[Bug-apl] inner product

Discussion:

e***@gmx.com

2017-03-17 16:17:19 UTC

Permalink

Hi

what am i missing here?

a←b←5 3⍴⍳8

a=b

+/a=b

a+.=b length error

Christian Robert

2017-03-17 16:26:42 UTC

Permalink

a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3

Post by e***@gmx.com
Hi
what am i missing here?
a←b←5 3⍴⍳8
a=b
+/a=b
a+.=b length error

e***@gmx.com

2017-03-17 17:14:46 UTC

Permalink

thanks

i have no clue how the result of a+.=⍉b as 5x5 result is obtained from a 5x3 array and 3x5 array shouldn't it be a 15x15 result ... but i guess i really just want to use +/a=b :)

On Fri, 17 Mar 2017 12:26:42 -0400

Post by Christian Robert
a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3

Post by e***@gmx.com
Hi
what am i missing here?
a←b←5 3⍴⍳8
a=b
+/a=b
a+.=b length error

Juergen Sauermann

2017-03-17 17:44:24 UTC

Permalink

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Hi, 
 
that would be more like an outer product: 
 
 a←b←5
3⍴⍳8 
 ⍴a∘.=b 
5 3 5 3 
 
The inner product reduces away the two middle axes of the outer
product. 
 
In the ⍉b case you have 
 
 
 ⍴a∘.=⍉b 
5 3 3 5 
 
 ⍴a+.=⍉b 
5 5 
 
That is, BTW, the reason while the last axis of a and the first axis
of b must match; otherwise 
the right function of . would get a length error. 
 
/// Jürgen 
 
 
 
<div class="moz-cite-prefix">On 03/17/2017 06:14 PM, <a class="moz-txt-link-abbreviated" href="mailto:***@gmx.com">***@gmx.com</a>
wrote: 
</div>
<blockquote cite="mid:***@gmx.com"
type="cite">
<pre wrap="">thanks

i have no clue how the result of a+.=⍉b as 5x5 result is obtained from a 5x3 array and 3x5 array shouldn't it be a 15x15 result ... but i guess i really just want to use +/a=b :)

On Fri, 17 Mar 2017 12:26:42 -0400
Christian Robert <a class="moz-txt-link-rfc2396E" href="mailto:***@polymtl.ca"><***@polymtl.ca></a> wrote:

</pre>
<blockquote type="cite">
<pre wrap=""> a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3

On 2017-03-17 12:17, <a class="moz-txt-link-abbreviated" href="mailto:***@gmx.com">***@gmx.com</a> wrote:
</pre>
<blockquote type="cite">
<pre wrap="">Hi

what am i missing here?

a←b←5 3⍴⍳8

a=b

+/a=b

a+.=b length error

</pre>
</blockquote>
<pre wrap="">
</pre>
</blockquote>
<pre wrap="">

</pre>
</blockquote>
 
</body>
</html>

Juergen Sauermann

2017-03-17 16:29:24 UTC

Permalink

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<meta content="text/html; charset=utf-8" http-equiv="Content-Type">
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Hi, 
 
the number of columns in a must match the number of rows
in b (or the other way around, 
I don't quite remember). 
 
Try: 
 
 a+.=⍉b 
3 0 0 0 0 
0 3 0 0 0 
0 0 3 0 0 
0 0 0 3 0 
0 0 0 0 3 
 
/// Jürgen 
 
 
<div class="moz-cite-prefix">On 03/17/2017 05:17 PM, <a class="moz-txt-link-abbreviated" href="mailto:***@gmx.com">***@gmx.com</a>
wrote: 
</div>
<blockquote cite="mid:***@gmx.com"
type="cite">
<pre wrap="">Hi

what am i missing here?

a←b←5 3⍴⍳8

a=b

+/a=b

a+.=b length error

</pre>
</blockquote>
 
</body>
</html>

e***@gmx.com

2017-03-17 17:05:09 UTC

Permalink

yes last dimension of a must be = to first dimension of b

but then

x←1 2 3 4
y←1 3 3 3

x+.=y works

On Fri, 17 Mar 2017 17:29:24 +0100

Hi,
the number of columns in a must match the number of rows in b (or the other way around,
I don't quite remember).
a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3
/// Jürgen
Hi
what am i missing here?
a←b←5 3⍴⍳8
a=b
+/a=b
a+.=b length error

Juergen Sauermann

2017-03-17 17:18:58 UTC

Permalink

<html>
<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type">
</head>
<body bgcolor="#FFFFFF" text="#000000">
Hi, 
 
yes. Because the first dimension of a is 4 and the last
dimension of b is also 4. 
 
/// Jürgen 
 
 
<div class="moz-cite-prefix">On 03/17/2017 06:05 PM, <a class="moz-txt-link-abbreviated" href="mailto:***@gmx.com">***@gmx.com</a>
wrote: 
</div>
<blockquote cite="mid:***@gmx.com"
type="cite">
<pre wrap="">
yes last dimension of a must be = to first dimension of b

but then

x←1 2 3 4
y←1 3 3 3

x+.=y works

On Fri, 17 Mar 2017 17:29:24 +0100
Juergen Sauermann <a class="moz-txt-link-rfc2396E" href="mailto:***@t-online.de"><***@t-online.de></a> wrote:

</pre>
<blockquote type="cite">
<pre wrap="">Hi,

the number of columns in a must match the number of rows in b (or the other way around,
I don't quite remember).

