Represent number N (N>1) as N = x + y, where x * y is the maximum possible value
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My problem is the following: I have to find such x and y (x+y=N) where x * y is the maximum possible value.
I have written an algorithm (which I think works well) and I am looking for more efficient and faster way to solve this problem.
This is my algorithm:
static void Find(int N)
int multiplication = 1;
int maxMultiplication = 1;
int desiredNumber = 1;
for (int i = 1; i <= N / 2; i++)
multiplication = i * (N - i);
if(multiplication > maxMultiplication)
maxMultiplication = multiplication;
desiredNumber = i;
Console.WriteLine("Desired presentation of number N is --> N = 0 + 1",
desiredNumber, N - desiredNumber);
c# algorithm
asked Jun 5 at 14:39
Beginner
896
add a comment |Â
up vote
0
down vote
favorite
My problem is the following: I have to find such x and y (x+y=N) where x * y is the maximum possible value.
I have written an algorithm (which I think works well) and I am looking for more efficient and faster way to solve this problem.
This is my algorithm:
static void Find(int N)
int multiplication = 1;
int maxMultiplication = 1;
int desiredNumber = 1;
for (int i = 1; i <= N / 2; i++)
multiplication = i * (N - i);
if(multiplication > maxMultiplication)
maxMultiplication = multiplication;
desiredNumber = i;
Console.WriteLine("Desired presentation of number N is --> N = 0 + 1",
desiredNumber, N - desiredNumber);
c# algorithm
asked Jun 5 at 14:39
Beginner
896
Do you know any differential calculus?
– Peter Taylor
Jun 5 at 15:49
I have done some courses. I think I will be able to understand the explanation...
– Beginner
Jun 5 at 15:51
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My problem is the following: I have to find such x and y (x+y=N) where x * y is the maximum possible value.
I have written an algorithm (which I think works well) and I am looking for more efficient and faster way to solve this problem.
This is my algorithm:
static void Find(int N)
int multiplication = 1;
int maxMultiplication = 1;
int desiredNumber = 1;
for (int i = 1; i <= N / 2; i++)
multiplication = i * (N - i);
if(multiplication > maxMultiplication)
maxMultiplication = multiplication;
desiredNumber = i;
Console.WriteLine("Desired presentation of number N is --> N = 0 + 1",
desiredNumber, N - desiredNumber);
c# algorithm
asked Jun 5 at 14:39
Beginner
896
My problem is the following: I have to find such x and y (x+y=N) where x * y is the maximum possible value.
I have written an algorithm (which I think works well) and I am looking for more efficient and faster way to solve this problem.
This is my algorithm:
static void Find(int N)
int multiplication = 1;
int maxMultiplication = 1;
int desiredNumber = 1;
for (int i = 1; i <= N / 2; i++)
multiplication = i * (N - i);
if(multiplication > maxMultiplication)
maxMultiplication = multiplication;
desiredNumber = i;
Console.WriteLine("Desired presentation of number N is --> N = 0 + 1",
desiredNumber, N - desiredNumber);
c# algorithm
asked Jun 5 at 14:39
Beginner
896
asked Jun 5 at 14:39
Beginner
896
asked Jun 5 at 14:39
Beginner
896
asked Jun 5 at 14:39
Beginner
896
896
Do you know any differential calculus?
– Peter Taylor
Jun 5 at 15:49
I have done some courses. I think I will be able to understand the explanation...
– Beginner
Jun 5 at 15:51
add a comment |Â
Do you know any differential calculus?
– Peter Taylor
Jun 5 at 15:49
I have done some courses. I think I will be able to understand the explanation...
– Beginner
Jun 5 at 15:51
Do you know any differential calculus?
– Peter Taylor
Jun 5 at 15:49
Do you know any differential calculus?
– Peter Taylor
Jun 5 at 15:49
I have done some courses. I think I will be able to understand the explanation...
– Beginner
Jun 5 at 15:51
I have done some courses. I think I will be able to understand the explanation...
– Beginner
Jun 5 at 15:51
add a comment |Â
1 Answer
1
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up vote
4
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accepted
$$fractextdtextdx x(N-x) = N - 2x$$
At a maximum, $fractextdftextdx = 0$, so the maximum occurs when $x = fracN2$. (Left as an exercise: show that it's a maximum and not a minimum).
Therefore you can skip the loop: desiredNumber = N / 2; is all you need. Note that if N is odd this gives the pair $leftlfloor fracN2 rightrfloor, leftlceil fracN2 rightrceil$, which is the correct solution in integers.
answered Jun 5 at 16:01
Peter Taylor
14k2454
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
@paparazzo, what'ss? And if that should say2, where are we in disagreement?
– Peter Taylor
Jun 5 at 16:41
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
And where are we in disagreement?desiredNumberwill be assigned12.
– Peter Taylor
Jun 5 at 16:50
1
@paparazzo I missed the MathJax at first as well: the stated pair isfloor(N/2), ceil(N/2), which will give12, 13forN=25(and otherwise equivalent to your suggestion offloor(N/2), N-floor(N/2))
– VisualMelon
Jun 5 at 17:12
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$$fractextdtextdx x(N-x) = N - 2x$$
At a maximum, $fractextdftextdx = 0$, so the maximum occurs when $x = fracN2$. (Left as an exercise: show that it's a maximum and not a minimum).
Therefore you can skip the loop: desiredNumber = N / 2; is all you need. Note that if N is odd this gives the pair $leftlfloor fracN2 rightrfloor, leftlceil fracN2 rightrceil$, which is the correct solution in integers.
answered Jun 5 at 16:01
Peter Taylor
14k2454
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
@paparazzo, what'ss? And if that should say2, where are we in disagreement?
