'for loop' exit at first passage

cvp

@mibradoc Pythonista 3.3 too
You need also to uncomment 146

cvp

This

			for numero_carre in range ( (numcarre_altex // 3) * 3,((numcarre_altex//3)*3) + 3 ) :
				print ('x:',numero_carre) # BUG !!  sortie de la boucle for après le 1er passage !! 
				if numero_carre != numcarre_altex and key[0] not in carrint(numero_carre) :
					numero_carre_sans_chiffre = numero_carre
					
				colonnes_possibles=[]
				for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :

also crashes with

grille à résoudre ? libre

[[1 4 0 0 7 0 9 2 0]
 [3 0 0 0 0 2 8 5 4]
 [0 0 0 0 5 0 6 0 7]
 [0 0 0 2 0 0 0 7 0]
 [9 0 6 0 0 0 4 0 1]
 [0 5 0 0 0 6 0 0 0]
 [0 0 9 0 6 0 0 0 0]
 [5 3 0 7 0 0 0 0 8]
 [0 8 0 0 2 0 0 3 9]]
50
(1, 1) [3, 4] 2
x: 0
Traceback (most recent call last):
  File "/private/var/mobile/Containers/Shared/AppGroup/668A7D98-7216-47ED-917D-AA0B6173167E/Pythonista3/Documents/aa.py", line 147, in <module>
    for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :
NameError: name 'numero_carre_sans_chiffre' is not defined

cvp

Your loop on colonne uses numero_carre_sans_chiffre which is only defined in a if.

Are sure you don't need to indent more this loop?

				if numero_carre != numcarre_altex and key[0] not in carrint(numero_carre) :
					numero_carre_sans_chiffre = numero_carre
					
					colonnes_possibles=[]
					for colonne in range((numero_carre_sans_chiffre % 3)*3,(numero_carre_sans_chiffre % 3)*3 + 3) :

mibradoc

@cvp you're right, but the bug remains on my iPad
It's strange that the same code runs without error on your device.
Thanks a lot to you and to @stephen

cvp

@mibradoc As we have made some changes in your code to let it run without name error, I'm not sure we use exactly the same code.
I also have an iPad (Mini 4) and I don't think comes from that.
Could you post a code which runs without name error on your device but which stops after one loop

stephen

pas de problème @mibradoc

here is translated script incase anyone needed/wanted
gist

stephen

@cvp i beleive the error is the issue originaly. i say this only because it raises after one pass of for loop

cvp

@stephen it is possible but as you have also changed some lines (comment, loop on colonne indentation), I just want to be sure that he and us try exactly the same code.

mibradoc

The code on Dropbox stops after one loop. Whith a breakpoint on line 141 and then step by step you will see that the loop is executed only one time, not three as expected. The name error comes because the bypass over the lines 143 and 144 where the variable 'numero_carre_sans_chiffre' is assigned.

cvp

@mibradoc yes, but as the name error occurs in the loop, it is normal the loop is stopped, isn'it?

mibradoc

@stephen Thanks for the translation.

mibradoc

@stephen and your code runs without errors ! Have you made some change other than translate ?

cvp

@mibradoc it was like I said, a problem of indentation of you loop on colonne!

mibradoc

@cvp I think the errors occurs in line 147, after the loop. This cannot prevent the loop from executing three times.

cvp

@mibradoc the problem in your original code is that the loop on colonne was in the loop On numéro and this crashed in the first occurrence in loop.
If you indent left the colonne loop, there is no crash...

Your code

for numéro...
  ...
  For colonne..

correction

for numéro...
  ...
for colonne...

mibradoc

@cvp OK, I'll check indentation

cvp

@mibradoc

See my previous post, edited

mibradoc

@cvp Great, It'works. Merci les amis ;-)

stephen

@mibradoc

I CHANGED:

for colonne in range((numero_carre % 3)*3,(numero_carre % 3)*3 + 3) :
#colonnes_possibles=[] ⇒ colonnes_possibles=[]



for numero_carre in range ( (numcarre_altex // 3) * 3,((numcarre_altex//3)*3) + 3 ) :
                print (numero_carre) # BUG !!  sortie de la boucle for après le 1er passage !! 
                #if numero_carre != numcarre_altex and key[0] not in carrint(numero_carre) :
                    #numero_carre_sans_chiffre = numero_carre
                    
                colonnes_possibles=[]
                for colonne in range((numero_carre % 3)*3,(numero_carre % 3)*3 + 3) :
                        if puzzle[ ligne_a_analyser , colonne ] == 0 and key[0] not in puzzle[:9,colonne]:
                            colonnes_possibles.append(colonne)
                if len(colonnes_possibles)  == 1 :
                    fn_remplir_case ( ligne_a_analyser , colonnes_possibles[0] , key[0] )
                        #print()
                        #print (puzzle)
                        #print ()

mibradoc

@stephen All is fine now, thank you.