I found this intersting challenge while I was reading an article on Linus Åkesson site, It attracted my attention, and I spent hours trying to figure it out until I successfully did! It was worth it.

Here is my explanation to the program, and my thought process.

Try to figure this out by yourself first

I am not a fan of Magic Numbers so first thing that took my attention were these numbers 1073741824, 30, 53, 55.

In computer programming, the term Magic Number is the unique values with unexplained meaning or multiple occurrences which could (preferably) be replaced with named constants

1. 1073741824 is the value of $2^{30}$, (so I knew that I am going to deal with a lot of 0s and 1s).
2. 30, something related to #1.
3. 53, ASCII representation for the number 5.
4. 55, ASCII representation for the number 7.

5 and 7? hmmm, let’s just remember these numbers and take them into considration.

Now, let’s break this program into two major parts:

1. The recurse function
2. The for loop over the exception message bytes!

🤯 My mind at some point: Why would anyone in the world loop over every byte in an Exception message?!! What is this weirdo recursive function trying to do??!! This program must be drunk!

I wrote a lot of Java code, but to be honest, this is the first time I see this syntax, so it was something new for me.

for(byte b : s.toString().getBytes()) switch(b) {
case ..
case ..
}


But of course this is not the biggest problem now.

The program takes as input two numbers let’s say they are (a) and (b) respectively. We assigne to the vairable y the value of a and pass b to that weird recurse function.

At the first moment I thought there is something wrong with recurse function; the two conditional branches does the same thing! What is the value of this check if(0 != (x & 1073741824)) then?! What’s the difference between line 5 and 7 they have the same exact line of code?!

💡 Gotcha moment:
WAIT A MINUTE!! 5 and 7? I saw these numbers before!

Exactly they are what switch cases referes to. Apparently that weird part that loops over the exception bytes, checks every stack frame that contains a number indicating the line number the recurse function were called from. (If it was coming from line 5 or line 7).

Now the important question is: What exactly the recurse function trying to do?

The recurse function take as input two arguemnts the number b (second arguments to the program) and the value 30, which will help in reaching the stopping condition. The recursive function stopping condition here is not something normal, it’s an exception thrown by line 5 or 7 indicating a division by zero, that we catch in line 13.

💡 Gotcha moment:
That’s why n - n / n is not same as n - 1.
At least in this program :)

To understand the effect of passing x * 2 to the recurse function we need understand how multiplying by 2 changes the bits in the binary representation.

3 (0b0011) * 2 = 6  (0b0110)
4 (0b0100) * 2 = 8  (0b1000)
5 (0b0101) * 2 = 10 (0b1010)


When we multiply by 2 we left shift bits by 1.

The number 1073741824 in binary is

0b1000000000000000000000000000000


There are 30 0s and only the most significant bit is on, so if we AND any value smaller than 1073741824 with it we will get 0.

if we started by passing the value 3 to the recurse function the if condition will always be false until and the recurse call in line 7 will get called which we reach when we get a value bigger than 1073741824 and the bit number 31 is on, in this case the condition will be true and we will go to line 5.

How many calls we need to reach line 5?

• 29 calls exactly.

because

2^29 * 3 = 1610612736 > 1073741824 (and bit 31 is 1)
2^28 * 3 = 805306368 < 1073741824


The number 3 (0b0011) has two 1s, so when we shift them 29 times and AND them with 2^30 that has 1 at bit 31 it will make the condition true, and we will go to line 5.

And we will have a call stack like this:

(recurse:5) --> 1
(recurse:5) --> 1
(recurse:7) --> 0
(recurse:7) --> 0
(recurse:7) --> 0
(recurse:7) --> 0
(recurse:7)  ...
(recurse:7)  ...
(recurse:7)  ...
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)
(recurse:7)  ...
(recurse:7) --> 0


At the top of stack we will find the two calls from line 5 after we shifted the 1s bits in number 3 29 times.

💡 Gotcha moment:
The recurse function is trying to represent the number 3 in binary using the exception stack frames.

That now answers a lot of questions regarding the for loop over the exception bytes. This program is trying to loop over the bits of second number (b) starting form the righmost bit, and the switch cases is differentiating between the zeros and ones. It’s like saying

case 0: x += y;
case 1: y += y;


Now comes the last and hardest part, understanding what this program is trying to achieve.

if a = 3, and b = 7 we will have X = 0, Y = 3.

While looping over the bits of b, every time we encounter 0 we will double the value of Y (3, 6, 12, 24, 48, …) and every time we encounter 1 we will add previous value of Y to X “and” double the value of Y

🎗️ Recall:
In Java we need to break after each case to avoid executing the next case block.

initial values:

-----------------------------
|   X     | Y  |      b     |
-----------------------------
|   0     | 3  | 0b0111 (7) |
-----------------------------


iteration #1 (bit: 1):

X = 0 + 3 = 3, Y = 6


iteration #2 (bit: 1):

X = 3 + 6 = 9, Y = 12


iteration #3 (bit: 1):

X = 9 + 12 = 21, Y = 24


all coming iteration won’t change the value of X so thet doesn’t matter. The final value of X is 21 so when b = 7, Y = 3 X will equal 21, this looks like a multiplication!

let’s try another set of value to make sure our hypothesis is correct

-----------------------------
| X (res) | Y  |      b     |
-----------------------------
|   0     | 4  | 0b0111 (7) |
-----------------------------


iteration #1 (bit: 1):

X = 0 + 4 = 4, Y = 8


iteration #2 (bit: 1):

X = 4 + 8 = 12, Y = 16


iteration #3 (bit: 1):

X = 12 + 16 = 28, Y = 32


4 * 7 = 28 ✅

Let’s generalize this pattern:

4 (0b0100) * 3 ---> (    4    )  * 3
5 (0b0101) * 3 ---> (  4 + 1  )  * 3
6 (0b0110) * 3 ---> (  4 + 2  )  * 3
7 (0b0111) * 3 ---> (4 + 2 + 1)  * 3
8 (0b1000) * 3 ---> (    8    )  * 3


That’s exactly what the program is trying to do, it converts b to binary representation using the exception stack frames trick, then it loops over the binary bits (frames) and for every 1 bit it tries to solve a sub-problem (something like dynamic programming) depending on the position of the bit. e.g If it was the third bit from the right then 2^(position-1) 2^2=4, let a = 3 then (4 * 3) = 12, it stores this value in X. Now if it encountered another 1 bit in the forth bit i.e (0b1100) which is binary representation of 12 then it sums the previous value of X (12) with the value of (2^3=8) * 3 which is equal to 24, then we have the final result = 24 + 12 = 36 and that’s the value of 3 * 12.

It turns out that this way of multiplication is called Ancient Egyptian multiplication ;)

That was very intersting indeed, hope you enjoyed as much as I did.

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