Moore's law is just an observation and not a real law like the Law of Gravity.
Over the years, developers have taken the increasing CPU speeds and...
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Bitwise operators for the win.
This is a rather beautiful "bit" of problem solving :)
It will work for both integers and floating points, positive and negative, and there is no issue with overflow. If someone can figure this out on their own, I think that's quite impressive. Most of us just know it from having seen it before though.
That's true. My initial solution to it was using addition/subtraction and once I got the concept of diffs, I was able to deduce a solution using bit manipulation.
JavaScript, by nature, stores all numbers as double-precision IEEE 754 floating point numbers. The 52 bits of the fractional portion is used to store integers. This results in 'maximum safe integer' of 2^53 - 1. Further complicating things is that JavaScript's bitwise operators only deal with the lowest 32 bits of JavaScript's 54-bit integers. So the bitwise approach will not work on big integers in JavaScript.
I guess BigInt new primitive proposal will do the trick :)
The swap with bitwise operators will be the fastest at the hardware level. :)
Rather disappointingly, in most modern hardware it's the one with the temp variable that is fastest.
Depending on the architecture and compiler, they end up using three registers to do a swap under the hood I think.
Yes, but the idea is that you do this in assembly. It was a common trick because it saved a register and (at worst) it ran in the same number of cycles.
When I started going through the comments, I was actually thinking "what about bitwise xor?" And sure enough, here it is.
^,^Yes. Whenever you can achieve something using bitwise operations I always prefer that.
x = x + y
y = x - y
x = x - y
Tricky bit: It can cause integer overflow if
x + y > int.MAXassuming cyclic overflows, i.e. INTMAX + 1 = INTMIN:
if
INTMAX - y < x(equivalent tox + y > INTMAXwithout cycles)then
x + y = INTMIN + y - (INTMAX - x - 1)ergo, rewriting the equations as
x' = x + yy' = x' - yx'' = x' - y'we get:
x' = INTMIN + y - (INTMAX - x - 1)y' = x' - yy' = INTMIN - (INTMAX - x - 1)y' = INTMIN + 1 - INTMAX + xy' = xx'' = INTMIN + y - (INTMAX - x - 1) - y'x'' = INTMIN + y - (INTMAX - x - 1) - xx'' = INTMIN + y - INTMAX + x + 1 - xx'' = INTMIN + 1 + y - INTMAXx'' = yOverflows schmoverflows.
EDIT: Might be worth noting that this approach and the bitwise operator are essentially the same thing. Combine two things in a way that is reversible if you have either of the two originals.
Unfortunately that works in theory but in practice and most compilers do not guarantee that behavior. For example:
Java: Will back cycle to -2147483648
Javascript: will eventually print infinity if the value is too high
Go: Will not compile
Validity checks were not a part of the question. But yes, you are right.
That's the second step in becoming a computer scientist.
limit x and y to 16 bit or store 32-bit numbers in 64-bit. max int for 64-bit signed or unsigned is twice that of it's 32-bit counterpart.
You cannot overcome size problem if it's expressed in bits. Only higher-order structures like arrays would enable that and then you are absolutely storing more than two variables (possibly more than one per entry if it's a pointer).
That defeats the purpose of doing that plus you are making a difficult work of complicating things. I suspect is better if you use a temp variable after all!
I suspect it's better if you don't try to copy with only two memory areas, but if you're going to worry about overflowing, then doubling storage size is the least complex thing you could do, and compatible with various CPU operating modes.
Good thing the question was specific to integers. With floating point arithmetic this wouldn't be guaranteed to work in all cases. Especially in Java 😭
This is probably the simplest case that shows the space vs time choice for optimisation. Here you are saving memory but have more operations. 3 arithmetic and 3 assignments. If you used an extra variable you still have 3 assignments but lose the 3 arithmetic but also have to add/remove the variable to from the stack.
