– 1. Implement a rule spiralDown, that takes a positive integer n and produces a list that counts down from n to 0, including negative copies of the values along the way. For example: spiralDown(4,L) should respond with: L=[4,-4,3,-3,2,-2,1,-1,0] – 2. Implement a rule secondCousinOnceRemoved. For example, secondCousinOnceRemoved(scott,gail) would return true if gail is scotts second cousin, once removed. Assume that the only facts available, are parent facts of the form: parent(x,y), meaning that person x is a parent of person y. For a definition of what is meant by a second cousin, once removed, see the wikipedia page: http://en.wikipedia.org/wiki/Cousin – 3. Solve the following cryptarithmetic problem using Prolog: P I N G P O N G + F U N I G N I P Each of the 7 different letters stands for a different digit. The aim is to find a substitution of digits for the letters such that the resulting sum is arithmetically correct. Your program should find ALL answers, that do not have leading zeros. It should be possible to query your solution in this manner: ?- crypto(P,I,N,G,O,F,U). Your solution should then produce all of the combinations of the digits that satisfy the addition problem above. Dont get confused between the letter O and the number 0 (zero). make sure you never let P=N, or G=F, etc all of the distinct letters have to stand for distinct digits. Use generate-and-test. – 4. Implement a rule called stepWith, that takes three parameters, a list L and two integers i and j. The rule returns true, if the value i can be stepped into the value j, using only legal steps. The list L provides a set of integers that make up the legal steps. For example, stepWith([7,12,19],6,32) would return true, because, starting at 6, there exists at least one sequence of additions using only the numbers in the list (7, 3, and 12), producing 28. Namely: 6+7+7+12 = 32. By contrast, stepWith([7,12,19],6,31) would be an example that should return false, since there is no sequence of additions starting from 6, and using only the values 7, 12, and 19, that results in 31. Make sure that your rule works for various values, and various size lists for L. You can assume that all of the integers are positive, and that i is less than j. – 5. Convert firstMidLast from the homework#2 Scheme assignment, to Prolog. Of course, you will need to add an additional parameter to hold the answer. For example, firstMidLast([10,11,12,13,14,15,16],Answer). should respond: Answer = [10,13,16] You may use your Scheme solution (or the posted one) as a starting point (although that isnt necessary). – Submit your solutions in a single file.
Programming
[Solved] SOLVED: Homework #3 solution
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