Raul, We've been moving for the last several weeks and I haven't studied this
yet.
It seems odd that f is so simple and g is impossible.
f=: 1 :'y u / y'
a=:%
a f i.4
0 0 0 0
_ 1 0.5 0.333333
_ 2 1 0.666667
_ 3 1.5 1
(i.3)%/i:3
0 0 0 0 0 0 0
_0.333333 _0.5 _1 _ 1 0.5 0.333333
_0.666667 _1 _2 _ 2 1 0.666667
g=: 1 :'x u /y'
(i.3) a g i:3
|domain error: scriptd
| (i.3) a g i:3
|[-17] c:\users\owner\j801-user\temp\113.ijs
Is there an easy way to write g ?
Linda
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Raul Miller
Sent: Friday, November 15, 2013 7:32 AM
To: Programming forum
Subject: Re: [Jprogramming] Times Table Therapy
& is compose here
http://www.jsoftware.com/help/dictionary/d630v.htm
`:6 is evoke gerund as a train
http://www.jsoftware.com/help/dictionary/d612.htm
the following / is outer product (or "table")
http://www.jsoftware.com/help/dictionary/d420.htm
I imagine you are already familiar with that one?
Here's my session from when I wrote that, along with some notes on my thinking.
I am including my mistaken experiments. I think this took me between fifteen
minutes and half an hour, but I did not time it so I do not know for sure.
First, I wanted to make sure that the thing worked, so I did a literal copy and
paste of your definition of T and then I pasted in the values to try to make it
work.
T=: 1
:'(,.(<":u),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(":,>{.y),(u,,''/''),":,>}.y'
'%' 1
:'(,.(<":u),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(":,>{.y),(u,,''/''),":,>}.y'
(i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Next, I started introducing small changes. I wanted to make sure that I
understood what you wrote and sometimes the easiest way of doing that is
finding equivalent expressions. And the easiest way there sometimes is to make
small changes that seem equivalent. I also wanted a shorter expression so that
I did not have so much code to think about. (I prefer to think in
transformations of data, and a lot of steps might be easy to write but reading?)
The first thing I did was change ,>}.y on the right to >{:y. Using }.
on the pair leaves an extra leading dimension (1) which you eliminate through ,
so why not just use {: and avoid it?
'%' 1
:'(,.(<":u),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(":,>{.y),(u,,''/''),":>{:y'
(i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
The next thing I did was remove the ravel of the rightmost '/' - it is getting
appended to '%' and '%/' is already rank 1.
'%' 1
:'(,.(<":u),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(":,>{.y),(u,''/''),":>{:y'
(i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Next I changed u to m. u can be a verb and this code assumes you are using a
noun.
'%' 1
:'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(":,>{.y),(m,''/''),":>{:y'
(i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Next, I wanted to evoke the verb named in m. There's a 128!: foreign which I
probably should have used, but I was not sure if ~ could be used on a primitive
like this.
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(>{.y)(m~/)>{:y'
(i.3);i:3
|ill-formed name
| (,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(>{.y)( m~/)>{:y
No. So, ok, let's box it and use evoke gerund:
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(>{.y)((<m)`:6/)>{:y'
(i.3);i:3
|domain error
| (,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y), ".(>{.y)((<m)`:6/)>{:y
Oops, I am computing a numeric result, so the ". has to go.
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),(>{.y)((<m)`:6/)>{:y'
(i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Now let's try cleaning it up, using an outer product instead of taking apart
boxes individually.
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),((<m)`:6/)&>/>y'
(i.3);i:3
┌─┬──────────────────┐
│%│_3 _2 _1 0 1 2 3│
├─┼──────────────────┤
│0│0 _0.5 _2 0 0 0 0 │
│1│ │
│2│ │
└─┴──────────────────┘
Oops, I need to compose unbox with the outer product, so that leads us to the
/&> part you were asking about.
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),((<m)`:6/)&>/y'
(i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Now I can also get rid of the (>}.y), by putting a [, inside my outer product
verb. (This is wrong, by the way - can you see my mistake?)
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.[A=:([,(<m)`:6/)&>/y' (i.3);i:3
┌─┬──────────────────────────────────┐
│%│ 0 1 2 0 0 0 0│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Now let's get rid of the [ to the left of A=: since it is doing nothing useful
for us.
'%' 1 :'(,.(<":m),<":,.>{.y),.(<{.":A),:<}.A=:([,(<m)`:6/)&>/y' (i.3);i:3
┌─┬──────────────────────────────────┐
│%│ 0 1 2 0 0 0 0│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Now let's clean up the manipulations on the left side of A which build those
labels. I basically want the first row of A in a separate box from the rest of
A, right?
