Riddle 1 Answer

I appear to have left Riddle 1 sitting out there without an official answer for almost seven months now. Sorry about that.

The answer given by Peter Byrne was valid, and essentially the one I was thinking of: while his example dealt with the untyped empty list (), I had the typed empty list `boolean$() in mind.

The insight here is that any and all are forms of min and max; and that min x,y, the min of the concatenation of two lists, is equal to min(min x;min y), the min of their separate mins (and mutatis mutandis for max). For this to work consistently for empty lists, the min of an empty list must be the maximum possible value for that data type (and mutatis mutandis for max).

Quick Tip: Intra-Statement Breakpoints

A statement referencing a non-existent variable is a common way to add a breakpoint to a q function.

q)f:{x:x+1;break;x+2}

While this is handy, it only lets you break between statements. A simple extension lets you break within statements, inspecting values at arbitrary points in the code, and continuing with execution once you’re done:

q)f:{x:x+1;{break;x}x+2}

Now, the break will occur after the portion of the statement to the right of the break function has executed, and x within the break function will have the value returned by that code. Since the break statement itself has no effects, and x contains the value returned by the code executed so far, typing : to continue will allow the rest of the break function to execute, returning x leftwards and allowing the rest of the statement to execute as if nothing had interrupted it.

Note that this is entirely legal inside qsql queries; I’ve often found it of particular utility there, since you can’t create new local variables inside a query. Unfortunately no specific examples come to mind at the moment; I’ll try to post one later to make it clearer how this technique works.