.proc circle_help ;; This is because jmp cant jump that long!
.include "circle.inc"


;;We have named the parts of the circle as such.
;;               |
;;      q22      |      q12
;;               |
;;   q21         |         q11
;;---------------X-----------------> X
;;   q31         |         q41
;;               |
;;      q32      |      q42
;;               |
;;               v Y
;; The q stands for quarter, whe have 4 quarter, and each quarter is split into 2


draw_pixel:;;q11
        LDA byte_to_paint ;A byte containing a single 1. Coresponds to X position in the chunk.
        ORA (btp_mem_pos), Y
        STA (btp_mem_pos), Y

draw_pixel_inv_y:;;q41
        STY temp
        LDA #$07
        SEC
        SBC temp
        TAY
        LDA byte_to_paint
        ORA (btp_mem_pos_inv_y), Y
        STA (btp_mem_pos_inv_y), Y
        ;hihi:
        ;jmp hihi
        LDY temp

draw_pixel_inv_y_mirror:
        SEC
        LDA btp_mem_pos_center_two
        SBC btp_mem_pos_inv_y
        STA btp_mem_pos_inv
        LDA btp_mem_pos_center_two + 1
        SBC btp_mem_pos_inv_y + 1
        STA btp_mem_pos_inv + 1

        STX temp_
        LDX byte_to_paint
	LDA inverse_factor_value, X;; (see END.s)
	LDX temp_

	; A = byte_to_paint_inv
        ORA (btp_mem_pos_inv), Y
        STA (btp_mem_pos_inv), Y


draw_pixel_mirror:;;q31
        SEC
        ;;C = 1 beacause the branching to while_x_bigger_then_y and SEC on first ittteration.
        LDA btp_mem_pos_center_two
        SBC btp_mem_pos
        STA btp_mem_pos_inv
        LDA btp_mem_pos_center_two + 1
        SBC btp_mem_pos + 1
        STA btp_mem_pos_inv + 1

        ;; calculate the inverted y value Y = 7-Y
        STY temp
        LDA #$07
        ;; C = 1 because arithmatic above
        SBC temp
        TAY

        ;; calculate byte_to_paint_inv 00000001 --> 10000000, 00000010 --> 01000000 ... etc
        ;; uses a table!
        STX temp_
        LDX byte_to_paint
	LDA inverse_factor_value, X;; (see END.s)
	LDX temp_

	; A = byte_to_paint_inv
        ORA (btp_mem_pos_inv), Y
        STA (btp_mem_pos_inv), Y

        ;;Recover the Y value (we changed it because evrything is inverted)
        LDY temp
        RTS
.endproc