JavaScript, by nature, stores all numbers as double-precision IEEE 754 floating point numbers. The 52 bits of the fractional portion is used to store integers. This results in 'maximum safe integer' of 253 - 1. Further complicating things is that JavaScript's bitwise operators only deal with the lowest 32 bits of JavaScript's 54-bit integers. So the bitwise approach will not work on big integers in JavaScript.
This was my solution as well. I honestly never thought about this problem seriously and went for a third variable every time. I'm taking capacity for granted. I learned my lesson. u.u
Wait.. readability of your code is also important!
Sure, it is. That's when you modularize your code and make an inline method or macro out of this.
x = 1 + 1
y = 2 - 1
x = 2 - 1
Nope, still true.
Sorry perceived wrongly my mistake. Deleted the previous comment. But the possibility of overflow is still there in this method.
If that's the case, then why swap at all?
Sorting functions. So you can write something simple like
Which would be inefficient, thus not recommended.
The function usually gets inlined anyway (depending on your compiler) and if it starts with
if x == y { return }then it just compiles toif x != y { /* swap x and y */ }In fact, might as well use a temporary variable at that point if you're expecting for compiler optimizations :shrug:. At that point they'd just become temporary registers and wouldn't have anything in memory, and a single
xchgcallYou could just write the registers yourself then...
Can you please elaborate on that?
x, y = y, x #
To be a little bit sneaky, use part of the function itself as swap space. This would only work if that space could be after the return instruction.
In pseudo-C:
This isn't thread-safe though, and a decent operating system might throw exceptions if you try to have executable code modify itself.
The
NOPs don't actually give you any memory. ThoseNOPs are wherever the executable image was loaded in memory, and the variableyepis on the stack. Assuming your standard C memory model.I misread the post on the phone, but I see what the code is doing. Although, I'm not sure what assigning to NOP does?
C memory layout:
The function code itself is in the
textsegment. The variables for a particular invocation of the function are on the stack. Padding out the function withNOPcreates space within the function, which I'm using as a variable. Basically I'm interpreting the "working memory" part of the post to mean "variables allocated on the stack". Now that I think of it, what I'm doing is basically a static variable, which (according to the link) C puts in theinitialized datasegment if you do it the right way, but then it would legitimately be a variable so I'd lose.It's probably not a legit way to do things (and might be impossible if the text segment is read-only as the link says it might be), but people had already posted the legit ways.
Thanks for that explanation. It's been a while since I've written C/C++, but this approach makes sense. I love the thought process of cleverly using that available memory, even if it doesn't work out. 👏👏👏
The variable
yepis declared within a function body without astaticqualifier, which means it's what the C standard calls an automatic storage duration variable. This means its duration is between the curly braces, and most compilers will use a register or put it on the stack.Being completely nitpicky here, but the variable
yepis declared after its use and also without a type. I'm sure modern compilers wouldn't even let you do that.Modern operating systems load the
textsegment into a virtual memory page marked as read-only. So yes, even if this compiles, it will generate a segfault on modern operating systems.yepisn't a variable, though, it's a label, representing an offset from the start of the function.I agree compilers and OS's will tend to prevent this. I said as much before, and trying to create a proof of concept in assembly has born that out.
It wasn't a completely fruitless exercise though, as it was a chance to learn some assembly, which provided some insights about the challenge that I might make a post about. Basically, swapping using math might not actually be more memory-efficient (however you define that, and without the kind of cheating I've been talking about) than swapping using "a variable".
It creates more memory though.
Only two options, either using memory or CPU
like XOR swap (bullet-proof solution)
Or addition & subtraction (which is not bullet-proof when dealing with a 32-bit register that would need extra care with overflow)
XOR swap is NOT bullet proof, you need to check if x == y first. If x == y, then the function sets both values to zero.
No, it will not. Let set x=5, y=5.
It is a nop actually, but still correct.
Since xor and subtraction were already mentioned, may I chime in with a circular shift over the 64 bit value of the whole function memory?
This one is different, but how exactly would you implement it in a programming language?
Basically, it works like this (x64 assembler):
mov rax, [m64]
ror rax, 32
mov [m64], rax
No, I mean on a high level programming language which you have no idea where is stored.
