help me with my logic assignment

[karl184]

please help me with my logic assignment. kailngn ko kasi ito by wednesday. sana matulungan nyo ako. =)

ONLY DEFINITE ANSWERS
NO PROBABILITIES
DEFEND ANSWER

In a certain mythical community, all politicians never tell the truth, and all non-politicians always tell the truth. A stranger meets 3 of these natives and asks the 1st of them: “Are you a politician?” The 1st native answers the question. The 2nd native then reports that the 1st denied being a politician. The 3rd native says that the 1st native is a politician.
· HOW MANY OF THESE 3 NATIVES ARE POLITICIANS?


Of 3 prisoners in a certain jail, one had normal vision, the 2nd has only one eye, and the third was totally blind. The jailor told them that from 3 white hats and 2 red hats, he would select 3 and put them on the prisoners’ heads. None of them could see what hat he wore. The jailor offered freedom to the 1st prisoner if he could tell what color hat he wore. To prevent a lucky guess, the jailor threatened execution for any incorrect answer. The 1st prisoner could not tell what color hat he wore. Next, the jailor made the same offer to the one eyed prisoner, but he could not tell what color hat he wore either. The jailor did not bother making the same offer to the blind prisoner, but then agreed to extend the same terms to him when he made the request. The blind prisoner then said:

“I do not need to have my sight,
From what my friends w/ eyes have said,
I clearly see my hat is ____________”
· HOW DID HE KNOW?

“That Abraham sure is one rich Arab.” Says Isaac. “He owns a hundred or more camels!” Well,” says Jacob, “I know for a fact that Abraham owns less than a hundred camels.” “Lets put it this way”, say Ishmael. “Abraham owns at least one camel.”

· IF ONLY ONE OF THE 3 STATEMENTS ABOVE IS TRUE, HOW MANY CAMELS DOES ABRAHAM OWN?

[Oscar01]

ANSWER TO #1:
The first native must answer "no" to the question. If he answers yes, and he is a politician, he would be telling the truth. If he answers yes, and he is not a politician, he would be lying. Neither can happen.

The first native cannot be a politician.

The second native is thus telling the truth.

The second native cannot be a politician.

The third native cannot be telling the truth.

The third native is thus the only politician.

[Oscar01]

ANSWER TO #2:
Each prisoner can wear a red or a white hate. By the counting principle (2 x 2 x 2), there are eight possibilities.

R R R --> impossible, because there are only two red hats

W R R --> impossible, or #1 would answer white.

R W R --> impossible, or #2 would answer white.

W W R --> impossible, or #1 would answer white after #2 speaks.

W R W

R W W

W W W

R R W

Thus, #3 must be wearing a white hat.


ANSWER TO #3:
Let C be the number of camels.

If C >= 100, then C also >= 1.

If C >= 1, then C either >=100 or < 100.

Thus, C cannot be >= 1 or >= 100.

C cannot also be < 0, because that is absurd.

Thus, C must be zero, and Abraham is not "one rich Arab."

Therefore, C < 100.

[karl184]

oo nga noh! hehe! galing naman! thanks!!

[mac_bolan00]

Originally posted by Oscar01
ANSWER TO #2:

W R W

R W W

W W W

R R W

Thus, #3 must be wearing a white hat.

both the two-eyed and the one-eyed guys were wearing red which they failed to identify. only then could the blind man be sure he was wearing white.

oscar01, we used the same method: trial-and-error.

[Oscar01]

Uh... no.

That's not trial and error. Mapping out an entire probability tree is never trial and error.

[Oscar01]

Someone turn this into a brainteaser thread.

[mac_bolan00]

Originally posted by Oscar01
Uh... no.

That's not trial and error. Mapping out an entire probability tree is never trial and error.
point taken, bro. :)

but then, a probability tree will only point out the possible states of nature. a decision tree, on the other hand, will show you the only acceptable state of nature which will fulfill the ncessary condition: that the blind man was 100% sure. that state of nature is RRW.

[Oscar01]

Hmmmm...

By the counting principle, we get the following possibilities:

#3 wearing white: 4 x 1

#3 wearing red: 4 x 1


Listing the probabilities and eliminating the impossible conditions:

#3 wearing white: 4 x 1

#3 wearing red: 0 x 1


#1 can't distinguish between W R W and R R W.

