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


;;We have named the parts of the circle as such.
;;       *       |
;;      qbb      |      qab
;;               |
;;   qba         |         qaa
;;---------------X-----------------> X
;;   qca         |         qda
;;               |
;;      qcb      |      qdb
;;               |       *
;;               v Y
;; The q stands for quarter, whe have 4 quarter, and each quarter is split into 2


draw_qaa:
        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

;; something is wrong | well isue for another time
draw_qcb:; xy swaped
        STY temp
        STX temp_
        LDA byte_to_paint
        STA temp__

        LDY #$00
        jmp goto_loop
loop:
        INY
goto_loop:
        ASL byte_to_paint; lsr asl
        BCC loop
        STY temp___


        LDX temp
       	LDA binary_factor, X; (see END.s)


        LDY temp___
        ORA (btp_mem_pos_swop), Y
        STA (btp_mem_pos_swop), Y;

;; this one is buggy
draw_qdb:; xy swaped and y is inverted.

        LDX temp
       	LDA binary_factor, X; (see END.s)
        TAX
        LDA inverse_factor_value, X;; (see END.s)

        ORA (btp_mem_pos_swop_inv_y), Y
        STA (btp_mem_pos_swop_inv_y), Y;

        SEC
        LDA #$07
        SBC temp___
        TAY
draw_qab:; xy swoped + mirroring

        LDA btp_mem_pos_center_two
        SBC btp_mem_pos_swop
        STA btp_mem_pos_inv
        LDA btp_mem_pos_center_two + 1
        SBC btp_mem_pos_swop + 1
        STA btp_mem_pos_inv + 1

        LDA #$07
        SBC temp
        TAX
        LDA binary_factor, X; (see END.s)

        ORA (btp_mem_pos_inv), Y
        STA (btp_mem_pos_inv), Y


        LDY temp
;        LDX temp_
        LDA temp__
        STA byte_to_paint
draw_qda:; y is inverted
        ;; invert Y, this is shared with qca
        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

draw_qca: ;;mirror technique
        ;SEC
        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 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)
	TAX ;; A is saved to x because qba use this as well

	; 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

draw_qba: ;;mirror_technique
        ;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

        TXA
        LDX temp_

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

increment_y_pos:
        INC Y_rel ; y++
        DEY
        BPL increment_y_pos_end
move_8px_down:
        LDY #$07
        ;; Switch to chunk bellow
        ; So we subtract  #$0140
        ; C = 1 because branching!
        Sub_16 btp_mem_pos, btp_mem_pos + 1, #$40, #$01, ! ;-320

        ;; Y is inverted
        Add_16 btp_mem_pos_inv_y, btp_mem_pos_inv_y + 1, #$40, #$01 ;+320
        ;; X and Y has swopped
        Sub_16 btp_mem_pos_swop, btp_mem_pos_swop + 1, #$08, #$00
        ;; X and Y has swoped and Y has inverted
        Add_16 btp_mem_pos_swop_inv_y, btp_mem_pos_swop_inv_y +1, #$08, #$00
increment_y_pos_end:
        RTS
.endproc