Cryptography

[Fadetoblack]

For Da Vince Code fans...or generally just riddles and cryptics fans.

An Easy one to start with.

Can you decode the following phrases?

1) Ni lkna rfnim ajnebse xat dn ahta ed tp ecxenia, treceb otdia seb nacgn. ~ Ihtondlr Owsihtni

2) Nilkn arf nima j nebdae, de ram eh tfoo wtf iter. ~ Cesapeek Yameerht

3) No tg nihsaw eg roege no dabana htes u cxe onr. ~ Effoot Rettebsiti

[TheOldhamWhisper]

1. In this world nothing can be said to be certain except death and taxes - Benjamin Franklin
2. Three may keep a secret if two of them are dead - Benjamin Franklin
3. It is better to offer no excuse than a bad one - George Washington

[Fadetoblack]

Ok TOW, we're just getting warmed up. You didn't even break a sweat there.

Ok guys, this has now turned into the can anyone beat TOW....:laugh

Here are some quotes by famous international personalities about India. Can you decipher them?

Jr bjr n ybg gb gur vaqvnaf, jub gnhtug hf ubj gb pbhag, jvgubhg juvpu ab jbegujuvyr fpvragvsvp qvfpbirel pbhyq unir orra znqr! - Nyoreg Rvafgrva

Vs gurer vf bar cynpr ba gur snpr bs rnegu jurer nyy gur qernzf bs yvivat zra unir sbhaq n ubzr sebz gur irel rneyvrfg qnlf jura zna ortna gur qernz bs rkvfgrapr, vg vf vaqvn! - Ebznvar Ebyynaq

Vaqvn vf gur penqyr bs gur uhzna enpr, gur ovegucynpr bs uhzna fcrrpu, gur zbgure bs uvfgbel, gur tenaqzbgure bs yrtraq, naq gur terng tenaq zbgure bs genqvgvba. Bhe zbfg inyhnoyr naq zbfg nfgevpgvir zngrevnyf va gur uvfgbel bs zna ner gernfherq hc va vaqvn bayl! - Znex Gjnva

Vaqvn pbadhrerq naq qbzvangrq puvan phyghenyyl sbe 20 praghevrf jvgubhg rire univat gb fraq n fvatyr fbyqvre npebff ure obeqre! - Uh Fuvu

Vs v jrer nfxrq haqre jung fxl gur uhzna zvaq unf zbfg shyyl qrirybcrq fbzr bs vgf pubvprfg tvsgf, unf zbfg qrrcyl cbaqrerq ba gur terngrfg ceboyrzf bs yvsr, naq unf sbhaq fbyhgvbaf, v fubhyq cbvag gb vaqvn. - Znk Zhryyre

[vegyjones]

We owe a lot to the Indians, who taught us how to count, without which no worthwhile scientific discovery could have been made! - Albert Einstein

[Fadetoblack]

Well done Veg.

He knew that one alright but he was so fast I thought he was cheating.

No cheating now people, it doesn't help you in the long run, and makes baby jesus cry.

Riitcjgz vehz mtam: 'R vehz aej pzstjmz R wzzl aej.'
Itcjgz vehz mtam 'R wzzl aej pzstjmz R vehz aej.'"
-Zgrsu Ogeii

"Rw vehz cuz ytgtlek essjgm cutc cde pzrwqm pzseiz ewz twl azc gzitrw cde."
-Zgrsu Ogeii

[TheOldhamWhisper]

Immature love says: I love you because I need you
Mature love says: I need you because I love you - Erich Fromm

In love the paradox occurs that two beings become one and yet remain two - Erich Fromm

That one was a lot tougher - I had the double letter at the end as an 's' for a while (which didn't help!).

[Fadetoblack]

It really is amazing how some people's minds work. Some have a head for puzzles, like my gf and my ex gf, whereas I am intrigued by them, but really can't be pushed.

Anyway....

For the following three words, each letter has been replaced with a number. A number represents the same letter for all words. Can you determine what the words are?

a. 72442442882
b. 911909
c. 625526255236

[vegyjones]

Damn, I had the i's as Mm's but changed it specifically because of those end double letters :mad:

[vegyjones]

I think that first one is Mississippi

Hasn't given me much clue as to the other 2 though! :doh

[TheOldhamWhisper]

c. Kinnikinnick

[TheOldhamWhisper]

b. Doodad (I've used it in a game of scrabble but don't ask me what it means!)