#2 can't distinguish between R W W and R R W.

But it doesn't matter; #3 knows he's wearing white regardless of what the other two are wearing.

Saying "decision tree" is floating meaningless jargon, since there is no decision to be made in the entire problem. Using your own definition, there's nothing else to look at but possible states of nature.

[mac_bolan00]

this is a three-step probability generation and the probabilities available to the third man depend on the outcome of the first two. again, the condition must be the blind man was absolutely sure he was wearing white. if one of the first two guys was in fact wearing white, then there was a 33% probability that the blind man would get a red hat.

the above condition requires you to make a choice or a 'decision' for the third man which, for this problem, is uniquely RRW.

[Oscar01]

(I'm having a lot of fun nitpicking with you here Mac.)

There are no probabilities involved whatsoever:

R R R

R R W

W R R *

R W R *

W W R *

W R W *

R W W *

W W W *

Taking the entire thing together, 6 out of 8 outcomes have either or both of the first two wearing white. You can go into the specific probability of each of the 8 outcomes, but I don't see how it helps.

After all, this is an exercise in logic, not probabilities.

By showing that all outcomes where he is wearing red are impossible if the first two are rational, he must conclude that he is wearing white regardless of what the other two are wearing.

Check your math Mac! :p

[Me]

To question #1:

1. The first native answered “no” whether or not he/she is a politician. But whether or not he/she is we cannot say at this point.

2. The second native is telling the truth (non-pol) in saying that the 1st denied.

3. The 3rd native’s contention despite pt.1 above makes him/her a liar (pol). A liar saying that the 1st is a politician leads to the conclusion that the first is a non-politician. :D

[Me]

Originally posted by Oscar01

W W R --> impossible, or #1 would answer white after #2 speaks.

In this scenario, the #2 prisoner, and not the #1, should be the one to conclude that he is wearing a white hat.

The #2 prisoner knows that #1 must be seeing at least one white hat, or else #2 & #3 both have reds on making it a give away for #1. #2 with that knowledge and seeing #3 wearing a red can easily conclude that the one white hat #1 is seeing is his. :D

[mac_bolan00]

no heavy math here either. again, understand the basic requisite. the blind man had to be sure he was wearing white. being sure means the absence of a probability that he was wearing red. had the outcome for the first two been different, he wouldn't have been sure.

[Me]

Originally posted by mac_bolan00

both the two-eyed and the one-eyed guys were wearing red which they failed to identify. only then could the blind man be sure he was wearing white.

Originally posted by mac_bolan00

but then, a probability tree will only point out the possible states of nature. a decision tree, on the other hand, will show you the only acceptable state of nature which will fulfill the ncessary condition: that the blind man was 100% sure. that state of nature is RRW.

Even if the two- and the one- eyed persons are wearing reds the third prisoner won’t be able to see that because he is a blind man! :D

But if he is confident that the other two are very logical, he can eliminate all scenarios that should enable the others to make a definite conclusion. He must have, like Oscar01, eliminated these which left him with scenarios that have him wearing a white hat. :D

[mac_bolan00]

sight did not matter. :rolleyes:

the other two were logical??? they both failed to guess the color. how does that make them logical?

[advancement]

I wonder if your logic would enable you to survive my POW camp.

If I were the jailer, I might put a white hat on the first clueless prisoner, the other white hat on the second clueless prisoner and if upon putting the third hat on the third prisoner's head he says “I do not need to have my sight, From what my friends w/ eyes have said, I clearly see my hat is white. do you want me to show you the math?” I would just say "no thank you" then pull out my .45 caliber pistol and blow his head off. :glee:

[sweetchick]

hmmm.. who's ur prof? i'm guessing it's Raj (close kme eh..lolz!)

i checked my notes.. pero di ko gets, tagal na nun eh, frosh pa ko..

basta eto nakalagay..

politician:
1 t f t T
2 f t t F
3 t f f F

so..the last 2 are politicians..

prisoners:

normal vision: white
one-eyed: white
blind: red

abraham:
ishmael: at least one

sorry kung magulo, basta yan unmg answers..

