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Hi Ryan,

It was amusing to realize that I have stated the rule in a complicated manner.

Taking your hint, if the difference of 2 numbers is odd, then one of them is even and the other odd - which implies that their sum is also odd.

That combined with my logic simply means that the sum of all the digits in the number must be odd.

So here is the revised version -

You like numbers where the*sum of digits of the numbers is ***Odd**.

That makes me realize that this is almost identical to Seetharaman's logic, except that I stop only at the first pass of summing up the digits; and not summing up the digits of the sum itself.

Regards,

Anubrato

It was amusing to realize that I have stated the rule in a complicated manner.

Taking your hint, if the difference of 2 numbers is odd, then one of them is even and the other odd - which implies that their sum is also odd.

That combined with my logic simply means that the sum of all the digits in the number must be odd.

So here is the revised version -

You like numbers where the

That makes me realize that this is almost identical to Seetharaman's logic, except that I stop only at the first pass of summing up the digits; and not summing up the digits of the sum itself.

Regards,

Anubrato

5 years ago

My take is that Ryan likes numbers where the *difference of sum of the alternate digits is ***odd**.

Here is my logic -

Ryan's likes:

25 => 5 – 2 = 3**Odd**

144 => (1+4) – 4 = 1**Odd**

300 => (3+0) – 0 = 3**Odd**

Ryan's dislikes:

24 => 4 – 2 = 2**Even**

145 => (1+5) – 4 = 2**Even**

400 => (4+0) – 0 = 4**Even**

Also, as already declared,

8889 => (8+9) – (8+8) = 1**Odd**

97 => 9 -7 = 2**Even**

So given all the above, the available options are:

37 => 7 – 3 = 4**Even**

64 => 6 – 4 = 2**Even**

200 => (2+0) – 0 = 2**Even**

1024 => (4+0) – (1+2) = 1**Odd**

65535 => (6+5+5) – (3+5) = 8**Even**

That leaves 1024 as the only option

Here is my logic -

Ryan's likes:

25 => 5 – 2 = 3

144 => (1+4) – 4 = 1

300 => (3+0) – 0 = 3

Ryan's dislikes:

24 => 4 – 2 = 2

145 => (1+5) – 4 = 2

400 => (4+0) – 0 = 4

Also, as already declared,

8889 => (8+9) – (8+8) = 1

97 => 9 -7 = 2

So given all the above, the available options are:

37 => 7 – 3 = 4

64 => 6 – 4 = 2

200 => (2+0) – 0 = 2

1024 => (4+0) – (1+2) = 1

65535 => (6+5+5) – (3+5) = 8

That leaves 1024 as the only option

5 years ago

Hello All,

I am trying to come up with an algorithm for a Tree as follows -

I have a set of parent-child relationships defined (please see attached file). For each relationship, the parent objct is a node on the tree and all its children are immediate neighbours of this node, as in any tree.

Now, I want to validate a 'sequence' for the tree nodes, which works as following.

Arbitrary 'sequence' numbers are assigned to each node e.g. A=1, B=2, D=3, C=4, F=5, E=6. Now, I want to 'validate' this sequence. The sequence is valid if for every node, all its parent nodes have a lower sequence number than itself.

For example, if for node C, both B and A are parents, then these sequences are valid: A=1, B=2, C=3 or A=2, B=1, C=3.

In the attached diagram, for example, the node A has children B and C. B has children D and E while C has only one child F.

Now, the following sequences are examples of 'valid' sequences -

1. A=1, B=2, C=3, D=4, E=5, F=6

2. A=1, B=2, D=3, C=4, F=5, E=6

1. A=1, C=2, F=3, B=4, D=5, E=6

Can anyone please suggest an efficient algorithm to achieve this?

Regards,

Anubrato

I am trying to come up with an algorithm for a Tree as follows -

I have a set of parent-child relationships defined (please see attached file). For each relationship, the parent objct is a node on the tree and all its children are immediate neighbours of this node, as in any tree.

Now, I want to validate a 'sequence' for the tree nodes, which works as following.

Arbitrary 'sequence' numbers are assigned to each node e.g. A=1, B=2, D=3, C=4, F=5, E=6. Now, I want to 'validate' this sequence. The sequence is valid if for every node, all its parent nodes have a lower sequence number than itself.

For example, if for node C, both B and A are parents, then these sequences are valid: A=1, B=2, C=3 or A=2, B=1, C=3.

In the attached diagram, for example, the node A has children B and C. B has children D and E while C has only one child F.

Now, the following sequences are examples of 'valid' sequences -

1. A=1, B=2, C=3, D=4, E=5, F=6

2. A=1, B=2, D=3, C=4, F=5, E=6

1. A=1, C=2, F=3, B=4, D=5, E=6

Can anyone please suggest an efficient algorithm to achieve this?

Regards,

Anubrato

6 years ago