6 balls and a scaleTwo Sheriffs and EavesdroppersSnow White and the Secret Message (Secret Sharing Problem)Two Sheriffs and Eavesdroppers - 2The oldest wins the prize, but they won't tell their age
Does a difference of tense count as a difference of meaning in a minimal pair?
What will happen if my luggage gets delayed?
How do we create new idioms and use them in a novel?
I reported the illegal activity of my boss to his boss. My boss found out. Now I am being punished. What should I do?
Proving a statement about real numbers
Do I really need to have a scientific explanation for my premise?
Was it really inappropriate to write a pull request for the company I interviewed with?
Should I take out a loan for a friend to invest on my behalf?
What would be the most expensive material to an intergalactic society?
How can I manipulate the output of Information?
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
Doesn't allowing a user mode program to access kernel space memory and execute the IN and OUT instructions defeat the purpose of having CPU modes?
How to resolve: Reviewer #1 says remove section X vs. Reviewer #2 says expand section X
Doubts in understanding some concepts of potential energy
What is better: yes / no radio, or simple checkbox?
Is it a Cyclops number? "Nobody" knows!
How do electrons receive energy when a body is heated?
Has a sovereign Communist government ever run, and conceded loss, on a fair election?
Are all players supposed to be able to see each others' character sheets?
How to design an organic heat-shield?
Shifting between bemols and diesis in the key signature
School performs periodic password audits. Is my password compromised?
Confusion about Complex Continued Fraction
Specifying a starting column with colortbl package and xcolor
6 balls and a scale
Two Sheriffs and EavesdroppersSnow White and the Secret Message (Secret Sharing Problem)Two Sheriffs and Eavesdroppers - 2The oldest wins the prize, but they won't tell their age
$begingroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
New contributor
$endgroup$
add a comment |
$begingroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
New contributor
$endgroup$
add a comment |
$begingroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
New contributor
$endgroup$
Suppose you are given 6 balls all of which look identical. You’re told that 4 of the balls all weigh the same, but there are 2 balls that have unequal weights. Additionally, these two balls together weigh the same as any two of the 4 balls having the same weight. Determine which two balls have different weights, and which one weighs less and which one weighs more, using a balance scale with the fewest number of weighings possible.
Hint given in book: First, calculate how many questions you have to answer about the possible relations between the weight of the balls, and use this to determine how many weighings with the balance scale are needed.
I believe the mathematical way of going about it would be taking log3 of (6 choose 2), but I'm not sure.
information-theory
information-theory
New contributor
New contributor
New contributor
asked yesterday
user57753user57753
284
284
New contributor
New contributor
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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
);
);
user57753 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%2f80488%2f6-balls-and-a-scale%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
$endgroup$
add a comment |
$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
$endgroup$
add a comment |
$begingroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
$endgroup$
I believe you're almost right in your assessment. Instead of
$log_3 binom62$ I believe the minimum number of weighings is $lceil log_3 [binom62 times 2] rceil$
This is because each weighing
has three possible outcomes: either the scales balance, the right side is heavier or the left side is heavier. So if the number of weighings is $X$ then the maximum possible number of different scenarios you can obtain is $3^X$.
However, there are $binom62=15$ ways to choose 2 faulty balls out of 6, and 2 ways to have one be heavier and one lighter, for a total of 30 different ways. Since we are looking at worst case scenarios, this means that if you only have three weighings (27 max scenarios), there is a way for two different arrangements to give the same outcome and thus be indistinguishable from each other.
For this reason
The minimum number of weighings is $lceil log_3 (15 times 2) rceil=4$
It's trivial to come up with a solution for 4 weighings. Call the balls ABCDEF. Weigh AB, CD, EF. If two of the weighings are balanced then the other will give the heavy and light balls. If two of the weighings are unbalanced, then for the fourth weighing weigh the two heavy sides. The heaviest side is the heavy ball and the lighter side of the weighing that didn't include the heavy ball is the light ball. I do not think it can be done in less than 4 weighings.
answered yesterday
AmorydaiAmorydai
88512
88512
add a comment |
add a comment |
user57753 is a new contributor. Be nice, and check out our Code of Conduct.
user57753 is a new contributor. Be nice, and check out our Code of Conduct.
user57753 is a new contributor. Be nice, and check out our Code of Conduct.
user57753 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%2f80488%2f6-balls-and-a-scale%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