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Instead of a series of hasps with a lock and key for each, what if the treasure chest is locked with a chain-like configuration? Within the links of the chain are nine larger rings (1-9). Each of the four slaves (A,B,C,D) and the vizier (V) have their own lock and key. Lock A goes through rings 1,2,3,4,7,8. Lock B goes through rings 1,2,3,5,6,8,9. Lock C goes through rings 1,2,4,5,7,8,9. Lock D goes through rings 2,3,6,7,8,9. Lock V goes through rings 3,4,5,6,7.

1-ABC-2-ABD-3-AV-4-CV-5-BV-6-DV-7-ACD-8-BCD-9

So if the vizier and any one slave or if any three slaves unlock their respective locks, the chain is broken.

Can't fully visualize this but think it would work

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Posted · Report post

what if the vizer was also a slave. that why he/she would have a key for being the slave and also a key for being a slave. So if the Vizer has 2 keys he/she only needed one more slave to help open the slave room. otherwise the 3 remaining slaves can open the room by themselves. so to answer the question as to how many locks there are, i think the answer is 3.

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Posted · Report post

Instead of a series of hasps with a lock and key for each, what if the treasure chest is locked with a chain-like configuration? Within the links of the chain are nine larger rings (1-9). Each of the four slaves (A,B,C,D) and the vizier (V) have their own lock and key. Lock A goes through rings 1,2,3,4,7,8. Lock B goes through rings 1,2,3,5,6,8,9. Lock C goes through rings 1,2,4,5,7,8,9. Lock D goes through rings 2,3,6,7,8,9. Lock V goes through rings 3,4,5,6,7.

1-ABC-2-ABD-3-AV-4-CV-5-BV-6-DV-7-ACD-8-BCD-9

So if the vizier and any one slave or if any three slaves unlock their respective locks, the chain is broken.

Can't fully visualize this but think it would work

in this way, 2 slaves can break the chain together... but the question states that it is opened when 3 of them worked together.....

but gud try

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I interpreted the question the same way, that ANY combination of 3 slaves should be able to open the door.

If this were the problem, then I wonder what the answer would be?

Find a combination which allows any 3 of 6 slaves to unlock the chest, (if this is possible), then, stitch two slaves together (like a human centipede) and give them the job of vazir. Then you have a vazir who can open the chest with 1 more of 4 slaves.

sorry but your step is wrong

what you have to do is give every key to a different pair of salves e.g. give key 1 to slaves A and B... keep diving the keys until no more pair of slaves is left... now say you have divided n keys to the slaves.... now give (n+1)th key to every slave and give n keys to the vizier

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Posted · Report post

in this way, 2 slaves can break the chain together... but the question states that it is opened when 3 of them worked together.....

but gud try

Don't think two slaves can unlock the chain. With my solution the vizier's lock goes through rings 3,4,5,6,7 so effectively the links between those rings cannot be broken by two slaves.

So if two slaves were to break the chain, it would have to be between rings 1 and 2, 2 and 3, 7 and 8, or 8 and 9.

If slaves A & B work together; lock C secures rings 1 and 2, lock D secures rings 2 and 3, and locks C and D secure rings 7 and 8 and rings 8 and 9.

If slaves A & C work together; lock B secures rings 1 and 2, locks B and D secure rings 2 and 3, lock D secures rings 7 and 8, and locks B and D secure rings 8 and 9.

If slaves A & D work together; locks B and C secure rings 1 and 2, lock B secures rings 2 and 3, lock C secures rings 7 and 8, and locks B & C secure rings 8 and 9.

If slaves B & C work together; lock A secures rings 1 and 2, locks A and D secure rings 2 and 3 and rings 7 and 8, and lock D secures rings 8 and 9.

If slaves B & D work together; locks A and C secure rings 1 and 2, lock A secures rings 2 and 3, locks A and C secure rings 7 and 8, and lock C secures rings 8 and 9.

If slaves C & D work together; locks A and B secure rings 1 and 2 and rings 2 and 3, lock A secures rings 7 and 8 and lock B secures rings 8 and 9.

A third slaves cooperation is needed in every case.

To break the chain between rings 1 and 2, slaves A,B,C must work together

To break the chain between rings 2 and 3, slaves A,B,D must work together

To break the chain between rings 7 and 8, slaves A,C,D must work together

To break the chain between rings 8 and 9, slaves B,C,D must work together

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Posted (edited) · Report post

All credit to plainglazed...I just made it visual.

Here is the chain. I imagine a circuit in each lock running between each adjacent prong. There is no curcuit between prongs farther than one ring apart. To open the chest a current it run through each ring and it checks if it can be completed with the adjacent ring.

Here are the four arrangements of vizier with slave.

#1 - When running a current from ring 3 it fails to complete with ring 4.