– Peter Taylor
Jun 5 at 16:41
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
And where are we in disagreement?desiredNumberwill be assigned12.
– Peter Taylor
Jun 5 at 16:50
1
@paparazzo I missed the MathJax at first as well: the stated pair isfloor(N/2), ceil(N/2), which will give12, 13forN=25(and otherwise equivalent to your suggestion offloor(N/2), N-floor(N/2))
– VisualMelon
Jun 5 at 17:12
 |Â
show 1 more comment
up vote
4
down vote
accepted
$$fractextdtextdx x(N-x) = N - 2x$$
At a maximum, $fractextdftextdx = 0$, so the maximum occurs when $x = fracN2$. (Left as an exercise: show that it's a maximum and not a minimum).
Therefore you can skip the loop: desiredNumber = N / 2; is all you need. Note that if N is odd this gives the pair $leftlfloor fracN2 rightrfloor, leftlceil fracN2 rightrceil$, which is the correct solution in integers.
answered Jun 5 at 16:01
Peter Taylor
14k2454
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
@paparazzo, what'ss? And if that should say2, where are we in disagreement?
– Peter Taylor
Jun 5 at 16:41
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
And where are we in disagreement?desiredNumberwill be assigned12.
– Peter Taylor
Jun 5 at 16:50
1
@paparazzo I missed the MathJax at first as well: the stated pair isfloor(N/2), ceil(N/2), which will give12, 13forN=25(and otherwise equivalent to your suggestion offloor(N/2), N-floor(N/2))
– VisualMelon
Jun 5 at 17:12
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$$fractextdtextdx x(N-x) = N - 2x$$
At a maximum, $fractextdftextdx = 0$, so the maximum occurs when $x = fracN2$. (Left as an exercise: show that it's a maximum and not a minimum).
Therefore you can skip the loop: desiredNumber = N / 2; is all you need. Note that if N is odd this gives the pair $leftlfloor fracN2 rightrfloor, leftlceil fracN2 rightrceil$, which is the correct solution in integers.
answered Jun 5 at 16:01
Peter Taylor
14k2454
$$fractextdtextdx x(N-x) = N - 2x$$
At a maximum, $fractextdftextdx = 0$, so the maximum occurs when $x = fracN2$. (Left as an exercise: show that it's a maximum and not a minimum).
Therefore you can skip the loop: desiredNumber = N / 2; is all you need. Note that if N is odd this gives the pair $leftlfloor fracN2 rightrfloor, leftlceil fracN2 rightrceil$, which is the correct solution in integers.
answered Jun 5 at 16:01
Peter Taylor
14k2454
answered Jun 5 at 16:01
Peter Taylor
14k2454
answered Jun 5 at 16:01
Peter Taylor
14k2454
answered Jun 5 at 16:01
Peter Taylor
14k2454
14k2454
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
@paparazzo, what'ss? And if that should say2, where are we in disagreement?
– Peter Taylor
Jun 5 at 16:41
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
And where are we in disagreement?desiredNumberwill be assigned12.
– Peter Taylor
Jun 5 at 16:50
1
@paparazzo I missed the MathJax at first as well: the stated pair isfloor(N/2), ceil(N/2), which will give12, 13forN=25(and otherwise equivalent to your suggestion offloor(N/2), N-floor(N/2))
– VisualMelon
Jun 5 at 17:12
 |Â
show 1 more comment
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
@paparazzo, what'ss? And if that should say2, where are we in disagreement?
– Peter Taylor
Jun 5 at 16:41
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
And where are we in disagreement?desiredNumberwill be assigned12.
– Peter Taylor
Jun 5 at 16:50
1
@paparazzo I missed the MathJax at first as well: the stated pair isfloor(N/2), ceil(N/2), which will give12, 13forN=25(and otherwise equivalent to your suggestion offloor(N/2), N-floor(N/2))
– VisualMelon
Jun 5 at 17:12
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
If N is odd then n/2 + n/s != n. A better generalization is n/2, n - n/s. OP is using integer math.
– paparazzo
Jun 5 at 16:37
@paparazzo, what's
s? And if that should say 2, where are we in disagreement?– Peter Taylor
Jun 5 at 16:41
@paparazzo, what's
s? And if that should say 2, where are we in disagreement?– Peter Taylor
Jun 5 at 16:41
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
Yes 2. In integer math 25/2 = 12 so I think need 12, 13
– paparazzo
Jun 5 at 16:43
And where are we in disagreement?
desiredNumber will be assigned 12.– Peter Taylor
Jun 5 at 16:50
And where are we in disagreement?
desiredNumber will be assigned 12.– Peter Taylor
Jun 5 at 16:50
1
1
@paparazzo I missed the MathJax at first as well: the stated pair is
floor(N/2), ceil(N/2), which will give 12, 13 for N=25 (and otherwise equivalent to your suggestion of floor(N/2), N-floor(N/2))– VisualMelon
Jun 5 at 17:12
@paparazzo I missed the MathJax at first as well: the stated pair is
floor(N/2), ceil(N/2), which will give 12, 13 for N=25 (and otherwise equivalent to your suggestion of floor(N/2), N-floor(N/2))– VisualMelon
Jun 5 at 17:12
 |Â
show 1 more comment
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Do you know any differential calculus?
– Peter Taylor
Jun 5 at 15:49
I have done some courses. I think I will be able to understand the explanation...
– Beginner
Jun 5 at 15:51