Related
For example, if I want to find
1085912312763120759250776993188102125849391224162 = a^9+b^9+c^9+d
the code needs to brings
a=3456
b=78525
c=217423
d=215478
I do not need specific values, only that they comply with the fact that a, b and c have 6 digits at most and d is as small as possible.
Is there a quick way to find it?
I appreciate any help you can give me.
I have tried with nested loops but it is extremely slow and the code gets stuck.
Any help in VB or other code would be appreciated. I think the structure is more important than the language in this case
Imports System.Numerics
Public Class Form1
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim Value As BigInteger = BigInteger.Parse("1085912312763120759250776993188102125849391224162")
Dim powResult As BigInteger
Dim dResult As BigInteger
Dim a As Integer
Dim b As Integer
Dim c As Integer
Dim d As Integer
For i = 1 To 999999
For j = 1 To 999999
For k = 1 To 999999
powResult = BigInteger.Add(BigInteger.Add(BigInteger.Pow(i, 9), BigInteger.Pow(j, 9)), BigInteger.Pow(k, 9))
dResult = BigInteger.Subtract(Value, powResult)
If Len(dResult.ToString) <= 6 Then
a = i
b = j
c = k
d = dResult
RichTextBox1.Text = a & " , " & b & " , " & c & " , " & d
Exit For
Exit For
Exit For
End If
Next
Next
Next
End Sub
End Class
UPDATE
I wrote the code in vb. But with this code, a is correct, b is correct but c is incorrect, and the result is incorrect.
a^9 + b^9 + c^9 + d is a number bigger than the initial value.
The code should brings
a= 217423
b= 78525
c= 3456
d= 215478
Total Value is ok= 1085912312763120759250776993188102125849391224162
but code brings
a= 217423
b= 78525
c= 65957
d= 70333722607339201875244531009974
Total Value is bigger and not equal=1085935936469985777155428248430866412402362281319
Whats i need to change in the code to make c= 3456 and d= 215478?
the code is
Imports System.Numerics
Public Class Form1
Private Function pow9(x As BigInteger) As BigInteger
Dim y As BigInteger
y = x * x ' x^2
y *= y ' x^4
y *= y ' x^8
y *= x ' x^9
Return y
End Function
Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
Dim a, b, c, d, D2, n As BigInteger
Dim aa, bb, cc, dd, ae As BigInteger
D2 = BigInteger.Parse("1085912312763120759250776993188102125849391224162")
'first solution so a is maximal
d = D2
'a = BigIntegerSqrt(D2)
'RichTextBox1.Text = a.ToString
For a = 1 << ((Convert.ToInt32(Math.Ceiling(BigInteger.Log(d, 2))) + 8) / 9) To a > 0 Step -1
If (pow9(a) <= d) Then
d -= pow9(a)
Exit For
End If
Next
For b = 1 << ((Convert.ToInt32(Math.Ceiling(BigInteger.Log(d, 2))) + 8) / 9) To b > 0 Step -1
If (pow9(b) <= d) Then
d -= pow9(b)
Exit For
End If
Next
For c = 1 << ((Convert.ToInt32(Math.Ceiling(BigInteger.Log(d, 2))) + 8) / 9) To c > 0 Step -1
If (pow9(c) <= d) Then
d -= pow9(c)
Exit For
End If
Next
' minimize d
aa = a
bb = b
cc = c
dd = d
If (aa < 10) Then
ae = 0
Else
ae = aa - 10
End If
For a = aa - 1 To a > ae Step -1 'a goes down few iterations
d = D2 - pow9(a)
For n = 1 << ((Convert.ToInt32(Math.Ceiling(BigInteger.Log(d, 2))) + 8) / 9) To b < n 'b goes up
If (pow9(b) >= d) Then
b = b - 1
d -= pow9(b)
Exit For
End If
Next
For c = 1 << ((Convert.ToInt32(Math.Ceiling(BigInteger.Log(d, 2))) + 8) / 9) To c > 0 Step -1 'c must be search fully
If pow9(c) <= d Then
d -= pow9(c)
Exit For
End If
Next
If d < dd Then 'remember better solution
aa = a
bb = b
cc = c
dd = d
End If
If a < ae Then
Exit For
End If
Next
a = aa
b = bb
c = cc
d = dd
' a,b,c,d is the result
RichTextBox1.Text = D2.ToString
Dim Sum As BigInteger
Dim a9 As BigInteger
Dim b9 As BigInteger
Dim c9 As BigInteger
a9 = BigInteger.Pow(a, 9)
b9 = BigInteger.Pow(b, 9)
c9 = BigInteger.Pow(c, 9)
Sum = BigInteger.Add(BigInteger.Add(BigInteger.Add(a9, b9), c9), d)
RichTextBox2.Text = Sum.ToString
Dim Subst As BigInteger
Subst = BigInteger.Subtract(Sum, D2)
RichTextBox3.Text = Subst.ToString
End Sub
End Class
[Update]
The below code is an attempt to solve a problem like OP's, yet I erred in reading it.
