Power in Cornwall

An old folk riddle:

As I was going to St Ives
I met a man with seven wives
Every wife had seven sacks
Every sack had seven cats
Every cat had seven kits
Kits, cats, sacks, wives –
How many were going to St Ives?

One – no calculation required – everyone else was coming from St Ives! From which we see the riddle is older than the railways.

So, what if the narrator met the polygamist on a train? Let’s suppose the polygamist is to be counted too. We start with him. And his wives.

In a scalar language we can loop 5 times, multiplying by 7 each time an initial seed of 1, and adding it to an accumulator.

count = 1; // me
for(i=0;i<5;i++) {
    count+= 7**i;
}
printf(count);

An array solution starts by listing the travellers.

      1 7            ⍝ a man with 7 wives
1 7
      +/1 7          ⍝ count them
8
      +/1 7 49       ⍝ 7 wives had 7 sacks
57

Hang on. There’s a pattern here.

      7*0 1 2        ⍝ 7 raised to consecutive powers
1 7 49
      +/7*0 1 2
57

The APL * means raise to a power. (For multiplication, we use the multiply sign × – how crazy is that?)

      +/7*0 1 2 3 4  ⍝ wives, sacks, cats, kits
2801
      +/7*⍳5         ⍝ ie first 5 integers
2801
      1++/7*⍳5       ⍝ and me
2802

But for vectors u and v and functions f and g, f/u g v is a degenerate case of u f.g v, which leads the way to expressions that work for arrays with more dimensions, so we could equally write:

      1+7+.*⍳5
2802

(Crowded train, I’d say.)

Works in q as well, where the syntax requires parens around the reduction:

q)1+(+/)7 xexp til 5
2802f
q)

You can experiment with the APL expressions in this post at TryAPL.org.

St Ives, by the way, is a charming seaside town in Cornwall. More power to it.