Given A,B and n find the minimum value of Ax-By where x+y = n
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Given $A$, $B$, $n$ we're interested in non negative values $Ax-By$, where $A,B,n$ $in$ $mathbb Z$, n $le$ $10^5$, $A+B=n$
and the value of the equation is minimized in case of multiple possible values of $x,y$
My idea is nothing but plain bruteforce, checking each values from $i$ to $n$ and $n-i$,thus finding minimum value.
For example:
$A=5,B=6$
$6*5 - 4*6 = 6$
Need some efficient ideas. Thanks in advance.
number-theory elementary-number-theory
asked Jan 9 at 4:38
some usersome user
32
$endgroup$
add a comment |
$begingroup$
Given $A$, $B$, $n$ we're interested in non negative values $Ax-By$, where $A,B,n$ $in$ $mathbb Z$, n $le$ $10^5$, $A+B=n$
and the value of the equation is minimized in case of multiple possible values of $x,y$
My idea is nothing but plain bruteforce, checking each values from $i$ to $n$ and $n-i$,thus finding minimum value.
For example:
$A=5,B=6$
$6*5 - 4*6 = 6$
Need some efficient ideas. Thanks in advance.
number-theory elementary-number-theory
asked Jan 9 at 4:38
some usersome user
32
$endgroup$
$begingroup$
are you assuming $A,B>0$? are $x,y in mathbb{R}$ or are they integers?
$endgroup$
– gt6989b
Jan 9 at 4:59
$begingroup$
yes A,B>0 and x,y are integers as well.
$endgroup$
– some user
Jan 9 at 5:06
$begingroup$
the actual value should be minimized
$endgroup$
– some user
Jan 9 at 10:47
$begingroup$
Sorry, I overlooked that you clarified that $Ax-By$ should be non-negative, so it boils down to minimize $|Ax-By|$
$endgroup$
– Peter
Jan 9 at 11:54
$begingroup$
Did you try to minimize the expression $$|A(n-y)+By|=|An-(A-B)y|$$ which we get by inserting $x=n-y$ ?
$endgroup$
– Peter
Jan 9 at 11:57
add a comment |
$begingroup$
Given $A$, $B$, $n$ we're interested in non negative values $Ax-By$, where $A,B,n$ $in$ $mathbb Z$, n $le$ $10^5$, $A+B=n$
and the value of the equation is minimized in case of multiple possible values of $x,y$
My idea is nothing but plain bruteforce, checking each values from $i$ to $n$ and $n-i$,thus finding minimum value.
For example:
$A=5,B=6$
$6*5 - 4*6 = 6$
Need some efficient ideas. Thanks in advance.
number-theory elementary-number-theory
asked Jan 9 at 4:38
some usersome user
32
$endgroup$
Given $A$, $B$, $n$ we're interested in non negative values $Ax-By$, where $A,B,n$ $in$ $mathbb Z$, n $le$ $10^5$, $A+B=n$
and the value of the equation is minimized in case of multiple possible values of $x,y$
My idea is nothing but plain bruteforce, checking each values from $i$ to $n$ and $n-i$,thus finding minimum value.
For example:
$A=5,B=6$
$6*5 - 4*6 = 6$
Need some efficient ideas. Thanks in advance.
number-theory elementary-number-theory
number-theory elementary-number-theory
asked Jan 9 at 4:38
some usersome user
32
asked Jan 9 at 4:38
some usersome user
32
edited Jan 9 at 11:19
some user
asked Jan 9 at 4:38
some usersome user
32
asked Jan 9 at 4:38
some usersome user
32
asked Jan 9 at 4:38
some usersome user
32
32
$begingroup$
are you assuming $A,B>0$? are $x,y in mathbb{R}$ or are they integers?
$endgroup$
– gt6989b
Jan 9 at 4:59
$begingroup$
yes A,B>0 and x,y are integers as well.
$endgroup$
– some user
Jan 9 at 5:06
$begingroup$
the actual value should be minimized
$endgroup$
– some user
Jan 9 at 10:47
$begingroup$
Sorry, I overlooked that you clarified that $Ax-By$ should be non-negative, so it boils down to minimize $|Ax-By|$
$endgroup$
– Peter
Jan 9 at 11:54
$begingroup$
Did you try to minimize the expression $$|A(n-y)+By|=|An-(A-B)y|$$ which we get by inserting $x=n-y$ ?
$endgroup$
– Peter
Jan 9 at 11:57
add a comment |
$begingroup$
are you assuming $A,B>0$? are $x,y in mathbb{R}$ or are they integers?
$endgroup$
– gt6989b
Jan 9 at 4:59
$begingroup$
yes A,B>0 and x,y are integers as well.
