Hey everyone, I'm starting a new article series here on catonmat that will be published on Fridays. It's called "**Visual Math Friday**." The goal of this article series is to assemble all the wonderful mathematical proofs and identities that can be proven in a single picture or drawing. I have been collecting them for years from various books, math journals, websites and video lectures and have collected several hundred of them.

If you have been following my blog, you'll have noticed that I love to explain things. I'll use the same writing style in this series as well. I'll post the visual proof together with my explanation. In the explanations I'll walk you through the proof, suggesting how to think about the visual image. But it's best if you try to understand the proof without reading my explanation and only then read it.

The proofs come from algebra, number theory, geometry, calculus, complex analysis, topology and various other fields of mathematics. I'll do one proof a week and will try to alternate between really easy, moderate and difficult proofs every week so that readers with various mathematical backgrounds can enjoy the series.

I'll start with a really easy integer sum proof this week. Without further ado, here it is.

Now try to figure what it proves without reading further! The colors play an important role in the visual proofs, as do the shape of the visual drawing, the shapes on it and their arrangement!

## The Proof

This picture proves that the sum of first n odd numbers is equal to n². In other words:

1 + 3 + 5 + 7 + 9 + 11 + ... + (2n-1) = n².

## Explanation

Imagine that the blue and green dots are marbles. Now let's look at the lower left corner. One blue marble is located there. Next if we follow the diagonal to the upper right corner, we see that the 1 blue marble is kind-of wrapped in 3 green marbles, then they in turn are wrapped in 5 blue marbles again, then those are wrapped in 7 green ones, which in turn are wrapped in 9 blue ones, then 11 green ones, and finally 13 ones.

This suggests that we could look at the sum 1 + 3 + 5 + 7 + 9 + 11 + 13. Well what is it? It's 49. Also what does this sum involve? It involves only odd numbers. Interesting.

Now let's count how many marbles we have on each side. 7 on one side and 7 on the other. And what is 7x7? It's 49! And how many numbers did we sum together? Well, seven - 1, 3, 5, 7, 9, 11, 13. Interesting.

Could it be that the sum of the first n odd numbers is equal to n²? Or in other words, is 1 + 3 + 5 + 7 + ... + (2n-1) equal to n²?

Let's try to prove this hypothesis by going one number further and adding 15 green marbles - 13 filled marbles and 2 semi-filled marbles:

Adding 15 marbles created a square of marbles again - now we have 64 marbles, 8 on each side.

The 2 semi-filled marbles are very important here. Each time we go further from n odd marbles to n+1 odd marbles, the difference changes by 2. This difference of two marbles can always be put where the semi-filled ones are in this picture.

This actually concludes the proof. We showed that the proof worked for several marbles, and showed how it still works if we went one step further. Therefore it works for all future steps. This method of proof is called mathematical induction and it's a very powerful and important tool in mathematics for proving sums like these.

## Before we go

Can you figure out the two proofs hidden in the logo of this post?

**Let me know in the comments if you can!**

## Get ahead of me

By the way, if you wish to get ahead of me in this series, you can get these three amazing books on visual proofs:

- Proofs Without Words I: Exercises in Visual Thinking.
- Proofs Without Words II: More Exercises in Visual Thinking.
- Math Made Visual.

Here are these books on Amazon.com:

**51**Comments March 03, 2010

# Vim Plugins You Should Know About, Part VII: ragtag.vim (formerly allml.vim)

This is the seventh post in the article series "**Vim Plugins You Should Know About**". This time I am going to introduce you to a plugin called "**ragtag.vim**". A month ago it was still known as "**allml.vim**" but now it has been renamed to **ragtag.vim**.

The best parts of RagTag are mappings for editing HTML tags. It has a mapping for quickly closing open HTML tags, a mapping for quickly turning the typed word into a pair of open/close HTML tags, several mappings for inserting HTML doctype, linking to CSS stylesheets, loading JavaScript `<script src="...">...</script>`

and it includes mappings for wrapping the typed text in a pair of `<?php ... ?>`

tags for PHP, or `<% ... %>`

for ASP or eRuby, and `{% .. %}`

for Django.

RagTag is written by **Tim Pope**. He's the master of Vim plugin programming. I have already written about two of his plugins - surround.vim and repeat.vim and more articles about his plugins are coming!

