A case of the sniffles












6












$begingroup$


The cubicle farm



The cubicle farm at the Colla-R water treatment plant is laid out in a neat square: eight rows of eight cubicles with a narrow corridor between each cubicle (in the diagram the thin black lines are the corridors). Each cubicle is currently occupied by an employee, and no employees are on holiday.



The cubicles identified by being coloured red have ill employees in them: they have contracted some water-borne illness and are infectious. Due to the layout of the cubicles, a healthy employee only contracts the illness if they have two ill immediate neighbours in the four cardinal compass directions (you may take North to be pointing upwards relative to the page). For example, the cubicle at the end of the second row from the top contains an employee who is about to become ill. Ill employees do not go home, do not recover, are not allowed to leave their cubicle, but (luckily for them) do not die. As per company policy, the Colla-R HR department have now quarantined the cubicle farm, and no employee may leave until either everyone is ill, or everyone is well.



If, at the start of each hour, any healthy employee who has two ill neighbours as described becomes ill and immediately infectious, will all the employees fall ill? If not, what is the minimum number and location of ill employees that would ensure they do all fall ill? (The Colla-R HR department would of course like to avoid this happening.)










share|improve this question









$endgroup$








  • 2




    $begingroup$
    "no employee may leave until either everyone is ill, or everyone is well" - how would the latter be possible? You said ill employees do not recover.
    $endgroup$
    – Rand al'Thor
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor and now you know something about the HR practices here....
    $endgroup$
    – postmortes
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor less jokingly though, it's just to cover off all the edge cases
    $endgroup$
    – postmortes
    9 hours ago






  • 2




    $begingroup$
    Same second question.
    $endgroup$
    – noedne
    8 hours ago
















6












$begingroup$


The cubicle farm



The cubicle farm at the Colla-R water treatment plant is laid out in a neat square: eight rows of eight cubicles with a narrow corridor between each cubicle (in the diagram the thin black lines are the corridors). Each cubicle is currently occupied by an employee, and no employees are on holiday.



The cubicles identified by being coloured red have ill employees in them: they have contracted some water-borne illness and are infectious. Due to the layout of the cubicles, a healthy employee only contracts the illness if they have two ill immediate neighbours in the four cardinal compass directions (you may take North to be pointing upwards relative to the page). For example, the cubicle at the end of the second row from the top contains an employee who is about to become ill. Ill employees do not go home, do not recover, are not allowed to leave their cubicle, but (luckily for them) do not die. As per company policy, the Colla-R HR department have now quarantined the cubicle farm, and no employee may leave until either everyone is ill, or everyone is well.



If, at the start of each hour, any healthy employee who has two ill neighbours as described becomes ill and immediately infectious, will all the employees fall ill? If not, what is the minimum number and location of ill employees that would ensure they do all fall ill? (The Colla-R HR department would of course like to avoid this happening.)










share|improve this question









$endgroup$








  • 2




    $begingroup$
    "no employee may leave until either everyone is ill, or everyone is well" - how would the latter be possible? You said ill employees do not recover.
    $endgroup$
    – Rand al'Thor
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor and now you know something about the HR practices here....
    $endgroup$
    – postmortes
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor less jokingly though, it's just to cover off all the edge cases
    $endgroup$
    – postmortes
    9 hours ago






  • 2




    $begingroup$
    Same second question.
    $endgroup$
    – noedne
    8 hours ago














6












6








6





$begingroup$


The cubicle farm



The cubicle farm at the Colla-R water treatment plant is laid out in a neat square: eight rows of eight cubicles with a narrow corridor between each cubicle (in the diagram the thin black lines are the corridors). Each cubicle is currently occupied by an employee, and no employees are on holiday.



The cubicles identified by being coloured red have ill employees in them: they have contracted some water-borne illness and are infectious. Due to the layout of the cubicles, a healthy employee only contracts the illness if they have two ill immediate neighbours in the four cardinal compass directions (you may take North to be pointing upwards relative to the page). For example, the cubicle at the end of the second row from the top contains an employee who is about to become ill. Ill employees do not go home, do not recover, are not allowed to leave their cubicle, but (luckily for them) do not die. As per company policy, the Colla-R HR department have now quarantined the cubicle farm, and no employee may leave until either everyone is ill, or everyone is well.



