What is Y Combinator?
Now we can do a recursion on lambda function. (If not, read the previous entry.)
Looking back, we’ll see that we have to pass its own function name as an argument to one of its own parameter. (Whoa, what a mouthful to speak!)
lambda f, x: 1 if x == 0 else x * f(f, x1)
Can we not doing that, and just write f(x1)
like other normal functions?
Take one step back and rewrite the factorial function in a sanity manner.
f = lambda x: 1 if x == 0 else x * f(x1)
This function works. Since we already reserve the name f
for it (although it’s not what we want). That is the name f
must exists in this scope in order for this function to works properly. And there’s another way to make f
exists without allocating the name in the global scope.
lambda f: lambda x: 1 if x == 0 else x * f(x1)
However, this function will not work right away. To make it works, we have to feed the function as an argument repeatedly. For example, to find $5!$, we have to unroll everything into:
(lambda f: lambda x: 1 if x == 0 else x * f(x1))(
(lambda f: lambda x: 1 if x == 0 else x * f(x1))(
(lambda f: lambda x: 1 if x == 0 else x * f(x1))(
(lambda f: lambda x: 1 if x == 0 else x * f(x1))(
(lambda f: lambda x: 1 if x == 0 else x * f(x1))(
(lambda f: lambda x: 1 if x == 0 else x * f(x1))(
(lambda whatever: 42)
)
)
)
)
)
)(5) # i know, its a sin writing lispsy code with c style indentation
That is we have to write lambda f: ...
again and again by ourselves. In the case of $5!$, we need to write the function at least five (plus one) times. If the factorial is larger, then we need to write the function repeatedly as many times.
So we need something that can passing the definition of this recursive function infinitely many times. This is the core concept of fixedpoint combinator. Which– I failed to derived it^{1} –we have one of the most famous, the Y combinator, which can be implement in Python^{2} as:
Y = lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v)))
Or with the mathematical definition $Y = \lambda f.(\lambda x.f(x\;x))(\lambda x.f(x\;x))$ so we can ditch $v$.
Try applying a function $g$, we’ll see that
\[\begin{align} Y\;g &= \Big( {\color{blue}\lambda f}.\big(\lambda x.{\color{blue}f}(x\;x)\big) \; \big(\lambda x.{\color{blue}f}(x\;x)\big) \Big) \; {\color{red}g} \\ &= \big(\lambda x.{\color{green}g}(x\;x)\big) \; \big(\lambda x.{\color{green}g}(x\;x)\big) \\ &= \big({\color{blue}\lambda x}.g({\color{blue}x}\;{\color{blue}x})\big) \; {\color{red}\big(\lambda x.g(x\;x)\big)} \\ &= g\Big( {\color{green}\big(\lambda x.g(x\;x)\big)}\;{\color{green}\big(\lambda x.g(x\;x)\big)} \Big) \\ &= g\Big( \big(\lambda x.{\color{green}g}(x\;x)\big)\;\big(\lambda x.{\color{green}g}(x\;x)\big) \Big) \\ &= g\Big( \Big( {\color{blue}\lambda f}. \big(\lambda x.{\color{blue}f}(x\;x)\big)\;\big(\lambda x.{\color{blue}f}(x\;x)\big) \Big) \; {\color{red}g} \Big) \\ &= g \big( Y\;g \big) \\ &= g \big( g \big( Y\;g \big) \big) = g \big( g \big( g \big( Y\;g \big) \big) \big) = g \big( g \big( g \big( g \big( \cdots \big) \big) \big) \big). \end{align}\]To apply it in programming, just write:
(lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))( # Y
lambda factorial: # function name
lambda n: 1 if n == 0 else n * factorial(n1) # function definition
)(5) # applying argument
What about a Fibonacci number…
(lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))(
lambda fib:
lambda n: n if n <= 1 else fib(n1) + fib(n2)
)(5)
Isn’t it easy?

Still curious on how we get it? Read Sorawee’s article, which walkthrough us step by step until we finally deriving the Y combinator. ↩

Actually, this code is Z combinator, which applied
x(x)(v)
right away. ↩
Revision notes:
 December 12, 2022:
Add link to Sorawee
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