Newbie question to under (&.)

[Roland Senn]

In https://www.jsoftware.com/books/pdf/easyj.pdf section 3 we have:

up =: *10&^

down =: %10&^

rnd =: <. @ (0.5&+)

round =: rnd @ up down ]

5.123 up 2 NB. 512.3 up is dyadic

512.3 down 2 NB. 5.123 down is dyadic and inverse function of up

rnd 3.657 5.123 NB. 4 5 rnd is monadic

3.657 5.123 round 2 NB. 3.66 5.12

Now as 'down' is the inverse of 'up', it should be possible to use 'under' (&.):

roundu =: rnd &. up

Unfortunately this doesn't work:

3.657 5.123 roundu 2

|domain error in roundu, executing dyad rnd&.up

| 3.657 5.123 roundu 2

I think the issue is: The verb 'up' is dyadic, but 'rnd' is monadic. So I tried:

rounduu =: ([: rnd [) &. up

This gives the same error!

In https://code.jsoftware.com/wiki/Essays/Under I found:

Round to p decimal places: ([: <. +&0.5) &. (*&(10^p))

However here p (number of digits to round) is a noun variable and not an argument of the verb!

How do I write a working version of roundu using under (&.)? How can I tell the interpreter to call 'rnd' monadically?

[Clifford Reiter]

You can make the "p" into an adverb argument:

roundm=:{{([: <. +&0.5) &. (*&(10^m)) }}

2 roundm 51.1234

51.12

2 roundm 51.1264

51.13

Not exactly what you were seeking. Also, you might look at :. which allows you to define an inverse, but using dyads makes it hard to use.

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[Roland Senn]

Many thanks! The {{(...)}} way to define verbs is new to me. Where is this syntax documented/described?

[Gilles Kirouac]

https://code.jsoftware.com/wiki/Vocabulary/DirectDefinition

~ Gilles

[Henry Rich]

Dyads make inverses trickier. x u^_1 y is

x&u^_1 y

which underlies the J rule that 'x is control information, y is data'. 

If (up) and (down) has been defined to take the number of digits as x and the value as y, they would be be inverses. 

Henry Rich 

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[More Rice]

Yes, you've the valence of u and v reversed.

the language specifies:

   x u&.v y ↔ vi (v x) u (v y)  NB. Ref

in roundu, I would think 'up' survived the monadic execution (not the results you want), but when it hits rnd, J detects it is calling the monadic verb there with x and y and issues the domain error.

in rounduu case, the domain error is from the obverse portion of 'up'. (And that's the more interesting part ...)

Henry, but wouldn't J assume v is always monadic (in this case) and would attempt only to find its monadic (not dyadic) obverse instead?

thanks 

Maurice

[Roland Senn]

Many Thanks! This explains everything!

I clicked on the (under) link in the NuVoc and landed on the page https://code.jsoftware.com/wiki/Vocabulary/ampdot. There the 2 very important formulas

   u&.v y ↔ vi u v y

   x u&.v y ↔ vi (v x) u (v y)

are missing.

How do I navigate from the main  Wiki page to this excellent https://www.jsoftware.com/help/dictionary/d631.htm page?

@Henry: I suggest to add these 2 formulas to the page linked by NuVoc?

[trx2358...@yahoo.com]

On the Wiki page:  https://code.jsoftware.com/wiki/Vocabulary/ampdot 

There is a link “Through to Dictionary” near the top which should bring you to the page you want.

Sent from Yahoo Mail for iPhone

[Roland Senn]

Excellent! Thanks!

[Jan-Pieter Jacobs]

I agree a formula like that would be more clear, particularly for &.: .

For &. they are not entirely precise, as they do not take into account that &. imposes the rank of v on u:

   *:^:_1 +/ *: 3 4
5
   +/&.*: 3 4
3 4

   +/&.:*: 3 4

5

The latter result should, according to the formula, be the same, but is not, because in the later executes +/ at rank 0 (the rank of *:).

I think the formula should read (and similarly, the dyadic case):

u&.v y <->  v^:_1 u"rv v y, with rv=: 0{.v b.0, i.e. the monadic rank of v.

That said, the wiki is a wiki, you could add it yourself if you want to...

Jan-Pieter