; The quarter square multiply, against numbers whose products are known.
;
; a * b = qs[a+b] - qs[|a-b|], and the table of quarter squares is built by adding, so nothing
; in the whole arrangement needs a multiply to exist. What this checks is that the arrangement
; actually multiplies: the cases below cover nought, one, the commutation, a square, and the
; largest product two bytes can hold.
;
; Written by Anachronaut

#Include print.asm
#Include math.asm

#Program

start:
  CALL mulReady             ; The table, once, before anything asks for a product.

  RSTA
  CALL times                ; 0 x 0
  INIA 0d7
  SETD.0 MulA
  STA.0
  RSTA
  SETD.0 MulB
  STA.0
  CALL show                 ; 7 x 0, which is the other way round from the last one.

  INIA 0d1
  CALL both
  CALL show                 ; 1 x 1

  INIA 0d12
  CALL both
  CALL show                 ; 12 x 12, a square, which is the case the identity leans on:
                            ; the difference term is nought and the whole answer is one entry.

  INIA 0d3
  SETD.0 MulA
  STA.0
  INIA 0d200
  SETD.0 MulB
  STA.0
  CALL show                 ; 3 x 200

  INIA 0d200
  SETD.0 MulA
  STA.0
  INIA 0d3
  SETD.0 MulB
  STA.0
  CALL show                 ; and 200 x 3, which had better agree.

  INIA 0xFF
  CALL both
  CALL show                 ; 255 x 255, the largest a byte times a byte can be.

  HALT

; A in both operands, for the square cases.
both:
  SETD.0 MulA
  STA.0
  SETD.0 MulB
  STA.0
  RET

; A in both, then show it. The nought case wants this and nothing else does.
times:
  CALL both
  CALL show
  RET

; The product, high byte then low, which is how a sixteen bit number reads.
show:
  CALL mul8
  SETD.0 MulHigh
  LDA.0
  CALL printByteHex
  SETD.0 MulLow
  LDA.0
  CALL printByteHex
  CALL lineFeed
  RET