Ain't it funny how the knight moves?

This is a good advertisement for the approach I came up with eventually: if you don't see a trivial way to the next square, find a sequence that will lead to it that starts in the lower right quadrant, then go to that quadrant and get to that square. (The lower right quadrant is the center ring in this diagram.)

I'm used to a version with pawns that involves less backtracking, so it took me a little while to stop trying to be "smarter" and just be okay with continually returning to home base.

Yes, start with the end square and work your way backwards. It is often only a single path that leads you there for the final steps. Once you develop the path close enough back to where you are then it’s easier to see how you can get onto it. Then just follow it.

Incidentally, this is also a good strategy for planning and breaking down goals in real life: begin with the end goal, then work your way backwards.

here's a script that will highlight previously visited squares as red. as a nice bonus it replaces the instructions with more accurate ones

  // Get all square elements
  squareElements = [...document.querySelectorAll("[data-testid='white-square']"), ...document.querySelectorAll("[data-testid='black-square']")];
  
  // create array of elements that have been landed on
  // these need to be set to red every time because the board resets every move
  let modifiedElements = [];
  observer = new MutationObserver(function (mutations) {
    mutations.forEach(function (mutation) {
      if (mutation.addedNodes.length > 0) {
        mutation.addedNodes.forEach(function(addedNode) {
          key = addedNode.getAttribute('data-testid').split('-')[1];
          if (!modifiedElements.includes(key)) {
            modifiedElements.push(key);
          }
          for (let i = 0; i  {
              return e.getAttribute('data-squareid') === modifiedElements[i]
            })
            el.style.backgroundColor = 'red';
          }
        });
      }
    });
  });
  
  // observe all square elements
  for (let i = 0; i