The below is for 1085912312763120759250776993188102125849391224162 = a^9+b^9+c^9+d^9+e and to minimize e.
Just became too excite about OP's interesting conundrum and read too quick.
I review this more later.
OP's approach is O(N*N*N*N) - slow
Below is a O(N*N*log(N)) one.
Algorithm
Let N = 1,000,000. (Looks like 250,000 is good enough for OP's sum of 1.0859e48.)
Define 160+ wide integer math routines.
Define type: pow9
int x,y,
int160least_t z
Form array pow9 a[N*N] populated with x, y, x^9 + y^9, for every x,y in the [1...N] range.
Sort array on z.
Cost so far O(N*N*log(N).
For array elements indexed [0... N*N/2] do a binary search for another array element such that the sum is 1085912312763120759250776993188102125849391224162
Sum closest is the answer.
Time: O(N*N*log(N))
Space: O(N*N)
Easy to start with FP math and then later get a better answer with crafter extended integer math.
Try with smaller N and total sum targets to iron out implementation issues.
In case a,b,c,d might be zero I got an Idea for fast and simple solution:
First something better than brute force search of a^9 + d = x so that a is maximal (that ensures minimal d)...
let d = 1085912312763120759250776993188102125849391224162
find max value a such that a^9 <= d
this is simple as we know 9th power will multiply the bitwidth of operand 9 times so the max value can be at most a <= 2^(log2(d)/9) Now just search all numbers from this number down to zero (decrementing) until its 9th power is less or equal to x. This value will be our a.
Its still brute force search however from much better starting point so much less iterations are required.
We also need to update d so let
d = d - a^9
Now just find b,c in the same way (using smaller and smaller remainder d)... these searches are not nested so they are fast ...
b^9 <= d; d-=b^9;
c^9 <= d; c-=b^9;
To improve speed even more you can hardcode the 9th power using power by squaring ...
This will be our initial solution (on mine setup it took ~200ms with 32*8 bits uints) with these results:
x = 1085912312763120759250776993188102125849391224162
1085912312763120759250776993188102125849391224162 (reference)
a = 217425
b = 65957
c = 22886
d = 39113777348346762582909125401671564
Now we want to minimize d so simply decrement a and search b upwards until still a^9 + b^9 <= d is lower. Then search c as before and remember better solution. The a should be search downwards to meet b in the middle but as both a and bhave the same powers only few iterations might suffice (I used 50) from the first solution (but I have no proof of this its just my feeling). But still even if full range is used this has less complexity than yours as I have just 2 nested fors instead of yours 3 and they all are with lower ranges...
