What does the 8 stand for in the Cantor pairing function?
$begingroup$
- What does the "8" stand for in the Cantor pairing function (which assigns a natural number to each pair of natural numbers)?
- Can I change it, or is it constant?
- The other function for getting the unique number when you use the two inputs, doesn't use "8" in its calculations, so why did the 8 gives us the right answer?
Below is the formula to find x and y such that π(x, y) = 1432:
8 × 1432 = 11456,
11456 + 1 = 11457,
√11457 = 107.037,
107.037 − 1 = 106.037,
106.037 ÷ 2 = 53.019,
⌊53.019⌋ = 53,
so w = 53;
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
so t = 1431;
1432 − 1431 = 1,
so y = 1;
53 − 1 = 52,
so x = 52; thus π(52, 1) = 1432.
To calculate π(52, 1):
52 + 1 = 79,
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
1431 + 1 = 1432,
so π(52, 1) = 1432.
probability combinatorics number-theory functions algorithms
edited Dec 10 '18 at 23:36
Blue
49.2k870157
asked Dec 10 '18 at 23:31
king amadaking amada
125
$endgroup$
add a comment |
$begingroup$
- What does the "8" stand for in the Cantor pairing function (which assigns a natural number to each pair of natural numbers)?
- Can I change it, or is it constant?
- The other function for getting the unique number when you use the two inputs, doesn't use "8" in its calculations, so why did the 8 gives us the right answer?
Below is the formula to find x and y such that π(x, y) = 1432:
8 × 1432 = 11456,
11456 + 1 = 11457,
√11457 = 107.037,
107.037 − 1 = 106.037,
106.037 ÷ 2 = 53.019,
⌊53.019⌋ = 53,
so w = 53;
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
so t = 1431;
1432 − 1431 = 1,
so y = 1;
53 − 1 = 52,
so x = 52; thus π(52, 1) = 1432.
To calculate π(52, 1):
52 + 1 = 79,
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
1431 + 1 = 1432,
so π(52, 1) = 1432.
probability combinatorics number-theory functions algorithms
edited Dec 10 '18 at 23:36
Blue
49.2k870157
asked Dec 10 '18 at 23:31
king amadaking amada
125
$endgroup$
2
$begingroup$
As described on the Wikipedia "pairing function" entry, the algorithm for retrieving $x$ and $y$ involves solving $w^2+1w-2t=0$. By the quadratic formula, the solution involves the square root of the expression $$1^2-4cdot 1cdot(-2t) qquadtext{that is,}qquad 8t+1$$ So, the $8$ is definitely a constant. Whether there's some deep "meaning" to the value ... I don't know.
$endgroup$
– Blue
Dec 10 '18 at 23:57
add a comment |
$begingroup$
- What does the "8" stand for in the Cantor pairing function (which assigns a natural number to each pair of natural numbers)?
- Can I change it, or is it constant?
- The other function for getting the unique number when you use the two inputs, doesn't use "8" in its calculations, so why did the 8 gives us the right answer?
Below is the formula to find x and y such that π(x, y) = 1432:
8 × 1432 = 11456,
11456 + 1 = 11457,
√11457 = 107.037,
107.037 − 1 = 106.037,
106.037 ÷ 2 = 53.019,
⌊53.019⌋ = 53,
so w = 53;
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
so t = 1431;
1432 − 1431 = 1,
so y = 1;
53 − 1 = 52,
so x = 52; thus π(52, 1) = 1432.
To calculate π(52, 1):
52 + 1 = 79,
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
1431 + 1 = 1432,
so π(52, 1) = 1432.
probability combinatorics number-theory functions algorithms
edited Dec 10 '18 at 23:36
Blue
49.2k870157
asked Dec 10 '18 at 23:31
king amadaking amada
125
$endgroup$
- What does the "8" stand for in the Cantor pairing function (which assigns a natural number to each pair of natural numbers)?
- Can I change it, or is it constant?
- The other function for getting the unique number when you use the two inputs, doesn't use "8" in its calculations, so why did the 8 gives us the right answer?
