It is possible that I make some error and there is a number that generate one infinite loop because e variable in code is too little (for example I wrong the number 3.323 or if variable t or r have integer part > of input w even if t-r decrease)

It is possible that I make some error and there is a number of input that generate one infinite loop because e variable in code is too little (for example I wrong the number 3.323 or if variable t or r have integer part > of input w even if t-r decrease)

I have not consider the case the number of input w is too big for variable ⎕FPC, for example when digits of number are > ⎕FPC÷3.322 (float for me has to be fixed point float because this is too tricly)

APL(NARS), 83 chars

r←q w;t;e t←1v⋄e←÷10*⌊(⎕FPC-{1>∣w:0⋄1+2⍟∣w})÷3.322v →3×⍳e>∣t-r←t-3÷⍨t-w÷t×t⋄t←r⋄→2 

//10+42+31=83

I followed the formula written in the post: https://codegolf.stackexchange.com/a/21793/120560. I think I have found the formula for min error allowable from +-*/ operations in the float point lenght xBit, due the precision is affected from the lenght of the integer part of the input number...

The error would be less than e←÷10*⌊(⎕FPC-{1>∣w:0⋄1+2⍟∣w})÷3.322v where w is the input number, ⎕FPC is the lenght in bit of the float point 1+2⍟w would be the lenght in bit of integer part of w, log_2(10)=3.32192809 here approximated as 3.322 (I don't know if it is the right number). Test

 aaa←27 64 1 18.609625 3652264 0.001 7 2E¯13 0 q¨aaa 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.381590388E¯38 q¨-aaa ¯3 ¯4 ¯1 ¯2.65 ¯154 ¯0.1 ¯1.912931183 ¯0.00005848035476 1.381590388E¯38 

 q 2 1.25992105 q 2v 1.25992105 q 2x 1.25992105 

 ⎕fpc←512 512÷3.322 154.1240217 150⍕q 2v 1.259921049894873164767210607278228350570251464701507980081975112155299676513959483729396562436255094154310256035615665259399024040613737228459110304269 

It is possible that I make some error and there is a number that generate one infinite loop because e variable in code is too little  (for example if variable t or r have integer part > of input w even if t-r decrease)

APL(NARS), 46 chars

r←q w;t t←1 →3×⍳1e¯9>∣t-r←t-3÷⍨t-w÷t×t⋄t←r⋄→2 

//8+4+34=46

I followed the formula written in the post: https://codegolf.stackexchange.com/a/21793/120560. It seems is ok with test

 aaa←27 64 1 18.609625 3652264 0.001 7 2E¯13 0 q¨aaa 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.568328545E¯9 q¨-aaa ¯3 ¯4 ¯1 ¯2.65 ¯154 ¯0.1 ¯1.912931183 ¯0.00005848035476 1.568328545E¯9 

It seems work for input big float v but not work for rationals...It could be ok with the restriction of question so ok.

The problem would be find in this case cube root function, the min number e that can find all digits possibles of one float when the sys has float XXXbits=⎕FPC (without generate one never ending loop for some number of input w due the number e is too small)....

I propose this function that would break convension of return the same type of the input, it would return a v number a float according ⎕FPC variable

r←q3 w;t;e t←1v⋄e←÷10*⌊(⎕FPC-{1>∣w:0⋄3÷⍨1+2⍟∣w})÷3.323v →3×⍳e>∣t-r←t-3÷⍨t-w÷t×t⋄t←r⋄→2 

The error in this case would be less than e←÷10*⌊(⎕FPC-{1>∣w:0⋄3÷⍨1+2⍟∣w})÷3.323v where w is the input number, ⎕FPC is the lenght in bit of the float point 1+2⍟w would be the lenght in bit of integer part of w, but I have to consider the 1/3 of that number because the resul will have 1/3 of these digits...log_2(10)=3.32192809 here approximated as 3.323 (I don't know if it is the right number, 3.322 is not right because I remember generate one never ending loop with the input 18.609625v). Test

 q3¨aaa 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.381590388E¯38  q3¨-aaa ¯3 ¯4 ¯1 ¯2.65 ¯154 ¯0.1 ¯1.912931183 ¯0.00005848035476 1.381590388E¯38 q3¨aaa×1v 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.872107764E¯154 q3¨aaa×1x 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.872107764E¯154 

 q3 2 1.25992105 q3 2v 1.25992105 q3 2x 1.25992105  q3 18.609625x 2.65 

 ⎕fpc←512 512÷3.322 154.1240217 150⍕q 2v 1.259921049894873164779198323884500528076109625698508732184198175469013575303302345933958149409424814616937169595159013873478464862408338554282464069753 150⍕q3 2v 1.259921049894873164767210607278228350570251464701507980081975112155299676513959483729396562436255094154310256035615665259399024040613737228459110304269 

note how q3 wuould be the right result, but q is right until 19th digit.

