Compressing Rather Large Table in APL

Question

I am currently working on finding a palindrome of made from multiplying numbers of length 3 together (from 900 to 1000). I am able to generate a table of palindromes in binary (1/0). However, the compress function (/) is giving me an error regarding storage. I am not sure why compressing the given binary table would cause such an issue.

new_set ← (900 + ⍳100)
{(⊃⍵)≡⌽⊃⍵}¨ ⊂∘⍕¨ (new_set ∘.× new_set)
{(⊃⍵)≡⌽⊃⍵}¨ ⊂∘⍕¨ (new_set ∘.× new_set) / ∘⍕¨(new_set ∘.× new_set)
WS FULL: Maximum workspace is 512 kilobytes.

Edit: Changing new_set to (⍳10) produces the same solution on the third line as the line before it (binary).. Really need help with the compress function. Any help greatly appreciated.

Answer 1

Expanding on the answer by B. Wilson,

So, the problem - We want to return the palindromes found in a multiplication table of the integers in the range of 900 to 1000.

We can start with the range:

   n ← (900 + ⍳100)

Straight-forward and as stated APL is right-associative, meaning functions apply to everything on the right, and one argument to the left. Given that, those parenthesis are superfluous and we can remove them.

Next, to produce the table:

   n ∘.× n

Great. Everyone loves outer product and as this problem is inherently O(n^2), there's not much we can do to avoid WS FULLs on larger arguments.

Here's where a subtly is introduced - do we want the unique multiplications? For example, should we include 901×901, as well as both 902×901 and 901×902, if not - there are a few things we can do to reduce space.

   ⊃,/(1↓n)×,\¯1↓n

This expression is one approach to avoid precisely those cases.

Now for the palindromes. In APL, as identified, it's idiomatic to use boolean masks to filter elements.

Your approach was

   {(⊃⍵)≡⌽⊃⍵}¨ ⊂∘⍕¨

Works well! I'll note that the ⊂ is superfluous as we immediately perform the inverse operation upon function invocation.

So, we can write:

   {⍵≡⌽⍵}∘⍕¨

Finally, the filter. You noticed that each element of the mask isn't paired up with each element of the table. Right, For non-singleton masks / will filter along major cells. For example:

   1 0 1 / 3 3⍴⍳9
1 3
4 6
7 9      

So, to get around your issue, we can ravel (flatten) both sides beforehand:

   {(,{⍵≡⌽⍵}∘⍕¨⍵)/,⍵} n ∘.× n

The complete solution! I'll posit some alternatives (minor changes) for investigation.

   {⍵/⍨(≡∘⌽⍨⍕)¨⍵}⊃,/(1↓n)×,\¯1↓n

   {⍵/⍨⍥,(≡∘⌽⍨⍕)¨⍵}∘.×⍨