'%' 1 :'(,.(<":m),<":,.>{.y),.({.,:&<}.)A=:([,(<m)`:6/)&>/y' (i.3);i:3
┌─┬──────────────────────────────────┐
│%│0 1 2 0 0 0 0 │
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Uh... right about here, I notice that I've messed things up. My top row is just
wrong. So let's abandon the above line of thought and just try making a table
that contains the top row and left column. I can't use '%' here but I am just
trying to untangle my thoughts, so I'll use a 0 for now until I have figured
out how to manipulate the rest of the data.
'%' 1 :'(0,>{:y),([,(<m)`:6/)&>/y' (i.3);i:3
0 _3 _2 _1 0 1 2 3
0 1 2 0 0 0 0 0
0 0 0 0 0 0 0 0
_0.333333 _0.5 _1 _ 1 0.5 0.333333 0
_0.666667 _1 _2 _ 2 1 0.666667 0
Ok, that is wrong because I have no leftmost column and I have two label rows
on top. Right about here, I noticed that I should have been using ], inside my
outer product, instead of [, (I want the contents of the righthand box not the
lefthand box).
'%' 1 :'(0,>{.y),(],(<m)`:6/)&>/y' (i.3);i:3
0 0 1 2 0 0 0
_3 _2 _1 0 1 2 3
0 0 0 0 0 0 0
_0.333333 _0.5 _1 _ 1 0.5 0.333333
_0.666667 _1 _2 _ 2 1 0.666667
Here, I have reversed my two label columns. I think here I noticed that my left
column is going on top instead of on the left. Rather than reason about that, I
simplify my expression further and ignore the leftmost column for a moment.
(It's easier to just perform experiments and glance at them than it is to think
things through.
Probably not the smartest technique, but do not accuse me of being smart and we
will be ok.)
'%' 1 :'(],(<m)`:6/)&>/y' (i.3);i:3
_3 _2 _1 0 1 2 3
0 0 0 0 0 0 0
_0.333333 _0.5 _1 _ 1 0.5 0.333333
_0.666667 _1 _2 _ 2 1 0.666667
Good. Here I have the table I want (without the boxes and without the left
column). So let's grab out the boxes:
'%' 1 :'({.;}.)(],(<m)`:6/)&>/y' (i.3);i:3
┌────────────────┬──────────────────────────────────┐
│_3 _2 _1 0 1 2 3│ 0 0 0 0 0 0 0│
│ │_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│ │_0.666667 _1 _2 _ 2 1 0.666667│
└────────────────┴──────────────────────────────────┘
Good enough? Not quite but I have not noticed that yet, so let's add the left
column.
'%' 1 :'(m;,.>{.y),.({.;}.)(],(<m)`:6/)&>/y' (i.3);i:3
┌─┬──────────────────────────────────┐
│%│_3 _2 _1 0 1 2 3 │
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Oh, right, I need to format the top row or it will not retain spacing when
separated from the bottom row
'%' 1 :'(m;,.>{.y),.({.;}.)":(],(<m)`:6/)&>/y' (i.3);i:3
┌─┬──────────────────────────────────┐
│%│ _3 _2 _1 0 1 2 3│
├─┼──────────────────────────────────┤
│0│ 0 0 0 0 0 0 0│
│1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
│2│_0.666667 _1 _2 _ 2 1 0.666667│
└─┴──────────────────────────────────┘
Done.
For this case I did not need to format the bottom right box, but in obscure
irrelevant cases that might be a good idea (like imagine the top row was 0 0.5
1 1.5 2 and the left column was 0 2 4 and the value for m was '*').
That said, I should probably have instead used
'%' 1 :'(m;,.>{.y),.({.;}.)":(''(],('',m,'')/)&>/'')128!:2 y' (i.3);i:3
Because it makes more sense to think of m as a string representing a verb than
it does to think of m as representing an unboxed gerund.
Thanks,
--
Raul
On Fri, Nov 15, 2013 at 2:28 AM, Linda Alvord <[email protected]> wrote:
> What definition of & is applied here?
>
> s=: 4 :'(],(<x)`:6/)&>/y'
> '%' s (i.3);i:3
> _3 _2 _1 0 1 2 3
> 0 0 0 0 0 0 0
> _0.333333 _0.5 _1 _ 1 0.5 0.333333
> _0.666667 _1 _2 _ 2 1 0.666667
>
> How does it work? I did find an explanation for `6/ but I can't find it
> again.
