Mystery Master	Analysis of Atrophysics Conference	Member

This four-star logic puzzle has 5 noun types, 5 nouns per type, 1 link, 31 facts, and 5 rules. What makes this puzzle interesting is one noun type (Attendance) does not have all of its values given, so you have to calculate them. The program does this via its rules. It needs 250 marks and 10 grids.

Attendance

Since most of the clues refer to attendance, below are all of the clues for this logic puzzle.

1. The first talk of the day was on the big bang theory.
2. Laura drew 24 listeners, more than any other speaker.
3. No two men spoke consecutively.
4. As many people attended the talk on pulsars as attended both the third speech and Gray's talk combined.
5. Howe's talk was attended by twice as many people as Ed's.
6. The last two speakers drew (not necessarily respectively) the largest and smallest numbers of attendees.
7. The talk about black holes drew half as many listeners as the second speech.
8. The talk on stellar evolution (which wasn't the one given by Christa Flynn) didn't draw the fewest attendees of the day.
9. The number of people who attended Samuel's talk was as much less than the number who attended the talk on quasars as it was more than the number who attended Ives's talk.
10. A different number of people attended each talk.

Since only Laura's attendance is known, let's name the attendances A1, A2, A3, A4, and A5. We will also associate talk1 with A1, talk2 with A2, and so on.

We know Laura had the largest attendance with 24 (clue 2), and the last two talks are the largest and smallest, in some order (clue 6). So Laura gave the 4th or 5th talk. We also know the talk on stellar evolution was not the smallest (clue 8). And the attendance for each talk is different (clue 10).

When we know Laura's attendance, this will allow us to (1) set her attendance to 24, and (2), the other attendance, A4 or A5, will be the smallest one.

There is one more thing. Since clue 3 states that "No two men spoke consecutively", the men must have given the 1st, 3rd, and 5th talks, in some order. This means Laura gave the 4th talk, which has the largest attendance, and therefore the 5th talk had the smallest attendance.

If we solve the puzzle without assumptions, and rules 3, 4, and 5 are disabled, the program finds 167/250 marks and 19/50 pairs. All of the facts have been fully examined and disabled. Rules 2 was invoked twice. We already knew the 1st talk was on the Big Bang Theory (clue 1). It finds that Laura's last name is Howe, and the 5th talk with the smallest attendance (A5) is on black holes. And Christa Flynn gave the 2nd talk. Here are the clues concerning attendance.

4. As many people attended the talk on pulsars as attended both the third speech and Gray's talk (1st or 5th) combined.
5. Howe's talk (4th) was attended by twice as many people as Ed's (1st, 3rd, 5th).
7. The talk about black holes (5th) drew half as many listeners as the second speech.
9. The number of people who attended Samuel's talk (1st, 3rd) was as much less than the number who attended the talk on quasars (2nd) as it was more than the number who attended Ives's talk (1st, 3rd, 5th).
10. A different number of people attended each talk.

The following clues, using pseudo algebra, is what we know about the attendances.

• Laura drew 24 listeners
• pulsars = talk3 + gray
• howe = 2*ed
• talk2 = 2*blackHoles
• 2*samuel = quasars + ives

And this is what we know so far.

• Laura Howe's 4th talk was on pulsars.
• Gray's was either the 1st or 5th talk.
• Howe's talk had 24 in attendance, so Ed's talk had 12. Ed's talk was 1st, 3rd, or 5th.
• The 5th talk on black holes was the smallest (not yet known), so talk 2 will double it (clue 7).
• Samuel was either 1st or 3rd.
• Quasars was the 2nd talk.
• Ive's talk was either 1st, 3rd, or 5th.

Rule 3

Rule 3 states "If Howe has the largest talk, then Ed did not talk on black holes (clues 5,7,10)". If we solve the puzzle with rule 3 enabled, we have 180/250 marks and 22/50 pairs. We see that Tom gave the 5th talk on black holes.

Rule 4

Revisiting clue 7, Christa (talk 2) had exactly twice as many attendees as Tom (black holes). Clue 4 states the talk on pulsars equals the attendance of talk 3 and Gray's talk. Now if Christa gave the talk on pulsars, that would mean (in pseudo-algebra): 2*blackHoles = talk3 + gray. Since (a) neither talk can be less than the talk on blackholes, and (b) they can't have the same attendance (clue 10), Christa can not give the talk on pulsars.

Rule 4 states: "If Christa gave the 2nd talk, then it was not on pulsars (clues 4,7,10)." Enable this rule, and we now have 206/250 marks and 32/50 pairs. We now see that the 4th talk is on pulsars.

Rule 5

Revisiting clue 4, the talk on pulsars equals the attendance of talk 3 and Gray's talk. Since the talk on pulsars drew 24 people, neither the 3rd talk nor Gray's talk can have 12 attendees (clues 4 and 10). Since Ed has 12 attendees (clue 5), this means he didn't speak third and he isn't Gray.

Rule 5 states: "If talk on pulsars was highest, then Ed is not 3rd or Gray (clues 4,5,10)." Enable this rule, and the program will find the solution.

This puzzle is challenging because you need to systematically work out the attendance numbers. By applying rules to this puzzle, this puzzle can be solved without making assumptions.