Here small working C++ example (sorry do not code in BASIC for decades):
//---------------------------------------------------------------------------
typedef uint<8> bigint;
//---------------------------------------------------------------------------
bigint pow9(bigint &x)
{
bigint y;
y=x*x; // x^2
y*=y; // x^4
y*=y; // x^8
y*=x; // x^9
return y;
}
//---------------------------------------------------------------------------
void compute()
{
bigint a,b,c,d,D,n;
bigint aa,bb,cc,dd,ae;
D="1085912312763120759250776993188102125849391224162";
// first solution so a is maximal
d=D;
for (a=1<<((d.bits()+8)/9);a>0;a--) if (pow9(a)<=d) break; d-=pow9(a);
for (b=1<<((d.bits()+8)/9);b>0;b--) if (pow9(b)<=d) break; d-=pow9(b);
for (c=1<<((d.bits()+8)/9);c>0;c--) if (pow9(c)<=d) break; d-=pow9(c);
// minimize d
aa=a; bb=b; cc=c; dd=d;
if (aa<50) ae=0; else ae=aa-50;
for (a=aa-1;a>ae;a--) // a goes down few iterations
{
d=D-pow9(a);
for (n=1<<((d.bits()+8)/9),b++;b<n;b++) if (pow9(b)>=d) break; b--; d-=pow9(b); // b goes up
for (c=1<<((d.bits()+8)/9);c>0;c--) if (pow9(c)<=d) break; d-=pow9(c); // c must be search fully
if (d<dd) // remember better solution
{
aa=a; bb=b; cc=c; dd=d;
}
}
a=aa; b=bb; c=cc; d=dd; // a,b,c,d is the result
}
//-------------------------------------------------------------------------
The function bits() just returns number of occupied bits (similar to log2 but much faster). Here final results:
x = 1085912312763120759250776993188102125849391224162
1085912312763120759250776993188102125849391224162 (reference)
a = 217423
b = 78525
c = 3456
d = 215478
It took 1689.651 ms ... As you can see this is much faster than yours however I am not sure with the number of search iterations while fine tuning ais OK or it should be scaled by a/b or even full range down to (a+b)/2 which will be much slower than this...
One last thing I did not bound a,b,c to 999999 so if you want it you just add if (a>999999) a=999999; statement after any a=1<<((d.bits()+8)/9)...
[Edit1] adding binary search
Ok now all the full searches for 9th root (except of the fine tunnig of a) can be done with binary search which will improve speed a lot more while ignoring bigint multiplication complexity leads to O(n.log(n)) against your O(n^3)... Here updated code (will full iteration of a while fitting so its safe):
//---------------------------------------------------------------------------
typedef uint<8> bigint;
//---------------------------------------------------------------------------
bigint pow9(bigint &x)
{
bigint y;
y=x*x; // x^2
y*=y; // x^4
y*=y; // x^8
y*=x; // x^9
return y;
}
//---------------------------------------------------------------------------
bigint binsearch_max_pow9(bigint &d) // return biggest x, where x^9 <= d, and lower d by x^9
{ // x = floor(d^(1/9)) , d = remainder
bigint m,x;
for (m=bigint(1)<<((d.bits()+8)/9),x=0;m.isnonzero();m>>=1)
{ x|=m; if (pow9(x)>d) x^=m; }
d-=pow9(x);
return x;
}
//---------------------------------------------------------------------------
void compute()
{
bigint a,b,c,d,D,n;
bigint aa,bb,cc,dd;
D="1085912312763120759250776993188102125849391224162";
// first solution so a is maximal
d=D;
a=binsearch_max_pow9(d);
b=binsearch_max_pow9(d);
c=binsearch_max_pow9(d);
// minimize d
aa=a; bb=b; cc=c; dd=d;
for (a=aa-1;a>=b;a--) // a goes down few iterations
{
d=D-pow9(a);
for (n=1<<((d.bits()+8)/9),b++;b<n;b++) if (pow9(b)>=d) break; b--; d-=pow9(b); // b goes up
c=binsearch_max_pow9(d);
if (d<dd) // remember better solution
{
aa=a; bb=b; cc=c; dd=d;
}
}
a=aa; b=bb; c=cc; d=dd; // a,b,c,d is the result
}
//-------------------------------------------------------------------------
function m.isnonzero() is the same as m!=0 just faster... The results are the same as above code but the time duration is only 821 ms for full iteration of a which would be several thousands seconds with previous code.
I think except using some polynomial discrete math trick I do not know of there is only one more thing to improve and that is to compute consequent pow9 without multiplication which will boost the speed a lot (as bigint multiplication is slowest operation by far) like I did in here:
How to get a square root for 32 bit input in one clock cycle only?
but I am too lazy to derive it...
