# A Gentle Introduction to Non-determinism in Scheme

Some of the most crucial steps in mental growth are based not simply on acquiring new skills, but on acquiring new administrative ways to use what one already knows.

―Marvin Minsky

## Table of contents

## Introduction

Nondeterministic programming is a technique wherein the flow of an algorithm is not linear, and there exists multiple possible continuations. The behavior of a computation can also change with the same inputs. There are several methods to achieve nondeterminism. In this article the method that I’ll use is backtracking.

Additionally, I’ll use Scheme to do it. In Scheme, you are permitted to go back in an early computation and back, later, with ease.

I will also discuss the prerequisite topics to make nondeterminism in Scheme easier to understand.

## The current continuation

In Scheme, the current continuation or the remaining computation is the remaining work to evaluate the rest of the computation.

In the expression

```
0
```

the current continuation—the remaining computation—is

```
(lambda (v) v)
```

and the remaining operation is

```
((lambda (v) v) 0)
```

which returns

```
0
```

That is the result because the remaining computation on that level—the top-level—is the identity function.

Bear in mind, that the name of the argument of the lambda is not important. It can be `v`

, `x`

, or `dog-cat-mouse`

:

```
((lambda (dog-cat-mouse) dog-cat-mouse) 0)
```

### Examples

In the expression

```
(* 1 2)
```

the remaining computation for `2`

—the second argument of `*`

—is

```
(lambda (v) (* 1 v))
```

and the remaining operation is

```
((lambda (v) (* 1 v)) 2)
```

which, in that level of computation returns

```
2
```

Consequently, the remaining computation is

```
(lambda (v) v)
```

and the remaining operation is

```
((lambda (v) v) 2)
```

which finally returns, at the top-level:

```
2
```

In the expression

```
(+ (* 1 2) 3)
```

the flow of the computation is

```
(* 1 2)
(+ 3)
```

Firstly, `(* 1 2)`

will be evaluated then the result becomes the first argument for `(+ 3)`

. In that empty space, the remaining computation is

```
(lambda (v) (+ v 3))
```

and the remaining operation is

```
((lambda (v) (+ v 3)) (* 1 2))
```

## The call/cc operator

In Scheme, you can capture the current continuation—the next computation—as a variable. You can do that with call/cc. It is an operator which accepts only one argument exclusively—a lambda—an anonymous function, with one argument:

```
(call/cc (lambda (k) k))
```

If you don’t have `call/cc`

define it with:

```
(define call/cc call-with-current-continuation)
```

The phrase *call with current continuation* can also be treated as *call with next computation*. The function that it calls is the lambda. Additionally, `call/cc`

passes the current continuation—the next computation—to that anonymous function.

In this lambda:

```
(lambda (k) k)
```

`k`

is a function itself which accepts one argument. So, in

```
(call/cc (lambda (k) k))
```

`call/cc`

returns the current continuation—simply the function. However, in

```
(call/cc (lambda (k) (k 0)))
```

`call/cc`

simply returns `0`

because in that level—the top-level—`k`

is

```
(lambda (v) v)
```

—the identity function. Because of that, the function

```
(call/cc (lambda (k) (k 0)))
```

is functionally equivalent to

```
((lambda (v) v) 0)
```

### Capturing

Going back to the earlier example:

```
(+ (* 1 2) 3)
```

you can capture the current continuation of `(* 1 2)`

with `call/cc`

:

```
(+ (call/cc (lambda (k) (* 1 2))) 3)
```

However, you may notice that you did not apply the function `k`

to anything. The expression `(* 1 2)`

is evaluated, then the result goes to `(+ _ 3)`

. In other words, that expression is functionally equivalent to:

```
(+ (* 1 2) 3)
```

### Application

Using the same example, let’s apply the function `k`

:

```
(+ (call/cc (lambda (k) (k (* 1 2)))) 3)
```

Inside the call of `call/cc`

:

```
(lambda (k) (k (* 1 2)))
```

the variable `k`

is the current continuation—the remaining computation. What is the remaining computation? This:

```
(+ 3)
```

the addition to `3`

. In order to represent that computation as function, it becomes:

```
(lambda (v) (+ v 3))
```

Wherein `(+ _ 3)`

is going to be represented by this lambda.

