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May 10th, 2013, 01:16 PM   #1
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If A and W are two given constants and A= x+(y*W) then how c

I'm not sure if this is the right place to ask this question but forgive, this problem has been bugging me for the last few days, so I thought maybe I should ask for some guidance.

If A and W are two given constants and A= x+(y*W) then how can I find x and y?


If A and W are two given constants and

A = x + (y * W)

then we know that:

y = (A-x)/W

x = A-(y*W)

also

x and y can be any finite integer greater than 0

W is the maximum value of x plus 1

So for example if x was 4 then W will be 5

then how can I generate two formulas to find x and y?

lets assume A = 24 and W = 5

and lets assume a maximum value of 4 to both x and y

then we will have a table of values like this


x = 0 | x = 1 | x = 2 | x = 3 | x = 4
----------------------------------------------------------------------
y = 0 | A = 0 | A = 1 | A = 2 | A =3 | A = 4
----------------------------------------------------------------------
y = 1 | A = 5 | A =6 | A = 7 | A = 8 | A = 9
----------------------------------------------------------------------
y = 2 | A = 10 | A =11 | A = 12 | A = 13 | A = 14
----------------------------------------------------------------------
y = 3 | A = 15 | A =16 | A = 17 | A =18 | A = 19
----------------------------------------------------------------------
y = 4 | A = 20 | A =21 | A = 22 | A = 23 | A = 24

OK, x and y are two unknown variables. We have this table and we have A and W as data. You can see that also there is pattern in the table.
for example if you give the x value of 4 and y any value between 0 to 4 you will see that the A on the y row and x column ( which is 4 in this case) is always equal to A [x,y]+ 5 on the same column but previous raw, except for y = 0, which is obviously the starting raw.
What's more interesting that if you assume maximum value of x = 3, then W = 4 which is obviously the number of columns, then the pattern will change to A[x,y] on the next raw is equal to A[x,y] + 4 on the previous raw, but that same column.
There is also an obvious pattern on raws, but the pattern won't change if you increase or decrease the maximum value of y, and the A[x,y] on the next column is always equal to previous A [x,y]+ 1 but the same raw.

So there are patterns, I have A, and I have W, I need two different equations to calculate x and y which are unknown variables.

It is clear that A is a two dimensional Array in form of A[x,y] which holds a value of A[x,y] = x + (y * W). Clearly x and y are indexes that have some relations to patterns on the table depending on A and W's value, what are those relations and how can I use them to generate a formula to calculate x and y?
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May 10th, 2013, 01:23 PM   #2
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Re: If A and W are two given constants and A= x+(y*W) then h

For several integer values of A and W, there will be multiple pairs of values (x, y) that will satisfy the equation. Obviously, if W > A then y = 0 and x = A is the only solution in nonnegative integers. Are x and y allowed to be negative? If so, then there are infinitely many solutions.
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May 10th, 2013, 01:40 PM   #3
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Re: If A and W are two given constants and A= x+(y*W) then h

Quote:
Originally Posted by icemanfan
For several integer values of A and W, there will be multiple pairs of values (x, y) that will satisfy the equation. Obviously, if W > A then y = 0 and x = A is the only solution in nonnegative integers. Are x and y allowed to be negative? If so, then there are infinitely many solutions.
x , y >=0 and they are integers,

Yes when W >= A, x is always x = A, and y is always = 0, that is obvious, but obviously when W < A then x != A and y != 0,
also W also can't be 0 because A is a two dimensional array.

x and y are finite values depending on A's value, if you look at the table. you will see that for any given value of A there is an obviously a finite x and y value.
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May 10th, 2013, 02:02 PM   #4
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Re: If A and W are two given constants and A= x+(y*W) then h

if A >=0 and an integer number then:

Obviously A[x,y] = {(A +0), (A+1) , (A+2) ,..., (A+n)}

from observing the table It's clear that n = x * y

It is understandable that the pattern is like:

A+0 = x +(y*W)+0 , A+1 = x +(y*W)+1 , A + 2 = x +(y*W)+2 , .... , A+n = x + (y*W)+n
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May 10th, 2013, 02:33 PM   #5
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Re: If A and W are two given constants and A= x+(y*W) then h

it's difficult to create a table on a forum based on texts, I added a few more dashes to make the table a bit more readable:
-------|--x = 0---|--x = 1---|---x = 2--|--x = 3--|--x = 4
------------------------------------------------------------
y = 0--|--A = 0---|--A = 1---|---A = 2--|--A =3---|--A = 4
------------------------------------------------------------
y = 1--|--A = 5---|--A =6---|---A = 7--|--A = 8--|--A = 9
------------------------------------------------------------
y = 2--|--A = 10--|--A =11--|--A = 12--|--A = 13--|--A = 14
------------------------------------------------------------
y = 3--|--A = 15--|--A =16--|--A = 17--|--A =18--|--A = 19
------------------------------------------------------------
y = 4 --|--A = 20--|--A =21--|--A = 22--|--A = 23--|--A = 24
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May 10th, 2013, 05:29 PM   #6
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Re: If A and W are two given constants and A= x+(y*W) then h

assuming y = [0,4] and x =[0,4] then:
We know from our data that W = max(x) +1 ---> W = 5
We can observe from the table that the following pattern is true:

for A = 24 ---> x = int( A / W ) + 0 = 4
for A = 23 ---> x = int( A / W ) - 1 = 3
for A = 22 ---> x = int( A / W ) - 2 = 2
for A = 21 ---> x = int( A / W ) - 3 = 1
for A = 20 ---> x = int( A / W ) - 4 = 0

for A = 19 ---> x = int( A / W ) + 1 = 4
for A = 18 ---> x = int( A / W ) + 0 = 3
for A = 17 ---> x = int( A / W ) - 1 = 2
for A = 16 ---> x = int( A / W ) - 2 = 1
for A = 15 ---> x = int( A / W ) - 3 = 0

for A = 14 ---> x = int( A / W ) + 2 = 4
for A = 13 ---> x = int( A / W ) + 1 = 3
for A = 12 ---> x = int( A / W ) + 0 = 2
for A = 11 ---> x = int( A / W ) - 1 = 1
for A = 10 ---> x = int( A / W ) - 2 = 0

for A = 9 ---> x = int( A / W ) + 3 = 4
for A = 8 ---> x = int( A / W ) + 2 = 3
for A = 7 ---> x = int( A / W ) + 1 = 2
for A = 6 ---> x = int( A / W ) + 0 = 1
for A = 5 ---> x = int( A / W ) - 1 = 0

for A = 4 ---> x = int( A / W ) + 4 = 4
for A = 3 ---> x = int( A / W ) + 3 = 3
for A = 2 ---> x = int( A / W ) + 2 = 2
for A = 1 ---> x = int( A / W ) + 1 = 1
for A = 0 ---> x = int( A / W ) + 0 = 0

now I need to rap this up and make a formula in order to find x without having to look in A's value at the table to determine x and/or y.
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May 11th, 2013, 09:42 AM   #7
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Re: If A and W are two given constants and A= x+(y*W) then h

Studying the problem more closely it becomes clear that the following information are useful:

x , y >= 0 and are a positive integers.

W , H > 0 and are a positive integers.

A[x,y] >= 0 and is a positive integer.

A[x,y] = x - ( y * W )

W = max ( x ) + 1 and it represents the number of columns in the A[x,y] array
H = max ( y ) + 1 and it represents the number of raws in the A[x,y] array

A[x,y+1] = A[x,y] + W , A[x,y+2] = A [x,y+1] + W , A[x,y+3] = A[x,y+2] + W , A[x,y+n] = A[x,y+n-1] + W

Clearly W plays an important role here, so in order to find y the following is always true regardless of W or H's value:

y = int ( A[x,y] /w )

Note: The value of A[x,y] and W are given as problem's data.

the rest is just calculations to find x:

x = A[x,y] - ( y * W)
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