Split a list of 24 items into four parts, so you need three out of the four parts to recover the entire list
$begingroup$
Imagine you have an ordered list of 24 words, and you want to split the list of words into four parts. You want to set it up so that three out of the four parts are needed in order to recover the entire list. For security, you also want to ensure that the entire list cannot be recovered with just two of the parts.
To make this a bit clearer, allow me to show an example with three parts, where two out of the three are required:
| part 1 | part 2 | part 3 |
|--------------|--------------|--------------|
| one | | one |
| two | | two |
| three | | three |
| four | | four |
| five | | five |
| six | | six |
| seven | | seven |
| eight | | eight |
| nine | nine | |
| ten | ten | |
| eleven | eleven | |
| twelve | twelve | |
| thirteen | thirteen | |
| fourteen | fourteen | |
| fifteen | fifteen | |
| sixteen | sixteen | |
| | seventeen | seventeen |
| | eighteen | eighteen |
| | nineteen | nineteen |
| | twenty | twenty |
| | twenty-one | twenty-one |
| | twenty-two | twenty-two |
| | twenty-three | twenty-three |
| | twenty-four | twenty-four |
Notice that if you have any two of the parts, you have the whole list. Also, if one of the parts is lost, you're fine. You can still retrieve the entire list.
Returning to the original question, how do I accomplish the above, again for 24 words, but where three out of four parts are required, and you can't recover the list with just two parts. Each word must exist in at least two of the parts, in case one of the four parts is lost.
When I try to do it by hand, I keep ending up with something that does not meet the requirements. I intuited that there would be 12 words in each of the four parts, but have been unable to split it by hand in an orderly way. I did manage to brute force it with a software script, but what the script generates looks very haphazard:
| part 1 | part 2 | part 3 | part 4 |
|--------------|--------------|--------------|--------------|
| one | | | one |
| | two | | two |
| three | | | three |
| four | | four | |
| five | | | five |
| | six | | six |
| | seven | | seven |
| | | eight | eight |
| nine | | nine | |
| ten | ten | | |
| eleven | | eleven | |
| twelve | | | twelve |
| | thirteen | thirteen | |
| | fourteen | fourteen | |
| | fifteen | fifteen | |
| | sixteen | sixteen | |
| seventeen | seventeen | | |
| eighteen | eighteen | | |
| nineteen | nineteen | | |
| twenty | | twenty | |
| | | twenty-one | twenty-one |
| | twenty-two | | twenty-two |
| | | twenty-three | twenty-three |
| | | twenty-four | twenty-four |
While the above does meet the requirements, it looks so random. I was hoping for something that follows some sort of pattern that is easy to recognize with the human eye (like the solution for the two-out-of-three-parts problem). It feels like it should be easy to solve given the right approach. Any ideas on finding an orderly pattern?
By the way, I'm not looking for any computer script (although feel free to provide one at your discretion). Apologies if this would not be considered a math problem. It feels like it could be, but I'm not quite sure how to categorize it. If it happens that your brain just intuits an orderly answer, feel free to put that here and I will gladly accept it. I don't necessarily need to know how you got there. :-)
permutations pattern-matching
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
$endgroup$
add a comment |
$begingroup$
Imagine you have an ordered list of 24 words, and you want to split the list of words into four parts. You want to set it up so that three out of the four parts are needed in order to recover the entire list. For security, you also want to ensure that the entire list cannot be recovered with just two of the parts.
To make this a bit clearer, allow me to show an example with three parts, where two out of the three are required:
| part 1 | part 2 | part 3 |
|--------------|--------------|--------------|
| one | | one |
| two | | two |
| three | | three |
| four | | four |
| five | | five |
| six | | six |
| seven | | seven |
| eight | | eight |
| nine | nine | |
| ten | ten | |
| eleven | eleven | |
| twelve | twelve | |
| thirteen | thirteen | |
| fourteen | fourteen | |
| fifteen | fifteen | |
| sixteen | sixteen | |
| | seventeen | seventeen |
| | eighteen | eighteen |
| | nineteen | nineteen |
| | twenty | twenty |
| | twenty-one | twenty-one |
| | twenty-two | twenty-two |
| | twenty-three | twenty-three |
| | twenty-four | twenty-four |
Notice that if you have any two of the parts, you have the whole list. Also, if one of the parts is lost, you're fine. You can still retrieve the entire list.
