A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

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A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Elvish Pillager » Sun Apr 23, 2017 1:50 am UTC

(I just wrote this. I sure hope I did the logic right.) (EDIT: I made a couple of mistakes writing the original puzzle, but I think they are fixed now, with much thanks to Poker's feedback in the thread below. The original puzzle was:
Spoiler:
Jane, Emily, and Mike were perfect logicians. One day, Jane said, "I'm thinking of four nonnegative integers, x1, y1, x2, and y2, that obey the following conditions:

|x1 - x2| >= 1
|x1 - x2| >= min(y1, y2)
|x1 - x2| <= 1 + min(y1, y2)
|y1 - y2| <= 1

"I'm going to tell x1 and y1 to Emily, and I'm going to tell x2 and y2 to Mike." Jane did this.

Emily said, "I don't know x2, and I wouldn't know it even if I knew y2."
Mike said, "I don't know x1, and I wouldn't know it even if I knew y1."
Emily said, "I don't know x2, and I wouldn't know it even if I knew y2."
Mike said, "I don't know x1, and I wouldn't know it even if I knew y1."
Emily said, "I don't know x2, and I wouldn't know it even if I knew y2."
Mike said, "I don't know x1, and I wouldn't know it even if I knew y1."
Jane interrupted, "Stop! You two could go on forever like that!"
Emily said, "I didn't know that."
Mike said, "I didn't know Emily didn't know that. If Emily had said she knew that, I wouldn't know whether Emily knew whether x2 is greater than 10. But now, I do."
Emily said, "Before Mike said that, I didn't know whether Mike knew which of x1 or x2 is bigger."

What were the numbers?


Jane, Emily, and Mike were perfect logicians. One day, Jane said, "I'm thinking of four nonnegative integers, x1, y1, x2, and y2, that obey the following conditions:

|x1 - x2| >= 1
|x1 - x2| >= min(y1, y2)
|x1 - x2| <= 1 + min(y1, y2)
|y1 - y2| <= 1

"I'm going to tell x1 and y1 to Emily, and I'm going to tell x2 and y2 to Mike." Jane did this.

Emily said, "I don't know x2, and I wouldn't know it even if I knew whether y2 was the same as y1."
Mike said, "I don't know x1, and I wouldn't know it even if I knew whether y1 was the same as y2."
Emily said, "I don't know x2, and I wouldn't know it even if I knew whether y2 was the same as y1."
Mike said, "I don't know x1, and I wouldn't know it even if I knew whether y1 was the same as y2."
Emily said, "I don't know x2, and I wouldn't know it even if I knew whether y2 was the same as y1."
Mike said, "I don't know x1, and I wouldn't know it even if I knew whether y1 was the same as y2."
Jane interrupted, "Stop! You two could go on forever like that!"
Emily said, "I didn't know that."
Mike said, "I didn't know Emily didn't know that. If Emily had said she knew that, I wouldn't know whether Emily knew whether x2 is greater than 15. But now, I do."
Emily said, "Before Mike said that, I didn't know whether Mike knew which of x1 or x2 is bigger."

What were the numbers?
Last edited by Elvish Pillager on Mon Apr 24, 2017 12:48 am UTC, edited 6 times in total.
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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Poker » Sun Apr 23, 2017 1:17 pm UTC

Elvish Pillager wrote:(I just wrote this. I sure hope I did the logic right.)

Jane, Emily, and Mike were perfect logicians. One day, Jane said, "I'm thinking of four nonnegative integers, x1, y1, x2, and y2, that obey the following conditions:

|x1 - x2| >= 1
|x1 - x2| >= min(y1, y2)
|x1 - x2| <= 1 + min(y1, y2)
|y1 - y2| <= 1

"I'm going to tell x1 and y1 to Emily, and I'm going to tell x2 and y2 to Mike." Jane did this.