:spinstar:

[advancement]

Tama si Oscar sa 1 and 3. sa 2 malabo pa.

[advancement]

In a certain mythical community, all politicians never tell the truth, and all non-politicians always tell the truth. A stranger meets 3 of these natives and asks the 1st of them: “Are you a politician?” The 1st native answers the question. The 2nd native then reports that the 1st denied being a politician. The 3rd native says that the 1st native is a politician.
· HOW MANY OF THESE 3 NATIVES ARE POLITICIANS?

Native 1 answers:
Yes I am a politician = Impossible to have been said
No I am not a Politician = He is either a politician (who denies being one) or a non-politician (telling the truth about his status). No is the only possible answer but there is still a possibility that Native 1 is a politician just denying that he is one.

Native 2: reports that Native 1 denied that he is a politician. He is telling the truth therefore he is a non-politician.

Take note that Native 2 isn't saying that 1 is a politician or not. 2 is only saying that 1 denied being a politician. He is telling the truth about what 1 said, not concluding if 1 is a politician or not.

Native 3: Says that the first native is a politician.
If 3 is a non-politician, then he is telling the truth therefore Native 1 is a politician. (Pattern now is P N N where P = Politician, N = Non Politician)
If 3 is a politician, then what he's saying is false therefore Native 1 is a non-politician (Pattern in this case is N N P).

We have the following possibilities:

N=Non Politician
P=Politician

1. P N N
2. N N P

Therefore the answer to the question: HOW MANY OF THESE 3 NATIVES ARE POLITICIANS? is ONE

But when asked, you are NOT SUPPOSED TO SAY who among 1 or 3 is DEFINITELY a politician/non-pol dahil variable yun. Baka kasi maliin ka (dahil sa solution) kahit tama number na binigay mo! State only what is needed. You can only conclude that Native 2 is a non-politician (N) but if Native 1 is P then 3 is N and vice versa.

So just answer ONE (1) because the question only requires the number of politicians and THAT'S IT OK? Of course, you have to discuss the solution too.

O.K. ba ka Oscar? ;)

BTW, in the hats question, the status of their eyes have no bearing on the question coz they couldn't see the hats anyway.

Oscar's answer to 3 is correct and well presented.

[Oscar01]

Originally posted by advancement
there is still a possibility that Native 1 is a politician just denying that he is one.
Oops. Got me there. I got 1 and didn't look back.

[Oscar01]

Originally posted by Me
The #2 prisoner knows that #1 must be seeing at least one white hat, or else #2 & #3 both have reds on making it a give away for #1. #2 with that knowledge and seeing #3 wearing a red can easily conclude that the one white hat #1 is seeing is his. :D
Yeah, I think that's the better way to eliminate the last red possibility.

Mine was weird.

If #1 sees #2 wearing a white hat and #3 wearing a red hat, he knows he sees either:

W W R

R W R

But if #2 doesn't answer, it means it can't be R W R, so he can tell it's white.

The straightforward one says that #2 sees

W R R

W W R

And he eliminates the first based on #1's answer.

[Oscar01]

We had a lot of these back in grade school, but the only one I remember is:

You are in a POW camp, and at your execution, your captor goes up to you and asks for your last words.

If you tell the truth, you will be shot.

If you tell a lie, you will be beheaded.

If you say nothing, you will be hanged.

What do you tell him?

[advancement]

Originally posted by Oscar01
We had a lot of these back in grade school, but the only one I remember is:

You are in a POW camp, and at your execution, your captor goes up to you and asks for your last words.

If you tell the truth, you will be shot.

If you tell a lie, you will be beheaded.

If you say nothing, you will be hanged.

What do you tell him?

I would tell him that I would be beheaded.

Having said something, I would not be hanged.
If they behead me, it would be the truth and disqualifies me from the 2nd condition.
They cannot shoot me, that would make my statement a lie so it contradicts the condition for shooting.

[Oscar01]

Galing!

How about this one...

You have 12 iron balls. All look exactly alike. They weigh the same, except one which is slightly heavier or lighter.

You are given balancing scales with 2 pans (you know... the one Lady Justice has in hand), and are asked to tell which ball is the odd one out.

But, you have to use the scale only three times.

