• kdb+

  Flouring the loaf

  Posted by SJT on May 22, 2022 at 12:00 am

  Now heres a task that cries out for a simple solution: put a border round a matrix. My matrix is boolean and represents a QR code (yes, well come to that) but its the same problem as, say, putting a 1px border on an image.

  So lets examine it as wrapping a char matrix in spaces. Well use 3 4#"ABCDEGHIJKL".

  Amend At

  Our first strategy is to manipulate indexes, which is often an efficient approach in q. We make a larger blank matrix for the result and write the original matrix in the right place.

  Start with the shape of the matrix; that is, count the rows and columns. (Shape is a concept that q did not inherit from its ancestor APL, but is easy enough to calculate.) We use the Zen monks for a point-free expression.

  q)show M:3 4#"ABCDEFGHIJKL" "ABCD" "EFGH" "IJKL" 
  q)count each 1 first M / shape of M 3 4

  So the result shape is 5 6, and here is our blank template:

  q){n:2+count each 1 firstx; n#" "}M 
  " " " " " " " " " "

  For the last move we could use Amend Each .' to map each item of M to a row-column pair in the result. But it should be more efficient to raze M and use Amend At @ to map all its items to the vector prd[n]#" " and then reshape it. Something like

  q){n:2+s:count each 1 firstx; n#@[prd[n]#" "; ??? ;:;raze x]}M 
  " " " ABCD " " EFGH " " IJKL " " "

   

  Above, ??? is some expression that returns the target indices for the items of M. Lets start with an easy expression wrong, but easy. Well write the items of M into the first positions of the result.

  q){n:2+s:count each 1 firstx; n#@[prd[n]#" ";til prd s;:;raze x]}M 
  "ABCDEF" "GHIJKL" " " " " " "

   

  Next we come to an often-overlooked overload of vs and sv: they encode and decode different (and variable) arithmetic bases. English pounds have 100 pennies (once known as New Pence) but once had 240, of which 12 made a shilling; and 20 shillings a pound.

   

  q)240*4.50 / 4.50 in old pence 
  1080f 
  q)100 20 12 vs 240*4.50 / 4.50 was 4 10s 0d. 
  4 10 0f 
  q)%[;240]100 20 12 sv 4 10 0 / 4/10/- in decimal coinage 
  4.5 
  q)%[;240]100 20 12 sv 4 17 6 / not every sterling amount has an exact equivalent 
  4.875

   

  We can use vs and sv to convert between row-col pairs and equivalent vector indices.

  q){n:2+s:count each 1 firstx; n#@[prd[n]#" ";n sv flip 1 1+/:s vs/:til prd s;:;raze x]}M 
  " " " ABCD " " EFGH " " IJKL " " "

   

  The above has a certain elegance in that it is probably efficient for a large matrix, but it does seem a lot of code for a simple task. If our matrices are small, perhaps we can see a simpler way?

  Join

  Join , looks like an obvious candidate. (And it will lead us to something about flip we might not have known; but well come to that.) We have to apply it to each of four sides, but we have decided we dont necessarily need the fastest expression for this.

  Looks straightforward: Join for top and bottom, Join Each for the sides.

   

  q),[;" "] " ",'" ",M,'" " 
  " " " ABCD " " EFGH " " IJKL " " "

   

  Ah, not quite that straightforward. Joining an atom doesnt use scalar extension the same way Join Each does. We could count the first row

  q){row:enlist(count first x)#" ";" ",'(row,x,row),'" "}M 
  " " " ABCD " " EFGH " " IJKL " " "

   

  Better, but the refactoring itch remains.

  The simplest operation is the Join Each, which exploits scalar extension.

  When I flour an unbaked loaf, I dont daub flour over it, I roll it in the flour.

   

  q)reverse flip ,'[" "] M 
  "DHL" "CGK" "BFJ" "AEI" " " 
  q)reverse flip ,'[" "] reverse flip ,'[" "] M 
  "LKJI " "HGFE " "DCBA " " " 
  q)4{reverse flip ,'[" "] x}/M 
  " " " ABCD " " EFGH " " IJKL " " "

   

  I dont need the (admittedly tiny) overhead of a lambda to apply a series of unaries. I can use a composition.

  q)4(reverse flip ,'[" "]@)/M 
  " " " ABCD " " EFGH " " IJKL " " "

   

  Now heres a surprise: we dont need the Each.

  q)4(reverse flip ,[" "]@)/M 
  " " " ABCD " " EFGH " " IJKL " " "

   

  How does that work? It turns out that flip uses scalar extension. The items of its argument must conform; that is, they must be same-length lists or atoms. But the result will have same-length lists.

   

  q)flip M 
  "AEI" "BFJ" "CGK" "DHL" 
  q)flip M,enlist "XYZ" / must conform! 
  'length [0] flip M,enlist "XYZ" ^ 
  q)flip M,"X" 
  "AEIX" "BFJX" "CGKX" "DHLX" 
  q)

   

  Loaf floured! And the QR code? Watch this space.

  SJT replied 8 months ago 1 Member · 0 Replies
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