Emily said, "I don't know x2, and I wouldn't know it even if I knew y2."
Mike said, "I don't know x1, and I wouldn't know it even if I knew y1."
Emily said, "I don't know x2, and I wouldn't know it even if I knew y2."
Mike said, "I don't know x1, and I wouldn't know it even if I knew y1."
Emily said, "I don't know x2, and I wouldn't know it even if I knew y2."
Mike said, "I don't know x1, and I wouldn't know it even if I knew y1."
Jane interrupted, "Stop! You two could go on forever like that!"
Emily said, "I didn't know that."
Mike said, "I didn't know Emily didn't know that. If Emily had said she knew that, I wouldn't know whether Emily knew whether x2 is greater than 10. But now, I do."
Emily said, "Before Mike said that, I didn't know whether Mike knew which of x1 or x2 is bigger."

What were the numbers?


Not sure if this is unsolvable or just me missing something, but here's what I've got:

Spoiler:
From the conditions: y1 and y2 are two numbers that are either identical or adjacent. x1 and x2 are two nonidentical numbers whose difference is the smaller of y1 and y2, possibly plus one. As a corollary, if either y1 or y2 is 0, x1 and x2 must be adjacent.

Statement 1: This statement eliminates (x1,y1) = (0,0) and (0,1): if Emily knew that min(y1,y2) = 0, and x1 = 0, Emily could prove that x2 = 1 from the above corollary. This eliminates (0,0) directly, and it eliminates (0,1) because Emily could not make that statement for sure without knowing y2 is not 0.

Statement 2: This statement eliminates (x2,y2) = (0,0) and (0,1) by the same logic as above, and in addition eliminates (1,0) and (1,1): if Mike knew that min(y1,y2)=0, and x2=1, then since Mike knows x1 is not 0, x1 must equal 2.

Statements 3-6: Proceeding in a similar fashion, Emily cannot have (0,0) through (4,1) and Mike cannot have (0,0) through (5,1).

Statement 7: The only number pairs that get eliminated in the above logic are (x,0) and (x,1). This statement proves that y1,y2 >= 2.

Statement 8: If y1 > 2, then y2 > 1, and so Emily would have known the loop of statements 1-6 would have been unending. The only way she couldn't have known that is if it was possible for y2 to be 1, so that one of Mike's (x,1) pairs could not be eliminated. Therefore, y1 = 2, and furthermore, from Emily's perspective, some (x2,1) pair where x2 >= 6 was possible, meaning x1 >= 4.

Statement 9: Mike not knowing that Emily didn't know proves nothing, because whether y2 = 2 or 3, Mike could not eliminate the possibility that y2 = 3 (or even 4), in which case Emily would have known. Now, given a value for y1, Emily would know that the difference between x1 and x2 is either y1 - 1, or y, or y + 1. Therefore, Emily would not know whether x2 was greater than 10 or not if her numbers were (8,2) to (13,2), (7,3) to (14,3), or (6,4) to (15,4); and would know in all other circumstances. Now that Mike knows Emily's numbers are (4,2) and above, Mike either knows x1 < 8, which would make Mike's numbers (4,2) and less or (4,3) and less; or Mike knows 8 <= x1 <= 13, which is impossible since Mike sees a difference of 3 as possibke; or Mike knows x > 13, which would make Mike's numbers (17,2) and greater or (17,3) and greater. In either case, before Emily's statement, Mike saw at least one "do not know" value as a possibility. Therefore if y2 is 2, some (x2+-3,3) must have been possible, meaning Mike's numbers could only be (4,2) or (17,2); and if y2 is 3, some (x2+-4,4) must have been possible, meaning Mike's numbers could only be (2,3) to (4,3) or (17,3) to (19,3).