I'm trying to write a simple function in VBA that will test a real value and output a string result if it's a perfect cube. Here's my code:
Function PerfectCubeTest(x as Double)
If (x) ^ (1 / 3) = Int(x) Then
PerfectCubeTest = "Perfect"
Else
PerfectCubeTest = "Flawed"
End If
End Function
As you can see, I'm using a simple if statement to test if the cube root of a value is equal to its integer portion (i.e. no remainder). I tried testing the function with some perfect cubes (1, 8, 27, 64, 125), but it only works for the number 1. Any other value spits out the "Flawed" case. Any idea what's wrong here?
You are testing whether the cube is equal to the double supplied.
So for 8 you would be testing whether 2 = 8.
EDIT: Also found a floating point issue. To resolve we will round the decimals a little to try and overcome the issue.
Change to the following:
Function PerfectCubeTest(x As Double)
If Round((x) ^ (1 / 3), 10) = Round((x) ^ (1 / 3), 0) Then
PerfectCubeTest = "Perfect"
Else
PerfectCubeTest = "Flawed"
End If
End Function
Or (Thanks to Ron)
Function PerfectCubeTest(x As Double)
If CDec(x ^ (1 / 3)) = Int(CDec(x ^ (1 / 3))) Then
PerfectCubeTest = "Perfect"
Else
PerfectCubeTest = "Flawed"
End If
End Function
#ScottCraner correctly explains why you were getting incorrect results, but there are a couple other things to point out here. First, I'm assuming that you are taking a Double as input because the range of acceptable numbers is higher. However, by your implied definition of a perfect cube only numbers with an integer cube root (i.e. it would exclude 3.375) need to be evaluated. I'd just test for this up front to allow an early exit.
The next issue you run into is that 1 / 3 can't be represented exactly by a Double. Since you're raising to the inverse power to get your cube root you're also compounding the floating point error. There's a really easy way to avoid this - take the cube root, cube it, and see if it matches the input. You get around the rest of the floating point errors by going back to your definition of a perfect cube as an integer value - just round the cube root to both the next higher and next lower integer before you re-cube it:
Public Function IsPerfectCube(test As Double) As Boolean
'By your definition, no non-integer can be a perfect cube.
Dim rounded As Double
rounded = Fix(test)
If rounded <> test Then Exit Function
Dim cubeRoot As Double
cubeRoot = rounded ^ (1 / 3)
'Round both ways, then test the cube for equity.
If Fix(cubeRoot) ^ 3 = rounded Then
IsPerfectCube = True
ElseIf (Fix(cubeRoot) + 1) ^ 3 = rounded Then
IsPerfectCube = True
End If
End Function
This returned the correct result up to 1E+27 (1 billion cubed) when I tested it. I stopped going higher at that point because the test was taking so long to run and by that point you're probably outside of the range that you would reasonably need it to be accurate.
For fun, here is an implementation of a number-theory based method described here . It defines a Boolean-valued (rather than string-valued) function called PerfectCube() that tests if an integer input (represented as a Long) is a perfect cube. It first runs a quick test which throws away many numbers. If the quick test fails to classify it, it invokes a factoring-based method. Factor the number and check if the multiplicity of each prime factor is a multiple of 3. I could probably optimize this stage by not bothering to find the complete factorization when a bad factor is found, but I had a VBA factoring algorithm already lying around:
Function DigitalRoot(n As Long) As Long
'assumes that n >= 0
Dim sum As Long, digits As String, i As Long
If n < 10 Then
DigitalRoot = n
Exit Function
Else
digits = Trim(Str(n))
For i = 1 To Len(digits)
sum = sum + Mid(digits, i, 1)
Next i