>
> Linda
>
>
>
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]] On Behalf Of Raul
> Miller
> Sent: Wednesday, November 13, 2013 7:38 AM
> To: Programming forum
> Subject: Re: [Jprogramming] Times Table Therapy
>
> Would
> '%' 1 :'(m;,.>{.y),.({.;}.)":(],(<m)`:6/)&>/y' (i.3);i:3 or
>
> '%' 4 :'(x;,.>{.y),.({.;}.)":(],(<x)`:6/)&>/y' (i.3);i:3
>
>
> be acceptable?
>
> The part to the right of the ": is essentially the same thing as your A.
>
> But note that I prefer to leave both the top and bottom boxes on the right
> formatted (where you only left the top box formatted).
>
> Thanks,
>
> --
> Raul
>
>
> On Wed, Nov 13, 2013 at 3:27 AM, Linda Alvord <[email protected]>wrote:
>
>> Thanks Raul. Here's the finished function. I would like not to
>> include A . Any ideas:
>>
>> a=:'%'
>> b=:i:3
>> c=:i.3
>> d=:c;b
>>
>> T=: 1
>> :'(,.(<":u),<":,.>{.y),.(<{.":A),:<}.[A=:(>}.y),".(":,>{.y),(u,,''/''),":,>}.y'
>> a T d
>> ┌─┬──────────────────────────────────┐
>> │%│ _3 _2 _1 0 1 2 3│
>> ├─┼──────────────────────────────────┤
>> │0│ 0 0 0 0 0 0 0│
>> │1│_0.333333 _0.5 _1 _ 1 0.5 0.333333│
>> │2│_0.666667 _1 _2 _ 2 1 0.666667│
>> └─┴──────────────────────────────────┘
>>
>> '>.' T (i:4);i:4
>> ┌──┬─────────────────────┐
>> │>.│_4 _3 _2 _1 0 1 2 3 4│
>> ├──┼─────────────────────┤
>> │_4│_4 _3 _2 _1 0 1 2 3 4│
>> │_3│_3 _3 _2 _1 0 1 2 3 4│
>> │_2│_2 _2 _2 _1 0 1 2 3 4│
>> │_1│_1 _1 _1 _1 0 1 2 3 4│
>> │ 0│ 0 0 0 0 0 1 2 3 4│
>> │ 1│ 1 1 1 1 1 1 2 3 4│
>> │ 2│ 2 2 2 2 2 2 2 3 4│
>> │ 3│ 3 3 3 3 3 3 3 3 4│
>> │ 4│ 4 4 4 4 4 4 4 4 4│
>> └──┴─────────────────────┘
>>
>>
>> Linda
>>
>> -----Original Message-----
>> From: [email protected] [mailto:pro
>> [email protected]] On Behalf Of Raul Miller
>> Sent: Saturday, November 09, 2013 8:45 AM
>> To: Programming forum
>> Subject: Re: [Jprogramming] Times Table Therapy
>>
>> First, g seemed a bit overly ornate, so I took the liberty of putting
>> it through a weight loss program:
>>
>> % 1 :'<}.":(>}.y),(>{.y)u/,>}.y' (i.3) ,&< i:3
>>
>> % 1 :'<}.":(>}.y),(>{.y)u/,>{:y' (i.3) ,&< i:3
>>
>> % 1 :'<}.":(>}.y),(>{.y)u/>{:y' (i.3) ,&< i:3
>>
>> % 1 :'<}.":(>}.y),u/&>/y' (i.3) ,&< i:3
>>
>> % 1 :'<}.":(],u/)&>/y' (i.3) ,&< i:3
>>
>>
>> (these all have the same result as a g d)
>>
>>
>> The phrase U&>/Y would apply the verb U between the contents of the
>> two boxes of Y (if Y is a pair of boxes). And, in this case, U would
>> be a verb with the result: "contents of the second box and a u table".
>>
>>
>> I am more comfortable with the short form than the long form because
>> the short form leaves me with extra space on the line so I can
>> inspect the rest of the sentence.
>>
>>
>> Anyways, note that we can get the top row back by getting rid of the
>> behead, like this:
>>
>> % 1 :'<":(],u/)&>/y' (i.3) ,&< i:3
>>
>> ┌──────────────────────────────────┐
>>
>> │ _3 _2 _1 0 1 2 3│
>>
>> │ 0 0 0 0 0 0 0│
>>
>> │_0.333333 _0.5 _1 _ 1 0.5 0.333333│
>>
>> │_0.666667 _1 _2 _ 2 1 0.666667│
>>
>> └──────────────────────────────────┘
>>
>>
>> If you want the top row in a different box from the rest of the rows,
>> you can replace that leading < with ,.@({.;}.)
