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. 

 
 

Puzzle : And Escape Story of Robbers Continues

Where story begins?

Babylas, Hilary, and Sosthenes have escaped the tower and divided their treasure into three bags. But now they must cross a river, and the boat can accommodate only two men at a time, or one man and a bag. None will trust another with his bag on the shore, but they agree that a man in the boat can be trusted to drop or retrieve a bag at either shore, as he’ll be too busy to tamper with it.

 How can they cross the river?

Source. 


This is how they can cross the river! 

 

Solution: Robbers' Planned Journey Across the River

Why they need a PLAN?

Let's recall that the boat can accommodate only two men at a time, or one man and a bag.

1. Sosthenes takes his bag across the river leaves it at other shore & comes back.

2. Sosthenes takes Hilari's bag to the other shore & leaves it there where his own bag is already there. 

3. Now, Hilari takes Sosthenes to the other shore, leaves him there & come back after recollecting own bag.

4. Hilari drops own bag at near shore & takes Babylas to other shore & returns back.

5. Next, he takes Babylas's bag & drops it at other shore where Babylas is waiting for his bag. And Hilary returns once again.

6. Finally, he collects his own bag and takes it to other shore.

The Love Island

Four men and four women are in Love Island. Each one falls in love with another and is himself/herself loved by someone else in a complete 8 person loop.

Peter falls in love with a girl who is unfortunately in love with Albert. Joseph loves a girl who loves the man who loves Elaine. Liza is loved by the guy who is loved by the girl who is loved by Paolo. Linda hates Paolo and is hated by the man whom Amy loves.

No guy/guy or girl/girl in Love Island.

Who loves Joseph and whom does Joseph love?

Know here who loves Joseph and Joseph loves whom! 

The Love Loop on the Love Isaland

What was the puzzle?

There are few hints given in the question.

-----------------------------------------------------------


1. Peter falls in love with a girl who is unfortunately in love with Albert. 

2. Joseph loves a girl who loves the man who loves Elaine. 

3. Liza is loved by the guy who is loved by the girl who is loved by Paolo. 

4. Linda hates Paolo and is hated by the man whom Amy loves.

----------------------------------------------------------- 

STEPS :

STEP 1 :

Let's name all 4 men as M1, M2, M3 and M4 while girls as G1, G2, G3 and G4.

STEP 2 :

Let's start with Peter. He can be at any position M1, M2, M3 or M4. Let's assume that he is at M1. As (1) points M2 must be Albert.

STEP 3 :

As per (2), there are 2 persons in between Joseph and Elaine. That is Joseph can be at M3 or M4 while Elaine can be at G4 or G1 respectively. Exactly, same are possible locations for Paolo and Liza respectively in the loop as (3) suggests.

STEP 4 :

In short, at M3 and M4, we have Joseph and Paolo and G1, G4 are occupied by Liza and Elaine but exact location yet to be known.

STEP 5 :

For The statement (4) to be TRUE, Linda must be hating Paolo. 

 👉 CASE 5.1 : If Paolo is at M3 then Linda can't be at G2 hence must be at G3. With that, Joseph will be at M4, Elaine at G1, Liza at G4 & only location for Amy as G2.

 👉 CASE 5.2 : If Paolo is at M4 then Linda can't be at G3 hence must be at G2. With that, Joseph will be at M3, Elaine at G4, Liza at G1 & only location for Amy as G3.

CASE 5.1 make other part of the statement (4) false i.e. Linda is hated by man (Paolo here) whom Amy loves is false in the case.

Hence, CASE 5.2 is valid and the final loop looks as - 

Finally, turning to the question asked.

Q. Who loves Joseph and whom does Joseph love?

A. Linda loves Joseph and Joseph loves Amy.

Who is the President of Logitopia?

Larry, Matt, and Nick live in the strange country of Logitopia. 

This country is inhabited by three races of people: the type A people who always tell the truth, the type B people who always lie, and the type C people who alternately tell the truth and lie. One of the three is the president of Logitopia.

Larry makes these two statements:

1. "The president is of a different race from the other two."
2. "Matt is not the president."

Matt makes these two statements:

1. "The president is a type B person."
2. "Larry is not the president."

Nick makes these two statements:

1. "Exactly two of us are of the same race."
2. "I am not the president."

Who is the president?

Know the NAME of the President!
 

Larry is the President of Logitopia

What was the puzzle?

First thing to note that it's not mentioned any where that three people are three different types; there may be 2 who are the same type.

Let's have a look at the statements made by 3.