Returning to the original question, how do I accomplish the above, again for 24 words, but where three out of four parts are required, and you can't recover the list with just two parts. Each word must exist in at least two of the parts, in case one of the four parts is lost.
When I try to do it by hand, I keep ending up with something that does not meet the requirements. I intuited that there would be 12 words in each of the four parts, but have been unable to split it by hand in an orderly way. I did manage to brute force it with a software script, but what the script generates looks very haphazard:
| part 1 | part 2 | part 3 | part 4 |
|--------------|--------------|--------------|--------------|
| one | | | one |
| | two | | two |
| three | | | three |
| four | | four | |
| five | | | five |
| | six | | six |
| | seven | | seven |
| | | eight | eight |
| nine | | nine | |
| ten | ten | | |
| eleven | | eleven | |
| twelve | | | twelve |
| | thirteen | thirteen | |
| | fourteen | fourteen | |
| | fifteen | fifteen | |
| | sixteen | sixteen | |
| seventeen | seventeen | | |
| eighteen | eighteen | | |
| nineteen | nineteen | | |
| twenty | | twenty | |
| | | twenty-one | twenty-one |
| | twenty-two | | twenty-two |
| | | twenty-three | twenty-three |
| | | twenty-four | twenty-four |
While the above does meet the requirements, it looks so random. I was hoping for something that follows some sort of pattern that is easy to recognize with the human eye (like the solution for the two-out-of-three-parts problem). It feels like it should be easy to solve given the right approach. Any ideas on finding an orderly pattern?
By the way, I'm not looking for any computer script (although feel free to provide one at your discretion). Apologies if this would not be considered a math problem. It feels like it could be, but I'm not quite sure how to categorize it. If it happens that your brain just intuits an orderly answer, feel free to put that here and I will gladly accept it. I don't necessarily need to know how you got there. :-)
permutations pattern-matching
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
$endgroup$
1
$begingroup$
I don't know how this may help solve the entire thing, but in your code generated example, across the rows, each phrase appears exactly twice (e.g. 'one' appears in #1,4 only, 'two' appears in #2,4 only, etc.) and that there are indeed twelve per part.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:36
1
$begingroup$
I figured out the solution (I think) so here's the hint: $_2C_4 = 6$, and use the fact that each phrase appears only twice per row.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:40
$begingroup$
Why don't you just renumber so that the twelve in column one are one thought twelve, the next picks in column two start at thirteen etc. There is more renumbering that could be done. But randomness in a list is sometimes a good thing - structure is exploitable by an enemy and might compromise security, if that is what os in view.
$endgroup$
– Mark Bennet
Dec 27 '18 at 7:33
add a comment |
$begingroup$
Imagine you have an ordered list of 24 words, and you want to split the list of words into four parts. You want to set it up so that three out of the four parts are needed in order to recover the entire list. For security, you also want to ensure that the entire list cannot be recovered with just two of the parts.
To make this a bit clearer, allow me to show an example with three parts, where two out of the three are required:
| part 1 | part 2 | part 3 |
|--------------|--------------|--------------|
| one | | one |
| two | | two |
| three | | three |
| four | | four |
| five | | five |
| six | | six |
| seven | | seven |
| eight | | eight |
| nine | nine | |
| ten | ten | |
| eleven | eleven | |
| twelve | twelve | |
| thirteen | thirteen | |
| fourteen | fourteen | |
| fifteen | fifteen | |
| sixteen | sixteen | |
| | seventeen | seventeen |
| | eighteen | eighteen |
| | nineteen | nineteen |
| | twenty | twenty |
| | twenty-one | twenty-one |
| | twenty-two | twenty-two |
| | twenty-three | twenty-three |
| | twenty-four | twenty-four |
Notice that if you have any two of the parts, you have the whole list. Also, if one of the parts is lost, you're fine. You can still retrieve the entire list.