>>
>>
>> Does this help?
>>
>>
>> Thanks,
>>
>>
>> --
>>
>> Raul
>>
>>
>>
>>
>>
>>
>> On Sat, Nov 9, 2013 at 3:15 AM, Linda Alvord <[email protected]
>> >wrote:
>>
>> > Thanks. I’m starting to understand things better. Here is my next
>> > problem.
>> >
>> > a=:%
>> > b=:i:3
>> > c=:i.3
>> > d=:(<c),<b
>> >
>> > ]M=:":(>}.d),(>{.d)a/,>}.d
>> > _3 _2 _1 0 1 2 3
>> > 0 0 0 0 0 0 0
>> > _0.333333 _0.5 _1 _ 1 0.5 0.333333
>> > _0.666667 _1 _2 _ 2 1 0.666667
>> > (<{.M),:<}.M
>> > ┌──────────────────────────────────┐
>> > │ _3 _2 _1 0 1 2 3│
>> > ├──────────────────────────────────┤
>> > │ 0 0 0 0 0 0 0│
>> > │_0.333333 _0.5 _1 _ 1 0.5 0.333333│
>> > │_0.666667 _1 _2 _ 2 1 0.666667│
>> > └──────────────────────────────────┘
>> >
>> > f=: 1 :'":(>}.y),(>{.y)u/,>}.y'
>> > a f d
>> > _3 _2 _1 0 1 2 3
>> > 0 0 0 0 0 0 0
>> > _0.333333 _0.5 _1 _ 1 0.5 0.333333
>> > _0.666667 _1 _2 _ 2 1 0.666667
>> >
>> > g=: 1 :'<}.":(>}.y),(>{.y)u/,>}.y'
>> > a g d
>> > ┌──────────────────────────────────┐
>> > │ 0 0 0 0 0 0 0│
>> > │_0.333333 _0.5 _1 _ 1 0.5 0.333333│
>> > │_0.666667 _1 _2 _ 2 1 0.666667│
>> > └──────────────────────────────────┘
>> >
>> > I want to modify g so that it attaches the top row with correct
>> > spacing as shown above.
>> >
>> > Linda
>> >
>> >
>> > -----Original Message-----
>> > From: [email protected] [mailto:
>> > [email protected]] On Behalf Of Raul Miller
>> > Sent: Friday, November 08, 2013 7:57 AM
>> > To: hProgramming forum
>> > Subject: Re: [Jprogramming] Times Table Therapy
>> >
>> > http://www.jsoftware.com/help/dictionary/cret.htm explains that return.
>> > exits an explicit definition.
>> >
>> > http://www.jsoftware.com/help/dictionary/d521.htm explains that the
>> > result of {. is the leading item of an array (which means one
>> > dimension less than the table).
>> >
>> > I'll presume that I do not need to document "behead" but just in
>> > case some of the younger readers are curious:
>> > http://www.jsoftware.com/help/dictionary/d531.htm
>> >
>> > Thanks,
>> >
>> > --
>> > Raul
>> >
>> >
>> >
>> > On Fri, Nov 8, 2013 at 3:26 AM, Linda Alvord
>> > <[email protected]
>> > >wrote:
>> >
>> > > What does return. Mean?
>> > >
>> > > Also:
>> > >
>> > > a=:*
>> > > b=:i:5
>> > > c=:i.3
>> > > d=:(<c),<b
>> > >
>> > > c */b
>> > > 0 0 0 0 0 0 0 0 0 0 0
>> > > _5 _4 _3 _2 _1 0 1 2 3 4 5
>> > > _10 _8 _6 _4 _2 0 2 4 6 8 10
>> > >
>> > > (>{.d)
>> > > 0 1 2
>> > > >}.d
>> > > _5 _4 _3 _2 _1 0 1 2 3 4 5
>> > >
>> > > (>{.d) */ >}.d
>> > > 0 0 0 0 0 0 0 0 0 0 0
>> > >
>> > > _5 _4 _3 _2 _1 0 1 2 3 4 5
>> > >
>> > > _10 _8 _6 _4 _2 0 2 4 6 8 10
>> > >
>> > > $ (>{.d) */ >}.d
>> > > 3 1 11
>> > >
>> > > So:
>> > >
>> > > (>{.d) */, >}.d
>> > > 0 0 0 0 0 0 0 0 0 0 0
>> > > _5 _4 _3 _2 _1 0 1 2 3 4 5
>> > > _10 _8 _6 _4 _2 0 2 4 6 8 10
>> > >
>> > > Why is ({.c) a list and (}.d) a table?