[Fadetoblack]

Well done guys.

Bob enters a hotel. "How many rooms are there here?", asks Bob. "Twenty", replies the receptionist, "We only have one room left - a single".

"That will do me perfectly", says Bob.

"Here you go, room number twenty-one."

Bob looked at the key and, sure enough, it said 21 in bold numbers on the keyring. Yet the receptionist said there were only twenty rooms. Bob found his room easily and realised how silly he had been.

Why was his room not number 20?

[TheOldhamWhisper]

A couple of possibles - some hotels don't have a room 9 because of the possibility of mix ups with the room 6 key or the hotel is one of many which uphold superstition and do not have a room 13.

[Fadetoblack]

I thought if I came at ya from a different direction I might catch you out.

I was wrong.

no 13 due to superstition is the answer. I'm not giving up yet.

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. BOTH SWITCHES ARE IN THEIR OFF POSITIONS NOW.* The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."

"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."

"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

*note - the only difference from Scenario B, the original position of the 2 switches are known.

[vegyjones]

What's the question?

[Fadetoblack]

The question though not stated in the given is implied, so I don't think it to be a big deal that it's missing.

[TheOldhamWhisper]

OK, this might take a few posts while I work out the finer details but we start off with 22 of the 23 using just one switch to show they have visited the room and using the other switch as a 'dummy' and the other guy taking a count of how many times the switch is in a certain position. Watch this space....

[vegyjones]

My thoughts.

The 23 prisoners select one spokesperson! The spokesperson will decide "when" they have completed the task!

On the spokespersons first visit, he switches Swith A ON!

When a prisoner goes to the room for the first time,
and he sees Switch A on, he must switch it OFF!

When a prisoner goes to the room for the first time,
and sees Switch A off, he must toggle switch B

Anyone not visiting for the first time, must toggle switch B.

Therefore, the spokesperson will know when a new person has been in the room because he will have to turn switch A ON

So, he can safely tell the warden on the 23rd time he has had to flick the switch!

:D How's that??

[TheOldhamWhisper]

I've typed it, so I might as well post it!

Right, prisoner 1 is given the job of counting.

When a prisoner enters the room, he switches A to the on position (if possible). If it is already in that position, he just uses switch B.

Whenever prisoner 1 goes into the room, he puts switch A to the off position and adds one to his count.

When he reaches 22, he can safely tell the warden that all prisoners have visited the room at least once.

[TheOldhamWhisper]

And - Vegy's is more accurate because I forgot to mention that the prisoners could only use switch A once! :)

[vegyjones]

Was it right about being the 23rd time?

I wasn't sure whether he had to bother waiting for his first time before starting the count since the switch would already be OFF!

Therefore, if it was on when he went in for his first time, it would only be
his 22nd visit when he knew everyone had visited!

But if he was the first in, then I think it would be 23!

Will have to wait for Fade to give the correct answer I suppose! :)

[Fadetoblack]

Hint
Assuming that:

A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.

B) There is no restriction on the number of time each prisoner can visit the switch room

C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.

[vegyjones]

Does that mean we're wrong?

[Fadetoblack]

No, just giving to time to think again. Whether you needed it or not.

Answer
1. Appoint a "scorekeeper" and the other 22 prisoners as "transmitters".

2. When a transmitter enter the switch room for the first time, he will flick switch B ON if he sees it in the OFF position. Alternatively, he will flick switch A if he sees switch B is in the ON position and waits for his next visit. The transmitter's mission is to find ONE opportunity to flick switch B to the ON position. After he completes his mission, he will flick switch A for all subsequent visits. In short, each transmitter's role is to send a signal to the scorekeeper that he has been to the switch room.

3. The scorekeeper's role is to flick off switch B whenever he sees it in the ON position. Each time he flicks off switch B he adds 1 to his score. When he sees switch B is already OFF, he flicks switch A instead and does not add to the score. As soon as he accumulates 22, he reports to the Warden.

[TheOldhamWhisper]

We were both using the wrong switch :laugh

[vegyjones]

We were both using the wrong switch :laugh:D

[Fadetoblack]

I was watching the thread and gave you extra time...but it was one of those things where if you think you're right, then you're not going to question anything. I must applaud you though, excellent the both of you. Give me a bell when ya want more :wink

[Fadetoblack]

In fact...