DigitalRoot = DigitalRoot(sum)
End If
End Function
Sub HelperFactor(ByVal n As Long, ByVal p As Long, factors As Collection)
'Takes a passed collection and adds to it an array of the form
'(q,k) where q >= p is the smallest prime divisor of n
'p is assumed to be odd
'The function is called in such a way that
'the first divisor found is automatically prime
Dim q As Long, k As Long
q = p
Do While q <= Sqr(n)
If n Mod q = 0 Then
k = 1
Do While n Mod q ^ k = 0
k = k + 1
Loop
k = k - 1 'went 1 step too far
factors.Add Array(q, k)
n = n / q ^ k
If n > 1 Then HelperFactor n, q + 2, factors
Exit Sub
End If
q = q + 2
Loop
'if we get here then n is prime - add it as a factor
factors.Add Array(n, 1)
End Sub
Function factor(ByVal n As Long) As Collection
Dim factors As New Collection
Dim k As Long
Do While n Mod 2 ^ k = 0
k = k + 1
Loop
k = k - 1
If k > 0 Then
n = n / 2 ^ k
factors.Add Array(2, k)
End If
If n > 1 Then HelperFactor n, 3, factors
Set factor = factors
End Function
Function PerfectCubeByFactors(n As Long) As Boolean
Dim factors As Collection
Dim f As Variant
Set factors = factor(n)
For Each f In factors
If f(1) Mod 3 > 0 Then
PerfectCubeByFactors = False
Exit Function
End If
Next f
'if we get here:
PerfectCubeByFactors = True
End Function
Function PerfectCube(n As Long) As Boolean
Dim d As Long
d = DigitalRoot(n)
If d = 0 Or d = 1 Or d = 8 Or d = 9 Then
PerfectCube = PerfectCubeByFactors(n)
Else
PerfectCube = False
End If
End Function
Fixed the integer division error thanks to #Comintern. Seems to be correct up to 208064 ^ 3 - 2
Function isPerfectCube(n As Double) As Boolean
n = Abs(n)
isPerfectCube = n = Int(n ^ (1 / 3) - (n > 27)) ^ 3
End Function
I have a problem here... if I have a table with few repeated string results. I want to know the value am the ammount of each.
For example. A function return an unknown "letters" and with unknown quantities in quantity
Function () return Table end
Table ={'a','a','c','b','b','a',...}
And I want to get this.
table.a={'a','a','a'}
table.b={'b','b'}
table.c={'c'}
....
....
I have no clue how to solve it...
Write a function, which creates a hash map of these things:
function RepetitionCounter(tInput)
local tCounter = {}
for i, v in ipairs(tInput) do
tCounter[v] = (tCounter[v] or 0) + 1
end
return tCounter
end
which you'll use as follows:
local tData = {'a','a','c','b','b','a',...}
local tCounts = RepetitionCounter(tData)
and the table tCounts would be as follows:
tCounts.a = 3
tCounts.b = 2
tCounts.c = 1
Modifying the function above by just a little, you can get the desired output. Replace the following line:
tCounter[v] = (tCounter[v] or 0) + 1
with
if not tCounter[v] then
tCounter[v] = {}
else
table.insert(tCounter[v], v)
end
I have been following books and PDFs on writing in FORTRAN to write an integration program. I compile the code with gfortran and get several copies of the following errors.
1)Unexpected data declaration statement at (1)
2)Unterminated character constant beginning at (1)
3)Unclassifiable statement at (1)
4)Unexpected STATEMENT FUNCTION statement at (1)
5)Expecting END PROGRAM statement at (1)
6)Syntax error in data declaration at (1)
7)Statement function at (1) is recursive
8)Unexpected IMPLICIT NONE statement at (1)
I do not know hat they truly mean or how to fix them, google search has proven null and the other topics on this site we about other errors. for Error 5) i put in Program main and end program main like i might in C++ but still got the same result. Error 7) makes no sense, i am trying for recursion in the program. Error 8) i read implicit none was to prevent unnecessary decelerations.
Ill post the code itself but i am more interested in the compiling errors because i still need to fine tune the array data handling, but i cant do that until i get it working.