Even for an assembly, the two variables could potentially be stored in two different registers. But we're going to assume not!
Cool idea anyways
I love you
Very contrived, only works on ARM, not even acceptable by all C compilers, but you never constrained the problem that way!
I went to bed, and realized that my answer is wrong. Exchanging the values in register is completely invisible to the function caller. Here's the correct answer.
Further, as
R0-R3are all caller-save registers in the ARM ABI, the function body, excluding the function entry stack shenanigans, uses zero memory.Oh and as prplz mentioned, any self-respecting compiler nowadays will use only registers any way you decide to implement this little stub of a function. Yes yes, the question is very contrived too, but hey.
Aren't you using a third register though?
Are you're thinking that you can do that because it's not in RAM?
The way the question is formulated, you're limited by having a certain amount of memory. Any register use doesn't count against that at all.
That's awesome. I did not know about it. I wonder if it swaps values instead of pointters.
Found this on SO @ stackoverflow.com/questions/445179....
So now we've reduced the problem to "How do you swap the two topmost stack items?" :)
As funny as it is, you declare a new pointer to the
swapfunction allocating new memory.I've written my answer with C in mind, and it does explain how it works if you try the calculation by hand: onlinegdb.com/B1IY6hFK7
When both
aandbhas the same value: onlinegdb.com/r1WeypFF7Result may vary depending on the language compiler, you might have to adjust the snippet according to your language's nuances.
The math trick works for numbers, but in Go:
(I think at the assembly level it uses a 3rd registry, but I'm not sure)
Same in Ruby.
I wrote the code based on the title only :D, i.e. swapping any 2 variables' values (any type) without using a third one:
Having $x and $y:
$x .= "-" . $y;
$y = explode("-", $x)[0];
$x = explode("-", $x)[1];
You only have 8 bytes remember?, this string manipulation takes more than that
I like defining one of these macros when swapping variables in C:
The second one works on floating point numbers too.
That's a good one. Although it won't work in Javascript I think.
x=10;y=5;console.log(x,y);(x=(x-(y = (x=x+y)-y)));console.log(x,y)
I think it doesn't like the double assignment on same expression. In JS you can do:
or
although less readable.
[] allocates new space I think so this solution won't be in constant space.
No need to go very far... and it will never cause overflow
x,y = y,x;
Syntaxic sugar :)
No idea if this counts
I'm sure this can be improved, but trying to get other stuff done ATM :D
working demo: dotnetfiddle.net/hP6T94
Thanks for the working demo! That is the correct solution. :)
Let's say the two values are in the
EAXandEBXregisters. Here is the fastest :In python
am i right ?
oh shorter,
So now, what is the second step ? :)
Love your thinking. Nobody ever said you can't write to file and read it back :D
This is a nice puzzle to work out.
Another interesting(?) lesson is that beyond saving space, once this solution used to really be faster than the "naive method", but on modern CPUs that is no longer the case.
I've seen this question come up a lot, along with fizzbuzz or "write a sorting function" and I'm not sure it's useful anymore.
It's a purely academic problem that touches on several different concepts, none of which are going to be used by 99% of programmers these days.
Why do you ask this question? What do people's different answers tell you, or are you more looking for how people attempt to solve it (if they've not heard it before) or how quickly they regurgitate a stock solution (if they have?)
If I asked this, and someone answered quickly (they already know why the dead man in the desert is lying next to an unopened pack...) then I'd follow up with something asking when they had ever had to do this in real life, or even if they could come up with three examples of situations where it would be a useful thing to do in real life. If they couldn't, then I don't think I would have learned anything useful about them.
What do you get out of it?
Good question! I used to ask this as a warm-up question back in the days at the beginning of an interview. (However, I've also used it in recent interviews after this post). I've never used this question to influence the outcome of the interview. It's merely conversational and a segue into academics to pique their interest into a deeper technical conversation.