Statement 10: After Emily made Statement 8, Mike, in most circumstances, would not know which of x1 or x2 is greater. Emily must have believed some other circumstance was possible. Now, whatever x2 Mike has, Emily could have had a greater x1. However, if x2 < 6, then x1 could not be less than x2 since their difference is at least 2, and x1 >= 4. Therefore, in order for Emily to not know if Mike knew, Emily must see some (5,2) or less or (5,3) or less as possible. Therefore, x1 <= 8. Therefore, Mike's upper range is eliminated, so Mike's numbers are (4,2), (2,3), (3,3), or (4,3); and Emily's numbers are (4,2), (5,2), (6,2) or (7,2) - (8,2) is eliminated since x2 is now <= 4.

And that's where I'm stuck. Granted, 13 possibilities is a lot better than an infinite number, but it's still a (non-baker's) dozen too many.

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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Elvish Pillager » Sun Apr 23, 2017 2:09 pm UTC

Poker wrote:Not sure if this is unsolvable or just me missing something, but here's what I've got:

Spoiler:
From the conditions: y1 and y2 are two numbers that are either identical or adjacent. x1 and x2 are two nonidentical numbers whose difference is the smaller of y1 and y2, possibly plus one. As a corollary, if either y1 or y2 is 0, x1 and x2 must be adjacent.

Statement 1: This statement eliminates (x1,y1) = (0,0) and (0,1): if Emily knew that min(y1,y2) = 0, and x1 = 0, Emily could prove that x2 = 1 from the above corollary. This eliminates (0,0) directly, and it eliminates (0,1) because Emily could not make that statement for sure without knowing y2 is not 0.

Statement 2: This statement eliminates (x2,y2) = (0,0) and (0,1) by the same logic as above, and in addition eliminates (1,0) and (1,1): if Mike knew that min(y1,y2)=0, and x2=1, then since Mike knows x1 is not 0, x1 must equal 2.

Statements 3-6: Proceeding in a similar fashion, Emily cannot have (0,0) through (4,1) and Mike cannot have (0,0) through (5,1).

Statement 7: The only number pairs that get eliminated in the above logic are (x,0) and (x,1). This statement proves that y1,y2 >= 2.

Statement 8: If y1 > 2, then y2 > 1, and so Emily would have known the loop of statements 1-6 would have been unending. The only way she couldn't have known that is if it was possible for y2 to be 1, so that one of Mike's (x,1) pairs could not be eliminated. Therefore, y1 = 2, and furthermore, from Emily's perspective, some (x2,1) pair where x2 >= 6 was possible, meaning x1 >= 4.

Statement 9: Mike not knowing that Emily didn't know proves nothing, because whether y2 = 2 or 3, Mike could not eliminate the possibility that y2 = 3 (or even 4), in which case Emily would have known. Now, given a value for y1, Emily would know that the difference between x1 and x2 is either y1 - 1, or y, or y + 1. Therefore, Emily would not know whether x2 was greater than 10 or not if her numbers were (8,2) to (13,2), (7,3) to (14,3), or (6,4) to (15,4); and would know in all other circumstances. Now that Mike knows Emily's numbers are (4,2) and above, Mike either knows x1 < 8, which would make Mike's numbers (4,2) and less or (4,3) and less; or Mike knows 8 <= x1 <= 13, which is impossible since Mike sees a difference of 3 as possibke; or Mike knows x > 13, which would make Mike's numbers (17,2) and greater or (17,3) and greater. In either case, before Emily's statement, Mike saw at least one "do not know" value as a possibility. Therefore if y2 is 2, some (x2+-3,3) must have been possible, meaning Mike's numbers could only be (4,2) or (17,2); and if y2 is 3, some (x2+-4,4) must have been possible, meaning Mike's numbers could only be (2,3) to (4,3) or (17,3) to (19,3).

Statement 10: After Emily made Statement 8, Mike, in most circumstances, would not know which of x1 or x2 is greater. Emily must have believed some other circumstance was possible. Now, whatever x2 Mike has, Emily could have had a greater x1. However, if x2 < 6, then x1 could not be less than x2 since their difference is at least 2, and x1 >= 4. Therefore, in order for Emily to not know if Mike knew, Emily must see some (5,2) or less or (5,3) or less as possible. Therefore, x1 <= 8. Therefore, Mike's upper range is eliminated, so Mike's numbers are (4,2), (2,3), (3,3), or (4,3); and Emily's numbers are (4,2), (5,2), (6,2) or (7,2) - (8,2) is eliminated since x2 is now <= 4.

