///////////////////////////////////////////////////////////////////////////////////////// // // Title: Pro and Con // Author: Monica Tenniel // Publication: Dell Logic Puzzles // Issue: April, 1998 // Page: 12 // Stars: 2 // // At the last meeting of the local city council, each member (Mr. Akerman, Ms. Baird, // Mr. Chatham, Ms. Duval, and Mr. Etting) had to vote on five motions, number 1 to 5 // in the clues below. Can you discover how each one voted on each motion? // // Note: a motion may have received zero or one yes vote, even though in real // life it's unlikely that both the maker and seconder of the motion would // change their minds before the motion came up for a vote. Each member voted // either yes or no on each motion; no one abstained from voting on any motion. // // Voting Chart // // 1 2 3 4 5 // Mr. Akerman // Ms. Baird // Mr. Chatham // Ms. Duval // Mr. Etting // // 1. Each motion got a different number of yes votes. // 2. In all, the five motions got three more yes votes than no votes. // 3. No two council members voted the same way on all five motions. // 4. The two women disagreed in their voting more often than they agreed. // 5. Mr. Chatham never made two yes votes on consecutive motions. // 6. Mr. Akerman and Ms. Baird both voted in favor of motion 4. // 7. Motion 1 received two more yes votes than motion 2 did. // 8. Motion 3 received twice as many yes votes as motion 4 did. // // Determine: fill in the chart (Yes/No) for each motion // ///////////////////////////////////////////////////////////////////////////////////////// // // Query: // all ProAndCon(x) // ///////////////////////////////////////////////////////////////////////////////////////// // // Result: // // x = [ {Akerman} [ {M1} 1, {M2} 1, {M3} 1, {M4} 1, {M5} 0], // {Baird} [ {M1} 1, {M2} 0, {M3} 1, {M4} 1, {M5} 0], // {Chatham} [ {M1} 1, {M2} 0, {M3} 1, {M4} 0, {M5} 0], // {Duval} [ {M1} 1, {M2} 1, {M3} 0, {M4} 0, {M5} 0], // {Etting} [ {M1} 1, {M2} 1, {M3} 1, {M4} 0, {M5} 0]] // ///////////////////////////////////////////////////////////////////////////////////////// // // Notes: // We use the number 1 for a "yes" vote, 0 for a "no" vote. // Although the number corresponding to a vote can only have one of these // two values, it was declared as "L" (large integer) as opposed to // "I" (integer). The constraints as specified in the code below are more // efficiently evaluated this way. In other words: in spite of using "L" instead // of "I" the puzzle is solved faster. // ///////////////////////////////////////////////////////////////////////////////////////// Name = Akerman | Baird | Chatham | Duval | Etting Motion = M1 | M2 | M3 | M4 | M5 Motions = Name->>Motion->L[0..1] {injection: Each member voted differently} pred ProAndCon(m::Motions) iff // 1. Each motion got a different number of yes votes. sums::Motion->>L[0..5] & {each different} sums(M1) = m(Akerman,M1) + m(Baird,M1) + m(Chatham,M1) + m(Duval,M1) + m(Etting,M1) & sums(M2) = m(Akerman,M2) + m(Baird,M2) + m(Chatham,M2) + m(Duval,M2) + m(Etting,M2) & sums(M3) = m(Akerman,M3) + m(Baird,M3) + m(Chatham,M3) + m(Duval,M3) + m(Etting,M3) & sums(M4) = m(Akerman,M4) + m(Baird,M4) + m(Chatham,M4) + m(Duval,M4) + m(Etting,M4) & sums(M5) = m(Akerman,M5) + m(Baird,M5) + m(Chatham,M5) + m(Duval,M5) + m(Etting,M5) & // 2. In all, the five motions got three more yes votes than no votes. no_votes::L & yes_votes::L & yes_votes + no_votes = 25 & {25 votes total: 5 members * 5 votes each} yes_votes = no_votes + 3 & sums(M1) + sums(M2) + sums(M3) + sums(M4) + sums(M5) = yes_votes & // 3. No two council members voted the same way on all five motions. {this is guaranteed by the type declaration of "Motions"} // 4. The two women disagreed in their voting more often than they agreed. DisagreedMore(m(Baird),m(Duval)) & // 5. Mr. Chatham never made two yes votes on consecutive motions. NoTwoConsecutiveYes(m(Chatham)) & // 6. Mr. Akerman and Ms. Baird both voted in favor of motion 4. m(Akerman,M4) = 1 & m(Baird,M4) = 1 & // 7. Motion 1 received two more yes votes than motion 2 did. sums(M1) = sums(M2) + 2 & // 8. Motion 3 received twice as many yes votes as motion 4 did. sums(M3) = 2*sums(M4) local pred DisagreedMore(m1::Motion->L[0..1],m2::Motion->L[0..1]) iff (diff1 = 1 & m1(M1) <> m2(M1) | diff1 = 0) & (diff2 = 1 & m1(M2) <> m2(M2) | diff2 = 0) & (diff3 = 1 & m1(M3) <> m2(M3) | diff3 = 0) & (diff4 = 1 & m1(M4) <> m2(M4) | diff4 = 0) & (diff5 = 1 & m1(M5) <> m2(M5) | diff5 = 0) & diff1 + diff2 + diff3 + diff4 + diff5 >= 3 local pred NoTwoConsecutiveYes(m::Motion->L[0..1]) iff m(M1) + m(M2) < 2 & m(M2) + m(M3) < 2 & m(M3) + m(M4) < 2 & m(M4) + m(M5) < 2

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