inverse method / formula for this?

weiss

Good day

i have this: x^1 + x^2 + x^3 + x^4

where x can be a decimal value of between 1 and 10

if i have a number: 2342 - what is the value of each x?

is there a multiple root finding algorithm for this?

Thomas_S_Hunterson

You could solve (for x) the equation:
x^4 + x^3 + x^2 + x^1 = 2342

What is it for? There are algebraic techniques or you could just use interpolation/estimation (hint, newton-raphson, (6<x<7), assuming it's just the first positive real root that you're looking for)

weiss

To tell the truth, i don't how to phrase the question which is why I asked a mod (before you replied) that the thread be removed.

i calculate total combinations for numbers 1 - 10, based on a set of numbers, lets say [2,3,4,2]

(10 ^ 0) * 2 = 2
(10 ^ 1) * 4 = 40
(10 ^ 2) * 3 = 300
(10 ^ 3) * 2 = 2000

total: 2 + 40 + 300 + 2000 = 2342

now i'd like to know an inverse method, where by i get [2,3,4,2] using 10 as the base.

so all i know is 10 and 2342, need to find what the roots are.

weiss

dr math probably better explains the calculation of total combinations.

http://mathforum.org/library/drmath/view/56114.html

weiss

should have added that the way i tried to do it was using division/modulus

2432 / (10 ^ 3) = 2
432 / (10 ^ 2) = 4
32 / (10 ^ 1) = 3
2 / (10 ^ 0) = 2

this is ok, but if i have a set of [10,1,10,1]

(10 ^ 0) * 10 = 10
(10 ^ 1) * 1 = 10
(10 ^ 2) * 10 = 1000
(10 ^ 3) * 1 = 1000

total: 10 + 10 + 1000 + 1000 = 2020

2020 / (10 ^ 3) = 2
20 / (10 ^ 2) = 0
20 / (10 ^ 1) = 2
0 / (10 ^ 0) = 0

[2,0,2,0]

i need the original set [10,1,10,1] if possible, must be a formula/method to do this?

Thomas_S_Hunterson

I don't think this is possible given the overflow in mod arithmetic.

What you want to do is somewhat akin to telling someone a number and asking them what two numbers you added to get that number. Obviously there are an infinitude of possibilities.

2020 could be given as [2,0,2,0], [10,1,10,1], [2020,0,0,0], [2010,1,0,0] etc

weiss

yeah..i'm not able to illustrate/describe it properly unfortunately.

possibly i need to look at multiple root finding algorithms or factoring algorithm..

Thomas_S_Hunterson

weiss wrote: »

yeah..i'm not able to illustrate/describe it properly unfortunately.

possibly i need to look at multiple root finding algorithms or factoring algorithm..

No I'm saying what you want to do is not possible.

As it stands, the function you are trying to apply to the vector is not injective and therefore no inverse exists.

weiss

hmm, no disrespect intended Sean_K but i don't believe its impossible, i've just failed to describe what it is i need to do in mathematical terms.

i see a way to use trial division.

if the result of a division is zero, this means the previous division was wrong.

As in previous example: 2020

In reverse order, start with 1000,decreasing by power of 10 for each element we want.

First result is 2

The second division results in zero, but during calculation of total number, its not possible to get result of zero..so it must be 1000 / (10 ^ 2) = 10

first is now 1, second is 10 ..continue for rest.

to be honest, i've not described what it is i want to do very well, apologies for that.

thank you for helping.

Thomas_S_Hunterson

Ok I think i see what you mean now; The set is comprised of integer values between 1 and 10? (sorry I assmued zero could be included).

This should be possible. Thinking about it quickly, subtract 1111 from the initial 4 digit integer (wrapping around from 0 to 9). Then split it out using the modulus/division method above and then add 1 to each of the numbers you end up with. (works in my head, i think)

/edit: sorry about the confusion, I tend to scan posts.

Thomas_S_Hunterson

Alternatively, working iteratively through the 4 digit number starting at the least significant digit:
Take it's value as the element of the set except if it's zero, in which case, the element of the set will be 10 and then subtract one from the next digit.

So on a computer you might repeatedly integer-divide the number by 10.

Take the remainder as your element of the set, except, as above of it is zero, in which case 10 goes in your set and 1 gets subtracted from the quotient.

GuanYin

I'm guessing you want to keep this now?

Sean K has obviously put time and effort into answering, it would be kinda rude to delete but it's up to you guys.

weiss

no delete, its ok now

mark renton

weiss wrote: »

should have added that the way i tried to do it was using division/modulus

2432 / (10 ^ 3) = 2
432 / (10 ^ 2) = 4
32 / (10 ^ 1) = 3
2 / (10 ^ 0) = 2

this is ok, but if i have a set of [10,1,10,1]

(10 ^ 0) * 10 = 10
(10 ^ 1) * 1 = 10
(10 ^ 2) * 10 = 1000
(10 ^ 3) * 1 = 1000

total: 10 + 10 + 1000 + 1000 = 2020

2020 / (10 ^ 3) = 2
20 / (10 ^ 2) = 0
20 / (10 ^ 1) = 2
0 / (10 ^ 0) = 0

[2,0,2,0]

i need the original set [10,1,10,1] if possible, must be a formula/method to do this?

your logic seems ok, but if your using base 10 then you should never have an integer greater than 9 in your set

weiss

what i should have seen so simple was to subtract the power from total before doing any modulus/division...:o

the result in code is roughly..

[PHP]
unsigned long long power = 10;

// get number of elements used to calculate value, must be atleast 1

for(elements = 1;value >= power;elements++) {
value -= power; // subtract
power *= 10; // multiply
}

for(int index = 0;index < elements;index++) {
num_set[index] = (value % 10); // modulus
value /= 10; // divide
}
[/PHP]

really simple thing.. thank you all for helping