## Previous articles in the series.

- Part I: surround.vim - find and edit surrounding parens, quotes, html tags, etc.
- Part II: repeat.vim - repeat the previous surrounding command.
- Part III: matchit.vim - extends the functionality of % key.
- Part IV: snipmate.vim - the best snippet plugin for vim
- Part V: a.vim - quickly change between source and header files (.c/.h).
- Part VI: nerd_tree.vim - awesome filesystem exploration plugin.

Here are examples of using the RagTag plugin.

## Quickly closing an open HTML tag.

Suppose you have typed `<div>`

and you want to close it without typing the whole closing tag `</div>`

yourself.

The quick way to do it with RagTag is to press

. This mapping automatically closes the last open HTML tag.**CTRL+X /**

**Extra tip**: If you didn't have this plugin you could quickly close the tag by typing `</`

and pressing

. The default Vim mapping **CTRL+X CTRL+O**

guesses the item that may come after the cursor. In case of an open HTML tag it's the close tag.**CTRL+X CTRL+O**

## Creating a pair of open/close HTML tags from a word.

Suppose you want to quickly create a pair of `<div></div>`

and place the cursor between the two tags.

The quick way to do it with RagTag is to type `div`

and press

. This mapping takes the typed word and creates a pair of HTML tags, one closing tag and one open tag on the same line.**CTRL+X SPACE**

However, if you wish to create open/close tag pair separated by a newline, type

.**CTRL+X ENTER**

Here is an example, if you just typed `div`

and then press `CTRL+X ENTER`

it will produce the following output:

<div>|</div>

indicates the position of cursor.**|**

## Insert HTML doctype.

If you type

RagTag will display a list of HTML doctypes to choose from. Defaults to HTML 4.01 Strict.**CTRL+X !**

This mapping is not that useful, given that I already introduced snipmate.vim plugin for creating snippets. Using snipmate.vim you can create a snippet "h" that would insert the whole HTML structure, including doctype, html, body, title, meta tags, etc.

## Link to a CSS stylesheet.

Typing

inserts the snippet for linking to a CSS stylesheet.**CTRL+X @**

<link rel="stylesheet" type="text/css" href="/stylesheets/|.css">

This is again not that useful, given that we have snipmate.vim.

The mapping is easy to remember because `@`

is used for importing in CSS.

## Insert meta content-type tag.

Typing

inserts the HTML meta tag for document's content type and encoding.**CTRL+X #**

<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">

The charset depends on document's charset. If it's utf-8, the mapping will set the charset to utf-8 in the meta tag.

## Load JavaScript document.

Typing

links to a JavaScript file.**CTRL+X $**

<script type="text/javascript" src="/javascripts/|.js">

The mapping is easy to remember because `$`

is a valid char in identifiers in many languages.

## Wrap the typed text in PHP, Django, eRuby template tags.

There are several different mappings for wrapping text in template tags. It's best to summarize them in the following table. The table assumes you had just typed "foo" and you are editing an eRuby document:

Mapping Result --------- ----------- CTRL+X = foo<%=|%> CTRL+X + <%= foo|%> CTRL+X - foo<%|%> CTRL+X _ <% foo|%> CTRL+X ' foo<%#|%> CTRL+X " <%# foo|%>

What this table shows us is, for example, that if you have typed "foo" and press

, the plugin will wrap "foo" in **CTRL+X _**`<% %>`

tags and place the cursor after "foo".

## Summary of all the RagTag mappings.

CTRL+X / Close the last open HTML tag CTRL+X SPACE Create open/close HTML tags from the typed word CTRL+X CR The same as CTRL+X SPACE but puts a newspace in between CTRL+X ! Insert HTML doctype CTRL+X @ Insert CSS stylesheet CTRL+X # Insert meta content-type meta tag CTRL+X $ Load JavaScript document For the following mappings, suppose that you have typed "foo". Mapping Result --------- ----------- CTRL+X = foo<%=|%> CTRL+X + <%= foo|%> CTRL+X - foo<%|%> CTRL+X _ <% foo|%> CTRL+X ' foo<%#|%> CTRL+X " <%# foo|%>

## How to install ragtag.vim?

To get the latest version:

- 1. Download ragtag.zip.
- 2. Extract ragtag.zip to
`~/.vim`

(on Unix/Linux) or`~\vimfiles`

(on Windows). - 3. Run
`:helptags ~/.vim/doc`

(on Unix/Linux) or`:helptags ~\vimfiles\doc`

(on Windows) to rebuild the tags file (so that you can read help`:help ragtag`

.) - 4. Restart Vim or source ragtag.vim by typing
`:so ~/.vim/plugin/ragtag.vim`

(on Unix/Linux) or`:so ~\vimfiles\plugin\ragtag.vim`

## Have Fun!

Have fun closing those HTML tags and until next time!