Returning to the original question, how do I accomplish the above, again for 24 words, but where three out of four parts are required, and you can't recover the list with just two parts. Each word must exist in at least two of the parts, in case one of the four parts is lost.
When I try to do it by hand, I keep ending up with something that does not meet the requirements. I intuited that there would be 12 words in each of the four parts, but have been unable to split it by hand in an orderly way. I did manage to brute force it with a software script, but what the script generates looks very haphazard:
| part 1 | part 2 | part 3 | part 4 |
|--------------|--------------|--------------|--------------|
| one | | | one |
| | two | | two |
| three | | | three |
| four | | four | |
| five | | | five |
| | six | | six |
| | seven | | seven |
| | | eight | eight |
| nine | | nine | |
| ten | ten | | |
| eleven | | eleven | |
| twelve | | | twelve |
| | thirteen | thirteen | |
| | fourteen | fourteen | |
| | fifteen | fifteen | |
| | sixteen | sixteen | |
| seventeen | seventeen | | |
| eighteen | eighteen | | |
| nineteen | nineteen | | |
| twenty | | twenty | |
| | | twenty-one | twenty-one |
| | twenty-two | | twenty-two |
| | | twenty-three | twenty-three |
| | | twenty-four | twenty-four |
While the above does meet the requirements, it looks so random. I was hoping for something that follows some sort of pattern that is easy to recognize with the human eye (like the solution for the two-out-of-three-parts problem). It feels like it should be easy to solve given the right approach. Any ideas on finding an orderly pattern?
By the way, I'm not looking for any computer script (although feel free to provide one at your discretion). Apologies if this would not be considered a math problem. It feels like it could be, but I'm not quite sure how to categorize it. If it happens that your brain just intuits an orderly answer, feel free to put that here and I will gladly accept it. I don't necessarily need to know how you got there. :-)
permutations pattern-matching
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
$endgroup$
Imagine you have an ordered list of 24 words, and you want to split the list of words into four parts. You want to set it up so that three out of the four parts are needed in order to recover the entire list. For security, you also want to ensure that the entire list cannot be recovered with just two of the parts.
To make this a bit clearer, allow me to show an example with three parts, where two out of the three are required:
| part 1 | part 2 | part 3 |
|--------------|--------------|--------------|
| one | | one |
| two | | two |
| three | | three |
| four | | four |
| five | | five |
| six | | six |
| seven | | seven |
| eight | | eight |
| nine | nine | |
| ten | ten | |
| eleven | eleven | |
| twelve | twelve | |
| thirteen | thirteen | |
| fourteen | fourteen | |
| fifteen | fifteen | |
| sixteen | sixteen | |
| | seventeen | seventeen |
| | eighteen | eighteen |
| | nineteen | nineteen |
| | twenty | twenty |
| | twenty-one | twenty-one |
| | twenty-two | twenty-two |
| | twenty-three | twenty-three |
| | twenty-four | twenty-four |
Notice that if you have any two of the parts, you have the whole list. Also, if one of the parts is lost, you're fine. You can still retrieve the entire list.
Returning to the original question, how do I accomplish the above, again for 24 words, but where three out of four parts are required, and you can't recover the list with just two parts. Each word must exist in at least two of the parts, in case one of the four parts is lost.