[sweetchick]

to the person who visited my blog and posted in my tagboard:

i was only trying to help..kaya nga i said i wasn't sure!!! hello?! pero thanks na rin for saying that, at least i know na mali pala notes ko back then..


:spinstar:

[raven23]

Nasagot na pala ni Oscar01 yung original questions. Pinagisipan ko pa and I even prepared a nicely worded set of answers. Anyway...

[diet_sarsi_pls]

Oscar01

step 1 : divide into 2 sets (6 balls on each) and put on scale the 1st time. whichever set has the odd ball tips the scale lower (or higher) depending on which criteria (heavier or lighter ball respectively)

step 2: use this set and divide into 2 (3 balls each) and for the 2nd time see which set again tips the scale, which of course would have the odd ball

step 3: of that set of 3 balls, balance out 2 balls. 3rd time. if they weigh the same, you surtely have the odd ball in hand (3rd ball). If not, the scale tells you which one it is.

[Oscar01]

Believe it or not, I haven't given the problem a go myself, but when I got the problem, it didn't say whether the ball was in fact heavier or lighter than the other 11. :)

[advancement]

Originally posted by diet_sarsi_pls
Oscar01

step 1 : divide into 2 sets (6 balls on each) and put on scale the 1st time. whichever set has the odd ball tips the scale lower (or higher) depending on which criteria (heavier or lighter ball respectively)

step 2: use this set and divide into 2 (3 balls each) and for the 2nd time see which set again tips the scale, which of course would have the odd ball

step 3: of that set of 3 balls, balance out 2 balls. 3rd time. if they weigh the same, you surtely have the odd ball in hand
(3rd ball). If not, the scale tells you which one it is.

I would detail the flaws in your solution:

Let
T=Odd ball
Y=Regular Ball

Step 1 (flaw starts here because the weight of T in comparison with Y [heavier or lighter] was not taken into account. It is not given, this automatically means that it has to be determined): Weigh 2 sets of 6 balls each. This requires using the scale once - but how would you know which set to eliminate? Set A or Set B?We have to determine that using...

Step 2: Requires using the scale twice -
Step 2A: weigh set A of 3 balls each (Subset A1 and Subset A2) if it tips the scales (Let's assume A1 > A2), proceed to step 3A, if not, eliminate this set and go to step 2B.
Step 2B: We now have eliminated the set that clearly doesn't have T which leaves us six balls. Go to step 3B.

Step 3A: Retaining Subset A1 on the scales, weigh it against set B1. If A1 = B1 then T < Y. If A1 > B1 then T > Y ... This is the third time the scale was used. We now have determined the weight of T in comparison with Y.
Step 3B: weigh set B of 3 balls each (Subset B1 and Subset B2) if it tips the scales (Let's assume B1 > B2), proceed to step 3A, if not, eliminate this set and go to step 3A switching the variables B and A. The problem is...this is the third time the scale was used so this constrains us from making any further measurements.

The number of steps already exceeded the parameter! hmmm... As for my solution, I'm getting close already. Hang in there :D

[advancement]

I got one part of the odd ball problem... feel free to point out any flaws and make corrections...

Let T = Odd Ball
Let W = Number of measurements in process
W<=3

I. Divide the balls into three equal parts we shall call Set A, B and C.

II. Weigh sets A and B on the scales (W=1)

A. If the scales don’t tip, the odd ball is in set C so isolate it. We now have the balls C1, C2, C3 and C4.

Weigh C1 and C2 (W=2)

A1.If the scales tip, (let C1 > C2) then remove C2 and place C3 (W=3)
a. If C1 = C3 then T = C2 (Lighter)
b. If C1 > C3 then T = C1 (Heavier)

A2. If the scales don’t tip, (C1=C2) then remove C2 and place C3 (W=3)
a. If C3 > C1, then T=C3 (Heavier)
b. If C3 < C1, then T=C3 (Lighter)
c. IF C3=C1 then T = C4

B. If the scales tip - ewan ko pa... (to be continued…esep esep muna tayo. :D )

[raven23]

Originally posted by Oscar01
Believe it or not, I haven't given the problem a go myself, but when I got the problem, it didn't say whether the ball was in fact heavier or lighter than the other 11. :) Believe me, I've been going at it since last night. I even posted what I thought was a solution but quickly recanted.