I was reading Knuth's "Selected Papers on Computer Science" and in chapter 13 "The IBM 650: An Appreciation from the Field" he had included a photo of a young himself from 1958 at his first computer.

Young Donald Knuth, age 20, at his first IBM 650 computer in 1958.

Don Knuth's first computer, as the chapter's title suggests, was the IBM 650. At that time Knuth had a part-time job helping the statisticians at Case Institute of Technology to draw graphs and do keypunching. Soon a strange new machine was installed across the hall, that the student newspaper called a "giant brain". It was actually an IBM 650.

One afternoon someone at the institute explained some of the machine's internal code to a bunch of freshmen, including Knuth. It all sounded mysterious to him, but it seemed to make a bit of sense, so he got ahold of some manuals. His first chance to try the machine came a few weeks later, when one of the upperclassmen at the fraternity needed to know the five roots of a particular fifth degree equation. Knuth decided to compute the roots using the 650. This was his first program on the 650.

Later Knuth realized how lucky he was to have had such a good first encounter with computers. The polynomial problem was well matched to his mathematical knowledge and interests, and he had a chance for hands-on experience, pushing buttons on the machine and seeing it punch the cards containing the answers.

Knuth's first large program was a tic-tac-toe that learned to play by remembering the relative desirability or undesirability of each position that it had ever encountered. The hardest part, as Knuth writes, was figuring out how to keep one digit of memory for each possible configuration of the board. The machine had just 2000 words of memory, each 10 digits long, plus a sign bit.

Next Knuth proceeded to writing a program that would find prime factors. The idea was that a person could set up any 10-digit number in the console switched and start his program, which would punch the corresponding prime factors on a card and stop. Knuth spent several weeks on this problem, rewriting his program several times. The final program took 11 minutes to determine that the number 9999999967 was prime.

The next major Knuth's program was the improvement of SOAP (Symbolic Optimal Assembly Program) and SOAP II assembly languages. He wrote SOAP III and learned about "creeping featurism," where each of his friends would suggest different things they wanted in an assembler. The program used all 2000 words of memory. Knuth doesn't think he could have gotten by with only 1999 words, because he had spent considerable time finding every last bit of space by using terrible tricks, such as using a single instruction on a specific address that would cause 4 side effects.

Along the way Knuth wrote a lot fun programs. One of the competitions between the students was to do as much as possible with programs that would fit on a single punch card - which had room for only eight instructions. One of the unsolved problems was to take the 10-digit number on the console and to reverse its digits from left to right, then display the answer and stop; nobody could figure out how to do this on a signle card. But one day Knuth proudly marched up to the machine and made a demonstration: he read in a card, then dialled the number 0123456789 on the console, and started the machine. Sure enough, it stopped, displaying the number 9876543210. Everybody applauded. He didn't explain until later that his card would display the number 9876543210 regardless of what number appeared on the console switches. :)

But there is more to the story. One day the IBM 650 machine got an extra set of console switches, that were called register 8004 (top row of switches in the picture). It turned out that nine instructions on an extended 650 were sufficient to reverse the digits of a number, and the ninth instruction could be put into the switches. Therefore he was able to solve the problem without cheating.

Knuth was very close with IBM 650. One night he missed a date with his wife-to-be, because he was so engrossed in debugging that he had forgotten all about the time.

The 650 provided Knuth with solid instruction in the art of programming. It was directly related to the topics of the first two technical articles he ever submitted for publication. Therefore it's not surprising at all that he decided to dedicate The Art of Computer Programming books to the IBM 650 computer. "This series of books is affectionately dedicated to the Type 650 computer once installed at Case Institute of Technology, in remembrance of many pleasant evenings."

**The End.**

If you liked this story, you may want to get Knuth's book "Selected Papers on Computer Science". The collection focuses on Knuth's publications that are addressed primarily to a general audience than to specialists.