Larry makes these two statements:

1. "The president is of a different race from the other two."
2. "Matt is not the president."

Matt makes these two statements:

1. "The president is a type B person."
2. "Larry is not the president."

Nick makes these two statements:

1. "Exactly two of us are of the same race."
2. "I am not the president."
  
We'll refer Larry's statements as L1 & L2, Matt's as M1 & M2 and those of Nick's as N1 & N2.

ANALYSIS : 

1] Suppose Matt's first statement M1 is TRUE. So, he is not a TYPE B person for sure & hence not the president. Therefore, L2 must be TRUE making sure Larry is not TYPE B person nor a president. So, Nick must be the president & TYPE B person whose both statements are FALSE. Since, N1 is turning out to be FALSE, neither Matt nor Larry can be TYPE B or both of them can't be of the same type i.e. TYPE A or TYPE C. 

However, since M2 turns out to be TRUE, Matt must be TYPE A person and hence leaving Larry as a TYPE C person. In that case, since L2 is TRUE, L1 must be FALSE. But in the case the president (TYPE B) is really different from the other two (TYPE A and TYPE C) making L1 TRUE. So, the assumption that M1 is TRUE goes wrong here in the case.

2] Now, let's suppose that Matt himself is the president. Then, L2 must be FALSE and M2, N2 would be TRUE. Since, M2 is TRUE, Matt (the president) can't be TYPE B & we know M1 is FALSE,  Matt must be TYPE C person & president.

Anyhow, Larry's first statement L1 can't be TRUE as in that case, he too will be TYPE C person as president Matt which is against the statement L1 itself. Since, his other statement L2 is FALSE, he must be TYPE B person.

Now, with N2 to be TRUE, if N1 is assumed to be TRUE then Nick would be TYPE A person. So, all three would belong to 3 different races which contradicts statement N1 itself. Hence, N1 must be FALSE. 

And if N1 FALSE and N2 TRUE, Nick would be TYPE C person as president Matt while Larry being TYPE B person. But this makes statement N1 TRUE which again contradicts our conclusion above. 

Therefore, Matt too can't be the president.

3] Suppose Nick is the president. That would make N2 FALSE and L2, M2 TRUE.
As concluded in STEP 1 above, we know M1 has to be FALSE. Then the President Nick can't be TYPE B person. With one of his false statement N2, Nick must be TYPE C person. Therefore, N1 must be TRUE. So, both Matt and Nick are TYPE C persons. 

That indicates president isn't of a different race than other two. That is L1 turns out to be FALSE. With L2 proved TRUE already, Larry would be TYPE C person. So all three would be TYPE C which contradicts TRUE statement N1.

4] Therefore, Larry must be the president.

That is L2, N2 must be TRUE and M2 must be FALSE. With M1 proved FALSE already in [1] above, Matt must be TYPE B person. Nick can't be TYPE B person with TRUE N2. 

If N1 is TRUE (i.e. Nick is TYPE A person) then Larry must be the other person with TYPE A (since Matt is TYPE B person) along with Nick for N1 to be TRUE. So, L1 too has to be TRUE. But, the president Larry is of the same race as that of Nick which is against L1 itself.

Hence, N1 must be FALSE and Nick must be TYPE C person. And therefore, L1 has to be TRUE making president Larry as a TYPE A person.

CONCLUSION : 

The President of Logitopia is Larry who is TYPE A person. Matt is TYPE B person and Nick is TYPE C person. 

Sequel : Story of Distribution of Loot

The five pirates mentioned previously are joined by a sixth, then plunder a ship with only one gold coin.

After venting some of their frustration by killing all on board the ship, they now need to divvy up the one coin. They are so angry, they now value in priority order:

1. Their lives
2. Getting money
3. Seeing other pirates die.

How can the captain save his skin?

This is how he should save himself! 

 The Prequel of the story!

Captain's Life Saving Proposal in Sequel

First read the story of sequel!

Let's name all the pirates as Pirate 6,5,4,3,2,1 as per their seniority. Now, the captain should respond with the logic below to save his skin.

Let's consider the cases where there are different number of pirates left on the ship after getting rid of seniors one by one.

----------------------------------------------------------------------------------------------

CASE 1 : 2 Pirates

The captain i.e. Pirate 2 can keep coin with him & obviously vote for himself (1/2 = 50% vote) to approve the proposal.

------------------------------------------------------------------------------------------------

CASE 2: 3 Pirates.

