Level-index is a system of representing numbers which has been proposed as an alternative to floating-point formats. It claims to virtually eliminate overflow (and underflow, in its symmetric form) from the vast majority of computations.

The usual floating-point format internally represents a number \$x\$ in terms of a mantissa \$m\in[1,2)\$ and an integer exponent \$e\$, where \$x=m\times2^e\$. In contrast, level-index represents a number by an integer level \$l\ge0\$ and an index \$i\in[0,1)\$, defined by
\$l+i=\psi(x)=\begin{cases} x,&x<1\\ 1+\psi(\log x),&\text{otherwise.} \end{cases}\$
Equivalently, \$l\$ and \$i\$ together represent the power tower number \$x=\psi^{-1}(l+i)=\operatorname{exp}^l(i)\$. Accordingly, \$\psi^{-1}\$ grows very fast; \$\psi^{-1}(5)\$ has more than a million digits to the left of the decimal point.


Given two numbers \$\psi(a),\psi(b)\$, compute their level-index sum \$\psi(a+b)\$. You may choose to input or output the level and index together or separately.

Your solution should be accurate to at least six digits after the decimal point and terminate in a reasonable time for all \$\psi(a+b)<8\$, and may not assume your floating-point representation has infinite precision.

Test cases

ψ(a)    ψ(b)    ψ(a+b)              a           b           a+b
0       0       0.0000000000000000  0           0           0
1       1       1.6931471805599453  1           1           2
2       2       2.5265890341390445  2.718281828 2.718281828 5.436563657
3       3       3.2046791426805201  1.5154262e1 1.5154262e1 3.0385245e1
4       4       4.0161875057657443  3.8142791e6 3.8142791e6 7.6285582e6
5       5       5.0000000044114734  2.3e1656520 2.3e1656520 4.7e1656520
1.70879 2.44303 2.6490358380192970  2.031531618 4.746554831 6.778086449
0.8001  3.12692 3.1367187876590184  0.8001      2.2470167e1 2.3270267e1
2.71828 3.14159 3.2134989633416356  7.774915755 2.368508064 3.1459996e1
4.6322  4.6322  4.6322790036002559  1.79455e308 1.79455e308 3.58910e308
4.80382 4.80382 4.8038229252940133  5.9037e4932 5.9037e4932 1.1807e4933
  • \$\begingroup\$ Why was it never adopted more widely? \$\endgroup\$
    – Jonah
    Aug 26, 2022 at 12:52
  • 2
    \$\begingroup\$ @Jonah Honestly, I have no idea. There is a related SO question from some years ago, but I don't think it's particularly comprehensive. \$\endgroup\$
    – att
    Aug 26, 2022 at 18:49
  • 2
    \$\begingroup\$ @Jonah Quite simply speed and memory. The advantages of systems like IEEE 754 in theses practicalities dwarf Level-index's strengths in preventing overflow and underflow. \$\endgroup\$
    – Noodle9
    Aug 27, 2022 at 12:39

2 Answers 2


SageMath, 93 bytes

lambda x,y:h(p(x)+p(y))
h=lambda x:x>1 and 1+h(log(x))or x
p=lambda x:x>1 and exp(p(x-1))or x

Try it online!


Python NumPy, 178 bytes

def f(I):
 if 4+i>j<5.5:
  while j>=1:l+=1;j=log(j)
 return j
p=lambda x:x>1 and exp(p(x-1))or x
from numpy import*

Attempt This Online!

Saved a bit by borrowing @Noodle9's p function.

Previous Python NumPy, 202 bytes

def f(I):
 if 4+i>j<5.5:
  for y in i,j:
   while y>=1+l:r=exp(r);y-=1
  while j>=1:l+=1;j=log(j)
 return l+j
from numpy import*

Attempt This Online!

I'm sure this can be golfed quite a bit.


With the given accuracy requirements we need something to bridge the gap between small values (<4.5, say) where we can use exp without overflowing and large (>5.5) where we can simply use the max. Luckily, NumPy has logaddexp built in which saves us one level of exponentiating.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.