**46**Comments February 24, 2010

# MIT Linear Algebra, Lecture 6: Column Space and Null Space

This is the sixth post in an article series about MIT's course "Linear Algebra". In this post I will review lecture six on **column spaces** and **null spaces** of matrices. The lecture first reviews vector spaces and subspaces and then looks at the result of intersect and union of vector subspaces, and finds out when Ax=b and Ax=0 can be solved.

Here is a list of the previous posts in this article series:

- Lecture 1: Geometry of Linear Equations
- Lecture 2: Elimination with Matrices
- Lecture 3: Matrix Multiplication and Inverse Matrices
- Lecture 4: A=LU Factorization
- Lecture 5: Vector Spaces and Subspaces

## Lecture 6: Column Space and Null Space

Lecture six starts with a reminder of what the vector space requirements are. If vectors **v** and **w** are in the space, then the result of adding them and multiplying them by a number stays in the space. In other words, all linear combinations of **v** and **w** stay in the space.

For example, the 3-dimensional space **R**^{3} is a vector space. You can take any two vectors in it, add them, and multiply them by a number and they will still be in the same **R**^{3} space.

Next, the lecture reminds subspaces. A subspace of some space is a set of vectors (including the **0** vector) that satisfies the same two requirements. If vectors **v** and **w** are in the subspace, then all linear combinations of **v** and **w** are in the subspace.

For example, some subspaces of **R**^{3} are:

- Any plane P through the origin (0, 0, 0).
- Any line L through the origin (0, 0, 0).

See the previous, lecture 5, on more examples of spaces on subspaces.

Now suppose we have two subspaces of **R**^{3} - plane P and line L. Is the union P∪L a subspace? **No.** Because if we take some vector in P and some vector in L and add them together, we go outside of P and L and that does not satisfy the requirements of a subspace.

What about intersection P∩L? Is P∩L a subspace? **Yes.** Because it's either the zero vector or L.

In general, given two subspaces S and T, union S∪T is a not a subspace and intersection S∩T is a subspace.

The lecture now turns to **column spaces of matrices**. The notation for a column space of a matrix A is C(A).

For example, given this matrix,

The column space C(A) is all the linear combination of the first (1, 2, 3, 4), the second (1, 1, 1, 1) and the third column (2, 3, 4, 5). That is, C(A) = { a·(1, 2, 3, 4) + b·(1, 1, 1, 1) + c·(2, 3, 4, 5) }. In general, **the column space C(A) contains all the linear combinations of columns of A**.

A thing to note here is that C(A) is a subspace of **R**^{4} because the vectors contain 4 components.

Now the key question, does C(A) fill the whole **R**^{4}? **No.** Because there are only three columns (to fill the whole **R**^{4} we would need exactly 4 columns) and also because the third column (2, 3, 4, 5) is a sum of first column (1, 2, 3, 4) and second column (1, 1, 1, 1).

From this question follows another question, the most important question in the lecture - Does Ax=b have a solution for every right-hand side vector b? **No.** Because the columns are not linearly independent (the third can be expressed as first+second)! Therefore the column space C(A) is actually a two-dimensional subspace of **R**^{4}.

Another important question arises - For which right-hand sides b can this system be solved? The answer is: **Ax=b can be solved if and only if b is in the column space C(A)**! It's because Ax is a combination of columns of A. If b is not in this combination, then there is simply no way we can express it as a combination.

That's why we are interested in column spaces of matrices. They show when can systems of equation Ax=b be solved.

Now the lecture turns to **null spaces of matrices**. The notation for a null space of a matrix A is N(A).

Let's keep the same matrix A:

The null space N(A) contains something completely different than C(N). **N(A) contains all solutions x's that solve Ax=0**. In this example, N(A) is a subspace of **R**^{3}.

Let's find the null space of A. We need to find all x's that solve Ax=0. The first one, obviously, is x = (0, 0, 0). Another one is x = (1, 1, -1). In general all x's (c, c, -c) solve Ax=0. The vector (c, c, -c) can be rewritten as c·(1, 1, -1).

Note that the null space c·(1, 1, -1) is a line in **R**^{3}.

The lecture ends with a proof that solutions x to Ax=0 always give a subspace. The first thing to show in the proof is that if x is a solution and x' is a solution, then their sum x + x' is a solution:

We need to show that if Ax=0 and Ax' = 0 then A(x + x') = 0. This is very simple. Matrix multiplication allows to separate A(x + x') into Ax + Ax' = 0. But Ax=0 and Ax'=0. Therefore 0 + 0 = 0.

