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[Robert Yarger]

Derek Bosch inquired recently on why the Stikman No 7 Puzzlebox was called the beast.  The answer is that Nick Baxter produced a mathmatical equation that proved that there were 666 different difficulty combinations based on the postion and direction that drawers were placed in this box.

Obviously, you guys enjoy (and are very good at) producing algebra equations, so I thought I would provide an opportunity for everyone to double check Nick's work.  Personally, when he explained it to me, I did not understand it.

This box has four drawers.  Each drawer is identical, but has a different maze pattern on each of its 4 lengthwise sides.  (The 4 maze patterns are also in the same order and direction on each drawer, so they are entirely identical). 

The middle pull tabs that you see in the picture have a dowel running through them that insersect with the maze pattens on each side of the wall it is embedded in.  As such, at any point in time, only 2 sides (the inner touching sides) of any particular drawer effects the puzzle solution.  The solution will vary based upon the which 2 maze sides face the interior of the box, as well as the direction they are facing.

Basically, you have

4 Drawers with 8 possible ways to insert each of them in the box (this includes inserting them backwards)

Only the two (touching inner) sides of any drawer effect the solution.  The outer maze patterns are inert.

Each different drawer position variation produces a new possible solution.

Don't forget that each drawer is identical, so there is the possibility of a lot of duplicate maze combinations that should not be counted. 

Don't forget that insertion of drawers relative to each other in the box is also a factor, but there can also be duplicates of this.

Also, there are some mathmatical combinations that have to be excluded because there is no way for the maze pattern existing on the opposite side of a drawer to affect a possible solution.  Only combinations of maze patterns that directly touch eachother as sides are a valid variant because the opposite pattern is by default always facing the exterior.

It is a caculation that is too complex for me, so I wish you all good luck.  We will see if Nick's original calculations will stand!!! 

Hey, I just build them.  I don't solve them.  By the way, I want to see the equation.  That should be a work of art in itself.  
   

[Robert Yarger]

I see there were no replies to this one.  Yeah, I would be scared to approach that myself.  That is why I just took Nick at his word.  I wouldn't know the first thing about disproving him, even if he were wrong.

[nick.baxter]

There is a very slick mathematical technique to count combinations in cases where symmetry makes some cases appear the same. Sometimes such problems can be solved by inclusion-exclusion, but frequently even this technique can get too complicated.

I wrote a paper for A Tribute to a Mathemagician, a Martin Gardner tribute book published by AK Peters. I recommend you buy the book, but the article is also on my puzzle pages at http://baxterweb.com/puzzles/burnside5.pdf. The article is written for puzzle designers with some mathematical interest, and describes the mathematics is terms a little bit more understandable for puzzle people. It was a difficult goal, and some say I actually succeeded!

Counting the number of "essentially different" arrangements of the drawers for the Beast is a perfect example of this technique.

There are two primary steps. One is to understand the ways to reorient the entire puzzle. There are eight such permutations: rotate parallel to the drawers: 90,180, 270; rotate 180 at the center of the side or top, rotate 180 along a diagonal (two ways), and one way is to not move it at all.

Then for each of these cases, count the number of ways to arrange the drawers so that the puzzle doesn't appear to change when transformed using that permutation. Counting these cases turns out to be a whole lot easier than for the original problem.

I know it doesn't make sense that this should work, but it does, and that is the magic of this mathematical technique. Read the article for some simpler cases, then come back here to see how the final count works out.

The eight cases:

• no move: each drawer goes in 8 ways, so there are 8⁴ total ways.
• rotate 90 (two identical cases): once one drawer is placed, then are the other drawers are fixed. That's 8 cases twice.
  
• rotate 180: here I can place two drawers independently, fixing the other two (the one's that are diagonally opposite). That's 8² cases.
• face rotation (two identical cases): Again, I can place two drawers independently, fixing the other two. That's 8² cases twice.
• diagonal rotation (two identical cases): the two drawers along the axis of rotation can be set independently; of the remaining two, one can be set freely, fixing the other. That's 8³ cases twice.

So the total is (8⁴+2*8+8²+2*8²+2*8³)/8 = 666!

