Chapter 9: Trains of Verbs
In this chapter we continue the topic of trains of verbs begun in Chapter 3. Recall that a train is an isolated sequence of functions, written one after the other, such as (+ * -).
Preparations
It will be convenient to have a few definitions ready to hand for the examples to come.
x =: 1 2 3 4 NB. a list of numbers sum =: + / NB. verb: sum of a list min =: <. / NB. verb: smallest item of a list max =: >. / NB. verb: largest item of a list
9.1 Review: Monadic Hooks and Forks
Recall from Chapter 3 the monadic hook, with the scheme:
(f g) y means y f (g y)
Here is an example, as a brief reminder: a whole number is equal to its floor:
| y =: 2.1 3 | <. y | y = <. y | (= <.) y |
2.1 3 |
2 3 |
0 1 |
0 1 |
Recall also the monadic fork, with the scheme:
(f g h) y means (f y) g (h y)
For example: the mean of a list of numbers is the sum divided by the number-of-items:
mean =: sum % #
| x | sum x | # x | (sum x) % (# x) | mean x |
1 2 3 4 |
10 |
4 |
2.5 |
2.5 |
Now we look at some further variations.
9.2 Dyadic Hooks
3 hours and 15 minutes is 3.25 hours. A verb hr, such that (3 hr 15) is 3.25, can be written as a hook. We want x hr y to be x + (y%60) and so the hook is:
hr =: + (%&60) 3 hr 15 3.25
The scheme for dyadic hook is:
x (f g) y means x f (g y)
with the diagram:
9.3 Dyadic Forks
Suppose we say that the expression "10 plus or minus 2" is to mean the list 12 8. A verb to compute x plus-or-minus y can be written as the fork (+,-):
| (10+2) , (10-2) | 10 (+,-) 2 |
12 8 |
12 8 |
The scheme for a dyadic fork is:
x (f g h) y means (x f y) g (x h y)
Here is a diagram for this scheme:
9.4 Review
There are four basic patterns of verb-trains. It may help to fix them in the memory by recalling these four verbs:
mean =: sum % # NB. monadic fork plusminus =: + , - NB. dyadic fork wholenumber =: = <. NB. monadic hook hr =: + (%&60) NB. dyadic hook
9.5 Longer Trains
Now we begin to look at ways in which to broaden the class of functions which can be defined as trains.
In general a train of any length can be analysed into hooks and forks. For a train of 4 verbs, e f g h, the scheme is that
e f g h means e (f g h)
that is, a 4-train (e f g h) is a hook, where the first verb is e and the second is the fork (f g h). For example, if y a list of numbers:
y =: 2 3 4
then the "norm" of y is (y - mean y), where mean is defined above as (sum % #). We see that the following expressions for the norm of y are all equivalent:
y - mean y _1 0 1 (- mean) y NB. as a hook _1 0 1 (- (sum % #)) y NB. by definition of mean _1 0 1 (- sum % #) y NB. as 4-train _1 0 1
A certain amount of artistic judgement is called for with long trains. This last formulation as the 4-train (- sum % #) does not bring out as clearly as it might that the key idea is subtracting the mean. The formulation ( - mean) is clearer.
For a train of 5 verbs d e f g h the scheme is:
d e f g h means d e (f g h)
That is, a 5-train (d e f g h) is a fork with first verb d, second verb e and third verb the fork (f g h) For example, if we write a calendar date in the form day month year:
date =: 28 2 1999
and define verbs to extract the day month and year separately:
Da =: 0 & {
Mo =: 1 & {
Yr =: 2 & {
the date can be presented in different ways by 5-trains:
| (Da , Mo , Yr) date | (Mo ; Da ; Yr) date |
28 2 1999 |
+-+--+----+ |2|28|1999| +-+--+----+ |
The general scheme for a train of verbs (a b c ...) depends upon whether the number of verbs is even or odd:
even: (a b c ...) means hook (a (b c ...))
odd : (a b c ...) means fork (a b (c ...))
9.6 Identity Functions
There is a verb [ (left bracket) which gives a result identical to its argument.
| [ 99 | [ 'a b c' |
99 |
a b c |
There is a dyadic case, and also a similar verb ]. Altogether we have these schemes
[ y means y
x [ y means x
] y means y
x ] y means y
| [ 3 | 2 [ 3 | ] 3 | 2 ] 3 |
3 |
2 |
3 |
3 |
The expression (+ % ]) is a fork; for arguments x and y it computes
(x+y) % (x ] y)
that is, (x+y) % y
| 2 ] 3 | (2 + 3) % (2 ] 3) | 2 (+ % ]) 3 |
3 |
1.66667 |
1.66667 |
Another use for the identity function [ is to cause the result of an assignment to be displayed. The expression foo =: 42 is an assignment while the expression [ foo =: 42 is not: it merely contains an assignment.
foo =: 42 NB. nothing displayed
[ foo =: 42
42
Yet another use for the [ verb is to allow several assignments to be combined on one line.
| a =: 3 [ b =: 4 [ c =: 5 | a,b,c |
3 |
3 4 5 |
Since [ is a verb, its arguments must be nouns, (that is, not functions). Hence the assignments combined with [ must all evaluate to nouns.