I don't think the problem can be solved without knowing first-hand whether the odd ball is lighter or heavier. Given that fact, the problem becomes trivial.

Without that fact, the best solution I could come up with still leaves one possibility (out of all other certainties) that after 3 uses of the scale, we're still left with two balls not knowing which one was lighter or heavier.

[mac_bolan00]

a 50% probability of solving it if you do the 6-3-1 process and go with either assumption (whether the odd ball is heavier or heavier).

here's an easier one:

a drug user named mac_bolan00 needs six roaches to make one joint. if mac_bolan00 had 36 roaches at the start, how many joints can he make?

:)

[advancement]

Tell you what people, you take care of your balls while I let these pretty babes here take care of mine ;)

Now THAT's logical!

[Me]

Using advancement’s variables:

I. Weigh 1,2,3,4 against 5,6,7,8 (W=1);

......A. Balance. therefore Y = 1,2,3,4,5,6,7,8 and therefore T is in 9,10, 11, 12
......Weigh 1,2,3 against 9,10,11 (W=2)

............1. Balance. therefore T=12
............We can stop here or proceed to find out if T is light or heavy.
............We proceed: Weight 12 against 1 (W=3)

..................a. Tips to 12. therefore T = 12 (heavy)

..................b. Tips to 1. therefore T = 12 (light)

............2. Tips to 9,10,11. therefore T is heavy and not 12.
............Weigh 9 against 10 (W=3)

..................a. Balance. therefore T = 11 (heavy)

..................b. Tips to 9. therefore T = 9 (heavy)

..................c. Tips to 10. therefore T = 10 (heavy)

............3. Tips to 1,2,3. therefore T is light and not 12.
............Weigh 9 against 10 (W=3)

..................a. Balance. therefore T = 11 (light)

..................b. Tips to 9. therefore T = 10 (light)

..................c. Tips to 10. therefore T = 9 (light)

......B. Tips to 1,2,3,4. therefore Y = 9,10,11,12 and therefore T is either heavy in
......1,2,3,4 or light in 5,6,7,8.
......Weigh 1,2,5 against 3,6,9 (W=2)

............1. Balance. therefore Y = 1,2,5,6 and therefore T is either a heavy 4 or
............ a light 7 or 8.
............Weigh 7 against 8 (W = 3)

..................a. Balance. therefore T = 4 (heavy)

..................b. Tips to 7. therefore T = 8 (light)

..................c. Tips to 8. therefore T = 7 (light)

............2. Tips to 1,2,5. therefore Y = 3,5 and therefore T is either a heavy 1 or 2
............ or a light 6.
............Weigh 1 against 2 (W = 3)

..................a. Balance. therefore T = 6 (light)

..................b. Tips to 1. therefore T = 1 (heavy)

..................c. Tips to 2. therefore T = 2 (heavy)

............3. Tips to 3,6,9. therefore Y = 1,2,6 and therefore T is either a heavy 3
............or a light 5.
............Weigh 3 against 9 (W = 3)

..................a. Balance. therefore T = 5 (light)

..................b. Tips to 3. therefore T = 3 (heavy)

......C. Tips to 5,6,7,8. therefore Y = 9,10,11,12 and therefore T is either light in
......1,2,3,4 or heavy in 5,6,7,8.
......Weigh 1,2,5 against 3,6,9 (W=2)

............1. Balance. therefore Y = 1,2,5,6 and therefore T is either a light 4
............or a heavy 7 or 8.
............Weigh 7 against 8 (W = 3)

..................a. Balance. therefore T = 4 (light)

..................b. Tips to 7. therefore T = 7 (heavy)

..................c. Tips to 8. therefore T = 8 (heavy)

............2. Tips to 1,2,5. therefore Y = 1,2,6 and therefore T is either a heavy 5
............ or a light 3.
............Weigh 5 against 9 (W = 3)

..................a. Balance. therefore T = 3 (light)

..................b. Tips to 5. therefore T = 5 (heavy)

............3. Tips to 3,6,9. therefore Y = 3,5 and therefore T is either a light 1 or 2
............ or a heavy 6.
............Weigh 1 against 2 (W = 3)