And that's where I'm stuck. Granted, 13 possibilities is a lot better than an infinite number, but it's still a (non-baker's) dozen too many.


Spoiler:
You're very close! Unfortunately, you missed something near the beginning that threw off the later numbers somewhat. (Tell me if you want a bigger hint.) EDIT: Wait, no, I think the puzzle is messed up. Give me a minute... EDIT 2: Yeah, originally I thought that the loop would only eliminate 0, not 1, as possibility for the y values. Now I see that it also eliminates 1, so I have increased the 10 to 13 in the problem statement, which I believe should narrow it down to one answer. EDIT: Wait, it needs to be 15!No, 14! Looks like you reasoned it out somewhat better than I did. EDIT: Wait, I think it's even worse than that! I'll have to think about this a lot more.

Looks like you're right that "I didn't know Emily didn't know that" doesn't add anything. It's left over from an older version where the restrictions on y1 and y2 were slightly different. I wonder if I can adjust the restrictions to make it a useful statement again.
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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Elvish Pillager » Sun Apr 23, 2017 3:48 pm UTC

About why the puzzle is broken:
Spoiler:
I believe the infinite sequence currently eliminates EVERY possible choice of values, although it's very complex to see why.

Figuring that out is a pretty interesting puzzle in its own right. I wonder what would be a good way to make that into an explicit puzzle that actually has an answer.
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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Elvish Pillager » Sun Apr 23, 2017 4:04 pm UTC

Trying to rehabilitate the original puzzle:

Spoiler:
My intention was to make the infinite loop only eliminate the y1=0 and y2=0 cases. I'd be grateful if anyone can come up with a wording that does that, or an explanation for why it's too hard to come up with such a wording.
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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Poker » Sun Apr 23, 2017 10:52 pm UTC

Elvish Pillager wrote:About why the puzzle is broken:
Spoiler:
I believe the infinite sequence currently eliminates EVERY possible choice of values, although it's very complex to see why.

Figuring that out is a pretty interesting puzzle in its own right. I wonder what would be a good way to make that into an explicit puzzle that actually has an answer.


I do see my mistake in my reasoning:

Spoiler:
My analysis for when y=0 and y=1 is still correct. Emily's statement eliminates up to (0,1), Mike's up to (1,1), and so on.

Case y=2: For Statement 1, Emily cannot eliminate anything about any (x,2). For statement 2, Mike cannot eliminate anything about any (x,2), for Mike could not tell between Emily having (x+1,1) and (x+2,1). For statement 3, however, since Mike eliminates (1,1), if Emily learns y2 = 1, Emily will learn x2 = 2, so Emily eliminates (0,2) as well as (1,0), (1,1), (2,0) and (2,1). Similarly, for statement 4, Mike eliminates (1,2), for statement 5, Emily eliminates (2,2), for statement 6, Mike eliminates (3,2), and so on. But what about the (2x+1,2) cases for Emily, and the (2x,2) cases for Mike? Well, on statement 6, if a (0,2) Mike were to learn that y1 = 2, Mike would know that x1=3 since (2,2) is eliminated. So (0,2) gets eliminated then along with (3,2). Similarly, Emily eliminates (1,2) and (4,2) on the theoretical statement 7, Mike eliminates (2,2) and (5,2), and so on.

Case y=3: On statement 6, after Emily has eliminated (2,2), a (0,3) Mike would learn x1 upon hearing y1 = 2, so Mike eliminates (0,3). Similarly, on Statement 7, Emily would eliminate (1,3). On Statement 8, Mike eliminates (2,3) as Elizabeth has eliminated (0,2) and (4,2). On Statement 9, Elizabeth eliminates (3,3) as Mike has eliminated (0,2), (1,2), and (5,2). On Statement 10, Mike eliminates (4,3), and so on Statement 12, Elizabeth eliminates (0,3) and (5,3), and so on.