When I try to do it by hand, I keep ending up with something that does not meet the requirements. I intuited that there would be 12 words in each of the four parts, but have been unable to split it by hand in an orderly way. I did manage to brute force it with a software script, but what the script generates looks very haphazard:
| part 1 | part 2 | part 3 | part 4 |
|--------------|--------------|--------------|--------------|
| one | | | one |
| | two | | two |
| three | | | three |
| four | | four | |
| five | | | five |
| | six | | six |
| | seven | | seven |
| | | eight | eight |
| nine | | nine | |
| ten | ten | | |
| eleven | | eleven | |
| twelve | | | twelve |
| | thirteen | thirteen | |
| | fourteen | fourteen | |
| | fifteen | fifteen | |
| | sixteen | sixteen | |
| seventeen | seventeen | | |
| eighteen | eighteen | | |
| nineteen | nineteen | | |
| twenty | | twenty | |
| | | twenty-one | twenty-one |
| | twenty-two | | twenty-two |
| | | twenty-three | twenty-three |
| | | twenty-four | twenty-four |
While the above does meet the requirements, it looks so random. I was hoping for something that follows some sort of pattern that is easy to recognize with the human eye (like the solution for the two-out-of-three-parts problem). It feels like it should be easy to solve given the right approach. Any ideas on finding an orderly pattern?
By the way, I'm not looking for any computer script (although feel free to provide one at your discretion). Apologies if this would not be considered a math problem. It feels like it could be, but I'm not quite sure how to categorize it. If it happens that your brain just intuits an orderly answer, feel free to put that here and I will gladly accept it. I don't necessarily need to know how you got there. :-)
permutations pattern-matching
permutations pattern-matching
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
asked Dec 27 '18 at 6:31
Nate CookNate Cook
1083
1083
1
$begingroup$
I don't know how this may help solve the entire thing, but in your code generated example, across the rows, each phrase appears exactly twice (e.g. 'one' appears in #1,4 only, 'two' appears in #2,4 only, etc.) and that there are indeed twelve per part.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:36
1
$begingroup$
I figured out the solution (I think) so here's the hint: $_2C_4 = 6$, and use the fact that each phrase appears only twice per row.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:40
$begingroup$
Why don't you just renumber so that the twelve in column one are one thought twelve, the next picks in column two start at thirteen etc. There is more renumbering that could be done. But randomness in a list is sometimes a good thing - structure is exploitable by an enemy and might compromise security, if that is what os in view.
$endgroup$
– Mark Bennet
Dec 27 '18 at 7:33
add a comment |
1
$begingroup$
I don't know how this may help solve the entire thing, but in your code generated example, across the rows, each phrase appears exactly twice (e.g. 'one' appears in #1,4 only, 'two' appears in #2,4 only, etc.) and that there are indeed twelve per part.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:36
1
$begingroup$
I figured out the solution (I think) so here's the hint: $_2C_4 = 6$, and use the fact that each phrase appears only twice per row.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:40
$begingroup$
Why don't you just renumber so that the twelve in column one are one thought twelve, the next picks in column two start at thirteen etc. There is more renumbering that could be done. But randomness in a list is sometimes a good thing - structure is exploitable by an enemy and might compromise security, if that is what os in view.
$endgroup$
– Mark Bennet
Dec 27 '18 at 7:33
1
1
$begingroup$
I don't know how this may help solve the entire thing, but in your code generated example, across the rows, each phrase appears exactly twice (e.g. 'one' appears in #1,4 only, 'two' appears in #2,4 only, etc.) and that there are indeed twelve per part.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:36
$begingroup$
I don't know how this may help solve the entire thing, but in your code generated example, across the rows, each phrase appears exactly twice (e.g. 'one' appears in #1,4 only, 'two' appears in #2,4 only, etc.) and that there are indeed twelve per part.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:36
1
1
$begingroup$
I figured out the solution (I think) so here's the hint: $_2C_4 = 6$, and use the fact that each phrase appears only twice per row.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:40
$begingroup$
I figured out the solution (I think) so here's the hint: $_2C_4 = 6$, and use the fact that each phrase appears only twice per row.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:40
$begingroup$
Why don't you just renumber so that the twelve in column one are one thought twelve, the next picks in column two start at thirteen etc. There is more renumbering that could be done. But randomness in a list is sometimes a good thing - structure is exploitable by an enemy and might compromise security, if that is what os in view.
