Jaap's Puzzle Page

Slide Rule Duel

Slide Rule Duel Slide Rule Duel Binary Bisect 5

This puzzle has moving pieces which in the top half of the frame form three overlapping semicircles, and in the bottom half only one semicircle. The bottom half can slide left and right so that its semicircle can line up with any of the three of the top half. In the solved position every quarter circle has 2 pieces. One piece in the shape of the overlapping area of semicircles, namely like a triangle with a flat base and two convex curved sides. The other piece is a thin vertical quadrangle, shaped a bit like the head of an axe. This latter piece comes in two types, mirror images of each other.
The puzzle is mixed up by lining up the lower semicircle with one of the upper ones to form a circle, and then rotating it a quarter or a half turn. Note that when quarter turns are used, some of the axe pieces will lie horizontal, and may block another circle from turning. The pieces of the top half are coloured, so that they have unique solved positions.

These puzzles were invented by Douglas A. Engel, who also invented various other puzzles such as:

There are two versions of this puzzle, both shown in the pictures above. The first is completely made of wood, and simply called the Slide Rule Duel. Douglas later used a different design, with plastic pieces and a transparent plastic front cover. He also made several other slide rule puzzles, so this puzzle got the more specific name "Binary Bisect 5".

The number of positions:

If you disregard the colours of the pieces, there are 36 possible configurations they can form. There is no easy way to see this, I simply counted them (or actually let my computer count them).
There are 8 axe pieces, 4 of each type. Given some configuration, any permutation of them is possible. This gives rise to 4!·4! possibilities.
There are 6 triangular pieces. It turns out that there are only 60 permutations that these can achieve instead of the 6!=720 that you might expect. It is tricky to show why this is the case, and I haven't been able to prove it. It is similar to what happens in the Bandaged Cube, however.
Putting all this together we get 36·60·4!·4! = 1,244,160 positions.

I have used a computer to calculate God's algorithm, and the results in the table below show that any position can be solved in at most 17 turns (13.156 on average), or 22 quarter turns (16.470 on average). These results were independently verified by Oriel Maxime.

Face turn metric




Let a clockwise quarter turn the left, middle, and right disks be denoted by the letters L, M, R respectively. Half turns will be denoted by L2, M2, and R2. Finally, anti-clockwise quarter turns will be L', M' , and R'.


Phase 1: Get the correct configuration.

  1. The diagram on the right shows all configurations that are possible. The lines connect configurations that differ by a single move. A green line is a turn of the M disc, red is an L turn, and blue an R turn. It is a simple matter to follow the lines in the diagram to the top and do the corresponding moves, so that the puzzle arrives at the standard configuration.

Phase 2: Solve the axe pieces.

  1. If the two white axe pieces are next to each other, then put them in place in the lower half of the puzzle using one of L2, M2, or R2. Then skip ahead to step d.
  2. Use L2, M2, and R2, moves to put the two white axe pieces at the far left and far right positions. This is quite easy to arrange.
  3. Bring the two white axes together in the lower half by doing L R L M2 R L R M2.
  4. If the coloured left-axe pieces are not in their correct order (red, blue, green), then put them in their correct positions using L2, M2 and R2 moves.
  5. If the orange right-axe piece is not in its correct position at the far right, then do L2 M2 L' R L M2 L' R' L'. If it is still not at the correct spot, then do the same sequence once more.
  6. If blue and green right-axe pieces need to be swapped, then do M2 L' R L M2 R' M L M' L' R L R'.

Phase 3: Solve the triangle pieces.

  1. If the triangle piece that belongs at the far left (the red one) lies at the far right, then do (M2 R2)3, i.e. repeat M2 R2 three times.
  2. Repeat (M2 L2)3 as often as is necessary to put the correct triangle piece (red) at the far left position.
  3. Repeat (M2 R2)3 as often as is necessary to put the correct triangle piece (orange) at the far right position.
  4. If the top triangles in the middle disk have to be swapped with the bottom triangles, then do M2 L' R L2 M L M2 R M R2 L R'.