Just for TOW....(well not just for him but you know what I mean)

I was playing a game of five card draw poker with a bunch of logicians. By the time we had finished bidding and were just about to reveal our cards, I was pretty confident I would win of the four of us remaining. I had three nines, some face card (I can't remember what suit or even whether it was a jack, queen, or king) and a four. (Or was it a five? I can't remember.)

I was even more sure when two of my opponents laid down their cards. One had a pair of fours and a pair of sevens, the other had a pair of twos and a pair of eights. My third opponent, however, laid down his five cards face down in a row. He said, "I have a straight, and the cards are, from lowest to highest: a ten, a jack, a queen, a king, and an ace. I have at least one card of each of the four suits: clubs, spades, hearts, and diamonds. I am fairly certain that this is the winning hand, but I'm feeling generous today, and I will give a third of the pot to whoever can determine which suit I have two cards of.

Now I know you can't figure it out without some clues. Here they are:

1. The king is next to at least one diamond.
2. The queen is next to exactly one heart.
3. The jack is next to at least one spade, but is not next to any hearts.
4. The ten is next to at least one club.
5. The ace does not border any black cards, nor does it border any diamonds.
6. My two cards of the same suit are not next to each other.
7. Of the ten possible pairings of cards, only one pair, when removed, leaves three cards in ascending order from left to right.
8. My ace is not the card on the far left."

There was a minute's silence. One of the other logicians said, "I give up! There's no way to figure that out!"
The other agreed. But I didn't. I had just figured out which suit he had two of.

Which suit is it?

[vegyjones]

CLUB

[TheOldhamWhisper]

OK, I've worked out the order of the cards is KJQTA and the J is a diamond and the T is a heart. The Queen is either a club or a spade. But now I am completely stuck :doh

And as my brain is now starting to rot, I'll pick this one up layer. Ooo

[Fadetoblack]

:laugh ok

[vegyjones]

I think your card order is wrong Oldham...

When you take out the 2 matching cards, the other 3 are in number order! :wink :D

[TheOldhamWhisper]

Nope - I'm pretty sure that is the order for No. 7 - The only pair you can remove is the King and Ten to leave JQA in order. It doesn't say anything about taking the matching cards out???

[TheOldhamWhisper]

Of course, I’m now in the position of the other two – because I need to use the picture card in my hand to determine the rest of the cards.

I can’t be holding the Jack – as I can’t deduce anything from doing so. It can’t be a Queen – I can only work out 4 of the cards suits whether it is a club or a spade. So I must be holding a King.

I know that the King (on the board) cannot be red – so I am holding either the spade or the club. If I am holding the Club, I still cannot say for certain what suit the Queen is so I must be holding the Spade.

That means the King on the board is a Club, the Queen must be a Spade (Rule 3) and the A must also be a Club (Rule 4).

Final Order:
K:clubs: J:diamonds: Q:spades: 10:hearts: A:clubs:

So the answer is – Clubs!

[Fadetoblack]

:D

Answer
The two cards on the ends and the card in the middle border a combined number of two suits. So at least one pair of them should be listed as bordering a card of the same suit in clues 1-5. Of the five, only the ace and the queen meet this criterion, so they must be two cards apart. One of them must be in the center, but it can't be the ace (the two cards on the ends would only border hearts - and one of them has to be bordering something else from clues 1-4), so it must be the queen. The ace is not on the left end from clue 8, so it must be on the right end, and the card next to it must be a heart. So far we have this(X represents unknown):
X X Q X A
X X X H X
At least one of the two cards to the left of the queen is lower than the queen (two of the remaining cards are lower, and only one can be right of the queen), so the triplet that is in ascending order is the queen, the ace, and the single card to the left of the queen that is lower than the queen. That means the king is left of the queen, and is also left of both the jack and the ten (otherwise, jack king ace or ten king ace would be a second triplet), so it must be on the far left. The jack must be left of the ten (or else ten jack ace would be a second triplet), so it must be second from the left, and the ten must be second from the right. The only card next to the king is the jack of diamonds (it's a diamond from clue 1). So far we have(T is ten):
K J Q T A
X D X H X
Now we seem to be stuck. We don't have any more clues that can be used. So how did I figure it out?
I held a face card (I said so myself in the intro). I must have known it cannot be that card and deduced the correct answer from there.
But which card did I have? We already know the suit of the jack, so my holding a jack would not help. Could I have held a queen? No, because I know the queen to either be a spade or club (clue 6), so one of the cards next to the queen would have its clue satisfied (see clues 3 and 4), and I could not determine the suit of the card on the other side of it.
Therefore, I must have held a king. But which king? I know the king is not a diamond (clue 6) or a heart (clue 3). If I held the king of clubs, then the king next to the jack would have satisfied the jack's clue, and I could not have determine the suit of either card next to the ten.
Therefore, I held the king of spades. The king of clubs must have been on the far left (only suit available), so the only card next to the jack that could be a spade is the queen, so the only card next to the ten that could be a club is the ace. In summary:
K J Q T A
C D S H C
Therefore there are two CLUBS.
Hide

