Puzzle : The Password Challenge by Evil Troll

A bridge was guarded by an evil troll. The troll was very intelligent, but he was also a coward. He was afraid of anyone smarter than him. So every time anyone tried to cross the bridge, the troll would set up a test. If the traveler passed the test, he would be allowed to cross. Otherwise, the troll would eat him.

Three travelers, Al, Ben and Carl, came across the bridge. 

The troll told them, "You may only cross my bridge if you know the password." 

He wrote five three-letter words on a rock. The five words were HOE, OAR, PAD, TOE, and VAT.

He then said, "I will tell each of you a different letter from the password. If you know what the password is, I will let you pass. But don't tell anyone else your letter." 

He then whispered a letter from the password to each traveler so that neither of the other two could hear him.

Then the troll asked Al, "Do you know what the password is?" "Yes," said Al, and the troll let him pass.

Then the troll asked Ben, "Do you know what the password is?" "Yes," said Ben, and the troll let him pass.
 
Then the troll asked Carl, "Do you know what the password is?" "Yes," said Carl, and the troll let him pass.

So, what is the password?

THIS is the correct password! 

One more such challenge by an evil troll! 

Solution : Intelligent Response to an Evil Troll

What was the challenge?

The list of words given by evil troll is - 

HOE, OAR, PAD, TOE, and VAT

Remember, all travelers i.e. Al, Ben and Carl knew the correct password straightaway as soon as evil troll whispered a letter from the password to each traveler.

STEPS : 

1] The unique letters (i.e. letters appearing only once) in above list of words are D, H, P, R and V. The evil troll must have one of these letters to Al, as Al could guess the correct password straightaway. 

The word TOE doesn't have any unique letter, hence, TOE is eliminated straightaway. 

Had Evil troll whispered any letter from TOE (i.e. T, O or E), then Al wouldn't have an idea whether the correct password is TOE or VAT, HOE or TOE or OAR,  HOE or TOE.

2] Now, Ben is smart enough to know that TOE is eliminated from the race after Al's response. He has to think about only four words i.e. about HOE, OAR, PAD, VAT.

The unique letters appearing in rest of words list are D, E, H, P, R, T, and V. One of these letters must be with Al and other must be with Ben. 

But the word OAR has only one unique letter i.e. R. If OAR was the password then only 1 of 3 travelers would have guessed the password correctly while other 2 would have been confused.

Therefore, OAR can't be the password.

3] Carl too smart enough to recognize that OAR and TOE are not the correct passwords. So, he has to think of only 3 words -  HOE, PAD, VAT.

Here, unique letters from the list of words are - D, E, H, O, P, T, and V.

Every traveler must have one unique letter from the above list. In fact, the password itself must be formed by only unique letters from the above list.

Words PAD and VAT has only 2 unique letters ( P & D, V & T respectively).

So, if PAD or VAT was the correct password then the one with letter A wouldn't have been able to guess the correct password.

Hence, HOE must be the correct password. 

4] The letter H must be with Al as T, O, E can't be with him. Similarly, Ben can't have letter O, so he must had letter E and he knows TOE is not the password after hearing Al's response. And the letter O must be with Carl and he knows TOE or OAR are not the passwords.

Puzzle : An Evil Troll on A Bridge

A bridge was guarded by an evil troll. The troll was very intelligent, but he was also a coward. He was afraid of anyone smarter than him. So every time anyone tried to cross the bridge, the troll would set up a test. If the traveler passed the test, he would be allowed to cross. Otherwise, the troll would eat him.

A traveler came across the bridge. 

The troll said, "You may only cross my bridge if you know the password." 

He then wrote thirteen pairs of letters on a rock:

A-V
B-W
C-Q
D-M
E-K
F-U
G-N
H-P
I-O
J-R
L-X
S-T
Y-Z

"These thirteen pairs consist of all 26 letters of the alphabet," said the troll. 

"The password contains thirteen letters, no two of which are the same. Each pair consists of one letter that is in the password and one other letter. If you wrote out the "other" letters in alphabetical order and then wrote each "password" letter under each one's corresponding "other" letter, you would have the correct spelling of the password."

Then the troll wrote five short words on the rock: FACE, QUEST, QUICK, SWITCH, and WORLD. 

"Each short word contains exactly the same number of letters with the password," he said.

So, what is the password? 

HERE is that PASSWORD!

Source

Solution : An Evil Troll on A Bridge Puzzle : Solution

What was the Puzzle?

The thirteen pairs of letters given by an evil troll are -

A-V
B-W
C-Q
D-M
E-K
F-U
G-N
H-P
I-O
J-R
L-X
S-T
Y-Z

And 5 short words given by troll are -  FACE, QUEST, QUICK, SWITCH, and WORLD.  

As described in the given details, we'll refer letter from password as PASSWORD letter & other as OTHER letter.

As per troll, those short words are having same number of PASSWORD letters.

STEPS :

1] Both S & T are appearing in the pair with each other. Hence, either S or T must be a PASSWORD letter but not both. Since, both letters are appearing in short word QUEST, that is QUEST having at least 1 PASSWORD letter for sure hence, all 5 must have at least 1 PASSWORD letter.

2] Suppose every short word has 1 PASSWORD letter. With S or T as 1 PASSWORD letter from QUEST, other letters Q, U, E can't be PASSWORD letters. 

If Q, U, E are not PASSWORD letters then C (from C-Q pair), F (from F-U pair) and K (from E-K pair) must be PASSWORD letters. 