Program main
implicit none
real, dimension(:,:), allocatable :: m, oldm
real a
integer io, nn
character(30) :: filename
real, dimension(:,:), allocatable :: alt, temp, nue, oxy
integer locationa, locationt, locationn, locationo, i
integer nend
real dz, z, integral
real alti, tempi, nuei, oxyi
integer y, j
allocate( m(0, 0) ) ! size zero to start with?
nn = 0
j = 0
write(*,*) 'Enter input file name: '
read(*,*) filename
open( 1, file = filename )
do !reading in data file
read(1, *, iostat = io) a
if (io < 0 ) exit
nn = nn + 1
allocate( oldm( size(m), size(m) ) )
oldm = m
deallocate( m )
allocate( m(nn, nn) )
m = oldm
m(nn, nn) = a ! The nnth value of m
deallocate( oldm )
enddo
! Decompose matrix array m into column arrays [1,n]
write(*,*) 'Enter Column Number for Altitude'
read(*,*) locationa
write(*,*) 'Enter Column Number for Temperature'
read(*,*) locationt
write(*,*) 'Enter Column Number for Nuetral Density'
read(*,*) locationn
write(*,*) 'Enter Column Number for Oxygen density'
read(*,*) locationo
nend = size(m, locationa) !length of column #locationa
do i = 1, nend
alt(i, 1) = m(i, locationa)
temp(i, 1) = log(m(i, locationt))
nue(i, 1) = log(m(i, locationn))
oxy(i, 1) = log(m(i, locationo))
enddo
! Interpolate Column arrays, Constant X value will be array ALT with the 3 other arrays
!real dz = size(alt)/100, z, integral = 0
!real alti, tempi, nuei, oxyi
!integer y, j = 0
dz = size(alt)/100
do z = 1, 100, dz
y = z !with chopped rounding alt(y) will always be lowest integer for smooth transition.
alti = alt(y, 1) + j*dz ! the addition of j*dz's allow for all values not in the array between two points of the array.
tempi = exp(linear_interpolation(alt, temp, size(alt), alti))
nuei = exp(linear_interpolation(alt, nue, size(alt), alti))
oxyi = exp(linear_interpolation(alt, oxy, size(alt), alti))
j = j + 1
!Integration
integral = integral + tempi*nuei*oxyi*dz
enddo
end program main
!Functions
real function linear_interpolation(x, y, n, x0)
implicit none
integer :: n, i, k
real :: x(n), y(n), x0, y0
k = 0
do i = 1, n-1
if ((x0 >= x(i)) .and. (x0 <= x(i+1))) then
k = i ! k is the index where: x(k) <= x <= x(k+1)
exit ! exit loop
end if
enddo
if (k > 0) then ! compute the interpolated value for a point not in the array
y0 = y(k) + (y(k+1)-y(k))/(x(k+1)-x(k))*(x0-x(k))
else
write(*,*)'Error computing the interpolation !!!'
write(*,*) 'x0 =',x0, ' is out of range <', x(1),',',x(n),'>'
end if
! return value
linear_interpolation = y0
end function linear_interpolation
I can provide a more detailed description of the exact errors, i was hoping that the error name would be enough since i have a few of each type.
I think I can spot a few serious errors in your code sample. The syntax error is that you have unbalanced parentheses in the exp(... statements. They should be like this:
tempi = exp(linear_interpolation(alt, temp, size(alt), alti) ) ! <- extra ")"
nuei = exp(linear_interpolation(alt, nue, size(alt), alti) )
oxyi = exp(linear_interpolation(alt, oxy, size(alt), alti) )
It's precisely things like this that can produce long strings of confusing errors like you're getting; therefore the advice Dave and Jonathan have given can't be repeated often enough.
Another error (the "unclassifiable statement") applies to your loops:
do(i=1, nend)
! ...
do(z=1, 100, dz)
! ...
These should be written without parentheses.
The "data declaration error" stems from your attempt to declare and initialise multiple variables like
real dz = size(alt)/100, z, integral = 0
Along with being positioned incorrectly in the code (as noted), this can only be done with the double colon separator:
real :: dz = size(alt)/100, z, integral = 0
I personally recommend always writing declarations like this. It must be noted, though, that initialising variables like this has the side effect of implicitly giving them the save attribute. This has no effect in a main program, but is important to know; you can avoid it by putting the initialisations on a separate line.
I need to create a 2 column array in ABAP so that a program can look up a record item (defined by the letters A - ZZZ) and then return the number associated with it.
For example:
A = 1
B = 2
C = 3
...
Z = 26
AA = 27
AB = 28
...