#2 - Ring 5 fails with ring 6

#3 - Ring 4 fails with ring 5

#4 - Ring 6 fails with ring 7

---Three slaves arrangements---

#1 - Ring 1 fails with ring 2

#2 - Ring 2 fails with ring 3

#3 - Ring 7 fails with ring 8

#4 - Ring 8 fails with ring 9

---Two slaves arrangements---

In each combination every ring is connected to the adjacent rings by at least one lock.

Did I miss anything?

Edited by curr3nt
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Posted (edited) · Report post

I think your ring count can be reduced to eight.

Switch the vizier to making a link between each adjacent rings except one and I think you can do it with 8 rings (seven circuits). I think we get back to the original solution of distribution using circuits instead of keys.

Would have to work it out to be sure but that is my hunch.

-edit-

Just noticed I kept misspelling circuit ><

Edited by curr3nt
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Posted · Report post

I think it can be done with less than 5 locks.

I say 3 locks and 4 key holes.

Each slave gets a copy of the same key and the v-guy gets a special key. The chest is set up as follows

L1 L2 L3

H1 H2 H3

H4

For the locks to disengage they must all be turned at the same time or else part of the mechanism will stop another from turning.

H1-3 are all the same kind of key hole and they match the slaves' keys.

H4 is a special key hole that can undo L1 and L2 simultaneously and only matches the v-guy's key.

Just my thought.

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Posted · Report post

All credit to plainglazed...I just made it visual.

Here is the chain. I imagine a circuit in each lock running between each adjacent prong. There is no curcuit between prongs farther than one ring apart. To open the chest a current it run through each ring and it checks if it can be completed with the adjacent ring.

Here are the four arrangements of vizier with slave.

#1 - When running a current from ring 3 it fails to complete with ring 4.

#2 - Ring 5 fails with ring 6

#3 - Ring 4 fails with ring 5

#4 - Ring 6 fails with ring 7

---Three slaves arrangements---

#1 - Ring 1 fails with ring 2

#2 - Ring 2 fails with ring 3

#3 - Ring 7 fails with ring 8

#4 - Ring 8 fails with ring 9

---Two slaves arrangements---

In each combination every ring is connected to the adjacent rings by at least one lock.

Did I miss anything?

nice attempt... but i dont have time to read all the stuff as im having my physics exam 2morrow.....

btw how did you draw all those diagrams?

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Posted · Report post

I think it can be done with less than 5 locks.

I say 3 locks and 4 key holes.

Each slave gets a copy of the same key and the v-guy gets a special key. The chest is set up as follows

L1 L2 L3

H1 H2 H3

H4

For the locks to disengage they must all be turned at the same time or else part of the mechanism will stop another from turning.

H1-3 are all the same kind of key hole and they match the slaves' keys.

H4 is a special key hole that can undo L1 and L2 simultaneously and only matches the v-guy's key.

Just my thought.

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Posted · Report post

The OP doesn't state that the locks be distinct, requiring different keys. This allows for the simple answer of 3 locks.

The stipulation"one can create copies of his key(s)" would prevent this.

The title is also fallacious given that no part of the riddle deals with any sort of impossibilities.

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Posted · Report post

The OP doesn't state that the locks be distinct, requiring different keys. This allows for the simple answer of 3 locks.

The stipulation"one can create copies of his key(s)" would prevent this.

The title is also fallacious given that no part of the riddle deals with any sort of impossibilities.

i know that.... i did it in 10 min. only . ... but the title needs to be a little catchy

it is understood that every lock is distinct and can have multiple keys... if locks are not distinct ,then whats their need

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Posted · Report post

I think your ring count can be reduced to eight.

Switch the vizier to making a link between each adjacent rings except one and I think you can do it with 8 rings (seven circuits). I think we get back to the original solution of distribution using circuits instead of keys.

Would have to work it out to be sure but that is my hunch.

-edit-

Just noticed I kept misspelling circuit ><

think you could be right. would require more keys though? also really admire the drawings - what program did you use?

ps - at least you spelled current correctly. gr3at analogy.

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Posted · Report post

uhm....I used MS Word 2007

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Posted · Report post

it is understood that every lock is distinct and can have multiple keys... if locks are not distinct ,then whats their need

Don't presume that what you believe is understood will be by others.

To give you an example for your question, my door (along with many others I'm sure) has a deadbolt and knob lock that both use the same key. Does this render its design useless? Also, consider a "superkey" or "masterkey" that locksmiths have that can open any lock. This removes the distinction for the keyholder, but not their need.

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Don't presume that what you believe is understood will be by others.

To give you an example for your question, my door (along with many others I'm sure) has a deadbolt and knob lock that both use the same key. Does this render its design useless? Also, consider a "superkey" or "masterkey" that locksmiths have that can open any lock. This removes the distinction for the keyholder, but not their need.

ok..... i accept my fault.. next time i'll explain the situation clearly

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