3 ms·

I think the whole approach with print5, print25 is the "correct" solution, but if I were interviewing I would give the best score to the person that gave system

by powera 16y ago

I think the whole approach with print5, print25 is the "correct" solution, but if I were interviewing I would give the best score to the person that gave system("/usr/bin/seq 1000"); as the answer.

acqq 16y ago
I wouldn't as that's clearly not a C or C++ solution but a Unix one.
tzs 16y ago
The idea of building up by having something that prints 5, and then using that to build a print 25, and so on, is good. That then inspires another question: given a function, foo(), that we wish to call exactly 1000 times, what is the way to do it in the least amount of code given that we are only allowed the following: 1. calling foo(). 2. defining new functions, whose bodies consist entirely of calls to previously defined functions or to foo. 3. calling those new functions. For computing amount of code, let's take it as 1 line per function call, and 1 line per function defined. So, for example (in pseudocode): function foo_50 foo foo ... foo foo_50 foo_50 ... foo_50 where there are 50 foo's in foo_50, and 20 calls to foo_50, would be a program of length 71. Breaking it down by powers of two: function foo_2 foo foo function foo_4 foo_2 foo_2 ... function foo_512 foo_256 foo_256 foo_512 foo_256 foo_128 foo_64 foo_32 foo_8 would have length 33. Is that the best we can do? How about for a general N? How about writing a program that given N writes the optimal program in the above form for N?
- RodgerTheGreat 16y ago
  This reminds me a bit of generating constants in BF: http://esolangs.org/wiki/Brainfuck_constants http://esolangs.org/wiki/Brainfuck_constants
- endtime 16y ago
  I think the final output of a simple solution to your last question can always be just four lines. def bar(baz): baz() baz() bar(bar(bar(bar(foo)))) and bar(bar(foo)) and foo() for N=21, for example. Actually, you can make it two lines if your goal is really to minimize LOC: def bar(baz): baz() and baz() bar(bar(bar(bar(foo)))) and bar(bar(foo)) and foo()
  - endtime 16y ago
    Whoops, `baz() and baz()` actually doesn't work, because foo may not return a truthy value (and bar definitely doesn't). One fix, which actually makes it even shorter, is just `baz(), baz()`. Same goes for `bar(bar(...)) and bar()`; all the `and`s should be commands instead in my solution above. Of course, this is Python-specific.
    - endtime 16y ago
      s/commands/commas/g

by powera 16y ago

I think the whole approach with print5, print25 is the "correct" solution, but if I were interviewing I would give the best score to the person that gave system("/usr/bin/seq 1000"); as the answer.

acqq 16y ago

I wouldn't as that's clearly not a C or C++ solution but a Unix one.

tzs 16y ago

The idea of building up by having something that prints 5, and then using that to build a print 25, and so on, is good. That then inspires another question: given a function, foo(), that we wish to call exactly 1000 times, what is the way to do it in the least amount of code given that we are only allowed the following: 1. calling foo(). 2. defining new functions, whose bodies consist entirely of calls to previously defined functions or to foo. 3. calling those new functions. For computing amount of code, let's take it as 1 line per function call, and 1 line per function defined. So, for example (in pseudocode): function foo_50 foo foo ... foo foo_50 foo_50 ... foo_50 where there are 50 foo's in foo_50, and 20 calls to foo_50, would be a program of length 71. Breaking it down by powers of two: function foo_2 foo foo function foo_4 foo_2 foo_2 ... function foo_512 foo_256 foo_256 foo_512 foo_256 foo_128 foo_64 foo_32 foo_8 would have length 33. Is that the best we can do? How about for a general N? How about writing a program that given N writes the optimal program in the above form for N?

RodgerTheGreat 16y ago
This reminds me a bit of generating constants in BF: http://esolangs.org/wiki/Brainfuck_constants http://esolangs.org/wiki/Brainfuck_constants
endtime 16y ago
I think the final output of a simple solution to your last question can always be just four lines. def bar(baz): baz() baz() bar(bar(bar(bar(foo)))) and bar(bar(foo)) and foo() for N=21, for example. Actually, you can make it two lines if your goal is really to minimize LOC: def bar(baz): baz() and baz() bar(bar(bar(bar(foo)))) and bar(bar(foo)) and foo()
- endtime 16y ago
  Whoops, `baz() and baz()` actually doesn't work, because foo may not return a truthy value (and bar definitely doesn't). One fix, which actually makes it even shorter, is just `baz(), baz()`. Same goes for `bar(bar(...)) and bar()`; all the `and`s should be commands instead in my solution above. Of course, this is Python-specific.
  - endtime 16y ago
    s/commands/commas/g