Basically, one side of the even-odd parity for each person fills out first, then the other.


Elvish Pillager wrote:Trying to rehabilitate the original puzzle:

Spoiler:
My intention was to make the infinite loop only eliminate the y1=0 and y2=0 cases. I'd be grateful if anyone can come up with a wording that does that, or an explanation for why it's too hard to come up with such a wording.


Using the original conditions:

Spoiler:
My first idea was simply:

Emily: "I don't know x2."
Mike: "I don't know x1."
etc.

Unfortunately, this only seems to eliminate the case where y1 = y2 = 0, and the case where one person has (0,0).

My second thought:

Emily: "I don't know x2, and I wouldn't know x2 even if I knew whether y2 was less than y1 or not."
etc.

But that still eliminates the y=1 cases, though I'm less certain if it eliminates everything as it did above.

Then I came upon this, which seems to do the trick:

Emily: "I don't know x2, and I wouldn't know x2 even if I knew whether y1 equals y2 or not."
etc.

The y = 0 cases get eliminated as normal - after all, their reasoning only involved supposing the other y is also 0, which is the same as saying the two y values are equal. And the only reason we could eliminate y = 1 in the original was that we learned that the other y = 0, so we would learn that the difference is 1. Here, that isn't the case: we learn that the other y = 0 or 2, so we don't know if the difference is 1 or 2.


Now that I know what you're aiming for, I'll try solving again, using my statement in its place, but I still run into problems:

Spoiler:
Statements 1-6: Emily does not have any of (0,0) through (4,0) and Mike does not have any of (0,0) through (5,0).

Statement 7: As shown above, this proves that y1, y2 > 0.

Statement 8: Emily did not know the loop was endless, so she must have thought the loop could end, as before. This is only possible if y1 = 1, and if x2 could still equal x1+1. In other words, x1 >= 5. So Emily's values are (5,1) or above.

Statement 9: y2 = 1 or 2, so Mike could only have as possibilities values for y1 of 1, 2, or 3. In these cases, the ranges for Emily's values for not knowing whether x2 is greater than 10 are (9,1) to (12,1), (8,2) to (13,2), and (7,3) to (14,3). Now that Mike knows Emily's true range, Mike either knows x1 < 9, meaning Mike's numbers are (6,1) and less or (6,2) and less; or Mike knows 9 <= x1 <= 12, which is impossible since Mike sees a difference of 2 as possible; or Mike knows x1 > 12, which would make Mike's numbers (15,1) and greater or (15,2) and greater. Whatever the case, before Emily's statement, Mike saw at least one "do not know" value as a possibility. Therefore, if y2 is 1, some (x2+-2,2) must have been possible, meaning Mike's numbers could only be (6,1) or (15,1); and if y2 is 2, some (x2+-3,3) must have been possible, meaning Mike's numbers could only be (4,2) to (6,2) or (15,2) to (17,2).

Statement 10: After Emily made Statement 8, Mike, in most circumstances, would not know which of x1 or x2 is greater. Emily must have believed some other circumstance was possible. Now, whatever x2 Mike has, Emily could have a greater x1. However, if x2 < 5, then x1 could not be less than x2 since their difference is at least 1, and x1 >= 4. Therefore, Emily must have seen some (4,1) or less or (4,2) or less as possible. Therefore, x1 <= 6, meaning Emily's numbers are either (5,1) or (6,1), and Mike's numbers are either (6,1), (4,2), (5,2) or (6,2). That leaves us with (5,1)(6,1), (5,1)(4,2), (5,1)(6,2), (6,1)(4,2), and (6,1),(5,2) as possibilities. 5 is better than 13, but it still isn't 1.


