Numbers of Armstong WP34s
11-30-2015, 07:02 PM
Post: #1
 ggauny@live.fr Senior Member Posts: 499 Joined: Nov 2014
Numbers of Armstong WP34s
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.

Gérard.
11-30-2015, 08:27 PM
Post: #2
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Numbers of Armstong WP34s
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
11-30-2015, 09:21 PM
Post: #3
 Dave Frederickson Senior Member Posts: 1,740 Joined: Dec 2013
RE: Numbers of Armstong WP34s
(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
12-01-2015, 06:35 AM
Post: #4
 ggauny@live.fr Senior Member Posts: 499 Joined: Nov 2014
RE: Numbers of Armstong WP34s
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.

Gérard.
12-01-2015, 10:11 AM
Post: #5
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Numbers of Armstong WP34s
(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
12-01-2015, 11:14 AM
Post: #6
 ggauny@live.fr Senior Member Posts: 499 Joined: Nov 2014
RE: Numbers of Armstong WP34s
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.

Gérard.
12-01-2015, 01:07 PM
Post: #7
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Numbers of Armstong WP34s
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
07-30-2018, 12:58 AM
Post: #8
 Albert Chan Senior Member Posts: 901 Joined: Jul 2018
RE: Numbers of Armstong WP34s
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.
 « Next Oldest | Next Newest »

User(s) browsing this thread: 1 Guest(s)