$endgroup$
– some user
Jan 9 at 5:06
$begingroup$
the actual value should be minimized
$endgroup$
– some user
Jan 9 at 10:47
$begingroup$
Sorry, I overlooked that you clarified that $Ax-By$ should be non-negative, so it boils down to minimize $|Ax-By|$
$endgroup$
– Peter
Jan 9 at 11:54
$begingroup$
Did you try to minimize the expression $$|A(n-y)+By|=|An-(A-B)y|$$ which we get by inserting $x=n-y$ ?
$endgroup$
– Peter
Jan 9 at 11:57
$begingroup$
are you assuming $A,B>0$? are $x,y in mathbb{R}$ or are they integers?
$endgroup$
– gt6989b
Jan 9 at 4:59
$begingroup$
are you assuming $A,B>0$? are $x,y in mathbb{R}$ or are they integers?
$endgroup$
– gt6989b
Jan 9 at 4:59
$begingroup$
yes A,B>0 and x,y are integers as well.
$endgroup$
– some user
Jan 9 at 5:06
$begingroup$
yes A,B>0 and x,y are integers as well.
$endgroup$
– some user
Jan 9 at 5:06
$begingroup$
the actual value should be minimized
$endgroup$
– some user
Jan 9 at 10:47
$begingroup$
the actual value should be minimized
$endgroup$
– some user
Jan 9 at 10:47
$begingroup$
Sorry, I overlooked that you clarified that $Ax-By$ should be non-negative, so it boils down to minimize $|Ax-By|$
$endgroup$
– Peter
Jan 9 at 11:54
$begingroup$
Sorry, I overlooked that you clarified that $Ax-By$ should be non-negative, so it boils down to minimize $|Ax-By|$
$endgroup$
– Peter
Jan 9 at 11:54
$begingroup$
Did you try to minimize the expression $$|A(n-y)+By|=|An-(A-B)y|$$ which we get by inserting $x=n-y$ ?
$endgroup$
– Peter
Jan 9 at 11:57
$begingroup$
Did you try to minimize the expression $$|A(n-y)+By|=|An-(A-B)y|$$ which we get by inserting $x=n-y$ ?
$endgroup$
– Peter
Jan 9 at 11:57
add a comment |
1 Answer
1
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$begingroup$
$y=n-x$ and you minimize $$Ax-B(n-x)=(A+B)x-Bn.$$
All values taken by this expression differ in multiples of $A+B$, and the smallest positive value is
$$(-Bn)bmod(A+B),$$ which is in $[0,A+B)$.
UPDATE:
The conditions in the title and in the body define completely different problems !
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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oldest
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$begingroup$
$y=n-x$ and you minimize $$Ax-B(n-x)=(A+B)x-Bn.$$
All values taken by this expression differ in multiples of $A+B$, and the smallest positive value is
$$(-Bn)bmod(A+B),$$ which is in $[0,A+B)$.
UPDATE:
The conditions in the title and in the body define completely different problems !
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
$y=n-x$ and you minimize $$Ax-B(n-x)=(A+B)x-Bn.$$
All values taken by this expression differ in multiples of $A+B$, and the smallest positive value is
$$(-Bn)bmod(A+B),$$ which is in $[0,A+B)$.
UPDATE:
The conditions in the title and in the body define completely different problems !
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
$y=n-x$ and you minimize $$Ax-B(n-x)=(A+B)x-Bn.$$
All values taken by this expression differ in multiples of $A+B$, and the smallest positive value is
$$(-Bn)bmod(A+B),$$ which is in $[0,A+B)$.
UPDATE:
The conditions in the title and in the body define completely different problems !
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
$endgroup$
$y=n-x$ and you minimize $$Ax-B(n-x)=(A+B)x-Bn.$$
All values taken by this expression differ in multiples of $A+B$, and the smallest positive value is
$$(-Bn)bmod(A+B),$$ which is in $[0,A+B)$.
UPDATE:
The conditions in the title and in the body define completely different problems !
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
edited Jan 9 at 12:49
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
answered Jan 9 at 12:41
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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$begingroup$
are you assuming $A,B>0$? are $x,y in mathbb{R}$ or are they integers?
$endgroup$
– gt6989b
Jan 9 at 4:59
$begingroup$
yes A,B>0 and x,y are integers as well.
$endgroup$
– some user
Jan 9 at 5:06
$begingroup$
the actual value should be minimized
$endgroup$
– some user
Jan 9 at 10:47
$begingroup$
Sorry, I overlooked that you clarified that $Ax-By$ should be non-negative, so it boils down to minimize $|Ax-By|$
$endgroup$
– Peter
Jan 9 at 11:54
$begingroup$
Did you try to minimize the expression $$|A(n-y)+By|=|An-(A-B)y|$$ which we get by inserting $x=n-y$ ?
$endgroup$
– Peter
Jan 9 at 11:57