Numbers of Armstong WP34s

[ggauny@live.fr]

Hi all,
I have made a little routine to know how are numbers of Armstrong, it is a curiosity arithmetic like this :

A one is when the sum of the digits of the number cube is egal to the number, as example :

1 is because, of course, 1^3 is 1,
153 is too because 1^3+5^3+3^ is 153.

In my knowledge only is 1, 153,370,371,407. I think it is all but not sure am-I.
For cumfortable use I merged 2 routines in one. Did you notice that I use goods ideas from forum like inc x and Dieter's 'DIVMOD' as if theses was built'in functions ?

Code:


fLBL'NAR'  ; stand for "Number of ARmstrong
2000   ; a limit
STO 03
0
STO 04
INC 04  ; it is 1,2,3,.....
RCL 04 
XEQ'TT1'
DSZ 03   ; to not over the 2000 limit
BACK 04
END

fLBL'TT1'
FILL
EXPT
INC X
STO 01
DROP
#010
RMDR
X^3
STO+ 02
DROP
#010
IDIV
ENTER
DSZ 01
BACK 009
RCL 02
X=? T
PSE 30  ; to have time to read and note on a sheet of paper
0
STO 01
0
STO 02
END.

Well, I am aware that serve for nothing, but it is funy, no ?

Like say Dieter : as usual comments are welcome.

[Thomas Klemm]

Instead of splitting the number into digits to calculate the sum of the cubes you can create both number and sum of the cubes from the digits. This leads to four nested loops to test 4-digit numbers.

Here's a Python program:

Code:

for a in range(10):
    p = a
    t = a**3
    for b in range(10):
        q = 10*p + b
        u = t + b**3
        for c in range(10):
            r = 10*q + c
            v = u + c**3
            for d in range(10):
                s = 10*r + d
                w = v + d**3
                if s == w:
                    print s

Result:
0
1
153
370
371
407


And this program is for the HP-42S:

Code:

{ 108-Byte Prgm }
LBL "NARCIS"
9E-3
STO 00      ; 0.009
STO 01      ; a
LBL 01      ; a-loop
RCL 01      ; a
IP
STO 05      ; p = a
3
Y↑X         ; a↑3
STO 06      ; t = a↑3
RCL 00      ; 0.009
STO 02      ; b
LBL 02      ; b-loop
RCL 02      ; b
IP
RCL 05      ; p
10
*           ; 10×p
RCL+ ST Y   ; b
STO 07      ; q = 10×p + b
X<>Y        ; b
3
Y↑X         ; b↑3
RCL+ 06     ; t
STO 08      ; u = t + b↑3
RCL 00      ; 0.009
STO 03      ; c
LBL 03      ; c-loop
RCL 03      ; c
IP
RCL 07      ; q
10
*           ; 10×q
RCL+ ST Y   ; c
STO 09      ; r = 10×q + c
X<>Y        ; c
3
Y↑X         ; c↑3
RCL+ 08     ; u
STO 10      ; v = u + c↑3
RCL 00      ; 0.009
STO 04      ; d
LBL 04      ; d-loop
RCL 04      ; d
IP
RCL 09      ; r
10
*           ; 10×r
RCL+ ST Y   ; s = 10×r + d
X<>Y        ; d
3
Y↑X         ; d↑3
RCL+ 10     ; w = v + d↑3
X=Y?        ; s = w ?
VIEW ST X
ISG 04      ; d
GTO 04      ; d-loop
ISG 03      ; c
GTO 03      ; c-loop
ISG 02      ; b
GTO 02      ; b-loop
ISG 01      ; a
GTO 01      ; a-loop
END

Cheers
Thomas

[Dave Frederickson]

(11-30-2015 07:02 PM)ggauny@live.fr Wrote:  A one is when the sum of the digits of the number cube is egal to the number ...

This is only true for 3-digit Armstrong Numbers. For an n-digit Armstrong Number, each digit is raised to the n-th power. So for 4-digit Armstrong Numbers, each digit is raised to the 4th power, not the 3rd.

1 is an Armstrong Number because 1^1=1, not 1^3=1.

Dave

[ggauny@live.fr]

Hi all,

I was ignoring that Armstrong's numbers was posssible after 3 digits, I am going to search for 4, 5...
But I am afraid the time to calcul solutions.

I appreciate your's intervention and also the program of Thomas for 42s, but I am a bit with difficulty to apprehend Python approach.

Good days for all.

[Thomas Klemm]

(12-01-2015 06:35 AM)ggauny@live.fr Wrote:  also the program of Thomas for 42s, but I am a bit with difficulty to apprehend Python approach.

You can just replace the exponent 3 by 4: Click run

Code:

for a in range(10):
    p = a
    t = a**4
    for b in range(10):
        q = 10*p + b
        u = t + b**4
        for c in range(10):
            r = 10*q + c
            v = u + c**4
            for d in range(10):
                s = 10*r + d
                w = v + d**4
                if s == w:
                    print s

Result:
0
1
1634
8208
9474
=> None


Cheers
Thomas

PS: If you want to go further with n you might extract that loop into a function that can be called recursively: Click run

Code:

def armstrong(n, i, a, b):
    if i == 0:
        if a == b:
            print a
    else:
        for d in range(10):
            armstrong(n, i-1, 10*a + d, b + d**n)
            
for n in range (1, 7):
    print "n = %d" % n
    armstrong(n, n, 0, 0)
    print

Result:
n = 1
0
1
2
3
4
5
6
7
8
9

n = 2
0
1

n = 3
0
1
153
370
371
407

n = 4
0
1
1634
8208
9474

n = 5
0
1
4150
4151
54748
92727
93084

n = 6
0
1
548834

=> None

[ggauny@live.fr]

Hallo Thomas,

Very thanks for your explanation, I will do an another routine to find with exposants up to three.
I have notice that if I am strangled it is permis to write a bit in french to be more clear, Marcus say the same thing of you.

Thanks again.

[Thomas Klemm]

There's another online tool to visualize Python.
Just copy this:

Code:

def armstrong(n, i, a, b):
    if i == 0:
        if a == b:
            print a
    else:
        for d in range(10):
            armstrong(n, i-1, 10*a + d, b + d**n)

armstrong(3, 3, 0, 0)

Paste it into the text-box and press [Visualize Execution].
Now you can step [Forward >] and [< Back] through the program just as you are used to from your calculators. However stepping back will undo the operation.

Cheers
Thomas

[Albert Chan]

I had solved Armstrong Numbers problem many years ago.

It searched all Armstrong Numbers (digits > 1) in 6 minutes on my laptop.
That is a huge 10^60 search space !

This is the gist of my routine:

If pattern 543 is checked, other permutations (345, 354, 435, 453, 534) can be skipped.
All permutations is going to produce the same sum anyway. So, 1 check = 3! = 6 tests

Digits pattern not reached 3 digits can be skipped: 100 110 111, 200, 210, 211, 220, ..., 432

Sum of over 3-digits can also be skipped: 964, 965, 966, 970, 971 ..., 988, 990, ... 999

Patterns to check = 433, 440, 441 ..., 963 (*)

The trick is to design how the pattern incremented.
You can guess the algorithm from above example.

Have you seen it ?
Combinatorial problem is fun ! :-)

(*) my code actually search to 988
To speed up the code, it need a simple test for endpoint.
So, the loop stop if the next pattern had too many 9's in it.