So, the role of `call/cc`

is to call a lambda, which accepts the remaining computation. Here, the remaining computation is called `k`

. Because of that, the consequence of:

```
(+ (call/cc (lambda (k) (k (* 1 2)))) 3)
```

is functionally equivalent to

```
((lambda (v) (+ v 3)) (* 1 2))
```

### Escaping

In

```
(define z #f)
(+ (call/cc (lambda (k) (set! z k) (* 1 2))) 3)
```

normally the result is

```
5
```

however, because we saved the current continuation—`k`

—in `z`

, we can return to that saved location with:

```
(z 1)
```

which returns

```
4
```

—another value. That mechanisms gives us the capability to escape computations, whether with a new value, an old value, or nothing.

For the reason that the current continuation is:

```
(lambda (v) (+ v 3))
```

the call

```
(z 1)
```

is functionally equivalent to

```
((lambda (v) (+ v 3)) 1)
```

instead of

```
((lambda (v) (+ v 3)) (* 1 2))
```

### Examples

In

```
(let ((x (call/cc (lambda (k) k)))) (x (lambda (o) "hi")))
```

the call

```
(x (lambda (o) "hi"))
```

becomes

```
((call/cc (lambda (k) k)) (lambda (o) "hi"))
```

there, the remaining computation is

```
(lambda (o) "hi")
```

which goes to the variable `k`

in the body of `call/cc`

. Like earlier, we apply `k`

to `(lambda (o) "hi")`

inside another lambda:

```
((lambda (v) (v (lambda (o) "hi"))) (lambda (o) "hi"))
```

which returns

```
"hi"
```

In

```
(((call/cc (lambda (k) k)) (lambda (x) x)) "hi")
```

the key is

```
((call/cc (lambda (k) k)) (lambda (x) x))
```

In the body of `call/cc`

, the remaining computation is

```
(lambda (x) x)
```

which goes to the variable `k`

in the body of `call/cc`

. Like earlier again, we apply it to the lambda:

```
((lambda (v) (v (lambda (x) x))) (lambda (x) x))
```

which returns

```
#<procedure x>
```

It means, that you can now apply this function to the remaining argument:

```
(((lambda (v) (v (lambda (x) x))) (lambda (x) x)) "hi")
```

which returns

```
"hi"
```

## The amb operator

One of the uses of the famous amb operator in Scheme is to implement non-deterministic programming by means of backtracking. With that mechanism, the computation can go to an earlier state; carry computations; change change; escape computations; and more.

In this article we’re going to use the amb operator to enable the backtracking mechanism.

### Definition

The definition that we’ll use is the one by shido.info/lisp. It is a macro to ensure that the arguments are not evaluated. Additionally, it uses lists internally.

```
(define call/cc call-with-current-continuation) ; 1
; 2
(define f #f) ; 3
; 4
(define-syntax amb ; 5
(syntax-rules () ; 6
((_) (f)) ; 7
((_ a) a) ; 8
((_ a b ...) ; 9
(let ((s f)) ; 10
(call/cc ; 11
(lambda (k) ; 12
(set! f (lambda () ; 13
(set! f s) ; 14
(k (amb b ...)))) ; 15
(k a))))))) ; 16
; 17
(define (really? x y) ; 18
(if (equal? x y) ; 19
(list x y) ; 20
(amb))) ; 21
; 22
(call/cc ; 23
(lambda (k) ; 24
(set! f (lambda () ; 25
(k 'no-choices))))) ; 26
```

These definitions work on MIT/GNU Scheme, Guile, Scheme48, Scsh, chibi-scheme, Gerbil, Chez Scheme, Chicken, Bigloo, Gauche, and Racket.