Possible correction:

Spoiler:
Trying to produce lower values for x2 still leaves us with an impossible choice for y2 between 1 and 2. Therefore, I agree that the 10 needs to be increased to 14. This would make Emily's "not knowing" ranges (13,1) to (16,1), (12,2) to (17,2), and (11,3) to (18,3), putting Mike's numbers at (10,1), (19,1), (8,2) to (10,2), or (19,2) to (21,2). The logic of Statement 10 is unchanged up to the point where Emily's numbers are determined to be either (5,1) or (6,1). In this case, since the difference between x1 and x2 is at most 2, the only possibilities are (6,1) and (8,2).

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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Elvish Pillager » Mon Apr 24, 2017 12:37 am UTC

Poker wrote:I do see my mistake in my reasoning:

Spoiler:
My analysis for when y=0 and y=1 is still correct. Emily's statement eliminates up to (0,1), Mike's up to (1,1), and so on.

Case y=2: For Statement 1, Emily cannot eliminate anything about any (x,2). For statement 2, Mike cannot eliminate anything about any (x,2), for Mike could not tell between Emily having (x+1,1) and (x+2,1). For statement 3, however, since Mike eliminates (1,1), if Emily learns y2 = 1, Emily will learn x2 = 2, so Emily eliminates (0,2) as well as (1,0), (1,1), (2,0) and (2,1). Similarly, for statement 4, Mike eliminates (1,2), for statement 5, Emily eliminates (2,2), for statement 6, Mike eliminates (3,2), and so on. But what about the (2x+1,2) cases for Emily, and the (2x,2) cases for Mike? Well, on statement 6, if a (0,2) Mike were to learn that y1 = 2, Mike would know that x1=3 since (2,2) is eliminated. So (0,2) gets eliminated then along with (3,2). Similarly, Emily eliminates (1,2) and (4,2) on the theoretical statement 7, Mike eliminates (2,2) and (5,2), and so on.

Case y=3: On statement 6, after Emily has eliminated (2,2), a (0,3) Mike would learn x1 upon hearing y1 = 2, so Mike eliminates (0,3). Similarly, on Statement 7, Emily would eliminate (1,3). On Statement 8, Mike eliminates (2,3) as Elizabeth has eliminated (0,2) and (4,2). On Statement 9, Elizabeth eliminates (3,3) as Mike has eliminated (0,2), (1,2), and (5,2). On Statement 10, Mike eliminates (4,3), and so on Statement 12, Elizabeth eliminates (0,3) and (5,3), and so on.

Basically, one side of the even-odd parity for each person fills out first, then the other.



Yep! That's the same conclusion I came to.

Poker wrote:Using the original conditions:

Spoiler:
My first idea was simply:

Emily: "I don't know x2."
Mike: "I don't know x1."
etc.

Unfortunately, this only seems to eliminate the case where y1 = y2 = 0, and the case where one person has (0,0).

My second thought:

Emily: "I don't know x2, and I wouldn't know x2 even if I knew whether y2 was less than y1 or not."
etc.

But that still eliminates the y=1 cases, though I'm less certain if it eliminates everything as it did above.

Then I came upon this, which seems to do the trick:

Emily: "I don't know x2, and I wouldn't know x2 even if I knew whether y1 equals y2 or not."
etc.

The y = 0 cases get eliminated as normal - after all, their reasoning only involved supposing the other y is also 0, which is the same as saying the two y values are equal. And the only reason we could eliminate y = 1 in the original was that we learned that the other y = 0, so we would learn that the difference is 1. Here, that isn't the case: we learn that the other y = 0 or 2, so we don't know if the difference is 1 or 2.


Now that I know what you're aiming for, I'll try solving again, using my statement in its place, but I still run into problems:

Spoiler:
Statements 1-6: Emily does not have any of (0,0) through (4,0) and Mike does not have any of (0,0) through (5,0).

