Credit: Hacker's Delight, Section 2-1: Manipulating Rightmost Bits, "A Novel Application"

Given a bitmask, gives the next bitmask, in lexicographic order (a.k.a. strictly incrementing order), with the same number of bits set (a.k.a. same population count). For example: 00100011 -> 00100101.

Modified here to wraparound correctly at the end of the word, which allows you to start with any given bitmask, then know you have tried all possible cases when the next bitmask is identical to the starting bitmask. This property avoids having to calculate n-choose-k (for a k-bit bitmask in an n-bit word) as a count, or have to detect and handle the highest-valued bitmask (i.e.: 11100000) as a special case.

Implementation, for x -> y:

- s = x & -x
- r = s + x
- c = carry(s + x)
- y = r | [(1 << (popcount(x^r)-2-2c))-1]

While this version uses the relatively large Population Count module instead of the much smaller Logarithm of Powers of Two module, its variable bit-shift is only for the fixed input value of 1, which requires much less hardware than the full-word shifter found in the ntz-based version. I don't yet know whether this is a good tradeoff for speed or area.

`default_nettype none module Bitmask_Next_with_Constant_Popcount_pop #( parameter WORD_WIDTH = 0 ) ( input wire [WORD_WIDTH-1:0] word_in, output reg [WORD_WIDTH-1:0] word_out );

First, let's define some constants used throughout. Rather than expect the simulator/synthesizer to get the Verilog spec right and extend integers correctly, we defensively specify the entire word.

localparam ZERO = {WORD_WIDTH{1'b0}}; localparam ONE = {{WORD_WIDTH-1{1'b0}},1'b1}; localparam TWO = {{WORD_WIDTH-2{1'b0}},2'b10}; initial begin word_out = ZERO; end

We find the least-significant bit set in the bitmask.

wire [WORD_WIDTH-1:0] smallest; Bitmask_Isolate_Rightmost_1_Bit #( .WORD_WIDTH (WORD_WIDTH) ) find_smallest ( .word_in (word_in), .word_out (smallest) );

Then add that smallest bit to the input, causing any run of consecutive 1 bits to ripple up into the next 0 bit. (e.g.: 1001100 -> 1010000)

We also save the carry-out to later deal with the case where the consecutive 1 bits were at the left end of the word and rippled into the carry out, so we can wraparound back to the right end without information loss.

wire [WORD_WIDTH-1:0] ripple; wire ripple_carry_out; Adder_Subtractor_Binary #( .WORD_WIDTH (WORD_WIDTH) ) calc_ripple ( .add_sub (1'b0), // 0/1 -> A+B/A-B .carry_in (1'b0), .A (word_in), .B (smallest), .sum (ripple), .carry_out (ripple_carry_out), // verilator lint_off PINCONNECTEMPTY .carries (), .overflow () // verilator lint_on PINCONNECTEMPTY );

Now we compute the number of bits which changed after the ripple addition. Any changed bits are on the right side of the ripple: the left side is always unchanged, except at the limit case where the carry out is set.

wire [WORD_WIDTH-1:0] changed_bits; Hamming_Distance #( .WORD_WIDTH (WORD_WIDTH) ) calc_changed_bits ( .word_A (word_in), .word_B (ripple), .distance (changed_bits) );

We need some corrections to the Hamming Distance, which are explained later. If we have not reached the left end of the word, then we need a correction of two, else zero.

reg [WORD_WIDTH-1:0] distance_adjustment = ZERO; always @(*) begin distance_adjustment = (ripple_carry_out == 1'b1) ? ZERO : TWO; end

If only one bit was rippled leftwards, then the Hamming Distance is necessarily two (e.g.: 010010 ripples to 010100, which changes two bits). So we subtract two from the Hamming Distance to bring it to zero and call that the normal case, as no set bits were lost. (Remember: we want to find the next bitmask with the same number of set bits.)

If there was a run of 1 bits, these would all have rippled leftwards into another bit. The Hamming Distance would therefore be the number of 1 bits in that run, plus the changed bit at the left (e.g.: 00111000 -> 01000000, for a Hamming Distance of 4). We also subtract two from this Hamming Distance. We have to rebuild the lost set bits at the far right end of the word, and the corrected Hamming Distance will allow us to do that later.

If we rippled all the way into the carry bit, then the Hamming Distance is necessarily equal to the number of set bits in the bitmask, as the carry bit is not included in the Hamming Distance calculation. We need that number to wraparound and create the first possible bitmask at the right end of the word (e.g.: 11100000 -> 00000111), so we subtract zero instead.

wire [WORD_WIDTH-1:0] adjusted_distance; Adder_Subtractor_Binary #( .WORD_WIDTH (WORD_WIDTH) ) calc_adjusted_distance ( .add_sub (1'b1), // 0/1 -> A+B/A-B .carry_in (1'b0), .A (changed_bits), .B (distance_adjustment), .sum (adjusted_distance), // verilator lint_off PINCONNECTEMPTY .carry_out (), .carries (), .overflow () // verilator lint_on PINCONNECTEMPTY );

To rebuild any lost set bits, we left-shift a 1 bit by the corrected Hamming Distance....

wire [WORD_WIDTH-1:0] shifted_one; Bit_Shifter #( .WORD_WIDTH (WORD_WIDTH) ) calc_shifted_ones ( .word_in_left (ZERO), .word_in (ONE), .word_in_right (ZERO), .shift_amount (adjusted_distance), .shift_direction (1'b0), // 0/1 -> left/right // verilator lint_off PINCONNECTEMPTY .word_out_left (), .word_out (shifted_one), .word_out_right () // verilator lint_on PINCONNECTEMPTY );

...and then subtract 1 to "reverse ripple" that bit into all the bits to the right, creating the lost run of set bits at the rightmost position. If there was only one bit rippled initially, then no set bit was lost, the corrected Hamming Distance was zero, the 1 bit is not shifted, and gets subtracted to zero, without affecting other bits.

wire [WORD_WIDTH-1:0] lost_ones; Adder_Subtractor_Binary #( .WORD_WIDTH (WORD_WIDTH) ) calc_lost_ones ( .add_sub (1'b1), // 0/1 -> A+B/A-B .carry_in (1'b0), .A (shifted_one), .B (ONE), .sum (lost_ones), // verilator lint_off PINCONNECTEMPTY .carry_out (), .carries (), .overflow () // verilator lint_on PINCONNECTEMPTY );

Finally, we OR the rippled bits (which contains the unchanged left-most part, plus the new set ripple bit) with the reconstructed bits lost to the initial ripple (if any). We now have the next bitmask with the same number of set bits, in strict incrementing order (a.k.a. lexicographic order).

always @(*) begin word_out = ripple | lost_ones; end endmodule