If the candidate has heard the question before, I will follow up with a question around if the solution handles overflow when numbers are large and also dive a bit into perf. We may dive into when and if this solution might be needed for specific use-cases. The reality is that the compiler optimizes the code better when using a temporary variable.
If they haven't heard the question before, it gives me a chance to assess how they work through the problem and helps me find an appropriate role for them if the interview is somewhat generic. If they dislike that problem or find it difficult to solve it, they are most likely not going to enjoy working on hardware/firmware. Most people who solve it really enjoy the problem and in my experience have performed quite well on the job.
Recently, I observed that this question is part of the prep material online and most people have seen it, so there's not much value in asking it. I still do it for fun sometimes.
Anyway, I agree with your comment though.
JAVA
int a=5;
int b=10;
a=a+b;
b=a-b;
a=a-b;
System.out.println(a);//10
System.out.println(b);//5
String s1="java1";
String s2="java2";
s1=s1+s2;
s2=s1.substring(0,(s1.length()-s2.length()));
s1=s1.substring(s2.length());
System.out.println(s1);//java2
System.out.println(s2);//java1
Here's my code snippet showing how we can swap two numbers without using temporary variable at coderseditor.com/?id=185
using System;
namespace CodeSample
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("Swap Number Sample");
int number1, number2;
Console.WriteLine("Enter the First Number : ");
number1 = int.Parse(Console.ReadLine());
Console.WriteLine("Enter the Second Number : ");
number2 = int.Parse(Console.ReadLine());
number1 = number1 - number2;
number2 = number1 + number2;
number1 = number2 - number1;
Console.WriteLine("First Number is " + number1);
Console.WriteLine("Second Number is " + number2);
Console.Read();
}
}
}
I think I must have skipped a few steps. The only occasion I've had to use such tricks is when playing Zachtronics games like TIS-100 or Shenzhen IO. I hope you are interviewing for embedded systems programmers. ;-)
The Law of Gravity is also just an observation :)
Someone has already posted my XOR solution 😄 this is one we occasionally ask in interviews, and it's one of my favorite abstract problems
Do you know how this is implemented internally?
let [x, y] = [y, x]
One extra reason to love JavaScript.
With non-blocking assignment to describe hardware in Verilog:
x <= y;
y <= x;
asic-world.com/tidbits/blocking.html
Not after you have replaced y in the second step.
[x, y] = [y, x]
Thanks for this post, I also take the modern "common" hardware for too granted.
Good reminder :)
Python a,b=b,a
Can I use pointers?
x:=1
y:=2
*(&x) = y
*(&y) = x
You can do anything as long as you swap the values and don't create more memory.
In your solution, x is overwritten by y so it isn't swapping both the values.
Ah, I thought it wouldn't work but at midnight computer tells you lies :).
The multiple assignment in Go ( assuming C/C++ are the same mechanism, and i see Python below) works though.
x, y = y,x
By using the ES6 destructuring assignment, that's how I would do it.
Does this solution meet memory constraints?
In general, it does unless you're talking about very large ones.
The entire swap ends up using three registers I think.
Move data to upper and lower RAX then ROL by 32
You could use the K combinator to swap any value, not just numbers!
Clearly I'm at step -1 🤓😂
if (y > x){
x += (y-x)/2;
y = 2*(y-x)
x += x-y
}
else if (y < x){
y += (y-x)/2;
x = 2*(y-x)
y += x-y
}
Probably not the best way ...
You might want to try it in a language of your choice. If it works, it's a very good first step. Then, you can think of optimizing it.
Wouldn't a bit swap work?
Your comments appear incorrect.
y = y-x is basically the difference between y and x and not -x.
You might want to try this solution in code for correctness.
Wouldn't that have upstream effects on the program?
If there are two variables a and b we could destructure them.
var a = 12;
var b = 21;
[a, b] = [b, a];
Values are interchanged :)
A = 00000000000000000000000000101010
B = 00000000000000000000000001111110
A = A or B = 00000000000000000000000001111110
B = A and B = 00000000000000000000000000101010