"long-frame rationing" is my term for what happens when the noun with the larger frame has to broken up into n-cells to pair up with the noun with the smaller frame.
It is demonstrated here: http://www.jsoftware.com/help/jforc/loopless_code_i_verbs_have_r.htm#_Toc141157994 For the example: 100 200 + i. 2 3 we see long-frame rationing occur when Henry says <QUOTE>The longer-frame operand (the right one) is broken up into operand 0-cells, each being paired with a copy of the shorter-frame operand cell. Each paired _1-cell becomes a row of paired operand cells:</QUOTE> so in my solution of 1 2 3 4 5 4 3 -. 2 4 I am at the long-frame rationing step where the frames are the same length, but I cannot pair them 1 for 1 because their shape is different: Write out the entire expression: 1 2 3 4 5 4 3 -. 2 4 Expand the verb rank(*1): -. b. 0 0 _ _ Rank Calculation: ---------------- left-noun-rank: 1 right-noun-rank: 1 left-verb-rank: _ right-verb-rank: _ lesser left rank, lr = 1 lesser right rank, rr = 1 The verb will be applied to 1-cells of the left operand and to 1-cells of the right operand. left-noun-shape: 7 right-noun-shape: 2 Frame Calulation(*2): -------------------- left-frame: lf = empty right-frame: rf = empty common frame: cf = empty length of this common frame: rcf = 0 -rcf = (_0) Operand Pairing: --------------- 1. Pair by common frame consisting of (-rcf)-cells: [1 2 3 4 5 4 3][2 4] 2. Pair by long-frame rationing. If the frames are the same length, then you simply pair them 1 by 1. *** HELP NEEDED HERE *** ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
