {$\mathbf{H_1}$}. It is a matrix and the subscript indicates that this is an operator for a single, or 1-base hit equivalent.

What we are trying to do here is find a matrix which we can use to multiply {$\left|Abbott \right\rangle$} so that the result is {$\left|Costello \right\rangle.$} But first let us see whether {$\mathbf{I_{19}},$} is in working order. We will test it by multiplying the general baseball state vector {$\left| baseball\right\rangle$} by it.

{$$\begin{bmatrix} 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1 \\\end{bmatrix} \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ b \\ k \\ P_B \\ P_{OD} \\ P_3 \\ P_4 \\ P_5 \\ P_6 \\ P_7 \\ P_8 \\ P_9 \end{pmatrix} = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ b \\ k \\ P_B \\ P_{OD} \\ P_3 \\ P_4 \\ P_5 \\ P_6 \\ P_7 \\ P_8 \\ P_9 \end{pmatrix}$$}

Well, we just lost all the mobile viewers and quite a number of the presbyopic ones too.

Just to recap how we multiply matrices, I'll work a few of these. A column vector, like {$\left| baseball\right\rangle$} a something by 1 matrix. In this case the something is 19. Matrices are Roman Catholic (thanks, Professor Matsen!), which means we mention the Rows first and the Columns second (that is RC in case anyone missed it. So {$\left| baseball\right\rangle$} is 19 by 1. In order to multiply matrices, the number of columns in the first one has to be equal to the number of rows in the second one. {$\mathbf{I_{19}},$} as identity matrices are wont, is square and 19 by 19. So we can multiply these matrices, whether we want to or not.

The resultant of matrix multiplication is a matrix with the same number of rows as the first matrix and the same number of columns as the second matrix. We are looking for 19 x1 here. The rule here is, an element in the resulting matrix is the sum of the products of the elements in the corresponding row of the first matrix with the elements of the corresponding column in the second matrix. So we go left to right in the first matrix an down in the second. Let us try a few.

{$$\begin{split} 1s_v \ + 0s_h \ + 0i \ + 0i_{TB} \ + 0o \ + 0r_1 \ + 0r_2 \ 0r_3 \ + 0b \ + 0k +& \cr 0P_B \ + 0P_{OD} \ + 0P_3 \ + 0P_4 \ + 0P_5 \ + 0P_6 \ + 0P_7 \ + 0P_8 \ + 0P_9 &= s_v \end{split}$$}

That went swimmingly, except perhaps when we got to the Ps because, you know, they are not real numbers and they are not even real integers. But we have a rule for multiplying whatever they are by zero. The result is zero, which is a number we can add.

We can do a couple of more.

{$$\begin{split} 0s_v \ + 1s_h \ + 0i \ + 0i_{TB} \ + 0o \ + 0r_1 \ + 0r_2 \ 0r_3 \ + 0b \ + 0k \ +& \cr 0P_B \ + 0P_{OD} \ + 0P_3 \ + 0P_4 \ + 0P_5 \ + 0P_6 \ + 0P_7 \ + 0P_8 \ + 0P_9 &= s_h \cr 0s_v \ + 0s_h \ + 1i \ + 0i_{TB} \ + 0o \ + 0r_1 \ + 0r_2 \ 0r_3 \ + 0b \ + 0k \ +& \cr 0P_B \ + 0P_{OD} \ + 0P_3 \ + 0P_4 \ + 0P_5 \ + 0P_6 \ + 0P_7 \ + 0P_8 \ + 0P_9 &= i \end{split}$$}

As we work through these we come to a place where we have to start dealing with Ps that have a non-zero coefficient. Here we remember we a rule for multiplying by 1 and a rule for adding zero.

So the question is what big matrix, 19 x 19 will change {$\left|Abbott \right\rangle$} to {$\left|Costello \right\rangle$}

{$$|Abbott \rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 60 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \end{pmatrix} \Longrightarrow \left|Costello \right\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 60 \\ 0 \\ 0 \\ 0 \\ 0 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \\ 60 \end{pmatrix}$$}

Basically, Hu's number 60 has to fill the spot at first base. All of the batting order has to move up one slot, and Hu's number (again) has to go to the bottom of the order.

Rotating the batting order seems the hardest thing, so we take it first.

