Fill two dimensional list by meeting given conditions
$begingroup$
Suppose you have a 2D array - a list that exits in both x and y axis - which is 4 x 4, and you need to fill the table by putting numbers in it obeying following restrictions.
Each cell is able to store a number that can be calculated according to its index. (Indexes start from [0,0])
If a cell's index is [i, j] it can store a number calculated as one of the following ways.
i x j + i - j
i x j + j - i
i ^ j
j ^ i
Each cell can carry at most multiplication of its adjacent cells. (horizonal and vertical)
Each cell should be colored in a way all odds are of same color and all evens are of same color and two same colored cell cannot stand side by side vertically or horizontally.
calculation-puzzle
asked Mar 29 at 9:54
DavisDavis
61
New contributor
Davis 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 |
$begingroup$
Suppose you have a 2D array - a list that exits in both x and y axis - which is 4 x 4, and you need to fill the table by putting numbers in it obeying following restrictions.
Each cell is able to store a number that can be calculated according to its index. (Indexes start from [0,0])
If a cell's index is [i, j] it can store a number calculated as one of the following ways.
i x j + i - j
i x j + j - i
i ^ j
j ^ i
Each cell can carry at most multiplication of its adjacent cells. (horizonal and vertical)
Each cell should be colored in a way all odds are of same color and all evens are of same color and two same colored cell cannot stand side by side vertically or horizontally.
calculation-puzzle
asked Mar 29 at 9:54
DavisDavis
61
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
How do you define $0^0$?
$endgroup$
– hexomino
Mar 29 at 9:59
$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
$endgroup$
– Rubio♦
yesterday
add a comment |
$begingroup$
Suppose you have a 2D array - a list that exits in both x and y axis - which is 4 x 4, and you need to fill the table by putting numbers in it obeying following restrictions.
Each cell is able to store a number that can be calculated according to its index. (Indexes start from [0,0])
If a cell's index is [i, j] it can store a number calculated as one of the following ways.
i x j + i - j
i x j + j - i
i ^ j
j ^ i
Each cell can carry at most multiplication of its adjacent cells. (horizonal and vertical)
Each cell should be colored in a way all odds are of same color and all evens are of same color and two same colored cell cannot stand side by side vertically or horizontally.
calculation-puzzle
asked Mar 29 at 9:54
DavisDavis
61
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Suppose you have a 2D array - a list that exits in both x and y axis - which is 4 x 4, and you need to fill the table by putting numbers in it obeying following restrictions.
Each cell is able to store a number that can be calculated according to its index. (Indexes start from [0,0])
If a cell's index is [i, j] it can store a number calculated as one of the following ways.
i x j + i - j
i x j + j - i
i ^ j
j ^ i
Each cell can carry at most multiplication of its adjacent cells. (horizonal and vertical)
Each cell should be colored in a way all odds are of same color and all evens are of same color and two same colored cell cannot stand side by side vertically or horizontally.
calculation-puzzle
calculation-puzzle
asked Mar 29 at 9:54
DavisDavis
61
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 9:54
DavisDavis
61
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 9:54
DavisDavis
61
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 29 at 9:54
DavisDavis
61
asked Mar 29 at 9:54
DavisDavis
61
61
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
How do you define $0^0$?
$endgroup$
– hexomino
Mar 29 at 9:59
$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
$endgroup$
– Rubio♦
yesterday
add a comment |
2
$begingroup$
How do you define $0^0$?
$endgroup$
– hexomino
Mar 29 at 9:59
$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
$endgroup$
– Rubio♦
yesterday
2
2
$begingroup$
How do you define $0^0$?
$endgroup$
– hexomino
Mar 29 at 9:59
$begingroup$
How do you define $0^0$?
$endgroup$
– hexomino
Mar 29 at 9:59
$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
$endgroup$
– Rubio♦
yesterday
$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
$endgroup$
– Rubio♦
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It turns out that this is
Impossible
Reasoning
The cell with index $[1,1]$ must contain the entry $ 1times 1 + 1 - 1 = 1^1 = 1$.