$endgroup$
– Mark Bennet
Dec 27 '18 at 7:33
$begingroup$
Why don't you just renumber so that the twelve in column one are one thought twelve, the next picks in column two start at thirteen etc. There is more renumbering that could be done. But randomness in a list is sometimes a good thing - structure is exploitable by an enemy and might compromise security, if that is what os in view.
$endgroup$
– Mark Bennet
Dec 27 '18 at 7:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So, given that the phrase appears in only two columns for its given row, let's find all the ways that the phrase can appear: Name the columns A, B, C, D. Then the 2/4 combinations are AD, BD, CD, AC, BC, AB.
Using these, 'one' appears only in A and D, 'two' appears only in B and D, 'three' appears only in C and D, and so on. For the image solution posted, I carried the pattern down all twenty-four phrases (I just used the numerals because I'm lazy and sleepy), but in reality, you only need to do it for the first six! Then for the rest, you can show it all. I'll edit in the two examples...
Image solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | | | 7 | Repeat from here
| | 8 | | 8 |
| | | 9 | 9 |
| 10| | 10| |
| | 11| 11| |
| 12| 12| | | end 2
| 13| | | 13|
| | 14| | 14|
| | | 15| 15|
| 16| | 16| |
| | 17| 17| |
| 18| 18| | | end 3
| 19| | | 19|
| | 20| | 20|
| | | 21| 21|
| 22| | 22| |
| | 23| 23| |
| 24| 24| | | end 4
In reality solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | 7 | 7 | 7 | The above cancellation keeps it limited to
| 8 | 8 | 8 | 8 | use three to find the entire thing, two won't do!
| 9 | 9 | 9 | 9 |
| 10| 10| 10| 10|
| 11| 11| 11| 11|
| 12| 12| 12| 12|
| 13| 13| 13| 13|
| 14| 14| 14| 14|
| 15| 15| 15| 15|
| 16| 16| 16| 16|
| 17| 17| 17| 17|
| 18| 18| 18| 18|
| 19| 19| 19| 19|
| 20| 20| 20| 20|
| 21| 21| 21| 21|
| 22| 22| 22| 22|
| 23| 23| 23| 23|
| 24| 24| 24| 24|
To be even clearer, with your three column example, you could easily do this:
Image solution:
| A | B | C |
-------------
| 1 | | 1 |
| | 2 | 2 |
| 3 | 3 | |
| 4 | 4 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 7 | 7 |
| 8 | 8 | 8 | Carry on list of numbers
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
$endgroup$
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
add a comment |
$begingroup$
Note that $4!=24$. Let the $24$ words be the permutations (or identify the given words wuth them) $1234,1243,1324,ldots, 4321$ and place $abcd$ in columns $a$ and $b$. Then three columns suffice because each word $abcd$ occurs (at most one of columns $a,b$ is missing), whereas two columns do not suffice (having only columns $a,b$ will miss the words ending in $ldots ab$ or $ldots ba$). Not ethat the method is somewhat wasteful - for example the twelve even permutations would be enough.
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
add a comment |
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2 Answers
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$begingroup$
So, given that the phrase appears in only two columns for its given row, let's find all the ways that the phrase can appear: Name the columns A, B, C, D. Then the 2/4 combinations are AD, BD, CD, AC, BC, AB.
Using these, 'one' appears only in A and D, 'two' appears only in B and D, 'three' appears only in C and D, and so on. For the image solution posted, I carried the pattern down all twenty-four phrases (I just used the numerals because I'm lazy and sleepy), but in reality, you only need to do it for the first six! Then for the rest, you can show it all. I'll edit in the two examples...
