Riddles and Puzzles

[ender516]

No problem Yours, just a small point, I didn't provide the impetus to your answer, I provided the answer.

Details, details...

Hurry up then with with the next one!

OK, a quick and simple one: {stolen from Raymond Smullyan}

The Island of Questioners
Somewhere in the vast reaches of the ocean, there is a very strange island known as the
Island of Questioners. It derives its name from the fact that its inabitants never make
statements, they only ask questions. The inhabitants ask only questions answerable by Yes or
No. Each inhabitant is one of two types, A and B. Those of type A ask only questions whose
correct answer is Yes; those of type B ask only questions whose correct answer is No. For
example, an inhabitant of type A could ask, "Does two plus two equal four?" But he could not
ask whether two plus two equals five. An inhabitant of type B could not ask whether two plus
two equals four, but he could ask whether two plus two equals five.
I once visited this island and met a couple named Ethan and Violet Russell. I heard Ethan
ask some, one, "Are Violet and I both of type B?' What type is Violet?

The answer to Ethan's question must be either Yes or No. If it were Yes, then Ethan would be saying in part that he can only ask questions whose correct answer is No, but this is self-contradictory. Therefore the answer to Ethan's question is No. This implies that one or both of them is of type A. But since the answer to Ethan's question is No, he is of type B. This implies that Violet is the type A of the pair.

[YoursFalsey]

As promised, solved quickly and simply! The ball is now in Ender's court!

[ender516]

Sorry to keep you all waiting. Here's another from Martin Gardner's book (same one as before):

Evaluate each of the 10 statements as to its truth or falsity:

1. Exactly one statement on this list is false.
2. Exactly two statements on this list are false.
3. Exactly three statements on this list are false.
4. Exactly four statements on this list are false.
5. Exactly five statements on this list are false.
6. Exactly six statements on this list are false.
7. Exactly seven statements on this list are false.
8. Exactly eight statements on this list are false.
9. Exactly nine statements on this list are false.
10. Exactly ten statements on this list are false.

[YoursFalsey]

Dang it. I'm pretty sure I've got this one- I love knight/knave puzzles and this falls into the same category of puzzle- but a) I feel like I've been hogging this thread lately and b) I don't have another puzzle handy at the moment. So I am going to hold off for a day or so while I find a good puzzle to pose and hope that someone else beats me to the punch....

[nippersean]

I have it as, 9. Nine statements are false. Is true all the others are false.

They are contradictory. Only one can be correct. Which leaves nine false statements - therefore 9) is the only true statement

[jonesthecurl]

I agree. what about if it said "at least n statements are false" rather than "exactly n" ?

[nippersean]

If it said "at least...." then first 5 T and 6-10 false?

[wercool]

Sorry to keep you all waiting. Here's another from Martin Gardner's book (same one as before):

Evaluate each of the 10 statements as to its truth or falsity:

1. Exactly one statement on this list is false.
2. Exactly two statements on this list are false.
3. Exactly three statements on this list are false.
4. Exactly four statements on this list are false.
5. Exactly five statements on this list are false.
6. Exactly six statements on this list are false.
7. Exactly seven statements on this list are false.
8. Exactly eight statements on this list are false.
9. Exactly nine statements on this list are false.
10. Exactly ten statements on this list are false.

i beileve that nine of these statments are false

Exactly ten statements on this list are false.this statment can not be true because to be true it must be false

[*]

Exactly two statements on this list are false.
[*]Exactly three statements on this list are false.
[*]Exactly four statements on this list are false.
[*]Exactly five statements on this list are false.
[*]Exactly six statements on this list are false.
[*]Exactly seven statements on this list are false.
[*]Exactly eight statements on this list are false.none of these statments can be true because only one statment can be true, for example number 5 cant be true because five others would have to be true which arent true

[*]Exactly one statement on this list is false.now this statment cant be true becuse 8 are false not 1

[*]Exactly nine statements on this list are false.this one is true

[ender516]

Sorry to be so slow getting here to judge the answer. We have had two entries with the same answer, both correct, with nippersean in first with a concise explanation which omitted a point (to be explained below) and with wercool coming in second with a more complete analysis which, I'm sorry, is a bit muddled from my point of view. I guess I will have to give it to nippersean, for a substantially complete and correct answer.

