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A Faster Radix Sort

by lft

This article describes an implementation of radix sort optimized for sprite multiplexers. The input is a set of actors numbered 0 to N-1, where each actor has a Y-position in the range 0–223 stored in an array (called ypos) on the zero-page. The routine will push the N actor numbers on the stack, in order of increasing Y-coordinates.

For a real multiplexer, it is actually preferable to either:

  • Push them in order of decreasing Y-coordinates, so they can be popped in top-to-bottom order, or
  • Not push them at all, but instead let the multiplexer traverse a linked list.

It is straightforward to modify the presented algorithm to meet the above requirements, and this will be discussed at the end. But the code is easier to understand if everything is kept in forward order during the sort.

The complete implementation is included at the end of the article, but the code is probably difficult to follow unless you read the text first.


The total execution time for 32 actors is 1970 cycles, which is 61.6 cycles per actor. Every additional actor adds 51 cycles to the execution time. The execution time is predictable, i.e. jitter-free.

The size of the code is about 2 kB, and it needs 60 bytes of zero-page storage in addition to the Y-positions. It is possible to reduce this to 32 bytes at the cost of a few extra cycles; doing so is left as an exercise for the reader.


We will perform a two-pass radix sort based on hexadecimal digits.

In the first pass, we place each actor in one of sixteen bins, based on the low nybble of their Y-position. We use linked lists to represent the bins, but the order of elements in each list is unimportant at this stage.

In the second pass, we take the actors out of the bins, starting with bin 0, then bin 1, etc., and place each actor at the end of one of fourteen lists, based on the high nybble. In this way, we end up with fourteen sorted lists: The first list contains every actor with a Y-position in the range $00–$0f, in order. The second list has all the actors with positions $10–$1f, in order. The fourteenth list has all the actors with positions $d0–$df, in order.

Finally, we traverse each of the fourteen lists in order, pushing the elements on the stack as we go. This completes the radix sort.

The linked lists

A linked list is typically represented by a global variable, indicating the first element of the list, and a next-pointer for each actor. End-of-list can be indicated by an invalid actor number, such as $ff.

Now, a common way of adding an element to a linked list is the following (in pseudo-code):

        next[i] = first
        first = i

This inserts the new element at the beginning of the list. If i is already in a register, this can be implemented with one load instruction and two store instructions.

But we will do something else: We introduce a tail-pointer, a variable to keep track of the address of the current end of the list. This variable might point to the variable called first, or it might point to one of the cells of the array called next. Now we can add an element to the end of the list like this:

        *tail = i
        tail = &next[i]

If we make sure that the next array and the first variable are located on the same page in memory, then we only need to update the LSB of the pointer. Furthermore, if we place the next array at the very beginning of a page, the pointer to element i becomes equal to i. Thus:

        *tail = i
        tail.lsb = i

If i is already in a register, this can be implemented with just two store instructions—eliminating the need for a load.

In our algorithm, the actors are never located in more than one list at the same time, so in principle we only need a single global array of next-pointers. But it turns out that we can save a couple of cycles if we use two arrays, one for each pass of the radix sort.

In the first pass, we need sixteen first-pointers and sixteen tail-pointers (one for each bin). In the second pass, we need fourteen first-pointers and fourteen tail-pointers.

Chaining lists together

After the first stage, we have sixteen lists. In the second stage, we traverse the elements of each of these lists in order. Observe that this is essentially the same as if we first concatenate the sixteen lists into one, and then traverse all the elements of that combined list.

Thanks to the tail-pointer representation, concatenating two lists is easy:

        *tail_A = first_B

To concatenate several lists, we have to start at the end and move towards the beginning of the final list:

        *tail_C = first_D
        *tail_B = first_C
        *tail_A = first_B

This is because some of the lists might be empty, in which case their first values would be uninitialized. Concatenating in the correct order ensures that each first value is valid just before we need it.

Afterwards, first_A indicates the first actor of the concatenated list, and we know that the list must contain exactly N elements (since all actors are present). We can traverse the list using an unrolled loop, stopping after the Nth element. Thus, we don't need to bother with an end-of-list marker at all. The final next-pointer can contain garbage.

The algorithm

The complete algorithm, in pseude-code, is as follows:

Stage 1: Reset the tail-pointers

        for j = 0 to 15:
                tail[j] = &first[j]

Stage 2: Collect actors by low-nybble

        for each actor i:
                j = ypos[i] & $0f
                *tail[j] = i
                tail[j] = &next[i]      // i.e. tail[j].lsb = i

Stage 3: Concatenate the lists

        for j = 14 down to 0:
                *tail[j] = first[j + 1]

        i = first[0]    // Prepare a list iterator for Stage 5

Stage 4: Reset the tail-pointers

        for j = 0 to 13:
                tail[j] = &first[j]

Stage 5: Traverse intermediate list, collect actors by high-nybble

        repeat as many times as there are actors:
                j = ypos[i] >> 4
                *tail[j] = i
                tail[j] = &next[i]      // i.e. tail[j].lsb = i
                i = next[i]

Stage 6: Concatenate the lists

        for j = 12 down to 0:
                *tail[j] = first[j + 1]

        i = first[0]    // Prepare a list iterator for Stage 7

Stage 7: Push the elements in order

        repeat as many times as there are actors:
                push i on stack
                i = next[i]

Indexed indirect mode

We need one more puzzle piece to see why this is efficient on the 6502. The indexed indirect addressing mode is rarely useful, but here it suddenly shines!

