RogerBW's Blog

The Weekly Challenge 192: Equal Flips For All 27 November 2022

I’ve been doing the Weekly Challenges. The latest involved a binary negation and list division. (Note that this is open until 27 November 2022.)

Task 1: Binary Flip

You are given a positive integer, $n.

Write a script to find the binary flip.

This turns out to be: take the binary representation (with no leading zeroes), then exchange all 0s and 1s to get an inverse value.

This is clearly intended to be a problem about binary representations. But we don't need to do that. Instead, I find the value of the leftmost set bit in the input by counting repeated shift rights until there's nothing left – e.g. 5 is processed as 101, 10, 1, 0, so three shifts. (I.e. log base 2.) Then shift 1 left that many times (getting 8). Subtract 1: 7, which gives me what in binary would be a series of "1" of the same length as the original number. Then subtract the input from that (gosh, an actual legitimate "take away the number you first thought of").

This sequence is in the OEIS and the graph makes this approach particularly obvious.

Some languages have access to CPU functions to count leading zeroes and thus get a binary magnitude that way, but I did this the same way throughout. JavaScript:

function binaryflip(n) {
    let m = n;
    let r = 0;
    while (m > 0) {
        m >>= 1;
        r += 1;
    return (1 << r) - 1 - n;

Task 2: Equal Distribution

You are given a list of integers greater than or equal to zero, @list.

Write a script to distribute the number so that each members are same. If you succeed then print the total moves otherwise print -1.

"Distribute" turns out to mean "subtract one from a list value, adding it to another adjacent list value". So this has to be done iteratively [0 3 0][1 2 0][1 1 1] = 2 steps , adding and subtracting 1 each time. If a number is higher than both its neighbours, the process is not immediately obvious, but this one seemed to work.


sub equaldistribution(@list) {

Check that an even distribution is possible.

    my $s = @list.sum;
    if ($s % @list.elems != 0) {
        return -1;

Get a target value for each element.

    my $m = $s div @list.elems;
    my $out = 0;
    my @w = @list;
    while (True) {
        for (0..@w.elems-2) -> $i {

Going through the list, if this value is higher than it needs to be, move any surplus into the next element.

            if (@w[$i] > $m) {
                my $v = @w[$i] - $m;
                @w[$i+1] += $v;
                $out += $v;
                @w[$i] = $m;

If it's lower, take as much as possible from the next element (without letting it go below zero.)

            } elsif (@w[$i] < $m) {
                my $v = min($m - @w[$i], @w[$i+1]);
                @w[$i+1] -= $v;
                $out += $v;
                @w[$i] += $v;

If the job's done, exit, otherwise loop again.

        if (@w.all == $m) {
    return $out;

Full code on github.

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