The answer found in Gardner's book covers any list of n such statements (provided n is greater than 1) follows:

At most one of the statements can be true because any two contradict each other. All the statements cannot be false, because this implies the list contains exactly zero false statements. Thus exactly one statement can be true. Thus exactly n - 1 are false and the (n - 1)st statement is true.

The bold statement is the point that nippersean missed, and if you do not get why that implication is a problem, ponder this: exactly zero false statements implies exactly n true statements, and with n greater than one, they will contradict each other.

Changing "exactly" to "at least" is pretty much the same as simply removing "exactly" (the "at least" would be understood). This variant was mentioned in Gardner's book and does in fact lead to 5 true and 5 false statements.

If you reduce the list to one statement, you get

1. Exactly one statement on this list is false.

This is equivalent to the old liar's paradox: "This statement is false."

You can get around this by adding the zero statement:

0. Exactly none of the statements on this list is false.

Interestingly, this moves the one true statement from position (n - 1) to position n.

So, you are up, nippersean. Let's have another.

[nippersean]

I've played a matchstick game - rules
21 matches placed on the pub table,
Player 1 takes either 1,2 or 3. Player 2 does likewise. The player who is left to pick up the last match loses.

How would you ensure victory? Can you explain it in maths terms?
What about if there were 22 matches at the begining 23,24 etc. What would happen if you could take between 1 and 4 etc..

[Ducttapers4JC]

I've played a matchstick game - rules
21 matches placed on the pub table,
Player 1 takes either 1,2 or 3. Player 2 does likewise. The player who is left to pick up the last match loses.

How would you ensure victory? Can you explain it in maths terms?
What about if there were 22 matches at the begining 23,24 etc. What would happen if you could take between 1 and 4 etc..

Ok, I am going to number the turns backwards, so that this makes sence. You need to leave your opponent with one match stick on the last turn (#1), so you want either 2,3, or 4 matches left the turn before (#2). You want your opponent left with 5 matches on turn #3, so that any number he takes will leave you with 2,3,or 4. To do that you need to have 6,7, or 8 on turn #4. You want your opponent left with 9 matches on turn #5 so that any number he takes will leave you with 6,7, or 8. If you follow the pattern back, you want you opponent left with 12 on #7, 15 on #9 and 19 on #11, and so if you go first, you can ensure victory by takeing 2 match sticks.

Follow the pattern farther for higher numbers.

For takeing more match sticks, simply subtract 1 (the number you want left at the end) and divide by the largest number of match sticks that can be taken + 1. the remainder is the number you should take at the begining to ensure victory.

[nippersean]

Nope. Though you're on the right lines with your first statement

[YoursFalsey]

For the game as stated- 21 matchsticks, picking 1, 2, or 3 sticks per turn, the key to victory is be player 2. However many sticks player 1 takes(x), if player 2 takes 4-x sticks, player 1 will eventually be forced to take the final stick. If on any turn player 2 follows suboptimal strategy (math-geek-speak for if player 2 f-s up) player 1 needs to take enough sticks to leave player 2 looking at 4k+1 matchsticks for some integer value of k.

If there are more matchsticks in the pile, but each turn is still one to three sticks, then each player still wants to leave 4k+1 matchsticks for the other player. (If there are 22 match sticks, player 1 will win as long as he picks 1 on his first turn and then follows optimal strategy. If there are 25 match sticks, it doesn't matter what player one picks, as player 2 can take enough to leave player 1 facing 21 matchsticks and the original problem...)

If each player is allowed to pick more sticks per turn, say one through n-1 for some number n, then each player now desires to leave the other player looking at a pile of nk+1 match sticks. (n=4 in the original case of 1 to 3 match sticks...) (Thus if each player was allowed 1-4 matchsticks or 1-9 matchsticks player 2 would still hold the edge in the 21 matchstick game, but if it was 1-6 matchsticks each turn, player 1 would hold the edge as long as he took 6 match sticks on his first turn.)

Matchstick games are a nice exercise for talking about modular arithmetic, as modular arithmetic is a very easy way to analyze such puzzles...

[nippersean]

Yours - absolutely brilliant. If only I'd asked you when this was set as my homework.

Your turn I believe

[YoursFalsey]

OK, Here goes.