Statement 7: As shown above, this proves that y1, y2 > 0.

Statement 8: Emily did not know the loop was endless, so she must have thought the loop could end, as before. This is only possible if y1 = 1, and if x2 could still equal x1+1. In other words, x1 >= 5. So Emily's values are (5,1) or above.

Statement 9: y2 = 1 or 2, so Mike could only have as possibilities values for y1 of 1, 2, or 3. In these cases, the ranges for Emily's values for not knowing whether x2 is greater than 10 are (9,1) to (12,1), (8,2) to (13,2), and (7,3) to (14,3). Now that Mike knows Emily's true range, Mike either knows x1 < 9, meaning Mike's numbers are (6,1) and less or (6,2) and less; or Mike knows 9 <= x1 <= 12, which is impossible since Mike sees a difference of 2 as possible; or Mike knows x1 > 12, which would make Mike's numbers (15,1) and greater or (15,2) and greater. Whatever the case, before Emily's statement, Mike saw at least one "do not know" value as a possibility. Therefore, if y2 is 1, some (x2+-2,2) must have been possible, meaning Mike's numbers could only be (6,1) or (15,1); and if y2 is 2, some (x2+-3,3) must have been possible, meaning Mike's numbers could only be (4,2) to (6,2) or (15,2) to (17,2).

Statement 10: After Emily made Statement 8, Mike, in most circumstances, would not know which of x1 or x2 is greater. Emily must have believed some other circumstance was possible. Now, whatever x2 Mike has, Emily could have a greater x1. However, if x2 < 5, then x1 could not be less than x2 since their difference is at least 1, and x1 >= 4. Therefore, Emily must have seen some (4,1) or less or (4,2) or less as possible. Therefore, x1 <= 6, meaning Emily's numbers are either (5,1) or (6,1), and Mike's numbers are either (6,1), (4,2), (5,2) or (6,2). That leaves us with (5,1)(6,1), (5,1)(4,2), (5,1)(6,2), (6,1)(4,2), and (6,1),(5,2) as possibilities. 5 is better than 13, but it still isn't 1.


Possible correction:

Spoiler:
Trying to produce lower values for x2 still leaves us with an impossible choice for y2 between 1 and 2. Therefore, I agree that the 10 needs to be increased to 14. This would make Emily's "not knowing" ranges (13,1) to (16,1), (12,2) to (17,2), and (11,3) to (18,3), putting Mike's numbers at (10,1), (19,1), (8,2) to (10,2), or (19,2) to (21,2). The logic of Statement 10 is unchanged up to the point where Emily's numbers are determined to be either (5,1) or (6,1). In this case, since the difference between x1 and x2 is at most 2, the only possibilities are (6,1) and (8,2).


Nice!
Spoiler:
I already thought of this wording, and I didn't think it worked, but I think I was making the same mistake I made in the first place. Looks like it works. I also originally didn't think of the x1>=5 thing because I only thought of Emily knowing they would repeat forever because y1 was high enough, but I didn't think of the fact that Emily could know from x1 that they were past the point of no return. I will update the puzzle with your corrections! I appreciate it! :-)
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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Elvish Pillager » Mon Apr 24, 2017 12:49 am UTC

A small additional correction:

Spoiler:
In your updated logic for the last step, you forgot to replace "x1 >= 4" with the newly computed "x1 >= 5", so you were off by one – the number in the problem needs to be 15, and the answers are (7, 1) and (9, 2).
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Re: A very fiendish one of those "figure out the answer based on logicians not knowing the answer" puzzles

Postby Poker » Mon Apr 24, 2017 1:56 am UTC

Elvish Pillager wrote:A small additional correction:

Spoiler:
In your updated logic for the last step, you forgot to replace "x1 >= 4" with the newly computed "x1 >= 5", so you were off by one – the number in the problem needs to be 15, and the answers are (7, 1) and (9, 2).


You're right, don't know how I didn't catch that myself. Glad I could help.


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