If, at the start of each hour, any healthy employee who has two ill neighbours as described becomes ill and immediately infectious, will all the employees fall ill? If not, what is the minimum number and location of ill employees that would ensure they do all fall ill? (The Colla-R HR department would of course like to avoid this happening.)










share|improve this question









$endgroup$




The cubicle farm



The cubicle farm at the Colla-R water treatment plant is laid out in a neat square: eight rows of eight cubicles with a narrow corridor between each cubicle (in the diagram the thin black lines are the corridors). Each cubicle is currently occupied by an employee, and no employees are on holiday.



The cubicles identified by being coloured red have ill employees in them: they have contracted some water-borne illness and are infectious. Due to the layout of the cubicles, a healthy employee only contracts the illness if they have two ill immediate neighbours in the four cardinal compass directions (you may take North to be pointing upwards relative to the page). For example, the cubicle at the end of the second row from the top contains an employee who is about to become ill. Ill employees do not go home, do not recover, are not allowed to leave their cubicle, but (luckily for them) do not die. As per company policy, the Colla-R HR department have now quarantined the cubicle farm, and no employee may leave until either everyone is ill, or everyone is well.



If, at the start of each hour, any healthy employee who has two ill neighbours as described becomes ill and immediately infectious, will all the employees fall ill? If not, what is the minimum number and location of ill employees that would ensure they do all fall ill? (The Colla-R HR department would of course like to avoid this happening.)







mathematics situation






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 9 hours ago









postmortespostmortes

508212




508212








  • 2




    $begingroup$
    "no employee may leave until either everyone is ill, or everyone is well" - how would the latter be possible? You said ill employees do not recover.
    $endgroup$
    – Rand al'Thor
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor and now you know something about the HR practices here....
    $endgroup$
    – postmortes
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor less jokingly though, it's just to cover off all the edge cases
    $endgroup$
    – postmortes
    9 hours ago






  • 2




    $begingroup$
    Same second question.
    $endgroup$
    – noedne
    8 hours ago














  • 2




    $begingroup$
    "no employee may leave until either everyone is ill, or everyone is well" - how would the latter be possible? You said ill employees do not recover.
    $endgroup$
    – Rand al'Thor
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor and now you know something about the HR practices here....
    $endgroup$
    – postmortes
    9 hours ago






  • 1




    $begingroup$
    @Randal'Thor less jokingly though, it's just to cover off all the edge cases
    $endgroup$
    – postmortes
    9 hours ago






  • 2




    $begingroup$
    Same second question.
    $endgroup$
    – noedne
    8 hours ago








2




2




$begingroup$
"no employee may leave until either everyone is ill, or everyone is well" - how would the latter be possible? You said ill employees do not recover.
$endgroup$
– Rand al'Thor
9 hours ago




$begingroup$
"no employee may leave until either everyone is ill, or everyone is well" - how would the latter be possible? You said ill employees do not recover.
$endgroup$
– Rand al'Thor
9 hours ago




1




1




$begingroup$
@Randal'Thor and now you know something about the HR practices here....
$endgroup$
– postmortes
9 hours ago




$begingroup$
@Randal'Thor and now you know something about the HR practices here....
$endgroup$
– postmortes
9 hours ago




1




1




$begingroup$
@Randal'Thor less jokingly though, it's just to cover off all the edge cases
$endgroup$
– postmortes
9 hours ago




$begingroup$
@Randal'Thor less jokingly though, it's just to cover off all the edge cases
$endgroup$
– postmortes
9 hours ago




2




2




$begingroup$
Same second question.
$endgroup$
– noedne
8 hours ago




$begingroup$
Same second question.
$endgroup$
– noedne
8 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

Answer 1




No, they will not all fall ill. In particular, none of the employees in the top (or bottom) row will fall ill as they need to have at least one infected neighbour in the same row. Since none are ill in the beginning, none will become ill.




Suggestion for the minimum




If all of the cubicles on a diagonal have ill employees then everybody will eventually fall ill. So this gives an upper bound of 8 for the minimum.




Proof that this is the minimum




One important thing to notice is that the total perimeter of the ill area never increases (this is due to the fact that the two cubicle walls providing the infection get absorbed into the infected area in the next step producing, at most, two new cubicle walls to the infected perimeter).

