-//. oblique adverb: last item bottom/right

[jerome fontaine]

NB. -//. oblique adverb: last item bottom/right

(9!:14'') NB. j9.7.0-beta4/j64/linux j9.5.2 j903

j9.7.0-beta4/j64/linux/commercial/www.jsoftware.com/2025-05-05T13:14:19/clang-14-0-0/SLEEF=1

c=:>: i.6

c

1 2 3 4 5 6

(c,0)(,:,-)0,c

1 2 3 4 5 6 0

0 1 2 3 4 5 6

1 1 1 1 1 1 _6

NB. last item -6 minus six

c,:c

1 2 3 4 5 6

1 2 3 4 5 6

-//. c,:c NB. last item 6 plus six

1 1 1 1 1 1 6

(],-//.) c,:c NB. last item 6 plus six

1 2 3 4 5 6 0

1 2 3 4 5 6 0

1 1 1 1 1 1 6

NB. I thank found _6 minus six

from France

[Henry Rich]

-/ is applied to each diagonal.  When the diagonal has only one atom, -/ doesn't change it.

   9://. c,:c
1 9 9 9 9 9 6

Perhaps you need
   +//. c,:-c

1 1 1 1 1 1 _6

Henry Rich

[jerome fontaine]

it computs a speed  with high line - low line

only the last column computs low line - high line

it appears with  minus not with add

[Henry Rich]

It's not computing low line - high line in the last diagonal.  It is passing the diagonal through unchanged.

Henry Rich

[jerome fontaine]

Ok  I shall remember to inverse   -  and  +  like you suggest

why  doesn't  comput the the last diagonale?

Thank you for your response

[Henry Rich]

It does compute the last diagonal.  That diagonal is the value

   ,6

and the computation is

   -/ ,6
6

That's just how u/ is defined.  You would have to work around that to create a verb that applies u monadically when given just one item.  Note that all applications of u in u/ are dyadic.

Henry Rich

[jerome fontaine]

]  c=:c,0

1 2 3 4 5 6 0

(],-//.) c,:c

1 2 3 4 5 6 0 0

1 2 3 4 5 6 0 0

1 1 1 1 1 1 _6 0

like this, my first idea works, avoid the last diag

with the new 0,  the car is really stopped

thank you for your help.