ofut@gatech.UUCP (12/05/83)
Off the top of my head, it seems that the first time I saw this puzzle,
you could not go outside of the box. That is, the outer boundary
was not to be crossed. With that, then maybe it is indeed reducible
to the Kroningsburg bridge problem, as SPAF said.
--
Jeff Offutt
School of ICS, Georgia Tech, Atlanta GA
CSNet: Ofut @ GATech ARPA: Ofut.GATech @ Csnet-Relay
uucp: ...!{akgua,allegra,rlgvax,sb1,unmvax,ulysses,ut-sally}!gatech!ofutleei@princeton.UUCP (12/06/83)
This is an interesting problem, but it has been wildly misrepresented on the net. In the original problem, CORNERS DO NOT ACT AS NODES. Thus, when we define an arc as any line connecting two nodes in the diagram, there is only one arc at each corner of the box. By a simple reduction, we can see that this reduces to an old, familiar graph tracing problem with a graph that looks like: /-------------------*--------------------\ | /---------------/|\----------------\ | | | +---------+----|----+---------+ | | | | | | | | | | | | \------*---------*---------*-------/ | | | | | / \ | | | | | +----|----+--/-+-\--+----|----+ | | | \ / | \ / | | \-----------*---/ | \---*------------/ | \-------------/ | +--------------+--------------+ As you can see, there are six nodes (*) in the graph and four of them have odd cardinality (? I don't think this is the right word). Since you can't trace (without lifting your pencil) any graph with more than two odd nodes (remember, if you enter a node, you have to leave by a different route) then you can't fulfill the requirements for the puzzle. It's impossible. Even if you count the corners as nodes, there are still four odd cardinality nodes in the corresponding graph. It cannot be done. Needless to say, the `solutions' that have come over the net are not only suspect but wrong. It was proven some three hundred years ago that these problems are unsolvable, and it is obvious why if you think about it for more than three seconds. SO STOP IT ALREADY!!! -- ------ Lee Iverson Princeton University ..!princeton!leei
leei@princeton.UUCP (12/06/83)
References: <2651@gatech.UUCP> Relay-Version: version B 2.10.1 6/24/83; site akgua.UUCP Posting-Version: version B 2.10.1 6/24/83; site princeton.UUCP Path: akgua!clyde!ihnp4!cbosgd!mhuxl!ulysses!princeton!leei Message-ID: <159@princeton.UUCP> Date: Tue, 6-Dec-83 03:28:57 EST Organization: Princeton Univ. EECS This is an interesting problem, but it has been wildly misrepresented on the net. In the original problem, CORNERS DO NOT ACT AS NODES. Thus, when we define an arc as any line connecting two nodes in the diagram, there is only one arc at each corner of the box. By a simple reduction, we can see that this reduces to an old, familiar graph tracing problem with a graph that looks like: /-------------------*--------------------\ | /---------------/|\----------------\ | | | +---------+----|----+---------+ | | | | | | | | | | | | \------*---------*---------*-------/ | | | | | / \ | | | | | +----|----+--/-+-\--+----|----+ | | | \ / | \ / | | \-----------*---/ | \---*------------/ | \-------------/ | +--------------+--------------+ As you can see, there are six nodes (*) in the graph and four of them have odd cardinality (? I don't think this is the right word). Since you can't trace (without lifting your pencil) any graph with more than two odd nodes (remember, if you enter a node, you have to leave by a different route) then you can't fulfill the requirements for the puzzle. It's impossible. Even if you count the corners as nodes, there are still four odd cardinality nodes in the corresponding graph. It cannot be done. Needless to say, the `solutions' that have come over the net are not only suspect but wrong. It was proven some three hundred years ago that these problems are unsolvable, and it is obvious why if you think about it for more than three seconds. SO STOP IT ALREADY!!! -- ------ Lee Iverson Princeton University ..!princeton!leei
cjp@vax135.UUCP (12/07/83)
Interesting. The last two puzzles have depended on psychological "tricks" or decoys. First you have the pirate treasure problem: it tries to dash your hopes of ever solving the problem by witholding information (which turns out to not be needed). Then, you have the LOOKS SIMPLE puzzle: when properly presented, it decoys you by masquerading as a non-trick puzzle while requiring a trick solution (like tracing down the length of a segment or crossing a segment at a corner). It's kind of fun to try to see through the smokescreen in these types of puzzle. Anyone got any more? Charles J. Poirier