We keep the sixteen tail-pointers on the zero-page. Suppose we have the current actor number (i) in the accumulator, and the desired list index (j) times two in the X register. Then:

        sta     (tail_pointers,x)       ; *tail[j] = i
        sta     tail_pointers,x         ; tail[j].lsb = i

As discussed previously, the next-pointers have to be located at the beginning of a page, and the first-pointers have to be located somewhere on the same page. But the first-pointers don't have to be consecutive. They will be immediate-operands in the code of Stages 3 and 6 (list concatenation).

6502 implementation

In order to truly minimize execution time, we will use separate memory areas for the two passes. This allows us to initialize both areas at once.

.var N_ACTOR = 32               // Must be even.
        // Zero-page variables:

ypos:   .fill   N_ACTOR, 0      // External input.

loptr:  .fill   32, 0
hiptr:  .fill   28, 0
        lda     #>lonext
.for(var i = 0; i < 16; i++) {
        sta     loptr+i*2+1
        lda     #>hinext
.for(var i = 0; i < 14; i++) {
        sta     hiptr+i*2+1
        // Stages 1 and 4:

        // The clean version:

        // .for(i = 0; i < 16; i++) {
        //      lda     #<join_lo+1+i*4
        //      sta     loptr+(15-i)*2
        // }
        // .for(i = 0; i < 14; i++) {
        //      lda     #<join_hi+1+i*4
        //      sta     hiptr+(13-i)*2
        // }

        // But instead we do the following:

        ldx     #$fb
.for(var i = 0; i < 8; i++) {
        lda     #<join_lo+1+i*8+4
        sax     loptr+(15-i*2)*2
        sta     loptr+(14-i*2)*2
        .if(i != 7) {
        sax     hiptr+(13-i*2)*2
        sta     hiptr+(12-i*2)*2
        // 2 + 8 * 14 - 6 = 108

        // Stage 2:

.for(var i = 0; i < N_ACTOR; i++) {
        ldy     ypos+i
        ldx     lobits,y
        lda     #i
        sta     (loptr,x)
        sta     loptr,x
        ldy     #0
        jmp     join_lo

        // 32 * 19 + 5 = 613

        .align  $100
        .fill   $100, (i & 15) * 2
        .fill   N_ACTOR, 0

        // Stage 3:

        .align  8
.for(var i = 14; i >= 0; i--) {
        lda     #0      // operand is first[i+1]
        sta     (loptr+2*i),y
        lda     #0      // operand is first[0]

        // 15 * 8 + 2 = 122

        // Stage 5:

.for(var i = 0; i < N_ACTOR; i++) {
        ldy     ypos,x
        ldx     hibits,y
        sta     (hiptr,x)
        sta     hiptr,x
        .if(i != N_ACTOR - 1) {
        lax     lonext,y
        ldy     #0
        jmp     join_hi

        // 2 + 32 * 24 - 6 + 5 = 769

        .align  $100
        .fill   $100, (i >> 4) * 2
        .fill   N_ACTOR, 0

        // Stage 6:

        .align  8
.for(var i = 12; i >= 0; i--) {
        lda     #0      // operand is first[i+1]
        sta     (hiptr+2*i),y
        lda     #0      // operand is first[0]

        // 13 * 8 + 2 = 106

        // Stage 7:
.for(var i = 0; i < N_ACTOR; i += 2) {
        lax     hinext,y
        .if(i != N_ACTOR - 2) {
        lda     hinext,x
        // 16 * 16 - 4 = 252

        // Total cycle count:
        // 108 + 613 + 122 + 769 + 106 + 252 = 1970

        // Note: Don't rts here, since there is data on the stack.


Having the actor numbers on the stack isn't all that useful. If you'd rather traverse the list during the visible portion of the display, as part of the multiplexer, simply replace all of Stage 7 by:

        sta     next_actor

And then, to obtain each successive actor:

        ldx     next_actor
        lda     hinext,x
        sta     next_actor

        // Do something with actor number X ...

Sometimes you do want the actor numbers on the stack, but you want them pushed in bottom-to-top order so the multiplexer can pick them up in top-to-bottom order. This can be achieved by modifying the look-up tables. Remember that the Y-coordinate is in the range $00–$df. Flip the range to reverse the sort:

        .fill   $100, (($df - i) & 15) * 2
        .fill   $100, (($df - i) >> 4) * 2
base/a_faster_radix_sort.txt · Last modified: 2020-04-08 00:03 by lft