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153
Advanced/Lambda Calculus/parts/00 intro.tex
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@ -67,7 +67,7 @@ $$
$$
This is very similar to the usual way we call functions: we usually write $f(5)$. \par
Above, we define our function $f$ \say{in-line} using lambda notation, \par
and we ommit the parentheses around 5 for the sake of simpler notation.
and we omit the parentheses around 5 for the sake of simpler notation.
\vspace{2mm}
@ -147,8 +147,13 @@ The first \say{pure} functions we'll define are $I$ and $M$:
\item $I = \lm x . x$
\item $M = \lm x . x x$
\end{itemize}
Both $I$ and $M$ take one function ($x$) as an input. \par
$I$ does nothing, it just returns $x$. \par
$M$ is a bit more interesting: it applies the function $x$ on a copy of itself.
Note that $I$ and $M$ don't have a meaning on their own. They are not formal functions. \par
\vspace{4mm}
Also, note that $I$ and $M$ don't have a meaning on their own. They are not formal functions. \par
Rather, they are abbreviations that say \say{write $\lm x.x$ whenever you see $I$.}
@ -163,11 +168,38 @@ Reduce the following expressions. \par
\begin{itemize}[itemsep=2mm]
\item $I~I$
\item $M~I$
\item $(I~I)~I$
\item $\Bigl(~ \lm a .(a~(a~a)) ~\Bigr) ~ I$
\item $\Bigl(~(\lm a . (\lm b . a)) ~ M~\Bigr) ~ I$
\end{itemize}
\begin{examplesolution}
\textbf{Solution for $(I~I)$:}\par
Recall that $I = \lm x.x$. First, we rewrite the left $I$ to get $(\lm x . x )~I$. \par
Applying this function by replacing $x$ with $I$, we get $I$:
$$
I ~ I =
(\lm x . \tzm{b}x )~\tzm{a}I =
I
\begin{tikzpicture}[
overlay,
remember picture,
out=-90,
in=-90,
distance=0.5cm
]
\draw[->,gray,shorten >=5pt,shorten <=3pt]
(a.center) to (b.east);
\end{tikzpicture}
$$
\vspace{0.5mm}
So, $I~I$ reduces to itself. This makes sense, since the identity
function doesn't change its input!
\end{examplesolution}
\vfill
@ -204,8 +236,8 @@ Remember that lambda calculus is left-associative.
\generic{Currying:}
In lambda calculus, functions are only allowed to take one argument.
However, we can emulate multivariable functions through \textit{currying}\footnotemark{}\hspace{-1ex}.
In lambda calculus, functions are only allowed to take one argument. \par
If we want multivariable functions, we'll have to emulate them through \textit{currying}\footnotemark{}\hspace{-1ex}.
\footnotetext{After Haskell Brooks Curry\footnotemark{}\hspace{-1ex}, a logician that contributed to the theory of functional computation.}
\footnotetext{
@ -216,9 +248,9 @@ However, we can emulate multivariable functions through \textit{currying}\footno
\vspace{1ex}
The idea behind currying is fairly simple: we make functions that return functions. \par
Consider the expression $A$ below. As before, it takes one input $f$ and returns the function $\lm a. f(f(a))$ (underlined).
Consider the expression $A$ below. It takes one input $f$ and returns the function $\lm a. f~(f~a)$ (underlined).
$$
A = \lm f .(\tzm{a} ~ \lm a . f(f(a)) ~ \tzm{b})
A = \lm f .(\tzm{a} ~ \lm a . f~(f~a) ~ \tzm{b})
\begin{tikzpicture}[
overlay,
remember picture
@ -230,11 +262,11 @@ A = \lm f .(\tzm{a} ~ \lm a . f(f(a)) ~ \tzm{b})
\end{tikzpicture}
$$
When we evaluate $A$ with one input, it constructs a new function with the argument we give it. \par
For example, let's apply $A$ to an arbirary function $N$:
For example, let's apply $A$ to an arbirary function $n$:
$$
A~N =
\lm \tzm{b}f.(~\lm a.f(f(a))~)~\tzmr{a}N =
\lm a.N(N(a))
A~n =
\lm f.(~\lm a.\tzm{b}f~(\tzm{c}f~a)~)~\tzmr{a}n =
\lm a.n~(n~a)
\begin{tikzpicture}[
overlay,
remember picture,
@ -244,13 +276,15 @@ $$
]
\draw[->,gray,shorten >=5pt,shorten <=3pt]
(a.center) to (b.center);
\draw[->,gray,shorten >=5pt,shorten <=3pt]
(a.center) to (c.center);
\end{tikzpicture}
$$
Above, $A$ replaced every $f$ in its definition with an $N$. \par
Above, $A$ replaced every $f$ in its definition with an $n$. \par
You can think of $A$ as a \say{factory} that constructs functions using the inputs we provide.
\problem{}<firstcardinal>
Let $C = \lm f. (\lm g. \lm x. (g(f(x))))$. What does $B$ do? \par
Let $C = \lm f. \Bigl[\lm g. \Bigl( \lm x. [~ g(f(x)) ~] \Bigr)\Bigr]$. What does $C$ do? \par
Evaluate $(C~a~b~x)$ for arbitary expressions $a$ and $b$. \par
\hint{Place parentheses first. Remember, function application is left-associative.}
@ -269,16 +303,16 @@ If we have an expression with repeated function definitions, we'll combine their
\vspace{1ex}
For example, $A = \lm f . (\lm a . f(f(a)))$ will become $A = \lm fa . f(f(a))$
For example, $A = \lm f .[ \lm a . f(f(a))]$ will become $A = \lm fa . f(f(a))$
\problem{}
Rewrite $C = \lm f. (\lm g. \lm x. (g(f(x))))$ from \ref{firstcardinal} using this shorthand.