Look at the last nine rows of the identity matrix. The first ten columns of all those rows are useless. They are zeros, and they can do nothing to affect the result of the matrix.

{$$\begin{bmatrix} 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 }\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0}\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0 }\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0 }\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0}\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0}\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0 }\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0 }\\ \style{background-color:yellow}{ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ }\style{color:red}{0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1 }\\\end{bmatrix} \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ b \\ k \\ \style{color:red}{ P_B \\ P_{OD} \\ P_3 \\ P_4 \\ P_5 \\ P_6 \\ P_7 \\ P_8 \\ P_9 }\end{pmatrix} = \begin{pmatrix} s_v \\ s_h \\ i \\ i_{TB} \\ o \\ r_1 \\ r_2 \\ r_3 \\ b \\ k \\ \style{color:red}{P_B \\ P_{OD} \\ P_3 \\ P_4 \\ P_5 \\ P_6 \\ P_7 \\ P_8 \\ P_9 }\end{pmatrix}$$}

Only the last 9 columns of the last nine rows of the big matrix and only the last nine rows of the baseball state vector have anything to do with producing the last nine rows of the resultant vector.

That being the case, we can peel those parts out to study the operations on the last nine rows of the Abbott state vector, to produce the last nine rows of the Costello vector.

{$$\begin{bmatrix} ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ?\ ? \end{bmatrix} \begin{pmatrix} 60 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \end{pmatrix} = \begin{pmatrix} 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \\ 60 \end{pmatrix}$$}

We are just copying numbers or not copying numbers, so all we need to fill in the unknown matrix is 1s and 0s. This is a good thing because only multiplication by 1 and multiplication by 0 are defined for uniform numbers. Moreover, only addition with 0 is defined for uniform numbers so each row of the big matrix had better have only one 1 and everything else on the row must be zero.

Looking at the first row, we want it to be eighteen. That is the number of the player who was on deck in the Abbott state. The first number in the row cannot be 1. That would just copy the 1st number of the partial Abbott state, and the rest of the row would have to be 0s because we cannot do anything else to a uniform number. But if the second element in the top row is 1, that copies 19 to the first row of the result, and that is what we want. This is pretty simple stuff, so we find:

{$$\begin{bmatrix} 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1 \\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \end{bmatrix} \begin{pmatrix} 60 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \end{pmatrix} = \begin{pmatrix} 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \\ 60 \end{pmatrix}$$}

So we now know how to rotate the batting order. We can paste this back over the identity matrix in the lower right corner. Since everything else stays the same, except for putting Hu on first, we only have one modification to make to the pasted up identity matrix, namely we have to copy Hu's number on to first base. I think this does it:

{$$\begin{bmatrix} 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1 \\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 0 \\\end{bmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 60 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ 0\\ 60 \\ 0 \\ 0 \\ 0 \\ 0 \\ 18 \\ 30 \\ 9 \\ 24 \\ 14 \\ 27 \\ 8 \\ 11 \\ 60 \end{pmatrix}$$}

{$$\mathbf{H_1} \left|Abbott \right\rangle = \left|Costello \right\rangle$$}

Okay. That was just the first batter of the game in the case that he got a base hit equivalent. And that is as far as I am going for now. There should be plenty to think about. Suppose Hu was put out instead. What would an operator look like that would increment the number of outs and move Hu to the bottom of the batting order? How would you handle players left on base at the end of the inning regarding batting order?

Homework

1. Verify that the 9x9 batter rotation matrix works.
2. Verity that {$\mathbf{H_1} \left|Abbott \right\rangle = \left|Abbott \right\rangle$} is true.
3. find the reverse operator {$\overleftarrow{\mathbf{H_1}}$} which produces {$\left|Abbott \right\rangle$} when it is applied to {$\left|Costello \right\rangle.$}
4. Find {$\mathbf{H_2}$} (The operator if Hu had hit a double instead.)
5. Assuming you have an 18x1 vector which is the batting order of both teams. Find an 18x18 operator that will move the fielding team from the bottom to the top of the batting order vector and move the number of the batter who just struck out to the bottom of his teams batting order. Assume there were no players left on base.
6. '''Extra Credit: " Explain the infield fly rule in 25 words or less. (Physics is not looking so hard now, is it?)

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This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.