Below is the formula to find x and y such that π(x, y) = 1432:
8 × 1432 = 11456,
11456 + 1 = 11457,
√11457 = 107.037,
107.037 − 1 = 106.037,
106.037 ÷ 2 = 53.019,
⌊53.019⌋ = 53,
so w = 53;
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
so t = 1431;
1432 − 1431 = 1,
so y = 1;
53 − 1 = 52,
so x = 52; thus π(52, 1) = 1432.
To calculate π(52, 1):
52 + 1 = 79,
53 + 1 = 54,
53 × 54 = 2862,
2862 ÷ 2 = 1431,
1431 + 1 = 1432,
so π(52, 1) = 1432.
probability combinatorics number-theory functions algorithms
probability combinatorics number-theory functions algorithms
edited Dec 10 '18 at 23:36
Blue
49.2k870157
asked Dec 10 '18 at 23:31
king amadaking amada
125
edited Dec 10 '18 at 23:36
Blue
49.2k870157
asked Dec 10 '18 at 23:31
king amadaking amada
125
edited Dec 10 '18 at 23:36
Blue
49.2k870157
edited Dec 10 '18 at 23:36
Blue
49.2k870157
edited Dec 10 '18 at 23:36
Blue
49.2k870157
49.2k870157
asked Dec 10 '18 at 23:31
king amadaking amada
125
asked Dec 10 '18 at 23:31
king amadaking amada
125
asked Dec 10 '18 at 23:31
king amadaking amada
125
125
2
$begingroup$
As described on the Wikipedia "pairing function" entry, the algorithm for retrieving $x$ and $y$ involves solving $w^2+1w-2t=0$. By the quadratic formula, the solution involves the square root of the expression $$1^2-4cdot 1cdot(-2t) qquadtext{that is,}qquad 8t+1$$ So, the $8$ is definitely a constant. Whether there's some deep "meaning" to the value ... I don't know.
$endgroup$
– Blue
Dec 10 '18 at 23:57
add a comment |
2
$begingroup$
As described on the Wikipedia "pairing function" entry, the algorithm for retrieving $x$ and $y$ involves solving $w^2+1w-2t=0$. By the quadratic formula, the solution involves the square root of the expression $$1^2-4cdot 1cdot(-2t) qquadtext{that is,}qquad 8t+1$$ So, the $8$ is definitely a constant. Whether there's some deep "meaning" to the value ... I don't know.
$endgroup$
– Blue
Dec 10 '18 at 23:57
2
2
$begingroup$
As described on the Wikipedia "pairing function" entry, the algorithm for retrieving $x$ and $y$ involves solving $w^2+1w-2t=0$. By the quadratic formula, the solution involves the square root of the expression $$1^2-4cdot 1cdot(-2t) qquadtext{that is,}qquad 8t+1$$ So, the $8$ is definitely a constant. Whether there's some deep "meaning" to the value ... I don't know.
$endgroup$
– Blue
Dec 10 '18 at 23:57
$begingroup$
As described on the Wikipedia "pairing function" entry, the algorithm for retrieving $x$ and $y$ involves solving $w^2+1w-2t=0$. By the quadratic formula, the solution involves the square root of the expression $$1^2-4cdot 1cdot(-2t) qquadtext{that is,}qquad 8t+1$$ So, the $8$ is definitely a constant. Whether there's some deep "meaning" to the value ... I don't know.
$endgroup$
– Blue
Dec 10 '18 at 23:57
add a comment |
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$begingroup$
As described on the Wikipedia "pairing function" entry, the algorithm for retrieving $x$ and $y$ involves solving $w^2+1w-2t=0$. By the quadratic formula, the solution involves the square root of the expression $$1^2-4cdot 1cdot(-2t) qquadtext{that is,}qquad 8t+1$$ So, the $8$ is definitely a constant. Whether there's some deep "meaning" to the value ... I don't know.
$endgroup$
– Blue
Dec 10 '18 at 23:57