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An Introduction to Artificial intelligence - Assignment 1 | CS 3600, Assignments of Computer Science

Material Type: Assignment; Professor: Starner; Class: Intro-Artificial Intell; Subject: Computer Science; University: Georgia Institute of Technology-Main Campus; Term: Spring 2006;

Typology: Assignments

Pre 2010

Uploaded on 08/05/2009

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Download An Introduction to Artificial intelligence - Assignment 1 | CS 3600 and more Assignments Computer Science in PDF only on Docsity! Solve the cryptarithmetic constraint satisfaction problem in the following figure using backtracking and forward checking, as well as the MRV and least-constraining-value heuristics. (The boxes represent constraints. The box at the top reflects the AllDiff constraint, while the other boxes represent the addition constraint. X1, X2, and X3 are carry values. You can find this problem as the example Figure 5.2 from your text.) There are some places where the algorithm still allows you to select one from several values for a variable. One possible solution would be: a. Choose the X 3 variable. Its domain is {0, 1}. b. Choose the value 1 for X 3 . (We can't choose 0; it wouldn't survive forward checking, because it would force F to be 0, and the leading digit of the sum must be nonzero.) c. Choose F , because it has only one remaining value. d. Choose the value 1 for F . e. Now X 2 and X 1 are tied for minimum remaining values at 2; let's choose X 2 . f. Either value survives forward checking, let's choose 0 for X 2 . g. Now X 1 has the minimum remaining values. h. Again, arbitrarily choose 0 for the value of X 1 . i. The variable O must be an even number (because it is the sum of T + T less than 5 (because O +O = R + 10 × 0). That makes it most constrained. j. Arbitrarily choose 4 as the value of O. k. R now has only 1 remaining value. l. Choose the value 8 for R. m. T now has only 1 remianing value. n. Choose the value 7 for T . o. U must be an even number less than 9; choose U . p. The only value for U that survives forward checking is 6. q. The only variable left is W . r. The only value left for W is 3. s. This is a solution. Since the problem is under-constrained, it was not actually necessary to use backtracking to reach a solution.
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