Image solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | | | 7 | Repeat from here
| | 8 | | 8 |
| | | 9 | 9 |
| 10| | 10| |
| | 11| 11| |
| 12| 12| | | end 2
| 13| | | 13|
| | 14| | 14|
| | | 15| 15|
| 16| | 16| |
| | 17| 17| |
| 18| 18| | | end 3
| 19| | | 19|
| | 20| | 20|
| | | 21| 21|
| 22| | 22| |
| | 23| 23| |
| 24| 24| | | end 4
In reality solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | 7 | 7 | 7 | The above cancellation keeps it limited to
| 8 | 8 | 8 | 8 | use three to find the entire thing, two won't do!
| 9 | 9 | 9 | 9 |
| 10| 10| 10| 10|
| 11| 11| 11| 11|
| 12| 12| 12| 12|
| 13| 13| 13| 13|
| 14| 14| 14| 14|
| 15| 15| 15| 15|
| 16| 16| 16| 16|
| 17| 17| 17| 17|
| 18| 18| 18| 18|
| 19| 19| 19| 19|
| 20| 20| 20| 20|
| 21| 21| 21| 21|
| 22| 22| 22| 22|
| 23| 23| 23| 23|
| 24| 24| 24| 24|
To be even clearer, with your three column example, you could easily do this:
Image solution:
| A | B | C |
-------------
| 1 | | 1 |
| | 2 | 2 |
| 3 | 3 | |
| 4 | 4 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 7 | 7 |
| 8 | 8 | 8 | Carry on list of numbers
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
$endgroup$
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
add a comment |
$begingroup$
So, given that the phrase appears in only two columns for its given row, let's find all the ways that the phrase can appear: Name the columns A, B, C, D. Then the 2/4 combinations are AD, BD, CD, AC, BC, AB.
Using these, 'one' appears only in A and D, 'two' appears only in B and D, 'three' appears only in C and D, and so on. For the image solution posted, I carried the pattern down all twenty-four phrases (I just used the numerals because I'm lazy and sleepy), but in reality, you only need to do it for the first six! Then for the rest, you can show it all. I'll edit in the two examples...
Image solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | | | 7 | Repeat from here
| | 8 | | 8 |
| | | 9 | 9 |
| 10| | 10| |
| | 11| 11| |
| 12| 12| | | end 2
| 13| | | 13|
| | 14| | 14|
| | | 15| 15|
| 16| | 16| |
| | 17| 17| |
| 18| 18| | | end 3
| 19| | | 19|
| | 20| | 20|
| | | 21| 21|
| 22| | 22| |
| | 23| 23| |
| 24| 24| | | end 4
In reality solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | 7 | 7 | 7 | The above cancellation keeps it limited to
| 8 | 8 | 8 | 8 | use three to find the entire thing, two won't do!
| 9 | 9 | 9 | 9 |
| 10| 10| 10| 10|
| 11| 11| 11| 11|
| 12| 12| 12| 12|
| 13| 13| 13| 13|
| 14| 14| 14| 14|
| 15| 15| 15| 15|
| 16| 16| 16| 16|
| 17| 17| 17| 17|
| 18| 18| 18| 18|
| 19| 19| 19| 19|
| 20| 20| 20| 20|
| 21| 21| 21| 21|
| 22| 22| 22| 22|
| 23| 23| 23| 23|
| 24| 24| 24| 24|
To be even clearer, with your three column example, you could easily do this:
Image solution:
| A | B | C |
-------------
| 1 | | 1 |
| | 2 | 2 |
| 3 | 3 | |
| 4 | 4 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 7 | 7 |
| 8 | 8 | 8 | Carry on list of numbers
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
$endgroup$
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
add a comment |
$begingroup$
So, given that the phrase appears in only two columns for its given row, let's find all the ways that the phrase can appear: Name the columns A, B, C, D. Then the 2/4 combinations are AD, BD, CD, AC, BC, AB.
