[code]

#define close_enough 0.01

main()

{

char facts[] = {"fat (grams)", "calcium (%)", "carbs (grams)", "vitamin D (%)", "cholesterol (%)" };

double eggs[] = {4, 0, 0, 1, 4};

double creamcheese[] = {9, 3, 4, 0, 0};

double sugar[] = {0, 0, 15, 0, 0};

int numbers_should_equal_this[] = { 15, 43, 4, 7 ,3};

int looper[3];

for (looper[0] = 0 ; looper[0] < 100 ; looper[0]++)

for (looper[1] = 0; looper[1] < 100 - looper[0]; looper[1]++)

for (looper[2] = 0; looper[2] < 100 - (looper[0] + looper[1]) ; looper[2]++)

if (eggs[0] * (looper[0] / 100.0) < close_enough)

if (creamcheese[0] * (looper[1] / 100.0) < close_enough)

if (sugar[0] * (looper[1] / 100.0) < close_enough)

[/code]

well you get the idea. Anyway I was wondering if you guys have any ideas on how to write a program like this (I got writers block) and also is there was a mathematical way to solve for an equation like this. Thanks alot guys

PS seems a shame to erase the above but I would have probably done this in real life:

[code]

#define close_enough 0.01

main()

{

char facts[] = {"fat (grams)", "calcium (%)", "carbs (grams)", "vitamin D (%)", "cholesterol (%)" };

double ingred[3][5] =

{

{ 0, 1, 2, 4, 3 },

{ 3, 4, 5, 0, 3 },

{ 0, 6, 0, 7, 8 }

};

int numbers_should_equal_this[] = { 15, 43, 4, 7 ,3};

int looper[3];

for (looper[0] = 0 ; looper[0] < 100 ; looper[0]++)

for (looper[1] = 0; looper[1] < 100 - looper[0]; looper[1]++)

for (looper[2] = 0; looper[2] < 100 - (looper[0] + looper[1]) ; looper[2]++)

if (ingred[0][0] * (looper[0] / 100.0) < close_enough)

if (inged[1][0] * (looper[1] / 100.0) < close_enough)

if (sugar[2][0] * (looper[1] / 100.0) < close_enough)

[/code]

you see what I mean

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## Comments

: ideas on how to write a program like this (I got writers block) and

: also is there was a mathematical way to solve for an equation like

: this. Thanks alot guys

:

Mathematically, say n ingredients, C being the final amount of calories, and c[leftbr]i[rightbr] being the amount of calories per unit of ingredient i (0 <= i < n).

We take x[leftbr]i[rightbr] to be the amount of ingredient i.

Then we have the n-dimensional linear equation:

[code] c[0] x[0] + c[1] x[1] + ... + c[n-1] x[n-1] = C [/code]

If we have only 2 ingredients, n=2 and this reduces to a well-known case:

[code]

//n = 2

c[0] x[0] + c[1] x[1] = C

//or more regularly written as:

a x + b y = c <-> y = -a/b x + c/b = p x + q

[/code]

In this case we already see that rather than a unique x and y (x[leftbr]0[rightbr] and x[leftbr]1[rightbr]) we get a straight line of possibilities.

In the general n>1 case we get the same result. The best we can do is to get a linear equation of n-1 variables. So there is no 'unique' solution.

I have forgot to mention here though that most of the solutions you'll get give a negative value for at least one of the ingredients.

One technique I can think of to find a ratio of ingredients is this:

Start at a certain ratio (I'll call it a point) with total calories added up smaller than C.

Then find an accompanying point with total calories added up larger than C

Next you take a point in between (for instance a ratio that is in the middle of the two other ratio's). Depending on whether or not the value of this point/ratio is greater than C, you take the Start point and this new point (new value > C), or this new point and the end point (new value < C).

Then repeat this process, until you are close enough to C as you want.

Mathematics guarantee that you approximate a point with value C.

There are some details about this approach that I haven't thought through fully yet So I'm afraid that in it's current form it won't work.

Maybe you can think of other, better methods? It's always fun to think about such challenges.

Best Regards,

Richard

The way I see it... Well, it's all pretty blurry

: : ideas on how to write a program like this (I got writers block) and

: : also is there was a mathematical way to solve for an equation like

: : this. Thanks alot guys

: :

:

: Mathematically, say n ingredients, C being the final amount of

: calories, and c[leftbr]i[rightbr] being the amount of calories per

: unit of ingredient i (0 <= i < n).

: We take x[leftbr]i[rightbr] to be the amount of ingredient i.

: Then we have the n-dimensional linear equation:

: [code]: c[0] x[0] + c[1] x[1] + ... + c[n-1] x[n-1] = C [/code]:

: If we have only 2 ingredients, n=2 and this reduces to a well-known

: case:

: [code]:

: //n = 2

: c[0] x[0] + c[1] x[1] = C

: //or more regularly written as:

: a x + b y = c <-> y = -a/b x + c/b = p x + q

: [/code]:

: In this case we already see that rather than a unique x and y

: (x[leftbr]0[rightbr] and x[leftbr]1[rightbr]) we get a straight line

: of possibilities.

:

: In the general n>1 case we get the same result. The best we can do

: is to get a linear equation of n-1 variables. So there is no

: 'unique' solution.

:

: I have forgot to mention here though that most of the solutions

: you'll get give a negative value for at least one of the ingredients.

:

: One technique I can think of to find a ratio of ingredients is this:

: Start at a certain ratio (I'll call it a point) with total calories

: added up smaller than C.

: Then find an accompanying point with total calories added up larger

: than C

: Next you take a point in between (for instance a ratio that is in

: the middle of the two other ratio's). Depending on whether or not

: the value of this point/ratio is greater than C, you take the Start

: point and this new point (new value > C), or this new point and the

: end point (new value < C).

: Then repeat this process, until you are close enough to C as you

: want.

: Mathematics guarantee that you approximate a point with value C.

:

: There are some details about this approach that I haven't thought

: through fully yet So I'm afraid that in it's current form it

: won't work.

:

: Maybe you can think of other, better methods? It's always fun to

: think about such challenges.

: Best Regards,

: Richard

:

: The way I see it... Well, it's all pretty blurry

can you explain a little more? I got lost. And can you give an example of using this system? Thank you