Now consider the cell with index $[0,1]$ which is adjacent to $[1,1]$. This must contain either
(i) $0 times 1 + 0 - 1 = -1$,
(ii) $0 times 1 + 1 - 0 = 1$
(iii) $0^1 = 0$
(iv) $1^0 = 1$
Now, (i), (ii) and (iv) are odd which would break rule 3.
However if we are in case (iii) then we break rule 2, since the cell with index $[1,1]$ can carry, at most, multiplication of its adjacent cells, which would be $0$ in this case.
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
$endgroup$
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
add a comment |
$begingroup$
My "solution" (a lateral thinking answer)
__0 NaN 1 NaNNaN 1 NaN 1__1 NaN 4 NaNNaN 1 NaN 9
Explanation
NaNmeaning not-a-number (according to the rules, a cell does not has to contain a number, but rather can do so). All other rules are obviously complied (if a cell contains a number, it is of the 4 permitted ones; NaNs (and their products) are never greater than any number; you should color all "odd" cells blue, "evens" green and NaNs grey, since the latter are neither even nor odd).
answered Mar 29 at 10:40
trolley813trolley813
1,14638
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
It turns out that this is
Impossible
Reasoning
The cell with index $[1,1]$ must contain the entry $ 1times 1 + 1 - 1 = 1^1 = 1$.
Now consider the cell with index $[0,1]$ which is adjacent to $[1,1]$. This must contain either
(i) $0 times 1 + 0 - 1 = -1$,
(ii) $0 times 1 + 1 - 0 = 1$
(iii) $0^1 = 0$
(iv) $1^0 = 1$
Now, (i), (ii) and (iv) are odd which would break rule 3.
However if we are in case (iii) then we break rule 2, since the cell with index $[1,1]$ can carry, at most, multiplication of its adjacent cells, which would be $0$ in this case.
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
$endgroup$
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
add a comment |
$begingroup$
It turns out that this is
Impossible
Reasoning
The cell with index $[1,1]$ must contain the entry $ 1times 1 + 1 - 1 = 1^1 = 1$.
Now consider the cell with index $[0,1]$ which is adjacent to $[1,1]$. This must contain either
(i) $0 times 1 + 0 - 1 = -1$,
(ii) $0 times 1 + 1 - 0 = 1$
(iii) $0^1 = 0$
(iv) $1^0 = 1$
Now, (i), (ii) and (iv) are odd which would break rule 3.
However if we are in case (iii) then we break rule 2, since the cell with index $[1,1]$ can carry, at most, multiplication of its adjacent cells, which would be $0$ in this case.
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
$endgroup$
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
add a comment |
$begingroup$
It turns out that this is
Impossible
Reasoning
The cell with index $[1,1]$ must contain the entry $ 1times 1 + 1 - 1 = 1^1 = 1$.
Now consider the cell with index $[0,1]$ which is adjacent to $[1,1]$. This must contain either
(i) $0 times 1 + 0 - 1 = -1$,
(ii) $0 times 1 + 1 - 0 = 1$
(iii) $0^1 = 0$
(iv) $1^0 = 1$
Now, (i), (ii) and (iv) are odd which would break rule 3.
However if we are in case (iii) then we break rule 2, since the cell with index $[1,1]$ can carry, at most, multiplication of its adjacent cells, which would be $0$ in this case.
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
$endgroup$
It turns out that this is
Impossible
Reasoning
The cell with index $[1,1]$ must contain the entry $ 1times 1 + 1 - 1 = 1^1 = 1$.
Now consider the cell with index $[0,1]$ which is adjacent to $[1,1]$. This must contain either
(i) $0 times 1 + 0 - 1 = -1$,
(ii) $0 times 1 + 1 - 0 = 1$
(iii) $0^1 = 0$
(iv) $1^0 = 1$
Now, (i), (ii) and (iv) are odd which would break rule 3.
However if we are in case (iii) then we break rule 2, since the cell with index $[1,1]$ can carry, at most, multiplication of its adjacent cells, which would be $0$ in this case.