The captain i.e. Pirate 3 offers coin to Pirate 1 to get his support (2/3 = 66%) on proposal. Since, Pirate 1 knows what is going to happen if Pirate 3 dies as crew reduced to 2 Pirates as in CASE 1.

------------------------------------------------------------------------------------------------

CASE 3: 4 Pirates.

The captain i.e. Pirate 4 offers coin to Pirate 2 thereby getting his support (2/4 =50% votes) to get approval on proposal. Again, here Pirate 2 is smart enough to agree on this proposal as he know what will happen if Pirate 4 is eliminated leaving behind 3 pirates on sheep as in CASE 3.

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CASE 4: 5 Pirates.
Now the captain i.e. Pirate 5 always will be in danger as he can give only coin to only 1 of remaining 4. So he can 'earn' only 1 vote in support of his proposal i.e. only 2/5 = 40% votes. Hence, there is no way his proposal get approval & he should be ready to die.

-------------------------------------------------------------------------------------------------

CASE 5: 6 Pirates.

Here, Pirate 5 has to agree whatever captain i.e. Pirate 6 offers as if Pirate 6 dies the case reduces to CASE 4 where Pirate 5 can't save himself. However, if both Pirate 6 & Pirate 5 die, then CASE 3 comes into reality where Pirate 2 gets coin. So Pirate 2 would never agree on proposal offered by Pirate 6.

However, if Pirate 6 offers coin to Pirate 4 then he would take it happily as he knows that what will happen of both Pirate 6 & 5 are killed as CASE 3 comes into picture where he has to give coin to Pirate 2 to save himself.

If Pirate 6 offers coin to Pirate 3 as well & earn his support as Pirate 3 also knows what is going to be the case of both 6 & hence 5 get eliminated. The case will be reduced to 4 Pirates as in CASE 3 where Pirate 4 will offer coin to Pirate 2.

The Pirate 6 even choose Pirate 1 to offer coin. Again, Pirate 1 smart to think that what will be the case if 6 & hence 5 are killed. It will be the scenario as in CASE 3 where Pirate 4 offers coin to Pirate 2.

In short, the Pirate 6 will always get support from Pirate 5 always in any case & any one from Pirate 4/3/1 if he offers coin to any one of these three. 

That's how he will get support of 50% (3/6) group to get approval for his proposal thereby saving his own life on approval of proposal offered.

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Story of Distiribution of 100 Coins Loot

Five ship’s pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).

The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.

If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.

What is the maximum number of coins the captain can keep without risking his life?

He can take away 98 coins! How? Read here! 

The Captain's Undeniable Proposal

What was the situation?  

He can keep 98 coins! Surprised? Read how.

Let's number 5 pirates as Pirate 5, Pirate 4.....Pirate 1 as per their descending order of seniority.

Pirate 5 keeps 98 coins with him and gives 1 coin each to Pirate 3 and Pirate 1.

Now Pirate 5 i.e Captain explains his decision -

CASE 1:

 If there were only 2 pirates then Pirate 2 would have taken all 100 coins after obtaining his own vote which accounts to 50% votes (50 % of 2 = 1).

CASE 2: 

 In case of 3 pirates, the Pirate 3 would have offered 1 coin to Pirate 1 & would have kept 99 coins with him. Now Pirate 1 wouldn't have any option other than agreeing on deal with Pirate 3. That's because if he doesn't agree then Pirate 3 would be eliminated & all coins would be with Pirate 2 as explained in above (case 1) of only 2 pirates. So votes of Pirate 1 and Pirate 3 which account to 66% (2 out of 3) votes of group locks this deal and Pirate 2 would be left without any coin.

CASE 3: 

 Now in case of 4 pirates, Pirate 4 would offer coin to Pirate 2 & would keep 99 coins with him. Now, Pirate 2 know what happens if Pirate 4 gets eliminated. Pirate 3 would offer 1 coin to Pirate 1 & will take away 99 coins. So Pirate 2 would definitely accept this deal. That's how votes of Pirate 4 and Pirate 2 makes 50% (2 out of 4) votes of group to pass the proposal. Pirate 3 and Pirate 1 can't do anything in this case.

By now, Pirate 3 and Pirate 1 realizes what happens of Pirate 5 gets eliminated. They won't be getting any coin if Pirate 4 becomes captain as explained above (case 3). So they have no option other than to vote for the proposal of Pirate 5. 

This way, Pirate 5, Pirate 3 and Pirate 1 (3/5 = 60% of crew) agree on proposal of Pirate 5 where he takes away 98 coins with 1 coin each to Pirate 3 and Pirate 1. 