It is possible that I make some error and there is a number that generate one infinite loop because e variable in code is too little  (for example I wrong the number 3.323 or if variable t or r have integer part > of input w even if t-r decrease)

APL(NARS), 83 chars

r←q w;t;e t←1v⋄e←÷10*⌊(⎕FPC-{1>∣w:0⋄1+2⍟∣w})÷3.322v →3×⍳e>∣t-r←t-3÷⍨t-w÷t×t⋄t←r⋄→2 

//10+42+31=83

I followed the formula written in the post: https://codegolf.stackexchange.com/a/21793/120560. I think I have found the formula for min error allowable from +-*/ operations in the float point lenght xBit, due the precision is affected from the lenght of the integer part of the input number...

The error would be less than e←÷10*⌊(⎕FPC-{1>∣w:0⋄1+2⍟∣w})÷3.322v where w is the input number, ⎕FPC is the lenght in bit of the float point 1+2⍟w would be the lenght in bit of integer part of w, log_2(10)=3.32192809 here approximated as 3.322 (I don't know if it is the right number). Test

 aaa←27 64 1 18.609625 3652264 0.001 7 2E¯13 0 q¨aaa 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.381590388E¯38 q¨-aaa ¯3 ¯4 ¯1 ¯2.65 ¯154 ¯0.1 ¯1.912931183 ¯0.00005848035476 1.381590388E¯38 

note that it seems ok even with rationals...

 q 2 1.25992105 q 2v 1.25992105 q 2x 1.25992105 

for big numbers it is possible too

 ⎕fpc←512 512÷3.322 154.1240217 150⍕q 2v 1.259921049894873164767210607278228350570251464701507980081975112155299676513959483729396562436255094154310256035615665259399024040613737228459110304269 

It is possible that I make some error and there is a number that generate one infinite loop because e variable in code is too little.

APL(NARS), 83 chars

r←q w;t;e t←1v⋄e←÷10*⌊(⎕FPC-{1>∣w:0⋄1+2⍟∣w})÷3.322v →3×⍳e>∣t-r←t-3÷⍨t-w÷t×t⋄t←r⋄→2 

//10+42+31=83

I followed the formula written in the post: https://codegolf.stackexchange.com/a/21793/120560. I think I have found the formula for min error allowable from +-*/ operations in the float point lenght xBit, due the precision is affected from the lenght of the integer part of the input number...

The error would be less than e←÷10*⌊(⎕FPC-{1>∣w:0⋄1+2⍟∣w})÷3.322v where w is the input number, ⎕FPC is the lenght in bit of the float point 1+2⍟w would be the lenght in bit of integer part of w, log_2(10)=3.32192809 here approximated as 3.322 (I don't know if it is the right number). Test

 aaa←27 64 1 18.609625 3652264 0.001 7 2E¯13 0 q¨aaa 3 4 1 2.65 154 0.1 1.912931183 0.00005848035476 1.381590388E¯38 q¨-aaa ¯3 ¯4 ¯1 ¯2.65 ¯154 ¯0.1 ¯1.912931183 ¯0.00005848035476 1.381590388E¯38 

note that it seems ok even with rationals...

 q 2 1.25992105 q 2v 1.25992105 q 2x 1.25992105 

for big numbers it is possible too

 ⎕fpc←512 512÷3.322 154.1240217 150⍕q 2v 1.259921049894873164767210607278228350570251464701507980081975112155299676513959483729396562436255094154310256035615665259399024040613737228459110304269 

It is possible that I make some error and there is a number that generate one infinite loop because e variable in code is too little (for example if variable t or r have integer part > of input w even if t-r decrease)