Using these, 'one' appears only in A and D, 'two' appears only in B and D, 'three' appears only in C and D, and so on. For the image solution posted, I carried the pattern down all twenty-four phrases (I just used the numerals because I'm lazy and sleepy), but in reality, you only need to do it for the first six! Then for the rest, you can show it all. I'll edit in the two examples...
Image solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | | | 7 | Repeat from here
| | 8 | | 8 |
| | | 9 | 9 |
| 10| | 10| |
| | 11| 11| |
| 12| 12| | | end 2
| 13| | | 13|
| | 14| | 14|
| | | 15| 15|
| 16| | 16| |
| | 17| 17| |
| 18| 18| | | end 3
| 19| | | 19|
| | 20| | 20|
| | | 21| 21|
| 22| | 22| |
| | 23| 23| |
| 24| 24| | | end 4
In reality solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | 7 | 7 | 7 | The above cancellation keeps it limited to
| 8 | 8 | 8 | 8 | use three to find the entire thing, two won't do!
| 9 | 9 | 9 | 9 |
| 10| 10| 10| 10|
| 11| 11| 11| 11|
| 12| 12| 12| 12|
| 13| 13| 13| 13|
| 14| 14| 14| 14|
| 15| 15| 15| 15|
| 16| 16| 16| 16|
| 17| 17| 17| 17|
| 18| 18| 18| 18|
| 19| 19| 19| 19|
| 20| 20| 20| 20|
| 21| 21| 21| 21|
| 22| 22| 22| 22|
| 23| 23| 23| 23|
| 24| 24| 24| 24|
To be even clearer, with your three column example, you could easily do this:
Image solution:
| A | B | C |
-------------
| 1 | | 1 |
| | 2 | 2 |
| 3 | 3 | |
| 4 | 4 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 7 | 7 |
| 8 | 8 | 8 | Carry on list of numbers
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
$endgroup$
So, given that the phrase appears in only two columns for its given row, let's find all the ways that the phrase can appear: Name the columns A, B, C, D. Then the 2/4 combinations are AD, BD, CD, AC, BC, AB.
Using these, 'one' appears only in A and D, 'two' appears only in B and D, 'three' appears only in C and D, and so on. For the image solution posted, I carried the pattern down all twenty-four phrases (I just used the numerals because I'm lazy and sleepy), but in reality, you only need to do it for the first six! Then for the rest, you can show it all. I'll edit in the two examples...
Image solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | | | 7 | Repeat from here
| | 8 | | 8 |
| | | 9 | 9 |
| 10| | 10| |
| | 11| 11| |
| 12| 12| | | end 2
| 13| | | 13|
| | 14| | 14|
| | | 15| 15|
| 16| | 16| |
| | 17| 17| |
| 18| 18| | | end 3
| 19| | | 19|
| | 20| | 20|
| | | 21| 21|
| 22| | 22| |
| | 23| 23| |
| 24| 24| | | end 4
In reality solution:
| A | B | C | D |
-----------------
| 1 | | | 1 |
| | 2 | | 2 |
| | | 3 | 3 |
| 4 | | 4 | |
| | 5 | 5 | |
| 6 | 6 | | | End of pattern
| 7 | 7 | 7 | 7 | The above cancellation keeps it limited to
| 8 | 8 | 8 | 8 | use three to find the entire thing, two won't do!
