Periodic table
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Display the row and column in the periodic table of the given atomic number.
- The periodic table
Let us consider the following periodic table representation.
__________________________________________________________________________
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
| |
|1 H He |
| |
|2 Li Be B C N O F Ne |
| |
|3 Na Mg Al Si P S Cl Ar |
| |
|4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr |
| |
|5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe |
| |
|6 Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn |
| |
|7 Fr Ra ° Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og |
|__________________________________________________________________________|
| |
| |
|8 Lantanoidi* La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu |
| |
|9 Aktinoidi° Ak Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr |
|__________________________________________________________________________|
- Example test cases;
-
1->1 1 -
2->1 18 -
29->4 11 -
42->5 6 -
57->8 4 -
58->8 5 -
72->6 4 -
89->9 4
Note that atomic values corresponding to Lantanoidi (57 to 71) and Aktinoidi (89 to 103) are to be mapped in their own rows.
- See also
- the periodic table
- This task was an idea from CompSciFact
- The periodic table in ascii that was used as template
ALGOL 68
<lang algol68>BEGIN # display the period and group number of an element, #
# given its atomic number #
INT max atomic number = 118; # highest known element #
# the positions are stored as: #
# ( group number * group multiplier ) + period #
INT group multiplier = 100;
[ 1 : max atomic number ]INT position;
STRING periodic table = "- ="
+ "-- -----="
+ "-- -----="
+ "-----------------="
+ "-----------------="
+ "--8--------------="
+ "--9--------------="
;
# construct the positions of the elements in the table #
# from the outline #
BEGIN
INT period := 1;
INT group := 1;
INT element := 1;
FOR t FROM LWB periodic table TO UPB periodic table DO
CHAR p = periodic table[ t ];
IF p = " " THEN
# no element at this position #
SKIP
ELIF p = "8" OR p = "9" THEN
# lantanoids or actinoids #
INT series period = IF p = "8" THEN 8 ELSE 9 FI;
INT series group := 4;
FOR e TO 15 DO
position[ element ] := ( group multiplier * series group ) + series period;
element +:= 1;
series group +:= 1
OD
ELSE
# there is a single element here #
position[ element ] := ( group multiplier * group ) + period;
element +:= 1;
IF p = "=" THEN
# final element of the period #
period +:= 1;
group := 0
FI
FI;
group +:= 1
OD
END;
# display the period and group numbers of test elements #
[]INT test = ( 1, 2, 29, 42, 57, 58, 72, 89, 113 );
FOR t FROM LWB test TO UPB test DO
INT e = test[ t ];
IF e < LWB position OR e > UPB position THEN
print( ( "Invalid element: ", whole( e, 0 ), newline ) )
ELSE
INT period = position[ e ] MOD group multiplier;
INT group = position[ e ] OVER group multiplier;
print( ( "Element ", whole( e, -3 )
, " -> ", whole( period, 0 ), ", ", whole( group, -2 )
, newline
)
)
FI
OD
END</lang>
- Output:
Element 1 -> 1, 1 Element 2 -> 1, 18 Element 29 -> 4, 11 Element 42 -> 5, 6 Element 57 -> 8, 4 Element 58 -> 8, 5 Element 72 -> 6, 4 Element 89 -> 9, 4 Element 113 -> 7, 13
Python
A solution trying hard not to encode too much data about the table.
<lang Python> def perta(atomic) -> (int, int):
NOBLES = 2, 10, 18, 36, 54, 86, 118 INTERTWINED = 0, 0, 0, 0, 0, 57, 89 INTERTWINING_SIZE = 14 LINE_WIDTH = 18
prev_noble = 0
for row, noble in enumerate(NOBLES):
if atomic <= noble: # we are at the good row. We now need to determine the column
nb_elem = noble - prev_noble # number of elements on that row
rank = atomic - prev_noble # rank of the input element among elements
if INTERTWINED[row] and INTERTWINED[row] <= atomic <= INTERTWINED[row] + INTERTWINING_SIZE: # lantanides or actinides
row += 2
col = rank + 1
else: # not a lantanide nor actinide
# handle empty spaces between 1-2, 4-5 and 12-13.
nb_empty = LINE_WIDTH - nb_elem # spaces count as columns
inside_left_element_rank = 2 if noble > 2 else 1
col = rank + (nb_empty if rank > inside_left_element_rank else 0)
break
prev_noble = noble
return row+1, col
- small test suite
TESTS = {
1: (1, 1), 2: (1, 18), 29: (4,11), 42: (5, 6), 58: (8, 5), 59: (8, 6), 57: (8, 4), 71: (8, 18), 72: (6, 4), 89: (9, 4), 90: (9, 5), 103: (9, 18),
}
for input, out in TESTS.items():
found = perta(input)
print('TEST:{:3d} -> '.format(input) + str(found) + (f' ; ERROR: expected {out}' if found != out else ))
</lang>