Try:

a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3

/// Jürgen

On 03/17/2017 05:17 PM, <a class="moz-txt-link-abbreviated" href="mailto:***@gmx.com">***@gmx.com</a> wrote:
Hi

what am i missing here?

a←b←5 3⍴⍳8

a=b

+/a=b

a+.=b length error

</pre>
</blockquote>
<pre wrap="">

</pre>
</blockquote>
 
</body>
</html>

e***@gmx.com

2017-03-17 19:00:10 UTC

Permalink

Hi

I assume you mean the vectors both have same length as their single dimension (no first or last - the same)

making them into 1 4⍴ arrays then requires the ⍉ use

don't you treat vectors differntly in the gnuapl code from 'true arrays' with 2 (or more dimensions)?

On Fri, 17 Mar 2017 18:18:58 +0100

Hi,
yes. Because the first dimension of a is 4 and the last dimension of b is also 4.
/// Jürgen
yes last dimension of a must be = to first dimension of b
but then
x←1 2 3 4
y←1 3 3 3
x+.=y works
On Fri, 17 Mar 2017 17:29:24 +0100
Hi,
the number of columns in a must match the number of rows in b (or the other way around,
I don't quite remember).
a+.=⍉b
3 0 0 0 0
0 3 0 0 0
0 0 3 0 0
0 0 0 3 0
0 0 0 0 3
/// Jürgen
Hi
what am i missing here?
a←b←5 3⍴⍳8
a=b
+/a=b
a+.=b length error

Nick Lobachevsky

2017-03-17 21:01:26 UTC

Permalink

The key to understanding inner product is that the inner dimensions of
the arguments have to be the same. The inner dimension here is 3.

a←2 3⍴⍳6
b←3 4⍴⍳12
a
0 1 2
3 4 5
b
0 1 2 3
4 5 6 7
8 9 10 11
a+.×b
20 23 26 29
56 68 80 92

To solve this, first transpose the right argument such that the inner
dimension goes to the back of the array and both arguments have the
same number of columns.

(¯1⌽⍳⍴⍴b)⍉b
0 4 8
1 5 9
2 6 10
3 7 11
a
0 1 2
3 4 5

Do the operations for every combination of rows in a and ⍉b. As we
are doing +.×
0 1 2 × 0 4 8 is 0 4 16, +/0 4 16 is 20. First element of the result
0 1 2 × 1 5 9 is 0 5 18, +/0 5 18 is 23. Second element of the result
And so on. Loop until done.

Shape of the result is (¯1↓⍴a),1↓⍴b or 2 4

For the vector and vector case, the lengths of both vectors have to be
the same. The result is simply +/ a × b

For higher order matrices, as before, the inner dimensions are
important. The others less so.
a←2 5 1 3⍴⍳30
b←3 4 2⍴⍳12
Here the idea is to collapse (i.e. multiply together) all but the
inner dimensions, then compute the result as if both arguments were
two dimensional matrices.
a←10 3⍴⍳30
b←3 8⍴⍳24
And as before, the shape of the product is (¯1↓⍴a),1↓⍴b or 2 5 1 4 2

See also http://www.dyalog.com/uploads/conference/dyalog16/presentations/U08_SIMD_Boolean_Array_Algorithms_slides.pdf
(The part about the STAR Inner Product Algorithm)

and

http://www.jsoftware.com/papers/innerproduct/ip1.htm

e***@gmx.com

2017-03-17 22:28:06 UTC

Permalink

very nice work - thanks you for spending the time .. everthing here holds when replacing +.x with +.= (which is of more interest to me)

On Fri, 17 Mar 2017 22:01:26 +0100

Post by Nick Lobachevsky
The key to understanding inner product is that the inner dimensions of
the arguments have to be the same. The inner dimension here is 3.
a←2 3⍴⍳6
b←3 4⍴⍳12
a
0 1 2
3 4 5
b
0 1 2 3
4 5 6 7
8 9 10 11
a+.×b
20 23 26 29
56 68 80 92
To solve this, first transpose the right argument such that the inner
dimension goes to the back of the array and both arguments have the
same number of columns.
(¯1⌽⍳⍴⍴b)⍉b
0 4 8
1 5 9
2 6 10
3 7 11
a
0 1 2
3 4 5
Do the operations for every combination of rows in a and ⍉b. As we
are doing +.×
0 1 2 × 0 4 8 is 0 4 16, +/0 4 16 is 20. First element of the result
0 1 2 × 1 5 9 is 0 5 18, +/0 5 18 is 23. Second element of the result
And so on. Loop until done.
Shape of the result is (¯1↓⍴a),1↓⍴b or 2 4
For the vector and vector case, the lengths of both vectors have to be
the same. The result is simply +/ a × b
For higher order matrices, as before, the inner dimensions are
important. The others less so.
a←2 5 1 3⍴⍳30
b←3 4 2⍴⍳12
Here the idea is to collapse (i.e. multiply together) all but the
inner dimensions, then compute the result as if both arguments were
two dimensional matrices.
a←10 3⍴⍳30
b←3 8⍴⍳24
And as before, the shape of the product is (¯1↓⍴a),1↓⍴b or 2 5 1 4 2
See also http://www.dyalog.com/uploads/conference/dyalog16/presentations/U08_SIMD_Boolean_Array_Algorithms_slides.pdf
(The part about the STAR Inner Product Algorithm)
and
http://www.jsoftware.com/papers/innerproduct/ip1.htm

Nick Lobachevsky

2017-03-18 06:57:49 UTC

Permalink

Thank you.

Forgot to mention scalar extension. When a scalar is supplied as an
argument, it's lengthened to match the inner dimension of the other
argument.
(3 4⍴'cattrattfatt') +.= 't'
2 2 2
(3 4⍴'cattrattfatt') +.= 'tttt'
2 2 2
'a' +.= 3 4⍴'catratfatbat'
1 1 1 1
'aaa' +.= 3 4⍴'catratfatbat'
1 1 1 1
If both arguments are scalar (a silly case) then it would be same as
if only the second function was used.
2+.×6
12