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- Steven Minsker
- J. Algorithms
- 1989

- Carole S. Klein, Steven Minsker
- Discrete Mathematics
- 1993

- Steven Minsker
- Inf. Process. Lett.
- 1991

- Steven Minsker
- SIGCSE Bulletin
- 2007

We propose a simple new variation of the Towers of Hanoi problem, in which there are three pegs arranged in a row and there are two stacks (black, white) of n rings each, initially located on the end pegs. The object is to exchange the stacks in accordance with the usual Hanoi rules, and with the additional restriction that rings cannot move directly from… (More)

- Steven Minsker
- SIGCSE Bulletin
- 2008

We propose another simple Towers of Hanoi variant, a hybrid between classical Hanoi and linear Hanoi, in which the rules governing movement depend on ring color. An optimal algorithm is presented. The problem and its heavily recursive solution are <i>not</i> difficult; perhaps one of its more interesting facets is that the optimality proof uses simultaneous… (More)

- Steven Minsker
- Inf. Process. Lett.
- 2005

The Little Towers of Antwerpen problem is a simple variant of the Towers of Hanoi problem requiring the interchanging of two stacks (black, white) of N rings using the third tower as a spare, subject to the usual rules governing ring movement. An optimal algorithm for the problem is presented and its performance is analyzed. 2005 Elsevier B.V. All rights… (More)

- Steven Minsker
- SIGCSE Bulletin
- 2009

We continue our study, begun in [3], of the classical/linear Towers of Hanoi "hybrid" problem, in which there are three pegs arranged in a row, and the rules governing ring movement depend on ring color. Whereas [3] dealt with perfect to perfect configuration problems in a very straightforward manner, the current paper discusses deterministic and dynamic… (More)

- Steven Minsker
- 2010

f3 real. (See, for instance, [1], p. 224, ex. 4.) We offer the following proof whic h, although it also uses Bieberbach's result, is considerably different. PROOF: Let cP (z) J~ . Then cP sends the open unit disk into itself. Let cP 1= cP and inductively 1 a? define cPn=cP OcPlI t. If cPll(Z)=An,1 z+AIl,z Z2+ . .. ,it is clear that A"'=M' A I ,2= M' AII ,I… (More)

- Steven Minsker
- Discrete Applied Mathematics
- 2014

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