A356480 a(n) is the minimal number of river crossings necessary to solve the missionaries and cannibals problem for n missionaries and n cannibals where the boat capacity is the minimum necessary to allow a solution.
1, 5, 11, 9, 11, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135
Offset: 1
Examples
Suppose n = 3 and that all the people must cross from the left river side to the right. Let m and c denote the number of missionaries and the number of the cannibals on the left bank of the river at any time. Let b=L if the boat is on the left bank, b=R if the boat is on the right bank. Then (m, c, b) fully captures the condition of the system. A solution of minimal length is then given by (3, 3, L)-->(2, 2, R)-->(3, 2, L)-->(3, 0, R)-->(3, 1, L)-->(1, 1, R)-->(2, 2, L)-->(0, 2, R)-->(0, 3, L)-->(0, 1, R)-->(1, 1, L)-->(0, 0, R).
References
- P. Norvig and S. J. Russell, Artificial Intelligence: A Modern Approach, Third Edition, 2010. Exercise 3.9.
Links
- R. Fraley, K. L. Cooke, and P. Detrick, Graphical Solution of Difficult Crossing Puzzles, Mathematics Magazine, Vol. 39 (3), pp. 151-157, (1966).
- Wikipedia, Missionaries and cannibals problem.
- Index entries for linear recurrences with constant coefficients, signature (2,-1).
Formula
G.f.: x*(4*x^6 - 4*x^5 + 4*x^4 - 8*x^3 + 2*x^2 + 3*x + 1)/(x-1)^2.
Comments