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
answered Mar 29 at 10:11
hexominohexomino
45.4k4139219
45.4k4139219
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
add a comment |
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
Actually it may be possible due to the fact that (according to the rules) rot13(rnpu pryy VF NOYR (be PNA) pbagnva n ahzore, ohg abg arprffnevyl FUBHYQ gb qb fb. Fbzr pryyf znl pbagnva ab ahzoref ng nyy (naq fb, abg orvat terngre guna nal bgure ahzoref, yvxr AnA va cebtenzzvat))
$endgroup$
– trolley813
Mar 29 at 10:20
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
@trolley813 That's quite an interesting take. Wouldn't it contravene the sentence "...you need to fill the table by putting numbers in it obeying following restrictions." in the opening paragraph?
$endgroup$
– hexomino
Mar 29 at 10:26
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
$begingroup$
maybe, but you can fill the table partially. I'm now writing my "solution".
$endgroup$
– trolley813
Mar 29 at 10:31
add a comment |
$begingroup$
My "solution" (a lateral thinking answer)
__0 NaN 1 NaNNaN 1 NaN 1__1 NaN 4 NaNNaN 1 NaN 9
Explanation
NaNmeaning not-a-number (according to the rules, a cell does not has to contain a number, but rather can do so). All other rules are obviously complied (if a cell contains a number, it is of the 4 permitted ones; NaNs (and their products) are never greater than any number; you should color all "odd" cells blue, "evens" green and NaNs grey, since the latter are neither even nor odd).
answered Mar 29 at 10:40
trolley813trolley813
1,14638
$endgroup$
add a comment |
$begingroup$
My "solution" (a lateral thinking answer)
__0 NaN 1 NaNNaN 1 NaN 1__1 NaN 4 NaNNaN 1 NaN 9
Explanation
NaNmeaning not-a-number (according to the rules, a cell does not has to contain a number, but rather can do so). All other rules are obviously complied (if a cell contains a number, it is of the 4 permitted ones; NaNs (and their products) are never greater than any number; you should color all "odd" cells blue, "evens" green and NaNs grey, since the latter are neither even nor odd).
answered Mar 29 at 10:40
trolley813trolley813
1,14638
$endgroup$
add a comment |
$begingroup$
My "solution" (a lateral thinking answer)
__0 NaN 1 NaNNaN 1 NaN 1__1 NaN 4 NaNNaN 1 NaN 9
Explanation
NaNmeaning not-a-number (according to the rules, a cell does not has to contain a number, but rather can do so). All other rules are obviously complied (if a cell contains a number, it is of the 4 permitted ones; NaNs (and their products) are never greater than any number; you should color all "odd" cells blue, "evens" green and NaNs grey, since the latter are neither even nor odd).
answered Mar 29 at 10:40
trolley813trolley813
1,14638
$endgroup$
My "solution" (a lateral thinking answer)
__0 NaN 1 NaNNaN 1 NaN 1__1 NaN 4 NaNNaN 1 NaN 9
Explanation
NaNmeaning not-a-number (according to the rules, a cell does not has to contain a number, but rather can do so). All other rules are obviously complied (if a cell contains a number, it is of the 4 permitted ones; NaNs (and their products) are never greater than any number; you should color all "odd" cells blue, "evens" green and NaNs grey, since the latter are neither even nor odd).
answered Mar 29 at 10:40
trolley813trolley813
1,14638
answered Mar 29 at 10:40
trolley813trolley813
1,14638
answered Mar 29 at 10:40
trolley813trolley813
1,14638
answered Mar 29 at 10:40
trolley813trolley813
1,14638
1,14638
add a comment |
add a comment |
Davis is a new contributor. Be nice, and check out our Code of Conduct.
Davis is a new contributor. Be nice, and check out our Code of Conduct.
Davis is a new contributor. Be nice, and check out our Code of Conduct.
Davis is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
How do you define $0^0$?
$endgroup$
– hexomino
Mar 29 at 9:59
$begingroup$
Has a correct answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
$endgroup$
– Rubio♦
yesterday