The Conquer Club cook, KeptainNoobie, decided to ask the guru on the hill how he could improve his ConquerClub game. At 9:00 A.M., he climbed the long trail up to the top of Somebigg Mountain (13.6 miles), finally reaching the top at 3:00 PM. He waited in line over night (A lot of us need to talk to the guru on the hill!), got some cryptic advice in the morning ("Think twice; click once!") and headed back down the long trail. By coincidence, he started once again at 9:00 AM and returned to the base of Somebigg Mountain at 3:00 PM again. Prove there is a time between 9:00 and 3:00 when KeptainNoobie was at the same spot in the trail on both days. Note he can walk at different speeds, rest, or even go backward whenever he wants.

Happy Thanksgiving!

[Beckytheblondie]

Isn't this inherent at passing between points A and B and back from B to A? Are you asking for a math proof?

[nippersean]

Yeah - don't get it Becky answerd it all ready a-b b-a you have to cross

[YoursFalsey]

Isn't this inherent at passing between points A and B and back from B to A? Are you asking for a math proof?

Sorry, my wife was pressuring me to finish up so we could get on the road for a dash to Grandmother's house. (Which was not over a river although we did go through some woods...) Yes, I was asking for a "proof" although I try to avoid that word since a) It causes many people's privates to shrink up and cringe within themselves and b) Proof tends to imply formality and I really am looking only for an informal demonstration that the answerer knows and can explain it so that someone who doesn't know understands. Remember, although we know the path's length and KeptainNoobie's start/end times, we know nothing about his pace, what breaks he took, whether he ever went backwards (and how far/how quickly/how often he did if so.) It is given that he passed through every point between A and B on both days- the difficulty is to prove that there is at least one point that he occupied at the same time on both days. (And like many such proofs, if you know the answer, it really does seem so blindingly obvious noone could miss it and if you are missing it, it is infuriating...)

[ender516]

Isn't this inherent at passing between points A and B and back from B to A? Are you asking for a math proof?

Sorry, my wife was pressuring me to finish up so we could get on the road for a dash to Grandmother's house. (Which was not over a river although we did go through some woods...) Yes, I was asking for a "proof" although I try to avoid that word since a) It causes many people's privates to shrink up and cringe within themselves and b) Proof tends to imply formality and I really am looking only for an informal demonstration that the answerer knows and can explain it so that someone who doesn't know understands. Remember, although we know the path's length and KeptainNoobie's start/end times, we know nothing about his pace, what breaks he took, whether he ever went backwards (and how far/how quickly/how often he did if so.) It is given that he passed through every point between A and B on both days- the difficulty is to prove that there is at least one point that he occupied at the same time on both days. (And like many such proofs, if you know the answer, it really does seem so blindingly obvious noone could miss it and if you are missing it, it is infuriating...)

Okay, then, we know that KeptainNoobie moved only along a unique path of length 13.6 miles. Imagine a graph of his displacement along this path with time on the horizontal axis, and distance from the starting point on the vertical axis. The origin of the graph represents the start of the trail at 9:00 A.M., and we know that the point representing his arrival at the summit (3:00 P.M., 13.6 miles) is on the graph. The function representing his ascent is continuous and single-valued (at any given time, he is in exactly one spot), and it is completely contained with the rectangle starting at the origin, running along the horizontal axis to 3:00 PM, then vertically to the summit point, then horizontally back to the vertical axis, and vertically down to the origin again. (He spends all his time between 9:00 and 3:00 on the first day somewhere on the trail.) If he travelled at a uniform speed, his function would trace the diagonal of the rectangle starting from the origin and ending at the summit point. Lacking knowledge of his speed, all we can say is that any time during his ascent, some of the path is above him and some below. So the function starts at the origin, ends at the summit point and divides the rectangle into upper and lower regions. The function of the descent on the second day is similar, but joins the other two corners of the rectangle. (He starts at (13.6 miles, 9:00 A.M.) and ends at (0, 3:00 P.M.).) So on the second day he starts in the upper region of the first day's graph and ends in the lower region. To get there, the graph of the descent must cross the graph of the ascent. At that point, he is at the same distance along the trail at the same time of day. Q.E.D.