[vegyjones]

That's how I had it

KQJTA

so that I had 3 cards in a row! Oldham had me worried that I had my crads wrong! :yikes:

But at least the answers were the same! :)

[Fadetoblack]

Eight and five, last name and given,
We are one six six six even;
The first in cow, the last in oxen
Three in damsel, three in vixen.

Question: What are we called?

[TheOldhamWhisper]

That's how I had it

KQJTA

so that I had 3 cards in a row! Oldham had me worried that I had my crads wrong! :yikes:

But at least the answers were the same! :)

But Condition No 2 is not fulfilled with the order you had them in Ooo

[vegyjones]

I had the Jack as a heart!

[TheOldhamWhisper]

But that would mean the 10 was a spade and the Ace cannot border a black card!

[vegyjones]

Oops Ooo

[TheOldhamWhisper]

I think I know the riddle but I'm looking for the 1666 connection :doh

Roman Numerals

.

[TheOldhamWhisper]

Can't believe I missed that...

write them in order of value to get 1666

.

[mathare]

Oldham is right and I think I have the connection. Maybe...

Take the C from 'cow' for 100, D, M and L from 'damsel' for a further 1550, X from 'oxen' for a further 10 and V, I and X from 'vixen' for 16 more. Making 1676 :doh So close, that has to be along the right lines

[Fadetoblack]

Very good.

The last name has eight letters (NUMERALS) while the first or given name has five (ROMAN).

Using each of the numerals once gives MDCLXVI, which is 1,666.

C ( which is found in Cow) is first of the seven in the alphabet while X ( which is found in oXen) is last.

Three of seven are found in DaMseL. Three are also found in VIXen.

Explanation of the hint:

I is the only one of five vowels among the seven Roman numerals.

C is found in Cat, M and I in MItten.

I and V are odd numbers. The rest are even numbers.

There are seven Roman numerals. One of them is found in seVen (the V).

[Fadetoblack]

A car thief, who had managed to evade the authorities in the past, unknowingly took the automobile that belonged to Inspector Detweiler. The sleuth wasted no time and spared no effort in discovering and carefully examining the available clues. He was able to identify four suspects with certainty that one of them was the culprit.

The four make the statements below. In total, six statements are true and six false.

Suspect A:
1. C and I have met many times before today.
2. B is guilty.
3. The car thief did not know it was the Inspector's car.

Suspect B:
1. D did not do it.
2. D's third statement is false.
3. I am innocent.

Suspect C:
1. I have never met A before today.
2. B is not guilty.
3. D knows how to drive.

Suspect D:
1. B's first statement is false.
2. I do not know how to drive.
3. A did it.

Which one is the car thief?

[vegyjones]

D

[TheOldhamWhisper]

.

The Answer is B and A's third question is irrelevant as it could be true or false. There are 2 contradictory statements and we thus need to find 4 more (proven) false statements. Taking each suspect in turn and assuming guilt, we can deduce that only if B is Guilty can we show 6 false statements!

.

[TheOldhamWhisper]

:ARsurrend

[Fadetoblack]

Consider the lilly. Not really. Consider that six statements are false. A's first statement and C's first statement contradict each other. One of them is false. C's and D's contradict each other. One of them is false. Therefore, there are four additional false statements.

Assume A is guilty. If so, A's second statement, B's second statement, and D's first statement are the additional false statements.

Assume D is guilty. If so, A's second statement, B's first statement, and D's third statement are false. This also only makes five false statements. D did not do it.

Assume C did it. If so, A's second statement, D's first and third statements are false. This again, makes only five false statements.

After ruling out suspects A, C and D, B is the culprit. B's third statement, C's second statement, and D's first and third statements are the additional false statements. This adds up to six.