Now, suppose there are just $7$ ill employees. Then, the total infected perimeter is at most $4 times 7 =28$. This can never increase, hence, the infection cannot cover all employees since the total perimeter is $32$.







share|improve this answer











$endgroup$













  • $begingroup$
    Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
    $endgroup$
    – postmortes
    9 hours ago












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f81449%2fa-case-of-the-sniffles%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









6












$begingroup$

Answer 1




No, they will not all fall ill. In particular, none of the employees in the top (or bottom) row will fall ill as they need to have at least one infected neighbour in the same row. Since none are ill in the beginning, none will become ill.




Suggestion for the minimum




If all of the cubicles on a diagonal have ill employees then everybody will eventually fall ill. So this gives an upper bound of 8 for the minimum.




Proof that this is the minimum




One important thing to notice is that the total perimeter of the ill area never increases (this is due to the fact that the two cubicle walls providing the infection get absorbed into the infected area in the next step producing, at most, two new cubicle walls to the infected perimeter).

Now, suppose there are just $7$ ill employees. Then, the total infected perimeter is at most $4 times 7 =28$. This can never increase, hence, the infection cannot cover all employees since the total perimeter is $32$.







share|improve this answer











$endgroup$













  • $begingroup$
    Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
    $endgroup$
    – postmortes
    9 hours ago
















6












$begingroup$

Answer 1




No, they will not all fall ill. In particular, none of the employees in the top (or bottom) row will fall ill as they need to have at least one infected neighbour in the same row. Since none are ill in the beginning, none will become ill.




Suggestion for the minimum




If all of the cubicles on a diagonal have ill employees then everybody will eventually fall ill. So this gives an upper bound of 8 for the minimum.




Proof that this is the minimum




One important thing to notice is that the total perimeter of the ill area never increases (this is due to the fact that the two cubicle walls providing the infection get absorbed into the infected area in the next step producing, at most, two new cubicle walls to the infected perimeter).

Now, suppose there are just $7$ ill employees. Then, the total infected perimeter is at most $4 times 7 =28$. This can never increase, hence, the infection cannot cover all employees since the total perimeter is $32$.







share|improve this answer











$endgroup$













  • $begingroup$
    Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
    $endgroup$
    – postmortes
    9 hours ago














6












6








6





$begingroup$

Answer 1




No, they will not all fall ill. In particular, none of the employees in the top (or bottom) row will fall ill as they need to have at least one infected neighbour in the same row. Since none are ill in the beginning, none will become ill.




Suggestion for the minimum




If all of the cubicles on a diagonal have ill employees then everybody will eventually fall ill. So this gives an upper bound of 8 for the minimum.




Proof that this is the minimum




One important thing to notice is that the total perimeter of the ill area never increases (this is due to the fact that the two cubicle walls providing the infection get absorbed into the infected area in the next step producing, at most, two new cubicle walls to the infected perimeter).

Now, suppose there are just $7$ ill employees. Then, the total infected perimeter is at most $4 times 7 =28$. This can never increase, hence, the infection cannot cover all employees since the total perimeter is $32$.







share|improve this answer











$endgroup$



Answer 1




No, they will not all fall ill. In particular, none of the employees in the top (or bottom) row will fall ill as they need to have at least one infected neighbour in the same row. Since none are ill in the beginning, none will become ill.




Suggestion for the minimum




If all of the cubicles on a diagonal have ill employees then everybody will eventually fall ill. So this gives an upper bound of 8 for the minimum.




Proof that this is the minimum




One important thing to notice is that the total perimeter of the ill area never increases (this is due to the fact that the two cubicle walls providing the infection get absorbed into the infected area in the next step producing, at most, two new cubicle walls to the infected perimeter).

Now, suppose there are just $7$ ill employees. Then, the total infected perimeter is at most $4 times 7 =28$. This can never increase, hence, the infection cannot cover all employees since the total perimeter is $32$.








share|improve this answer














share|improve this answer



share|improve this answer








edited 8 hours ago

























answered 9 hours ago









hexominohexomino

45.8k4140219




45.8k4140219












  • $begingroup$
    Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
    $endgroup$
    – postmortes
    9 hours ago


















  • $begingroup$
    Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
    $endgroup$
    – postmortes
    9 hours ago
















$begingroup$
Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
$endgroup$
– postmortes
9 hours ago




$begingroup$
Answer 1 is correct; if you can prove 8 as the minimum you get the tick :)
$endgroup$
– postmortes
9 hours ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f81449%2fa-case-of-the-sniffles%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

How to label and detect the document text images

Vallis Paradisi

Tabula Rosettana