The Wires UndergroundThe grades of the students in a classA puzzle of trust and lies, allies and spiesSink the SubmarineThe Magic Money MachineLabeling wires in a bundleMr. Hilbert and the Problem of the Rogue ResistorThe game of Shafiqa and HabibiA way to beat the system?A Strategy Game Involving Conquering of RegionsThe Alchemist's brewSolution or Origin of logic puzzle
Was the Boeing 2707 design flawed?
Why do banks “park” their money at the European Central Bank?
How is linear momentum conserved in case of a freely falling body?
When one problem is added to the previous one
To get so rich that you are not in need of anymore money
Prevent use of CNAME record for untrusted domain
Can an ISO file damage—or infect—the machine it's being burned on?
Why do proofs of Bernoulli's equation assume that forces on opposite ends point in different directions?
Does ostensible/specious make sense in this sentence?
Evaluated vs. unevaluated Association
"There were either twelve sexes or none."
When calculating a force, why do I get different result when I try to calculate via torque vs via sum of forces at an axis?
Are there any elected officials in the U.S. who are not legislators, judges, or constitutional officers?
Discussing work with supervisor in an invited dinner with his family
Can you cast bonus action and reaction spells while already casting a spell?
Foreign language movie, people enter a church but find they can't leave
How to gently end involvement with an online community?
What is the loud noise of a helicopter when the rotors are not yet moving?
Breaker Mapping Questions
How can I unambiguously ask for a new user's "Display Name"?
How were medieval castles built in swamps or marshes without draining them?
Ghidra: Prepend memory segment in assembly listing view
Boot Windows from SAN
Who was the most successful German spy against Great Britain in WWII, from the contemporary German perspective?
The Wires Underground
The grades of the students in a classA puzzle of trust and lies, allies and spiesSink the SubmarineThe Magic Money MachineLabeling wires in a bundleMr. Hilbert and the Problem of the Rogue ResistorThe game of Shafiqa and HabibiA way to beat the system?A Strategy Game Involving Conquering of RegionsThe Alchemist's brewSolution or Origin of logic puzzle
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?
strategy
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?
strategy
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago
$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago
add a comment |
$begingroup$
There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?
strategy
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
There are ten wires that run underground across a river. Thus on each of the banks, there are ten ends of the wires. Unfortunately, given an end of a wire on this bank, one doesn't know which end on the other bank it corresponds to. Suppose that you an ohmmeter and a bunch of wires. (You can then, for example, connect two ends on side A of the river, and go to the side B and measure the resistance between every pair of ends to figure out which pair on side B corresponds to the pair that is connected on side A). What is the minimum number of crossings you need to make in order to figure out the matching ends of all the ten wires?
strategy
strategy
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
athin
13.2k3 gold badges41 silver badges108 bronze badges
13.2k3 gold badges41 silver badges108 bronze badges
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
dvcdvc
211 bronze badge
211 bronze badge
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dvc is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago
$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago
add a comment |
$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago
$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago
$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago
$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago
$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago
$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To solve the problem, at least I can cross
$7$ times.
And the strategy is:
- Connect two wires, says wire $1$ and $2$, then cross.
- Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.
- Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.
- Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.
- Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.
- Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.
- Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.
- Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.
For improvement, if
I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.
$endgroup$
add a comment |
$begingroup$
The absolute minimum is
2 crossings
First some observations:
Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4
The actual strategy:
So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.
Next
Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.
Proof:
It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "559"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
dvc is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f87384%2fthe-wires-underground%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To solve the problem, at least I can cross
$7$ times.
And the strategy is:
- Connect two wires, says wire $1$ and $2$, then cross.
- Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.
- Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.
- Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.
- Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.
- Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.
- Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.
- Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.
For improvement, if
I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.
$endgroup$
add a comment |
$begingroup$
To solve the problem, at least I can cross
$7$ times.
And the strategy is:
- Connect two wires, says wire $1$ and $2$, then cross.
- Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.
- Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.
- Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.
- Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.
- Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.
- Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.
- Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.
For improvement, if
I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.
$endgroup$
add a comment |
$begingroup$
To solve the problem, at least I can cross
$7$ times.
And the strategy is:
- Connect two wires, says wire $1$ and $2$, then cross.
- Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.
- Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.
- Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.
- Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.
- Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.
- Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.
- Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.
For improvement, if
I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.
$endgroup$
To solve the problem, at least I can cross
$7$ times.
And the strategy is:
- Connect two wires, says wire $1$ and $2$, then cross.
- Find the corresponding connected wires of $1$ and $2$. Connect another two wires $3$ and $4$, then cross.
- Find the corresponding connected wires of $3$ and $4$. Connect another two wires $5$ and $6$, then cross.
- Find the corresponding connected wires of $5$ and $6$. Connect another two wires $7$ and $8$, then cross.
- Find the corresponding connected wires of $7$ and $8$. You also know which two lefts are $9$ and $10$. Now, connect wires corresponding to $1$ and $3$, then cross.
- Find the corresponding connected wires of $1$ and $3$. You also know for sure about $1$, $2$, $3$, and $4$. Now, connect wires corresponding to $5$ and $7$, then cross.
- Find the corresponding connected wires of $5$ and $7$. You also know for sure about $5$, $6$, $7$, and $8$. Now, connect wires corresponding to $2$ and $9$, then cross.
- Find the corresponding connected wires of $2$ and $9$. You also know for sure about $9$ and $10$.
For improvement, if
I can pair more than once, after the fourth crossing, I can just pair $1$ and $3$, $5$ and $7$, also $2$ and $9$ simultaneously. So I can cross only $5$ times.
answered 7 hours ago
athinathin
13.2k3 gold badges41 silver badges108 bronze badges
13.2k3 gold badges41 silver badges108 bronze badges
add a comment |
add a comment |
$begingroup$
The absolute minimum is
2 crossings
First some observations:
Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4
The actual strategy:
So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.
Next
Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.
Proof:
It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The absolute minimum is
2 crossings
First some observations:
Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4
The actual strategy:
So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.
Next
Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.
Proof:
It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The absolute minimum is
2 crossings
First some observations:
Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4
The actual strategy:
So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.
Next
Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.
Proof:
It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The absolute minimum is
2 crossings
First some observations:
Notice that when you connect ends of wires together, you are forming a partition. Notice that 10 is a triangular number, that is 10 = 1 + 2 + 3 + 4
The actual strategy:
So, first group wires into groups of one, two, three and four. Cross the river. You will be able to identify each group by continuity.
Next
Take one from each group, and form a group of four that way. Then take one from the remaining 3 groups, and so on. Once again, you have formed four differently sized groups, which will be identifiable once you cross the river. Additionally, since each group is formed by members of different groups, you will be able to tell which one is which.
Proof:
It obviously can't be solved with one crossing, as that would mean that you can connect the wires somehow on one end, and read something different on each wire.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
dratini0dratini0
211 bronze badge
211 bronze badge
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
dratini0 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
dvc is a new contributor. Be nice, and check out our Code of Conduct.
dvc is a new contributor. Be nice, and check out our Code of Conduct.
dvc is a new contributor. Be nice, and check out our Code of Conduct.
dvc is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f87384%2fthe-wires-underground%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
This must be a duplicate
$endgroup$
– Dr Xorile
7 hours ago
$begingroup$
What counts as one wiring? If I connect end A1 to A2 and B1 to B2 is that two and then measure the resistance between A1 and B1, is that one wiring or two?
$endgroup$
– Helena
7 hours ago