'Morning Melange' - Puzzle

This morning, the popular Bay area cable access TV show, "Morning Melange", featured six guests (including Francine and Evan). Each guest lives in a different town in the region (including Corte Madera), and each has a different talent or interest that was the focus of his or her segment. The segments began at 6:45, 7:00, 7:15, 7:30, 7:45 and 8:00. 

Discover, for each time, the full name of the featured guest, where he or she lives and his or her special interest.
 
1. The first three guests were, in some order: the person surnamed Ivens, the person from Berkeley and the antique car collector.

2. Damien's segment was sometime before Lautremont's segment.

3. Krieger's segment began at 7:45.

4. Alice appeared after the person from Oakland and before the person surnamed Morley.

5. The people from Berkeley and Daly City aren't of the same sex.

6. The six guests were: Cathy, the person whose first name is Damien, the person whose last name is Novak, the person from Daly City, the person from Palo Alto and the bungee jumper.

7. The last name of the financial adviser is either Lautremont or Novak.

8. Jaspersen's segment began exactly 45 minutes after the beekeeper's segment.

9. The crepe chef went on sometime before the person from Sausalito and sometime after Brandon.

10. The hypnotherapist's segment began at 7:15.

HERE is SOLUTION! 

'Morning Melgane' Puzzle - Solution

What was the puzzle?

Rewriting all the given clues once again.

--------------------------------------------------------------------------- 

1. The first three guests were, in some order: the person surnamed Ivens, the person from Berkeley and the antique car collector.

2. Damien's segment was sometime before Lautremont's segment.

3. Krieger's segment began at 7:45.

4. Alice appeared after the person from Oakland and before the person surnamed Morley.

5. The people from Berkeley and Daly City aren't of the same sex.

6. The six guests were: Cathy, the person whose first name is Damien, the person whose last name is Novak, the person from Daly City, the person from Palo Alto and the bungee jumper.

7. The last name of the financial adviser is either Lautremont or Novak.

8. Jaspersen's segment began exactly 45 minutes after the beekeeper's segment.

9. The crepe chef went on sometime before the person from Sausalito and sometime after Brandon.

10. The hypnotherapist's segment began at 7:15.

--------------------------------------------------------------------------- 


STEPS :

1] Let's make table like below for simplicity & easy understanding.

   
2] As per (10), hypnotherapist took segment starting at 7:15 & (3) suggest that Krieger started at 7:45.

3] As per (8), Jaspersen must be either at 7:30, 7:45 or 8:00 (i.e. in second half) while beekeeper must be at 6:45 or 7:00 or 7:15. But 7:15 segment is already occupied by hypnotherapist and Krieger is there already at 7:45.
So, Jaspersen can't be at 7:45. Hence, Jaspersen must be at 7:30 and beekeeper at 6:15 for (8) to be true.

With that, the antique car collector pointed by (1) must be at 6:45.

4] Since, as per (7), the financial adviser isn't Jaspersen or Kreiger ,therefore is on during the 8:00 segment. As per (4), Morley can't be in first 2 segments & as per (7) Morely is not a financial adviser. Thus, Morley needs to be at 7:15.

5] With that as (4) suggests, Alice must be on during segment started at 7:00 & person from Oakland took 6:45.

6] Now, as (1) suggests person having surname Ivans has to be at 6:45 & the person at Berkeley must be at 7:15.

7] The two guests appeared in the 6:45 and 7:15 segments can't be Novak, from Daly City or Palo Alto, or interested in bungee jumping (Hint 6). Therefore, these guests must be Cathy and Damien.

8] So for (9) to be Brandon must be at 7:30,  the chef must be at 7:45, and the guest from Sausalito at 8:00.

9] There are 6 different guest pointed by Hint (6). So, the bungee jumper who has to at 7:30 can't be from Daly City or Palo Alto. Hence, he must be from Corte Madera. 

10] Now, Alice has to be from Daly City or Palo Alto & hence can't have surname Novak as (6) points 6 different guests. Therefore, Novak must be at 8:00 and Alice at 7:00 must be Lautremont.

 11] With that, the hint (2) suggests that Damien must be at 6:45 and hence Cathy at 7:15.

12] Now it's clear that, Cathy is from Berkeley & hence as (5) suggests Alice can't be from Dale City and is therefore from Palo Alto. So Krieger must be from Dale City.

13] In the last, as per (5) itself, Kreiger, who is from Daly City, cannot be Francine, and must be named Evan. This leaves Francine at 8:00.