In that case, FACE will have 2 PASSWORD letters viz. C & E which goes against our assumption of having exactly 1 PASSWORD letter in each short word. 

3] Let's assume along with S or T the second PASSWORD letter is E i.e each short word has 2 PASSWORD letters. Again, Q, U can't be PASSWORD letters but C (from C-Q pair) & F (from F-U pair) must be. Still then FACE will have 3 PASSWORD letters which goes against our assumption of exactly 2 PASSWORD letter in each short word. 

4] Now, let's assume along with S or T the second PASSWORD letter is U. Again, Q, E can't be PASSWORD letters but C (from C-Q pair) & K (from E-K pair) must be. Still then QUICK will have 3 PASSWORD letters which goes against our assumption of exactly 2 PASSWORD letter in each short word. 

5] Let's assume there are 4 PASSWORD letters in each short word. So apart from S or T, the letters Q, U, E of short word QUEST must be PASSWORD letters. 

If Q, U, E are PASSWORD letters then C (from C-Q pair), F (from F-U pair) and K (from E-K pair) must NOT be the PASSWORD letters. 

In the case, the short word FACE will have maximum only 2 PASSWORD letters (not sure about A from A-V pair) which again goes against our assumption of exactly 4 PASSWORD letter in each short word. 

6] Hence, each short word must be having 3 PASSWORD letters. 

If Q, E are the PASSWORD letters with S or T in QUEST, then C & K can't be PASSWORD letters. With that, Q, U, I will be 3 PASSWORD letters in QUICK. And if U too is the PASSWORD letter then QUEST will have 4 PASSWORD letters. 

If Q, U are the PASSWORD letters with S or T in QUEST, then C & F can't be PASSWORD letters. With that, FACE can have maximum of only 1 PASSWORD letter. 

7] Hence, U & E must be other 2 PASSWORD letters apart from S or T in short word QUEST. So Q must not be the PASSWORD letter but C must be. Also, F and K can't be the PASSWORD letters.  Hence, FACE will have E, C and A as PASSWORD letters. 

If A is PASSWORD letter then V (from A-V pair) can't be the PASSWORD letter.

8] Next, from QUICK we will have, C, U and obviously I as 3 PASSWORD letters after Q, K are ruled out. If I is PASSWORD letter then O (from I-O pair) can't be the PASSWORD letter.

9] Just like QUEST, SWITCH too have either S or T as PASSWORD letter. Moreover, it has I & C as PASSWORD letters. Hence, H & W must not be the PASSWORD letters.

10] So, if W & O are not the PASSWORD letters then other 3 letters of WORLD i.e. R, L, D must be PASSWORD letters. With that M (from D-M pair), J (from J-R pair) and X (from L-X pair) are ruled out.

11] So far we have got - 

PASSWORD letters - U, E, C, A, I, R, L, D, Either S or T.

OTHER letters - Q, F, K, V, O, H, W, M, J, X  

12] Arranging every OTHER letter in alphabetical order & writing down corresponding PASSWORD letter below it -

OTHER :  F   H   J   K   M   O   Q   V   W   X
PASS.  :  U   P   R   E   D   I    C   A    B   L 

13] Now, S-T, G-N, Y-Z are the only 3 pairs left. And correct placement for these pairs must be like.

OTHER :  F   G   H   J   K   M   O   Q   S   V   W   X   Z
PASS.  :  U   N   P   R   E   D   I    C   T   A    B    L   Y

CONCLUSION : 

The PASSWORD that an evil troll has set must be UNPREDICTABLY. 

 
 

The Antisocial Club Challenge

The Antisocial Club meets every week at Jim's Bar. Since they are so antisocial, however, everyone always sits as far as possible from the other members, and no one ever sits right next to another member. Because of this, the 25-stool bar is almost always less than half full and unfortunately for Jim the members that don't sit at the bar don't order any drinks. 

Jim, however, is pretty smart and makes up a new rule: The first person to sit at the bar has to sit at one of two particular stools. If this happens, then the maximum number of members will sit at the bar. 

Which stools must be chosen? Assume the stools are numbered 1 to 25 and are arranged in a straight line.

Wise Selection of first 2 Seats made available! 

Logical Response to The Antisocial Club Challenge

What was the challenge?

The bar owner Jim put only 2 options for first person to seat and those options are Stool No. 9 and Stool No.17. We'll call seat for stool for convenience.

Choice of numbers 9 and 17 is because of symmetry explained below.

Suppose the first person choose seat no.9.

Then the second person has to choose the furthest seat i.e. seat no.25.

The third person will choose seat no.1 which is furthest from seat no.9 and 25.

The fourth person won't have option other than seat no.25 which is the furthest from rest of 3.

For the fifth person, seats numbered 5, 13 and 21 are available which are equidistant from 2 persons. We assume he chooses farthest from the seat occupied by person who entered just ahead of him. That is he chooses seat no.5.

Since, seat no.21 is furthest from seat no.5 and only seat equidistant from 17 & 25, the sixth person will choose seat no.21.

With that, 7th person won't have any option other than seat no.13 to maintain distance from others.

With the same logic, next 6 persons occupies seat numbered 3, 23, 7, 19, 11, 15.

So there will be 13 people maximum in the bar with no one seating next to each other.

The same seat numbers will be occupied even if the first person choose seat no.17. The second person will choose seat no.1, third will choose seat no.25, fourth will choose seat no.9 and so on.

But if the first person doesn't choose 9 or 17 then there can be less than or equal to 12 people maximum in the bar following their antisocial trait.