Rewrite $C = \lm f.\lm g.\lm x. (g(f(x)))$ from \ref{firstcardinal} using this shorthand.
\vspace{25mm}
Remember that this is only notation. \textbf{Curried functions are not multivariable functions, they are simply shorthand!}
Any function presented with this notation must still be evaluated one variable at a time, just like an un-curried function.
Substituting all curried variables at once may cause errors, as we'll see below.
Substituting all curried variables at once will cause errors.
\pagebreak
@ -311,63 +345,70 @@ Substituting all curried variables at once may cause errors, as we'll see below.
We say two functions are \textit{equivalent} if they differ only by the names of their variables:
$I = \lm a.a = \lm b.b = \lm \heartsuit . \heartsuit = ...$ \par
\vspace{2mm}
\begin{instructornote}
The idea behind this is very similar to the idea behind \say{equivalent groups} in group theory: \par
we do not care which symbols a certain group or function uses, we care about their \textit{structure}. \par
The idea behind this is very similar to the idea behind \say{equivalent groups} in group theory: \par
we do not care which symbols a certain group or function uses, we care about their \textit{structure}. \par
\vspace{2mm}
\vspace{2mm}
If we have two groups with different elements with the same multiplication table, we look at them as identical groups.
The same is true of lambda functions: two lambda functions with different variable names that behave in the same way are identical.
If we have two groups with different elements with the same multiplication table, we look at them as identical groups.
The same is true of lambda functions: two lambda functions with different variable names that behave in the same way are identical.
\end{instructornote}
\vspace{4mm}
%\iftrue
\iffalse
\generic{$\alpha$-Conversion:}
There is one more operation we need to discuss. Those of you that have worked with any programming language may
find that this section sounds like something you've seen before.
\generic{$\alpha$-Conversion:}
There is one more operation we need to discuss. Those of you that have worked with any programming language may
find that this section sounds like something you've seen before.
\vspace{2mm}
\vspace{2mm}
Variables inside functions are \say{scoped.} We must take care to keep separate variables separate.
Variables inside functions are \say{scoped.} We must take care to keep separate variables separate.
For example, take the functions \par
$A = \lm a b . a$ \par
$B = \lm b . b$
For example, take the functions \par
$A = \lm a b . a$ \par
$B = \lm b . b$
\vspace{2ex}
\vspace{2ex}
We could say that $(A~B) = \lm b . (\lm b . b)$, and therefore
$$
((A~B)~I)
= (~ (\lm \tzm{b}b . (\lm b . b))~\tzmr{a}I ~)
= \lm I . I
\begin{tikzpicture}[
overlay,
remember picture,
out=225,
in=315,
distance=0.5cm
]
\draw[->,gray,shorten >=5pt,shorten <=3pt]
(a.center) to (b.center);
\end{tikzpicture}
$$
We could say that $(A~B) = \lm b . (\lm b . b)$, and therefore
$$
((A~B)~I)
= (~ (\lm \tzm{b}b . (\lm b . b))~\tzmr{a}I ~)
= \lm I . I
\begin{tikzpicture}[
overlay,
remember picture,
out=225,
in=315,
distance=0.5cm
]
\draw[->,gray,shorten >=5pt,shorten <=3pt]
(a.center) to (b.center);
\end{tikzpicture}
$$
Which is, of course, incorrect. $\lm I . I$ is not a valid function.
This problem arises because both $A$ and $B$ use the input $b$.
However, each $b$ is \say{bound} to a different function:
One $b$ is bound to $A$, and the other to $B$. They are therefore distinct.
Which is, of course, incorrect. $\lm I . I$ is not a valid function.
Both $A$ and $B$ use the input $b$. However, each $b$ is \say{bound} to a different function:
One $b$ is bound to $A$, and the other to $B$. They are therefore distinct.
\vspace{2ex}
\vspace{2ex}
Let's rewrite $B$ as $\lm b_1 . b_1$ and try again: $(A~B) = \lm b . ( \lm b_1 . b_1) = \lm bb_1 . b_1$ \par
Now, we correctly find that $(A~B~I) = (\lm bc . c)~I = \lm c . c = B = I$.
Let's rewrite $B$ as $\lm b_1 . b_1$ and try again: $(A~B) = \lm b . ( \lm b_1 . b_1) = \lm bb_1 . b_1$ \par
Now, we correctly find that $(A~B~I) = (\lm bc . c)~I = \lm c . c = B = I$.
\fi
\problem{}
Let $Q = \lm abc.b$ \par
Reduce $(Q~a~c~b)$.
Let $Q = \lm abc.b$. Reduce $(Q~a~c~b)$. \par
\hint{
You may want to rename a few variables. \\
The $a,b,c$ in $Q$ are different than the $a,b,c$ in the expression!
}
\begin{solution}
I'll rewrite $(Q~a~c~b)$ as $(Q~a_1~c_1~b_1)$: