TWC 205: Exclusive Third Or First

In which we ponder the highest bit, and find a much faster Max_XOR.

TWC Task #1 - Third Highest


Given an array of integers, return the Third Highest if found, otherwise return the maximum.


  • The last example shows that duplicate values are to be suppressed; task1(5,4,4,3) should return 3.

  • The very efficient single-pass algorithm for third-highest-maximum is very inefficient, w.r.t. programmer time, so I just sorted.


use v5.36;
use List::Util qw<uniq>;
sub task1 ( $ns ) {
    my ( $x1, $x2, $x3 ) = sort { $b <=> $a } uniq @{$ns};
    return $x3 // $x1;


sub task1 ( @ns ) {
    (.elems == 3 ?? .head !! .tail) given @ns.sort.squish.tail(3);
  • .squish is more efficient than .unique, and can be used as an exact replacement when the list is already sorted.

  • given aliases the RHS expression into $_, allowing the concise (elems,head,tail) method calls.

TWC Task #2 - Maximum XOR


Given an array of integers, find the highest value obtained by XORing any two distinct elements.


use v5.36;
use ntheory    qw<forcomb>;
use List::Util qq<max>;
sub task2 ( @ns ) {
    my $r = 0;
    forcomb {
        $r = max( $r, $ns[$_[0]] ^ $ns[$_[1]] )
    } @ns, 2;
    return $r;


sub task2 { @_.unique.combinations(2).map({ [+^] .list }).max }

[+^] is the "reduction" form of the +^ XOR operator; it XORs all the things on its right, which in this case are both of the elements in the two-element list passed to map.

That simple .combinations(2) works fine, with it's N*(N-1)/2 performance, until the input array gets large.

With a bit of analysis, we can go much faster!

sub task2_fast ( @ns ) {
    sub hi-bit ( UInt $n ) { $n.log2.floor }

    my %grouped = @ns.unique.classify(&hi-bit);

    my ($top_group, @lesser_groups) = %grouped.sort(-*.key)

    if !@lesser_groups {
        my $removes_hi-bit = 1 +< hi-bit($top_group[0]);
        return task2_fast( $top_group.list »-» $removes_hi-bit );
    else {
        return [max] {
            ($top_group.list X+^ .list).max;

What we know:

  • XOR on single bits:

    • 0^0 == 0
    • 0^1 == 1
    • 1^0 == 1
    • 1^1 == 0
  • XOR is commutative (order does not matter): a^b == b^a

    • So, we can use .combinations(2) instead of @ns Xop @ns Cartesian product.
  • a^a==0, which is the minimum possible result, and is only achievable with a^a.

    • x^y == 0 implies x==y
    • So, .unique is safe (and helpful); duplicate elements XOR to zero.
    • Ack! Corner case not handled: @ns.elems > 1, but @ns.unique.elems == 1.
  • Raku has 3 XOR ops:

    • xor short-circuiting boolean XOR, low precedence
    • ^^ short-circuiting boolean XOR, high precedence
    • +^ numeric bitwise XOR (which is what this task needs)
  • The highest binary bit set in an integer can be found in many ways in C and assembler. Traditionally, the rightmost bit ("ones" position) is 0, next to the left ("twos" position) is 1, next ("fours" position) is 2, and so on.

    • In Perl, floor(log($N)/log(2)) works, where floor is from the POSIX module, or from the new Perl v5.36 "builtin".
    • In Raku, this is easily done with .log2.floor. e.g. the highest bit in 3 (0b11) is 1, and the highest bit in 3 (0b100) is 2:

      $ raku -e 'for sort classify *.log2.floor, ^1024 {
          say join "\t", .key, .value.minmax.gist,
                               .value.minmax.bounds.fmt("0b%b", "..")
        Hi-Bit   RangeInDecimal   RangeInBinary
          -Inf   0..0             0b0..0b0
           0     1..1             0b1..0b1
           1     2..3            0b10..0b11
           2     4..7           0b100..0b111
           3     8..15         0b1000..0b1111
           4    16..31        0b10000..0b11111
           5    32..63       0b100000..0b111111
           6    64..127     0b1000000..0b1111111
           7   128..255    0b10000000..0b11111111
           8   256..511   0b100000000..0b111111111
           9   512..1023 0b1000000000..0b1111111111
  • Even if all we know about two integers is the highest bit set in each, we still know something about their XOR. Given A > B:

    • If A and B have different highest bits, A has the higher of the two highest bits, and A XOR B will have that highest bit from A. Further, (A XOR B) will always be greater than B.
    • Otherwise, A and B have the same highest bit, so A XOR B will clear that bit. Further, (A XOR B) will always be less than A.

We will see concrete examples of these "different highest bits" and "same highest bits" later. For now, we posit that some benefit will be gained if we group (via .classify) the input array by the highest bit set.

    sub hi-bit ( UInt $n ) { $n.log2.floor }
    my %grouped = @ns.unique.classify(&hi-bit);

Now if we only had a way to re-combine all these groups, that guaranteed we did not miss any combinations of the original @ns elements...

Useful technique from combinatorics:

(or, The Whole Is Always Equal To The Sum Of Its Parts, But Only When Using The Correct Definition Of "Sum")

Classifying (sub-grouping) all of some list into groups, and then taking the cross of each .combinations(2) of those groups, plus the .combinations(2) within each group, will exactly equal a generic .combinations(2) of the initial list. Every two-element pairing will still occur exactly once. e.g:

$ raku -e '.say for (1..30).combinations(2)' | wc -l
435     # 30*29/2, as expected

$ raku -e '
    my @groups = (1..30).classify(* % 10)*.value); # 10 subgroups, by last-digit for fun
    for @groups.combinations(2) -> ($g1, $g2) { # All 2-combinations of groups,
        .say for $g1.list X $g2.list;           #   one group crossed with another group
    for @groups -> $g1 {                        # Each group,
        .say for $g1.combinations(2);           #   2-combinations of only itself
' | wc -l
435     # same count as the simple `.combinations(2)`, just sliced in a (potentially) more useful fashion.

Once grouped by highest set bit, we can deduce:

  • Same-bits become zero, and all in a group have the same high bit.
    • XOR of any X within a group with any Y within that same group always yield a cancellation of the high bit. The result will be in a lower group.
    • XOR of any X within a group with any Y within a lower group can never cause a cancellation of the X high bit. The result will be in the same group as X.
    • Therefore, the only time you have to cross (or .combinations(2)) within a group, is if that group is the only group.
    • Also, since no combination of any X or Y from any non-top groups can combine with XOR to generate the high bit, no groups need to be crossed with each other using X+^; only the top group needs to be crossed with every other group.
    • Whenever there is only one group, you still have a shortcut, because the max-XOR that they can generate is the same as the XOR of that group with their high-bit turned off; one or more rounds of recursion will get us to a multi-group solution, guaranteed by the initial .unique.

Extended example - "same highest bits"

Consider @ns = 876, 920, 930;. All are in "group 9" (512..1023).

876 == 0b1101101100, after bit-9 removed: 0b101101100 == 364
920 == 0b1110011000, after bit-9 removed: 0b110011000 == 408
930 == 0b1110100010, after bit-9 removed: 0b110100010 == 418

876 +^ 920 == 364 +^ 408 == 244
876 +^ 930 == 364 +^ 418 == 206
920 +^ 930 == 408 +^ 418 ==  58

So, max_XOR(876,920,930) will have the same result as max_XOR(364,408,418), because all the XOR-ed combinations of the former equal all the XOR-ed combinations of the latter. So, we recurse with the reduced numbers. Again, they are all in the same group, this time "group 8".

364 == 0b101101100, after bit-8 removed: 0b01101100 == 108
408 == 0b110011000, after bit-8 removed: 0b10011000 == 152
418 == 0b110100010, after bit-8 removed: 0b10100010 == 162

So, max_XOR(364,408,418) will have the same result as max_XOR(108,152,162), so we recurse again.

(108,152,162) are not all in one group, so they have "different highest bits", which I needed a bigger example to show.

Extended example - "different highest bits"

Consider @ns = flat 1003, 1001, 1..511;

When grouped by hi-bit, (1003,1001) are in "group 9", because their highest bit is 9: (0b1111101011, 0b1111101001). In this example, I call this group G9. All the rest are in groups between 0 and 8. In real code, they would be in separate groups, but for this example, I lump them all together, calling the super-group G0_8. How can we combine these?

G0_8 X+^ G0_8 always has bit 9 unset, because none of them have bit 9 set.
  G9 X+^   G9 always has bit 9 unset, because the XOR cleared it.
  G9 X+^ G0_8 always has bit 9   set, since the G9 element has bit 9, and nothing in G0_8 has bit 9 to counteract it.

More generally, for any case of groups with "different highest bits", we only need to check the top group against all the lower groups.

In this particular case, the simple .combinations(2) would require 513*512/2 = 131328 XORs, while this grouping only needs 2*511 = 1022 XORs. 131328/1022 == 128.500978. Even given the extra work of grouping, that will be quite a speed-up!

Worst case would be if the top group was full, and all lower groups were full. Even then, we cut the number of XORs in half.

Not done:

I expect that a further speed-up is possible, by noting that for any individual N in the top group, if a group exists whose hi-bit matches the first "zero" bit in N, no group lower than that needs searching. E.g. for 943 (group 9) (0b1110101111), the zeros are in bits 6 and 4. All elements of groups 8,7,6 must be XOR, but if group 6 had any elements, groups lower than 6 may be skipped. Otherwise, groups 5,4 must be XOR, but if group 4 had any elements, groups lower than 4 may be skipped. I did not have time to code this supposition into my solution above, nor did I verify my own logic. It might even result in a net loss of speed, given the work to find the positions of zeros in every N in the top group.

Bud: Well, let’s see, we have on the bags, Who’s on first, What’s on second, I Don’t Know is on third…
Lou: That’s what I want to find out.
-- Abbott & Costello Video

1 Comment

If modules are allowed, then the following gives pretty much optimal performance, slightly slower than O(N); heapifing an array is O(N), removing the max of the heap is O(Log2(N)). The heapify algorithm is pretty interesting, using the siftDown() algorithm the intitial pass is a surprising O(N)!


One thing I found confusing in your writeup, your perl solution takes a reference as the argument for the perl code, but then in your example you call it with a list. I chose to implement the list API.

use v5.34;
use Algorithm::Heapify::XS qw(max_heapify max_heap_shift);
sub task1 {
my @ret;
while (@_ and @ret<3) {
my $top = max_heap_shift(@_);
push @ret, $top if !@ret or $ret[-1] != $top;
return $ret[2] // $ret[0];
print task1(5,4,4,3);

Leave a comment

About Bruce Gray

user-pic "Util" on IRC and PerlMonks. Frequent speaker on Perl and Raku, but infrequent blogger.