| 9 | 9 | 9 | 9 |
| 10| 10| 10| 10|
| 11| 11| 11| 11|
| 12| 12| 12| 12|
| 13| 13| 13| 13|
| 14| 14| 14| 14|
| 15| 15| 15| 15|
| 16| 16| 16| 16|
| 17| 17| 17| 17|
| 18| 18| 18| 18|
| 19| 19| 19| 19|
| 20| 20| 20| 20|
| 21| 21| 21| 21|
| 22| 22| 22| 22|
| 23| 23| 23| 23|
| 24| 24| 24| 24|
To be even clearer, with your three column example, you could easily do this:
Image solution:
| A | B | C |
-------------
| 1 | | 1 |
| | 2 | 2 |
| 3 | 3 | |
| 4 | 4 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 7 | 7 |
| 8 | 8 | 8 | Carry on list of numbers
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
edited Dec 27 '18 at 7:04
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
answered Dec 27 '18 at 6:55
Christopher MarleyChristopher Marley
1,117115
1,117115
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
add a comment |
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
You can think about it the reverse way. You're selecting 3 columns of the four, meaning you're excluding one, which isn't enough to get rid of the fact that each phrase appears twice for its row. Selecting two columns means that one of the numbers won't show up just because that's part of the $_2C_4$ generated list. $AB not= CD$
$endgroup$
– Christopher Marley
Dec 27 '18 at 7:06
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
$begingroup$
Wow, perfect solution and excellent explanation. Thank you for generalizing it for the three column solution as well!
$endgroup$
– Nate Cook
Dec 27 '18 at 7:59
add a comment |
$begingroup$
Note that $4!=24$. Let the $24$ words be the permutations (or identify the given words wuth them) $1234,1243,1324,ldots, 4321$ and place $abcd$ in columns $a$ and $b$. Then three columns suffice because each word $abcd$ occurs (at most one of columns $a,b$ is missing), whereas two columns do not suffice (having only columns $a,b$ will miss the words ending in $ldots ab$ or $ldots ba$). Not ethat the method is somewhat wasteful - for example the twelve even permutations would be enough.
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
add a comment |
$begingroup$
Note that $4!=24$. Let the $24$ words be the permutations (or identify the given words wuth them) $1234,1243,1324,ldots, 4321$ and place $abcd$ in columns $a$ and $b$. Then three columns suffice because each word $abcd$ occurs (at most one of columns $a,b$ is missing), whereas two columns do not suffice (having only columns $a,b$ will miss the words ending in $ldots ab$ or $ldots ba$). Not ethat the method is somewhat wasteful - for example the twelve even permutations would be enough.
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
add a comment |
$begingroup$
Note that $4!=24$. Let the $24$ words be the permutations (or identify the given words wuth them) $1234,1243,1324,ldots, 4321$ and place $abcd$ in columns $a$ and $b$. Then three columns suffice because each word $abcd$ occurs (at most one of columns $a,b$ is missing), whereas two columns do not suffice (having only columns $a,b$ will miss the words ending in $ldots ab$ or $ldots ba$). Not ethat the method is somewhat wasteful - for example the twelve even permutations would be enough.
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
$endgroup$
Note that $4!=24$. Let the $24$ words be the permutations (or identify the given words wuth them) $1234,1243,1324,ldots, 4321$ and place $abcd$ in columns $a$ and $b$. Then three columns suffice because each word $abcd$ occurs (at most one of columns $a,b$ is missing), whereas two columns do not suffice (having only columns $a,b$ will miss the words ending in $ldots ab$ or $ldots ba$). Not ethat the method is somewhat wasteful - for example the twelve even permutations would be enough.
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
answered Dec 27 '18 at 7:34
Hagen von EitzenHagen von Eitzen
281k23272505
281k23272505
add a comment |
add a comment |
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$begingroup$
I don't know how this may help solve the entire thing, but in your code generated example, across the rows, each phrase appears exactly twice (e.g. 'one' appears in #1,4 only, 'two' appears in #2,4 only, etc.) and that there are indeed twelve per part.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:36
1
$begingroup$
I figured out the solution (I think) so here's the hint: $_2C_4 = 6$, and use the fact that each phrase appears only twice per row.
$endgroup$
– Christopher Marley
Dec 27 '18 at 6:40
$begingroup$
Why don't you just renumber so that the twelve in column one are one thought twelve, the next picks in column two start at thirteen etc. There is more renumbering that could be done. But randomness in a list is sometimes a good thing - structure is exploitable by an enemy and might compromise security, if that is what os in view.
$endgroup$
– Mark Bennet
Dec 27 '18 at 7:33