Yay! Another logic problem...this one might be harder

[jwvdcw]

There are 100 monks in a monastery who have sworn to a vow of silence. They are the most devoted monks in the world, and also the most logical.

On Christmas, God speaks to the monks at their daily meeting and says that some of you will be leaving the monastery to go on a mission*. "You may not communicate with one another in any way, you must simply leave the monastery at midnight of the night you realize you are selected and meet up on the mounain. Last night, I put a single red dot on the forehead of the monks who are selected. I expect you will know when to meet me."

God never told them how many monks were dotted, there were no reflective surfaces in the monastery so they couldn't tell if they were dotted themselves, and the monks made no verbal, written or non-verbal communication of any kind. But sure enough, all five monks who had been dotted met up on the mountain, just when God expected them to.

On what day did they go to the Mountain? And explain their thought process throughout each day that passes.

[pudley4]

<font color="white"> It was day 5.

Day 1 - If only one monk had been dotted, he would realize it immediately because he would see no other monks with dots; so he would leave that night.

Day - If there were 2 monks with dots, they would see each other and expect the other to leave if he didn't see any others with dots. When they awaken on the second day and the other monk is still there, each knows he has a dot on his own head, so he leaves that night.

The same reasoning can be used to calculate the 5 monks left on the fifth night </font>

[jwvdcw]

long explanation of the answer taken from Jericho at footballguys:

<font color="white"> The Diary of a Dotted Monk...

DAY ONE:

I see four dots. Therefore there must be either four dots total, and I am undotted, or five dots total, and I am dotted. So it is *possible* that there are only four dots. So IF there are only four dots then one of the dotted monks I see will see only three dots.

It is *possible* then that one monk sees only three dots. Therefore, that monk would know there are either three or four dots. IF that monk assumes there are only three dots, it is *possible* that he will assume that one of the dotted monks he sees will see only TWO dots (even though we know better).

It is therefore *possible* that a monk sees only three dots and that monk would consider it *possible* that a monk sees only two dots. A monk that sees two dots would know that there are either two or three dots. IF that monk assumes there are only two dots (a perfectly valid assumption if you only see two), it is *possible* that he would believe that one of those other monks sees only ONE dot (even though we know better).

It is therefore *possible* that a monk sees only three dots and that monk would consider it *possible* that a monk sees only two dots, and therefore it also *possible* that the second monk thinks it's *possible* for a monk to see only one dot. A monk that sees one dot would know that there are either one or two dots. IF that monk assumes there is only one dot (again, a perfectly valid assumption - these are hypothetical "possible" monks), it is *possible* that he would believe that the dotted monk he sees can see no dots.

So, it is *possible* that a monk sees only three dots and that monk would consider it *possible* that a monk sees only two dots, and therefore it is also *possible* that the second monk thinks it's *possible* for a monk to see only one dot, and therefore it's also *possible* for this third monk to think it's *possible* for a monk to see zero dots. A monk who sees zero dots would obviously leave tonight. I know that no monks will leave tonight, because I see four dots, but I must take into account all possibilities. I go to sleep.

DAY TWO:

As expected, no monks left. It is now no longer possible for any monk, hypothetical or otherwise, to believe that a monk saw zero dots. And therefore there are at least two dots. However, it is still *possible* that a monk sees only three dots and that monk would consider it *possible* that a monk sees only two dots, and therefore it's also *possible* that the second monk thinks it's *possible* for a monk to see one dot. A monk who sees only one dot now must assume that he also has a dot and would leave. I know that nobody is leaving yet because I see four dots, but I must take into account all possibilities. I go to sleep.

DAY THREE:

As expected, no monks left. However, it is now no longer possible for any monk, hypothetical or otherwise, to believe that any monk saw only one dot. And therefore there are at least three dots. However, it is still *possible* that a monk sees only three dots and that monk would consider it *possible* that a monk sees only two dots. A monk who sees two dots must assume that he also has a dot and would leave. I know that nobody is leaving yet because I see four dots, but I must take into account all possibilities. I go to sleep.

DAY FOUR:

As expected, no monks left. However, it is now no longer possible for any monk, hypothetical or otherwise, to believe that any monk saw only two dots. And therefore there are at least four dots. Hah, I knew that already. However, it is still *possible* that a monk sees only three dots. A monk who sees three dots must assume that he also has a dot and would leave. I am unsure whether people will leave tonight or not. I go to sleep.

DAY FIVE:

Interesting, no monks left. Therefore, it is no longer possible for any monk, hypothetical or otherwise, to believe that any monk saw only three dots. Therefore there are at least five dots. And since I can only see four, I leave tonight.

</font>

[GuyOnTilt]

This one was a little harder, but still doesn't pass the "Satisfy GoT's Need for a Difficult Logic Puzzle" test. /images/graemlins/wink.gif

My Answer:

<font color="white">The 5 monks each realized they had marks on the 5th day, including Chrismas Day, which means they went to the Mountain on the night of the 29th.

It's easiest to solve the puzzle by thinking of it this way:

If only one of them is marked, they will know they are marked because somebody has to be and nobody else is, therefore they will leave on the first night. If two of them are marked, the 2 who are marked will see one other mark and will know that either 1 or 2 are marked. The next morning, if the marked one is still there, then they will know that their forehead is marked, since if it wasn't that monk would've left the night before. Note that both marked monks will go through this same thought process. Therefore, the two marked ones will leave the 2nd night. If 3 are marked, then each will see 2 marked foreheads and know that it's between 2 and 3. They'll all wait the first two nights to see if the other two follow through with the thought process above. If the other 2 monks who are marked are still there the morning of the 3rd day, then you yourself must be marked. Same thought process for the 4th day, and the 5th day. Since it was a given in the logic puzzle that there were 5 marked monks, they will all arrive on the mountain the night of the 29th.

This puzzle would've been more difficult if it was given what night they arrived on instead of the number of monks marked I think... </font>

GoT

[GuyOnTilt]

long explanation of the answer taken from Jericho at footballguys:

What is this?! You cheated and looked up the answer on the internet?! That's weak sauce.

GoT

[jwvdcw]

[ QUOTE ]
long explanation of the answer taken from Jericho at footballguys:

What is this?! You cheated and looked up the answer on the internet?! That's weak sauce.

GoT

[/ QUOTE ]

Huh?? I was the one that asked the question! Its another message board, where we already went through tons of these logic problems. I got the answer correct, but couldn't type out the reasoning nearly as well as that post did, so I just copied and pasted.