>> > >
>> > > Linda
>> > >
>> > >
>> > > -----Original Message---
>> > > From: [email protected] [mailto:
>> > > [email protected]] On Behalf Of Raul
>> > > Miller
>> > > Sent: Thursday, November 07, 2013 10:07 AM
>> > > To: Programming forum
>> > > Subje.ct: Re: [Jprogramming] Times Table Therapy
>> > >
>> > > We can replace
>> > > g=: 1 :',.(":u),":,.>{.y'
>> > > with
>> > > g=: 1 :',.(":5!:5<''u''),":,.>{.y'
>> > >
>> > > A problem is that u is a verb in your example, and you want a
>> > > noun representation of it.
>> > >
>> > > But this runs into a problem:
>> > >
>> > > g=: 1 :',.(":5!:5<''u''),":,.>{.y'
>> > > * g (<i.3),<i.5
>> > > |value error: y
>> > > | ,.(":5!:5<'u'),":,.>{. y
>> > >
>> > > We need an unquoted reference to u (or one of the other such
>> > > names), or x and y are interpreted to mean u and v.
>> > >
>> > > So:
>> > >
>> > > g=: 1 :',.(":5!:5<''u''),":,.>{.y return. u'
>> > > * g (<i.3),<i.5
>> > > *
>> > > 0
>> > > 1
>> > > 2
>> > >
>> > > Does this make sense?
>> > >
>> > > --
>> > > Raul
>> > >
>> > >
>> > >
>> > > On Wed, Nov 6, 2013 at 11:07 PM, Linda Alvord
>> > > <[email protected]
>> > > >wrote:
>> > >
>> > > > This the data I want to use:
>> > > >
>> > > > a=:*
>> > > > b=:i:5
>> > > > c=:i.3
>> > > > ]d=:(<c),<b
>> > > > ┌─────┬──────────────────────────┐
>> > > > │0 1 2│_5 _4 _3 _2 _1 0 1 2 3 4 5│
>> > > > └─────┴──────────────────────────┘
>> > > >
>> > > > This this is the correct result with the wrong data:
>> > > >
>> > > > a=:'*'
>> > > > f=: 1 :',.(":u),":,.>{.y'
>> > > > a f d
>> > > > *
>> > > > 0
>> > > > 1
>> > > > 2
>> > > >
>> > > > Here is the error I can’t fix:
>> > > > a=:*
>> > > > g=: 1 :',.(":u),":,.>{.y'
>> > > > a g d
>> > > > |domain error: a
>> > > > | ,. (":u),":,.>{.y
>> > > >
>> > > >
>> > > > Is there a way to make g work correctly?
>> > > >
>> > > > Linda
>> > > >
>> > > >
>> > > > -----Original Message-----
>> > > > From: [email protected] [mailto:
>> > > > [email protected]] On Behalf Of Roger
>> > > > Hui
>> > > > Sent: Sunday, November 03, 2013 7:02 PM
>> > > > To: Programming forum
>> > > > Subject: Re: [Jprogramming] Times Table Therapy
>> > > >
>> > > > > My first adverb! Linda
>> > > >
>> > > > Given the topic and the person, it seems appropriate to point
>> > > > out that
>> > > Ken
>> > > > Iverson credited Linda Alvord for getting over a pedagogic hurdle.
>> > > > From *Kenneth
>> > > > E. Iverson <http://www.jsoftware.com/papers/autobio.htm>*,
>> > > > 2008,
>> > section
>> > > > 5:
>> > > >
>> > > > There were also surprises in the writing. Although the great
>> > > > utility
>> > of
>> > > > matrices was recognized (as in a 3-by-2 to represent a
>> > > > triangle), there
>> > > was
>> > > > a great reluctance to use them because the concept was
>> > > > considered to be
>> > > too
>> > > > difficult.
>> > > >
>> > > > Linda Alvord said to introduce the matrix as an outer product —
>> > > > an idea that the rest of us thought outrageous, until Linda
>> > > > pointed out that
>> > the
>> > > > kids already knew the idea from familiar addition and
>> > > > multiplication
>> > > tables.
>> > > >
>> > > >
>> > > >
>> > > >
>> > > > On Sun, Nov 3, 2013 at 7:04 AM, Linda Alvord
>> > > > <[email protected]
>> > > > >wrote:
>> > > >
>> > > > >
>> > > > > My first adverb! Linda
>> > > > >
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