[YoursFalsey]

Okay, then, we know that KeptainNoobie moved only along a unique path of length 13.6 miles. Imagine a graph of his displacement along this path with time on the horizontal axis, and distance from the starting point on the vertical axis. The origin of the graph represents the start of the trail at 9:00 A.M., and we know that the point representing his arrival at the summit (3:00 P.M., 13.6 miles) is on the graph. The function representing his ascent is continuous and single-valued (at any given time, he is in exactly one spot), and it is completely contained with the rectangle starting at the origin, running along the horizontal axis to 3:00 PM, then vertically to the summit point, then horizontally back to the vertical axis, and vertically down to the origin again. (He spends all his time between 9:00 and 3:00 on the first day somewhere on the trail.) If he travelled at a uniform speed, his function would trace the diagonal of the rectangle starting from the origin and ending at the summit point. Lacking knowledge of his speed, all we can say is that any time during his ascent, some of the path is above him and some below. So the function starts at the origin, ends at the summit point and divides the rectangle into upper and lower regions. The function of the descent on the second day is similar, but joins the other two corners of the rectangle. (He starts at (13.6 miles, 9:00 A.M.) and ends at (0, 3:00 P.M.).) So on the second day he starts in the upper region of the first day's graph and ends in the lower region. To get there, the graph of the descent must cross the graph of the ascent. At that point, he is at the same distance along the trail at the same time of day. Q.E.D.

Yep, that pretty much sums it up. Another way to look at it (if terms like continuous and singlevalued seem too mathematical for you) is imagine a 2nd person follows Keptains speed and path from the first day exactly on the second day. Since the Keptain starts out above the 2nd person and ends up below him, at some point they pass each other- that point is the place where Keptain occupied the same place at the same time both days.

If that is what Becky meant by inherant, I'm willing for the torch to pass to her next- I know it feels like to me like Ender and I have posed a disproportionate share of the recent puzzles...

[ender516]

Okay, then, we know that KeptainNoobie moved only along a unique path of length 13.6 miles. Imagine a graph of his displacement along this path with time on the horizontal axis, and distance from the starting point on the vertical axis. The origin of the graph represents the start of the trail at 9:00 A.M., and we know that the point representing his arrival at the summit (3:00 P.M., 13.6 miles) is on the graph. The function representing his ascent is continuous and single-valued (at any given time, he is in exactly one spot), and it is completely contained with the rectangle starting at the origin, running along the horizontal axis to 3:00 PM, then vertically to the summit point, then horizontally back to the vertical axis, and vertically down to the origin again. (He spends all his time between 9:00 and 3:00 on the first day somewhere on the trail.) If he travelled at a uniform speed, his function would trace the diagonal of the rectangle starting from the origin and ending at the summit point. Lacking knowledge of his speed, all we can say is that any time during his ascent, some of the path is above him and some below. So the function starts at the origin, ends at the summit point and divides the rectangle into upper and lower regions. The function of the descent on the second day is similar, but joins the other two corners of the rectangle. (He starts at (13.6 miles, 9:00 A.M.) and ends at (0, 3:00 P.M.).) So on the second day he starts in the upper region of the first day's graph and ends in the lower region. To get there, the graph of the descent must cross the graph of the ascent. At that point, he is at the same distance along the trail at the same time of day. Q.E.D.

Yep, that pretty much sums it up. Another way to look at it (if terms like continuous and singlevalued seem too mathematical for you) is imagine a 2nd person follows Keptains speed and path from the first day exactly on the second day. Since the Keptain starts out above the 2nd person and ends up below him, at some point they pass each other- that point is the place where Keptain occupied the same place at the same time both days.

If that is what Becky meant by inherant, I'm willing for the torch to pass to her next- I know it feels like to me like Ender and I have posed a disproportionate share of the recent puzzles...

Ah, yours is a much more intuitive solution, and I am also willing to let it go to someone else. (I had to take that book by Martin Gardner back to the library.)

[the.killing.44]

Easy CC one:

CC Player X has 1300 points. But he's yet to win a game. How?

[frankiebee]

Easy CC one:

CC Player X has 1300 points. But he's yet to win a game. How?

He scored 300 points in terminator games without finishing them

[ender516]

Easy CC one:
CC Player X has 1300 points. But he's yet to win a game. How?

He scored 300 points in terminator games without finishing them

That sounds right, but let me ask, since I don't play anything but standard: he may have finished the games and lost, because he still gets points for opponents that he has eliminated regardless of who wins, right? That's what I understand from reading the Game Options tab of the Instructions page, anyway.

[YoursFalsey]

Correct.

Also, for terminator games the winner is the last one standing. If player X joins an eight player terminator game, elimates 6 of his opponents, and then gets whacked by the player Y, player Y is the winner of the game, even though X's rating improved by more.