### Deconstruction

In this section, we are going to deconstruct the definitions of the amb operator and other functions. The function `really?`

technically is not part of the definition, however, we’ll use it to demonstrate the functionality of `amb`

.

Here are superficial steps of the definitions:

- In line 1
`call/cc`

is bound to`call-with-current-continuation`

mainly for implementations which doesn’t have that definition. - In line 3
`f`

is bound to a default value. - In lines 5 to 16 is the body of
`amb`

. - In lines 18 to 21 is an example function: if
`x`

and`y`

are equal, it returns a list; if not,`amb`

is called. - In lines 23 to 26,
`call/cc`

is called, in which the true initial value of`f`

is initialized with a lambda which returns`'no-choices`

. - Let’s go back to the body of
`amb`

in lines 5 to 16. In line 7, if`amb`

is called as such:

the function`(amb)`

`f`

is called, with whatever that`f`

can do. - In line 8, if
`amb`

is called as such:

the argument`(amb "dog")`

`"dog"`

is simply returned. - However, with more than one arguments, the behavior of
`amb`

changes. Firstly, in line 10, the current value of`f`

is bound to`s`

, next in line 11,`call/cc`

is called, capturing the current continuation to`k`

. - Inside the body of the lambda, the global variable
`f`

will have a new value—another lambda—which will not be evaluated until it is explicitly requested. In that body,`f`

will have the earlier value of`s`

, then`k`

will be called with the value`b ...`

, which are the remaining arguments for`amb`

. - Lastly, the
`k`

function—the remaining computation—will be applied to the value`a`

.

### Evaluation

Using `really?`

we will watch the evaluation steps in the use of `amb`

. If we have the following expression,

```
(really? (amb "dog" "cat" 9) (amb "mouse" 9))
```

here are the simplified evaluation steps:

`(amb "dog" "cat" 9)`

will be evaluated;- For the reason that more than one argument is passed to
`amb`

let’s go to line 9; - The local variable
`s`

will have the current value of`f`

in the top-level:`(lambda () (k 'no-choices))`

- The lambda in line 12 will be called with the remaining computation to
`k`

; - In line 13
`f`

will have a new value. The value will be a lambda; - When the lambda is called,
`k`

again will have the value of`s`

, which is the lambda in step 3. `k`

will be applied to`a`

, wherein,`(k a)`

is:

which consequently becomes`((lambda (v) (really? v (amb "mouse" 9))) "dog")`

because of the call`((lambda (v) ((lambda (w) (really? v w)) "dog")) "mouse")`

`(amb "mouse" 9)`

;- Because
`"dog"`

is not equal to`"mouse"`

the`amb`

operator will be called like:`(amb)`

- Because
`(amb)`

doesn’t have arguments, the expression in line 7 happens;

wherein the value of`(f)`

`k`

is the lambda in lines 13 to 15. The value of`f`

again becomes:`(lambda () (k 'no-choices))`

- Then, in line 15
`k`

will be called like`(k "cat" 9)`

. - The steps will be repeated, calling
`really?`

many times. The steps will look like:`(really? "dog" "mouse") (really? "dog" 9) (really? "cat" "mouse") (really? "cat" 9) (really? 9 "mouse") (really? 9 9)`

- Because
`9`

is equal to`9`

, the expression

will be evaluated in the body of`(list 9 9)`

`really?`

which finally returns`'(9 9)`

## Closing remarks

In this article, we noticed that with `call/cc`

we have easily achieved non-determinism through backtracking with `amb`

and basic Scheme functions. However, there are more elaborate and better methods of achieving this.

The type of continuations that we dealt with is the non-delimited continuation in contrast with a delimited continuation. I must also mention, that there’s an argument against the use of `call/cc`

.

Anyway, I hope that you learned something good from this post.