A bit of news: The basic Congruent partition problem has been listed at:
http://maven.smith.edu/~orourke/TOPP/P73.html
Question: Given a convex polygon P and a number N so that P can somehow be cut into N mutually congruent *non-convex* pieces with no wastage - ie. a 'perfect congruent partition' exists for P for that N. Then is it true that P necessarily also allows a partition into N mutually congruent convex pieces?
In other words, can we say: "there is no P and N such that P can be perfectly congruent-partitioned into N pieces only when the pieces are necessarily non-convex"?
For N = 2, the above question seems to have an affirmative answer - no convex polygon appears to have only a congruent 2-partition such that pieces are non-convex. Beyond N=2, nothing much seems to be known.
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