The second thing to show is that if x is a solution, then c·x is a solution:

We need to show that if Ax=0 then A(c·x)=0. This is again very simple. Matrix multiplication allows to bring c from A(c·x) outside c·A(x) = c·0 = 0.

That's it. We have proven that solutions x to Ax=0 always form a subspace.

Here is the video of the sixth lecture:

Direct link: http://www.youtube.com/watch?v=8o5Cmfpeo6g

Topics covered in lecture six:

- [01:00] Vector space requirements.
- [02:10] Example of spaces R
^{3}. - [02:40] Subspaces of spaces.
- [03:00] A plane P is a subspace of R
^{3}. - [03:50] A line L is a subspace of R
^{3}. - [04:40] Union of P and L.
- [07:30] Intersection of P and L.
- [09:00] Intersection of two subspaces S and T.
- [11:50] The column space C(A) of a matrix A.
- [16:20] Does Ax=b have a solution for every b?
- [19:45] Which b's allow Ax=b to be solved?
- [23:50] Can solve Ax=b exactly when b is in C(A).
- [28:50] The null space N(A) of a matrix A.
- [37:00] Why is the null space a vector space?
- [37:30] A proof that the null space is always a vector space.
- [41:50] Do the solutions to Ax=b form a subspace?

Here are my notes of lecture six:

Have fun with this lecture! The next post is going to be about general theory of solving equations Ax=0, pivot variables and special solutions.

PS. This course is taught from Introduction to Linear Algebra textbook. Get it here:

In this post I'll derive the Y-combinator and explain all the steps taken. The key to understanding the Y-combinator is knowing precisely what it does. I am sure many people never understand Y-combinator because of this very reason, or because the introduction is too lengthy to comprehend. Therefore I'll give the shortest possible explanation of Y-combinator before I derive it so that you know what the end result should be.

The Y-combinator allows an anonymous function to call itself, that is, **it allows anonymous function recursion**. Since anonymous functions don't have names, they can't refer to themselves easily. The Y-combinator is a clever construct helps them to refer to themselves. That's it. That's all the Y-combinator does. Remember that.

I'll use the Scheme programming language to derive the Y-combinator. My derivation is based on the one in "The Little Schemer" book. I took the examples from the book, filled in the missing steps and explained the steps in more details.

Here it is.

Suppose we want to write the `length`

function, which, given a list, returns the number of elements in it. It's really easy if we can give the function a name:

```
(define length
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
```

But now suppose that we can't give names - we can only use anonymous functions:

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list)))))))
```

Suddenly there is no way this anonymous function can refer to itself. What do we put in ** ???**? We can't refer to the name

`length`

anymore because it's an anonymous function, which doesn't have any name. One thing we can try is to put the function itself in the place of **:**

`???`

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list) ; the
(cond ;
((null? list) 0) ; function
(else ;
(add1 (??? (cdr list)))))) ; itself
(cdr list))))))
```

This is not much better, we are still left with ** ???**. But there is a bright side to it, this function can determine lengths of lists with one element and no elements. Let's try inserting the same anonymous function in place of

**again:**

`???`

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
(cdr list))))))
(cdr list))))))
```

Now this function can determine lengths of lists with 0, 1 and 2 elements. If we continue this way, we can construct the `length`

function for lists with a certain number of elements. But this is not what we want, we the real `length`

function that works for lists with any number of elements.

As we all know, repeating code is not a good thing. Let's try to factor out the repetitions and rewrite the original anonymous function slightly. Instead of leaving `???`

in the code, let's pass it to an anonymous function via an argument called `length`

.

```
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
???)
```

Notice that this code invokes the first `lambda (length)`

function and passes it `???`

as the `length`

argument. This function in turn returns the original anonymous function:

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
```

Now let's try constructing the previous two length functions that work for lists with a maximum of one and two elements. First the function for lists with a maximum of one element:

```
((lambda (f)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (f (cdr list)))))))
((lambda (g)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (g (cdr list)))))))
???))
```

Here `???`

first gets passed to `lambda (g)`

, which returns the original anonymous function, which then gets passed to `lambda (f)`

, which in turn returns the function that works for lists with one element:

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
(cdr list))))))
```

Similarly, the following code returns a function that works for lists with a maximum of two elements:

```
((lambda (f)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (f (cdr list)))))))
((lambda (g)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (g (cdr list)))))))
((lambda (h)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (h (cdr list)))))))
???)))
```

Since the argument names `f`

, `g`

,`h`

in `(lambda (f) ...)`

, `(lambda (g) ...)`

, `(lambda (h) ...)`

are independent, we can rename all of them to `length`

, to make it look more similar to the `length`

function:

```
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
((lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
???)))
```

We are still repeating code. The obvious thing we can do about it is create an anonymous function called `mk-length`

(make length) that creates this code for us:

```
((lambda (mk-length)
(mk-length ???))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```

This is pretty tricky. Observe how `(lambda (mk-length) ...)`

gets the `(lambda (length) ...)`

function passed as the `mk-length`

argument, which in turn accepts `???`

as an argument and returns our original anonymous function:

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
```

Now let's try constructing length functions for lists of length one and two. For the list of length one, we just need to apply `mk-length`

on itself:

((lambda (mk-length) (mk-length (mk-length ???))) (lambda (length) (lambda (list) (cond ((null? list) 0) (else (add1 (length (cdr list))))))))

Let's go through this code. First `(lambda (length) ...)`

gets passed to the `lambda (mk-length)`

function as the `mk-length`

argument. Then it applies the result of `(mk-length ???)`

(which is the original anonymous function) to the `mk-length`

. This produces our well known function that works on lists with one or none elements:

```
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (??? (cdr list))))))
(cdr list))))))
```

Similarly, by adding another call to `mk-length`

, we get the function that works for lists with two or less elements:

```
((lambda (mk-length)
(mk-length
(mk-length
(mk-length ???)))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```

Now, since we don't know what `???`

is, let's just call it `mk-length`

and see what happens:

```
((lambda (mk-length)
(mk-length mk-length))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```

Nothing bad actually happens, we still get the same original anonymous function, except in place of `???`

we have the function:

```
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list)))))))
```

Notice also that argument names `mk-length`

and `length`

in `lambda (mk-length)`

and `lambda (length)`

are independent. Therefore we can rename `length`

to `mk-length`

to remind that the first argument to `mk-length`

is `mk-length`

:

```
((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (mk-length (cdr list))))))))
```

And now comes the key trick. We replace `mk-length`

with a self-application `(mk-length mk-length)`

:

```
((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 ((mk-length mk-length) (cdr list))))))))
```

If you try this code out, you'll see that it returns the length of a list for any list. Here is an example of applying it to a list of 10 elements:

```
(((lambda (mk-length)
(mk-length mk-length))
(lambda (mk-length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 ((mk-length mk-length) (cdr list))))))))
'(a b c d e f g h i j))
```

The function works because it keeps adding recursive uses by passing `mk-length`

to itself, just as it is about to expire. This is not yet the Y-combinator, but we have successfully managed to recursively call an anonymous function. Now we need to massage the code a bit to separate the anonymous function from the self-applicative code.

The first step is to move the self-applicative code out as much as possible:

((lambda (mk-length) (mk-length mk-length)) (lambda (mk-length)((lambda (length) (lambda (list) (cond ((null? list) 0) (else (add1 (length (cdr list)))))))(lambda (x) ((mk-length mk-length) x)))))

Now the code in bold looks very much like the `length`

function we need! Let's move it outside as well:

((lambda (le) ((lambda (mk-length) (mk-length mk-length)) (lambda (mk-length) (le (lambda (x) ((mk-length mk-length) x))))))(lambda (length) (lambda (list) (cond ((null? list) 0) (else (add1 (length (cdr list))))))))

There we have it! The anonymous `length`

function has been separated from the self-applicative call. The code in **bold** is the `length`

function, and the code not-in-bold is the Y-combinator!

Now let's rename `mk-length`

to `f`

and join the lines to make it more compact:

```
((lambda (le)
((lambda (f) (f f))
(lambda (f)
(le (lambda (x) ((f f) x))))))
(lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
```

And we are done. We can now write down the definition of Y-combinator:

```
(define Y
(lambda (le)
((lambda (f) (f f))
(lambda (f)
(le (lambda (x) ((f f) x)))))))
```

And we can apply it to our anonymous `length`

function:

```
((Y (lambda (length)
(lambda (list)
(cond
((null? list) 0)
(else
(add1 (length (cdr list))))))))
'(a b c d e f g h i j))
; ==> 10
```

This is one of many derivations of the applicative-order Y-combinator. See the publication "The Why of Y" for another derivation.