..................a. Balance. therefore T = 6 (heavy)

..................b. Tips to 1. therefore T = 2 (light)

..................c. Tips to 2. therefore T = 1 (light)


whew! :D

[diet_sarsi_pls]

Oscar01: yeah, I thought my initial solution was too easy and so figured you meant that the oddball's weight is unknown (lighter or heavier). I just had to wait for you to clarify that :grinroll:

For me, in order of priority I should identify:
1. the set/subset in which the oddball belongs to then
2. identify if it is lighter or heavier, then lastly
3. the oddball itself
....all in 3 tries of using the scale

here goes:

divide first into 2 sets of 6 (A nd B). Each set will then be divided into 2 sets of 3 balls (subsets A1+A2 and B1+B2 respectively)

1st weighing (to identify which set the oddball is in) : weigh A1 against A2 (or B1 vs B2 doesnt matter) coz if it balances then the oddball is in the other set and if not, then the oddball is in any of the current subsets you're testing or vice-versa. Regardless, after this, you now know which main set (A or B) is normal. Right?

2nd weighing (to identify if the oddball is lighter or heavier): Weigh A against B. Since you now know which set is "not normal", the side to which the normal set/ or "abnormal" set tips will tell you if the oddball is lighter or heavier. Right?

3rd weighing (to identify the oddball itself - it's kinda tricky) :D The 3rd weighing "depends" on what happened in the 1st weighing from which you make inferences.

so.....

If in the 1st weighing the scales don't balance ( this is the much easier part) you can then infer which subset the oddball belongs to because just like in the 2nd weighing, where you were able to determine if the oddball is lighter or heavier, the side to which it tips over tells you which subset of 3 has the oddball. Example, in the 1st try A2 is lower than A1, so you're sure it has the oddball but are not certain yet if it's lighter or heavier....but if in the second weighing you determined the oddball to be lighter, recalling the results from the 1st weighing you can pinpoint that A1 has the lighter oddball. Correct? Again it doesn't matter if it's lighter or heavier but the key here is knowing the exact subset of 3 balls to use for the final testing. Having done so, weigh 2 balls from this subset against each other (in this case A1) 3rd weighing na. Naturally, kung balanced out the oddball is in your hand , if not, again the direction to which it tips tells you which ball is lighter or heavier so ayun identified na which ball.

but what if from the 1st weighing eh balanced ang subsets A1 and A2?Eto na yung "tricky part"......

weigh B1 against B2 coz you're sure nan duon ang odd ball in either subset 3rd weighing na. Then again based on the 2nd weighing you can tell if the oddball is lighter or heavier, so after the 3rd weighing you can identify the oddbal subset. Tapos from this subset put 1 ball in 1 hand and another 1 ball in the other hand. Close your eyes and CONCENTRATE coz I believe if you try REAL HARD you can tell which ball is lighter or heavier and thus is the odd ball. It's not difficult considering the density of iron cast is about 7000kg/m3. Kung feeling mo pareho ang weights, then yung oddball is the 3rd ball not in your hands. HEHEHEHE :grinroll: - Well?!?! I used the scale 3 times didn't I?!So what are you guys complaining about?

I'm so sorry but that's the best solution I could think of. :rotflmao:

hey mac_bolan00: can you spare me a joint? Nagka-headache ako tuloy sa kaisisp. and advancement: when the babes are finished with you, kindly remind them it's my turn naman.

[mac_bolan00]

Originally posted by diet_sarsi_pls
hey mac_bolan00: can you spare me a joint? Nagka-headache ako tuloy sa kaisisp. and advancement: when the babes are finished with you, kindly remind them it's my turn naman.
congratulations. i couldn't have figured it out. now, your invitation to a pot session is, unfortunately, one of the better hits i get here in PEX. PM me. :D

[straightshooter]

Originally posted by mac_bolan00
a 50% probability of solving it if you do the 6-3-1 process and go with either assumption (whether the odd ball is heavier or heavier).

here's an easier one:

a drug user named mac_bolan00 needs six roaches to make one joint. if mac_bolan00 had 36 roaches at the start, how many joints can he make?

:)

Seven joints. One roach. Matamis ba rolling paper mo?