Mike Simmons wrote:I think six is the minimum number of people involved. (Well, for, how shall I put it... traditional family trees. ) One solution is to take Salvin's answer and send E or F home. We need one or the other, but we don't need both. Or we can send them both home, if we can augment G with another cousin of opposite gender.
salvin francis wrote:
Mike Simmons wrote:I think six is the minimum number of people involved. (Well, for, how shall I put it... traditional family trees. ) One solution is to take Salvin's answer and send E or F home. We need one or the other, but we don't need both. Or we can send them both home, if we can augment G with another cousin of opposite gender.
They are both required actually.
If i am a guy - A
and i came with my sis - B
and my mom - C
and my Dad - D
and my Mon/Dad's Brother - E
[omit F]
and his son/daughter - G.
1. someone was there with their mother. A to C
2. someone was there with their father. A to D
3. someone was there with their sister. A to B
4. someone was there with their brother. B to A
5. someone was there with their son. C or D to A
6. someone was there with their daughter. C or D to B
7. someone was there with their aunt. G to C
8. someone was there with their uncle. A or B to E
9. someone was there with their neice. E to B
10. someone was there with their nephew. E to A
11. someone was there with their cousin. A or B to G
Total = 6
Mike Simmons wrote:I think six is the minimum number of people involved. (Well, for, how shall I put it... traditional family trees. ) One solution is to take Salvin's answer and send E or F home. We need one or the other, but we don't need both. Or we can send them both home, if we can augment G with another cousin of opposite gender.
Ryan McGuire wrote:Six is still higher than I'm looking for. And yes, we're talking about traditional family trees. i.e. nothing like I'm My Own Grandpa.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Mike Simmons wrote:
salvin francis wrote:
Ryan McGuire wrote:
salvin francis wrote:
There is a Titar in front of a Titar
There is a Titar behind a Titar
Titar in front,
Titar behind,
How many Titar in all ?
Let's see... I drew a diagram of the scenario described. I have a row of 15 titars side-by-side. In front of the fifth one is one more titar, so...
16. Final answer.
Answer = 3
[Titar][Titar][Titar]
Two seems sufficient.
fred rosenberger wrote:...I'll say 5.
The solution is obvious, and is left as an exercise for the reader.Ryan McGuire wrote:
fred rosenberger wrote:...I'll say 5.
Show your work.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
fred rosenberger wrote:
The solution is obvious, and is left as an exercise for the reader.Ryan McGuire wrote:
fred rosenberger wrote:...I'll say 5.
Show your work.
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors
Ryan McGuire wrote:And yes, we're talking about traditional family trees. i.e. nothing like I'm My Own Grandpa.
Ryan McGuire wrote:
But we all know that titar have an overly developed "personal" protection drive, bordering on paranoia, when just one other titar is present. Therefore, any time only two titar are in close proximity to each other it usually results in a frenetic battle to the death. This would result in a puzzler more like, "Titar triumphantly on top of Titar; Titar bleeding to death under Titar; How many Titar in all?" The only time there can be multiple (live) titar in a small area is when there is three or more, thus making Salvin's answer of three quite correct indeed.
There is a Titar in front of a Titar
There is a Titar behind a Titar
A Titar in front AND A
Titar behind,
How many Titar in all ?
Vinod Tiwari | Twitter | Shikshanirman | Guftgu
Vinod Tiwari wrote:I got 3.
Mike Simmons wrote:
salvin francis wrote:
Ryan McGuire wrote:
salvin francis wrote:
There is a Titar in front of a Titar
There is a Titar behind a Titar
Titar in front,
Titar behind,
How many Titar in all ?
Let's see... I drew a diagram of the scenario described. I have a row of 15 titars side-by-side. In front of the fifth one is one more titar, so...
16. Final answer.
Answer = 3
[Titar][Titar][Titar]
Two seems sufficient.
Vinod Tiwari | Twitter | Shikshanirman | Guftgu
Vinod Tiwari wrote:How you got 2?
There is one Titar in front of me and another behind me.
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