Shortest expression for {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4}

Question

Given list of integers {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4}. For those who interested these numbers are used in weekday calculation.

Weekday = (m[n] + d + y + y>>2 + y/400 - y/100) % 7;, where m[n] - expression I'm searching, d - day of month, y - year - (month <= 2).

Construct expression consisting of arithmetic, logic and bitwise operators, which will output for positive integer n integer m so that m % 7 equals n-th number in the list.

Branches, ternary operators, table lookups and pointers are not allowed.

Score:
1 - for | & ^ ~ >> << operators
1.1 - for + - < > <= >= == != ! && || operators
1.2 - for * operator
1.4 - for / % operators

Answer with lowest score wins.

Personally I have found:

(41*n)>>4+((n+61)>>4)<<2 with score 6.4. I thought this will be hard to find so provided own expression to start with.

Answer 1

2 2.2

I love arbitrary precision arithmetic.

0x4126030156610>>(n<<2)

Or, if you don't like hex,

1146104239711760>>(n<<2)

Test:

print([(0x4126030156610>>(n<<2))%7 for n in range(1,13)])
[0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]

Answer 2

2.0

(127004 >> i) ^ 60233

or (score 2.2) :

(i * 3246) ^ 130159

All found with brute force :-)

Answer 3

35.3

I suspect this may be the least efficient method to create the list:

1.7801122128869781e+003 * n - 
1.7215267321373362e+003 * n ^ 2 + 
8.3107487075415247e+002 * n ^ 3 - 
2.0576746235987866e+002 * n ^ 4 + 
1.7702949291688071e+001 * n ^ 5 + 
3.7551387326116981e+000 * n ^ 6 - 
1.3296432299817251e+000 * n ^ 7 + 
1.8138635864087030e-001 * n ^ 8 - 
1.3366764519057219e-002 * n ^ 9 + 
5.2402527302299116e-004 * n ^ 10 - 
8.5946393615396631e-006 * n ^ 11 -
7.0418841304671321e+002

I just calculated the polynomial regression. I'm tempted to see what other terrible method could be attempted.

Notably, I could save 3.3 points if the result was rounded. At this point, I don't think that matters.

Answer 4

3.2

Zero based solution:

7 & (37383146136 >> (i*3))

One based solution:

7 & (299065169088 >> (i*3))

I initially thought that the %7 operation would be counted as well and % being an expensive operation here, I tried to solve it without it.

I came to a result of 3.2 like this:

// Construction of the number
// Use 3 bits per entry and shift to correct place
long c = 0;
int[] nums = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
for (int i = nums.Length - 1; i >= 0; i--)
{
    c <<= 3;
    c += nums[i];
}
// c = 37383146136

// Actual challenge
for (int i = 0; i < 13; i++)
{
    Console.Write("{0} ",7 & 37383146136 >> i*3);
}

I'd be interested in optimizations using this approach (without %). Thanks.

Answer 5

Python (3)

Since there are quite a few of these questions these days, I decided to make a program to automatically solve them in 3 (or 2) tokens. Here's the result for this challenge:

G:\Users\Synthetica\Anaconda\python.exe "C:/Users/Synthetica/PycharmProjects/PCCG/Atomic golfer.py"
Input sequence: 0 3 2 5 0 3 5 1 4 6 2 4
f = lambda n: (72997619651120 >> (n << 2)) & 15
f = lambda n: (0x426415305230L >> (n << 2)) & 15
f = lambda n: (0b10000100110010000010101001100000101001000110000 >> (n << 2)) & 15

Process finished with exit code 0

Proof that this works:

f = lambda n: (72997619651120 >> (n << 2)) & 15

for i in range(12):
   print i, f(i)
 
0 0
1 3
2 2
3 5
4 0
5 3
6 5
7 1
8 4
9 6
10 2
11 4

Answer 6

4.3, 20 chars in python and yet 32bit assembly friendly

For completeness, I also add this python expression which was part of a quora answer some years ago using the so-called kefalonian constant [1], [2] :

• 0xd50e9994>>36-3*m&7

Test case:

• print([0xd50e9994>>(36-3*m)&7 for m in range(1,13)])

[1] https://www.quora.com/What-does-floor-division-in-Python-do/answer/Fotis-Georgatos

[2] https://qr.ae/psZnDr