9.6.1 Example: Hook as Abbreviation
The monadic hook (g h) is an abbreviation for the monadic fork ([ g h). To demonstrate, suppose we have:
g =: , h =: *: y =: 3
Then each of the following expressions is equivalent.
([ g h) y NB. a fork 3 9 ([ y) g (h y) NB. by defn of fork 3 9 y g (h y) NB. by defn of [ 3 9 (g h) y NB. by defn of hook 3 9
9.6.2 Example: Left Hook
Recall that the monadic hook has the general scheme
(f g) y = y f (g y)
How can we write, as a train, a function with the scheme
( ? ) y = (f y) g y
There are two possibilities. One is the fork (f g ]):
f =: *:
g =: ,
(f g ]) y NB. a fork
9 3
(f y) g (] y) NB. by meaning of fork
9 3
(f y) g y NB. by meaning of ]
9 3
For another possibility, recall the ~ adverb with its scheme (x f~ y) = (y f x). Our train can be written as the hook (g~ f).
(g~ f) y NB. a hook 9 3 y (g~) (f y) NB. by meaning of hook 9 3 (f y) g y NB. by meaning of ~ 9 3
9.6.3 Example: Dyad
There is a sense in which [ and ] can be regarded as standing for left and right arguments.
f =: 'f' & , g =: 'g' & ,
| foo =: (f @: [) , (g @: ]) | 'a' foo 'b' |
(f@:[) , (g@:]) |
fagb |
9.7 The Capped Fork
The class of functions which can be written as unbroken trains can be widened with the aid of the "Cap" verb [: (leftbracket colon)
The scheme is: for verbs f and g, the fork [: f g means f @: g. For example, let
f =: 'f' & , g =: 'g' & , y =: 'y'
then [: is illustrated by:
| f g y | (f @: g) y | ([: f g) y |
fgy |
fgy |
fgy |
Notice how the sequence of three verbs ([: f g) looks like a fork, but with this "capped fork" it is the MONADIC case of the middle verb f which is applied.
The [: verb is valid ONLY as the left-hand verb of a fork. It has no other purpose: as a verb it has an empty domain, that is, it cannot be applied to any argument. Its usefulness lies in building long trains. Suppose for example that:
h =: 'h'&,
then the expression (f , [: g h) is a 5-train which denotes a verb:
(f , [: g h) y NB. a 5-train
fyghy
(f y) , (([: g h) y) NB. by meaning of 5-train
fyghy
(f y) , (g @: h y) NB. by meaning of [:
fyghy
(f y) , (g h y) NB. by meaning of @:
fyghy
'fy' , 'ghy' NB. by meaning of f g h
fyghy
9.8 Constant Functions
Here we continue looking at ways of broadening the class of functions that we can write as trains of verbs. There is a built-in verb 0: (zero colon) which delivers a value of zero regardless of its argument. There is a monadic and a dyadic case:
| 0: 99 | 0: 2 3 4 | 0: 'hello' | 88 0: 99 |
0 |
0 |
0 |
0 |
As well as 0: there are similar functions 1: 2: 3: and so on up to 9: and also the negative values: _9: to _1:
| 1: 2 3 4 | _3: 'hello' |
1 |
_3 |
0: is said to be a constant function, because its result is constant. Constant functions are useful because they can occur in trains at places where we want a constant but must write a verb, (because trains of verbs, naturally, contain only verbs).
For example, a verb to test whether its argument is negative (less than zero) can be written as (< & 0) but alternatively it can be written as a hook:
negative =: < 0:
| x =: _1 0 2 | 0: x | x < (0: x) | negative x |
_1 0 2 |
0 |
1 0 0 |
1 0 0 |
9.9 The "Constant" Conjunction
The constant functions _9: to 9: offer more choices for ways of defining trains. Neverthless they are limited to single-digit scalar constants. We look now at at a more general way of writing constant functions. Suppose that k is the constant in question:
k =: 'hello'
An explicit verb written as (3 : 'k') will give a constant result of k:
| k | (3 : 'k') 1 | (3 : 'k') 1 2 |
hello |
hello |
hello |
Since k is explicit, its rank is infinite: to apply it separately to scalars we need to specify a rank R of 0:
| k | R =: 0 | ((3 : 'k') " R) 1 2 |
hello |
0 |
hello hello |
The expression ((3 : 'k') " R) can be abbreviated as (k " R) with the aid of the Constant conjunction " (double quote)
| k | R | ((3 : 'k') " R) 1 2 | 'hello' " R 1 2 |
hello |
0 |
hello hello |
hello hello |
Note that if k is a noun, then the verb (k"R) means: the constant value k produced for each rank-R cell of the argument. By contrast, if v is a verb, then the verb (v"R) means: the verb v applied to each rank-R cell of the argument.
The general scheme can be represented as:
k " R means (3 : 'k') " R
This is the end of Chapter 9.