(Unicode APL chars in this message! and a spoiler if you haven't yet found the solution!)
I love showing off the APL (from the original book) solution for this to people who have heard nothing of array-based languages: V +.× ∊ ∨.= N ∘.| V which finds the sum of the numbers in the vector V which are multiples of any number in the vector N. (You have to picture +.× as one above the other for full effect, just as they are here: http://www.jsoftware.com/papers/APL.htm <http://www.jsoftware.com/papers/APL.htm>; the epsilon has a bar above it as well, but I can't type that.) Anyhow, the previous solution is easily translated into ISO APL: V←⍳999 N←3 5 V+.×0∨.=N∘.|V or into tacit J: 3 5 (] +/ . * 0 +./ . = |/) i.1000 If you're just beginning to learn J, you should know that while it might have been elegant to write in APL, dot products like +./ . = and +/ . * should be avoided in J, and be replaced by the faster and easier-to-understand +./@:= and +/@:* (they are only interchangeable if the left and right arguments are vectors however!). If I "explicate" (render explicit) the previous tacit solution (and remove the dot products), it gives: V=: i.1000 N=: 3 5 +/ V * +./ 0 = N |/ V The |/ makes a table of remainders from the divisions of all the num- bers in V (along the rows) by all the numbers in N (vertically). 0 = … looks for 0-remainders, which correspond to multiples of numbers in N in V. +./ ORs the lines together, so you get a vector which contains a 1 iff the corresponding number in V is a multiple of ANY number in N, and a 0 otherwise (*./ would correspond to multiples of ALL the numbers in N). Multiply by V, you get a vector of 0s and of multiples of any number in N. All that is needed then is to take the sum +/ . Try playing around with this a little: instead of multiplying by the boolean vector that corresponds to the multiples (0 +./@:= |/) you can use # (copy or compress if x is boolean). You can also check for multiples by seeing if N = the GCD table of N and V ([ = +./) or by checking if V = the rows of the LCM table (] ="1 *./) and try out the rank conjunction. Best regards, Louis ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
