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Crack the padlock logic challenge - python brute force!
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| def rightplace(a,b,len=3): | |
| astr = str.zfill(str(a),len) | |
| bstr = str.zfill(str(b),len) | |
| return [1 if astr[x]==bstr[x] else 0 for x in range(len)] | |
| def wrongplace(a,b,len=3): | |
| astr = str.zfill(str(a),len) | |
| bstr = str.zfill(str(b),len) | |
| return [1 if astr[x] in bstr and astr[x] != bstr[x] else 0 for x in range(len)] | |
| for i in range(0,1000,1): | |
| ## 682: One number is correct and well placed. | |
| if sum(rightplace(i,682)) != 1: continue | |
| ## 614: One number is correct but wrongly placed. | |
| if sum(wrongplace(i,614)) != 1: continue | |
| ## 206: Two numbers are correct but wrongly placed. | |
| if sum(wrongplace(i,206)) != 2: continue | |
| ## 738: Nothing is correct. | |
| if sum(wrongplace(i,738)) != 0 or sum(rightplace(i,738)) != 0: continue | |
| ## 780: One number is correct but wrongly placed. | |
| if sum(wrongplace(i,380)) != 1: continue | |
| print(str.zfill(str(i),3)) | |
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