AZ =
BA =
...
BZ =
CA =
...
...
ZZZ =
Please can you suggest how I can code this.
Is there a better option than writing an array?
Thanks.
you don't need to lookup the value in a table. this can be calculated:
parameters: p_input(3) type c value 'AAA'.
data: len type i value 0,
multiplier type i value 1,
result type i value 0,
idx type i.
* how many characters are there?
len = strlen( p_input ).
idx = len.
* compute the value for every char starting at the end
* in 'ABC' the C is multiplied with 1, the B with 26 and the A with 26^2
do len times.
* p_input+idx(1) should be the actual character and we look it up in sy-abcde
search p_input+idx(1) in SY-ABCDE.
* if p_input+idx(1) was A then sy-fdpos should now be set to 0 that is s why we add 1
compute result = result + ( sy-fdpos + 1 ) * multiplier.
idx = idx - 1.
multiplier = multiplier * 26.
enddo.
write: / result.
i didn't test the program and it has pretty sure some syntax errors. but the algorithm behind it should work.
perhaps I'm misunderstanding, but don't you want something like this?
type: begin of t_lookup,
rec_key type string,
value type i,
end of t_lookup.
data: it_lookup type hashed table of t_lookup with unique key rec_key.
then once it's populated, read it back
read table it_lookup with key rec_key = [value] assigning <s>.
if sy-subrc eq 0.
" got something
else.
" didn't
endif.
unfortunately, arrays don't exist in ABAP, but a hashed table is designed for this kind of lookup (fast access, unique keys).
DATA: STR TYPE STRING, I TYPE I, J TYPE I, K TYPE I, CH TYPE C, RES
TYPE INT2, FLAG TYPE I.
PARAMETERS: S(3).
START-OF-SELECTION.
I = STRLEN( S ).
STR = S.
DO I TIMES.
I = I - 1.
CH = S.
IF CH CO '1234567890.' OR CH CN SY-ABCDE.
FLAG = 0.
EXIT.
ELSE.
FLAG = 1.
ENDIF.
SEARCH SY-ABCDE FOR CH.
J = I.
K = 1.
WHILE J > 0.
K = K * 26.
J = J - 1.
ENDWHILE.
K = K * ( SY-FDPOS + 1 ).
RES = RES + K.
REPLACE SUBSTRING CH IN S WITH ''.
ENDDO.
* RES = RES + SY-FDPOS.
IF FLAG = 0.
MESSAGE 'String is not valid.' TYPE 'S'.
ELSE.
WRITE: /, RES .
ENDIF.
Use this code after executing.
I did a similar implementation some time back.
Check this it it works for you.
DATA:
lv_char TYPE char1,
lv_len TYPE i,
lv_len_minus_1 TYPE i,
lv_partial_index1 TYPE i,
lv_partial_index2 TYPE i,
lv_number TYPE i,
result_tab TYPE match_result_tab,
lv_col_index_substr TYPE string,
lv_result TYPE i.
FIELD-SYMBOLS:
<match> LIKE LINE OF result_tab.
lv_len = strlen( iv_col_index ) .
lv_char = iv_col_index(1).
FIND FIRST OCCURRENCE OF lv_char IN co_char RESULTS result_tab.
READ TABLE result_tab ASSIGNING <match> INDEX 1.
lv_number = <match>-offset .
lv_number = lv_number + 1 .
IF lv_len EQ 1.
ev_col = ( ( 26 ** ( lv_len - 1 ) ) * lv_number ) .
ELSE.
lv_len_minus_1 = lv_len - 1.
lv_col_index_substr = iv_col_index+1(lv_len_minus_1) .
CALL METHOD get_col_index
EXPORTING
iv_col_index = lv_col_index_substr
IMPORTING
ev_col = lv_partial_index2.
lv_partial_index1 = ( ( 26 ** ( lv_len - 1 ) ) * lv_number ) + lv_partial_index2 .
ev_col = lv_partial_index1 .
ENDIF.
Here The algorithm uses a recursive logic to determine the column index in numbers.
This is not my algorithm but have adapted to be used in ABAP.
The original algorithm is used in Open Excel, cant find any links right now.