[Robert Yarger]

Nick,

Glad you finally found your way here to defend you numbers.  The debate has been going on for a while, but unfortunatly, most of the debate took place in a different area of the forum.  I had to find it by doing a search for "beast" from the home page.  Below is Jim's ending quote from elsewhere on the website, and Derek has his equation posted somewhere else.  Hopefully you have put the matter to rest, but I thought I would attempt to get all this stuff together in one place.

JIM'S RESPONSE-------

Hi Derek and Robert,

I feel somewhat gratified by Derek's number.  I spent a bit of time the other day using a standard counting argument to try to compute the number of assemblies of "The Beast" up to rotation and came up with:

4(16+15+14+...+1)=4(16)(17)/2=544.

I may have some duplications embedded in my solution somewhere.  (Disclaimer: I have not played with the puzzle in a while and so I may be overlooking something.)  In any event, I could not produce 666.   

Best,
Jim

[Canuck]

Nice to see you here Nick!  Is that a Lee Krasnow 'Third Stellation' as your avatar?  Very nice puzzle

[nick.baxter]

Actually it's Pelikan's version of Coffin's Superstar. One of the nicer pieces from my recent auction.

[nick.baxter]

Seeing that Derek got a different count by using BurrTools was compelling motivation to review my work. Unfortunately (for those who like how the Beast got its name), Derek's count is correct.

For those who care, the last case in my earlier post had the error--the count for that case should have been zero. Here are the cases again, done right this time.

The eight cases:

• no move: each drawer goes in 8 ways, so there are 8⁴ total ways.
• rotate 90 (two identical cases): once one drawer is placed, then are the other drawers are fixed. That's 8 cases twice.
  
• rotate 180: here I can place two drawers independently, fixing the other two (the one's that are diagonally opposite). That's 8² cases.
• face rotation (two identical cases): Again, I can place two drawers independently, fixing the other two. That's 8² cases twice.
• diagonal rotation (two identical cases): the two drawers along the axis of rotation must rotate to be identical with themselves, which cannot happen. So there are no instances for this case.

So the total is (8⁴+2*8+8²+2*8²+2*0)/8 = 538

Now that that's settled, the only remaining question is whether or not the Beast deserves to retain it's nickname!

[Canuck]

Actually it's Pelikan's version of Coffin's Superstar. One of the nicer pieces from my recent auction.

Wow, that is a very nice puzzle, thanks Nick

[jstrayer]

It is good to see that the 538 has been conclusively settled.  Number theoretically, 538 is relatively uninteresting.  Robert, what say you?

[Eric Fuller]

Maybe rename it the "Wanna be-ast" box?

[Robert Yarger]

And you said that you weren't witty?

[Ron Zezima]

Actually Jim, I find the number "538" very intriguing, both for the complex mathematical calculation and for the fact that the number 538 is in direct historical relation to the Beast: though perhaps somewhat less known than the number 666. Obviously, this is not a religious forum, so I will attempt to keep this entry on point. There are different views of the interpretation of Daniel and the Revelation.  Utilizing the historical interpretation, Papal Rome has long been identified as the Sea Beast of Revelation 13. There are 6 identification points. One of these 6 points contains the year 538 AD: the year in which Papal Rome enforced its civil authority (which it received from the Roman Emperor, Justinian, back in 533) to vanquish the Ostrogoths from Rome. The Ostrogoths were the last of three arian powers (Ostrogoths, Heruli, Vangals). The Papacy received unchallenged authority in western Europe in 538 AD.  According to scripture, this Beast power would rule for 1,260 prophetic days or literal years (538 + 1260 = 1798).  In 1798, Napolean's French general Berthia marched into Rome, dethroned the Pope, and took him captive. For those of us who like mathematics, the 1,260 day prophecy is also referred to as 42 months, and a time, times, and a dividing of time (see Daniel 7). Time was used to denote one year, times equals 2 years, and a dividing of time is half a year. A Biblical lunar month was 30 days. All methods total 1,260. It would seem the "538" puzzle is more interesting than ever and should retain it's original title: the Beast.  ;-)

[Canuck]

Very interesting Ron

[Kerry Verne]

538 is also the total number of electoral votes in the US election system.

[jstrayer]

Ok, ok, I stand corrected.  THE BEAST it remains!