- Joined
- Sep 22, 2010
I know what you're going to say--I'm preaching to the choir. But what prompted me to think about this is wallylutz's claim (in the 2011 Worlds forum) that a triple-triple's advantage is that it frees up jumping passes that a skater can fill with even more valuable jumps. But this proposition shows that there is no such advantage, and may even carry a disadvantage.
By the way, I credit this idea to Mathman and Serious Business.
Proposition: For any CoP-compliant seven-triple jump layout with two triple-triple combinations, one 3-jump combination and two double axels, there is another layout consisting of exactly the same jumps but with both triple-triples removed such that:
1. The base value of the jump layout is at least the same.
2. The GOE potential is higher.
(We disregard the 1/2 mark bonus of 10% without loss of generality)
The Proof is rather trivial. Substitute a double axel as the first jump in both triple-triple combinations. If the same type of jump is used as the first jump in both triple-triples, then substitute that triple jump into the 3-2-2 combo to comply with the Zayak rule. Now you have a seven-triple layout with no triple-triple combination at all, but exactly the same base value and higher GOE potential since every jumping pass has a triple.
One can even extend this idea to the ultimate layout for a ladies free skate (with no quad). Consider this 8-triple jump layout with two triple-triples and one triple-triple-double (!).
3A+3L
3A+3L
3Lz+3T+2L
3F
3S
2A
2A
Now rearrange it like this:
2A+3L
2A+3L
3A+3T+2L
3A
3Lz
3F
3S
Exact same jumps, exact same base value, and higher GOE potential. But it's much easier in comparison, since I don't think that we've ever seen a 3A+3L combo even from the men's competition.
Isn't it a travesty that first jump layout is not higher in base value than the second, and can earn less GOE points in total?
By the way, once can generalize the proposition to the following statement:
For any n-triple jump layout where n=7 or n<7 consisting of one 3-jump combination, two double axels and at least one triple-triple combination, there exists another layout such that:
1. It contains exactly the same jumps (hence identical base value).
2. It contains no triple-triple combination (hence easier).
3. It has greater GOE potential (hence a disincentive to perform the triple-triple).
I'm not going to bother to prove the case where n < 7 since it's even more trivial than the above.
By the way, I credit this idea to Mathman and Serious Business.
Proposition: For any CoP-compliant seven-triple jump layout with two triple-triple combinations, one 3-jump combination and two double axels, there is another layout consisting of exactly the same jumps but with both triple-triples removed such that:
1. The base value of the jump layout is at least the same.
2. The GOE potential is higher.
(We disregard the 1/2 mark bonus of 10% without loss of generality)
The Proof is rather trivial. Substitute a double axel as the first jump in both triple-triple combinations. If the same type of jump is used as the first jump in both triple-triples, then substitute that triple jump into the 3-2-2 combo to comply with the Zayak rule. Now you have a seven-triple layout with no triple-triple combination at all, but exactly the same base value and higher GOE potential since every jumping pass has a triple.
One can even extend this idea to the ultimate layout for a ladies free skate (with no quad). Consider this 8-triple jump layout with two triple-triples and one triple-triple-double (!).
3A+3L
3A+3L
3Lz+3T+2L
3F
3S
2A
2A
Now rearrange it like this:
2A+3L
2A+3L
3A+3T+2L
3A
3Lz
3F
3S
Exact same jumps, exact same base value, and higher GOE potential. But it's much easier in comparison, since I don't think that we've ever seen a 3A+3L combo even from the men's competition.
Isn't it a travesty that first jump layout is not higher in base value than the second, and can earn less GOE points in total?
By the way, once can generalize the proposition to the following statement:
For any n-triple jump layout where n=7 or n<7 consisting of one 3-jump combination, two double axels and at least one triple-triple combination, there exists another layout such that:
1. It contains exactly the same jumps (hence identical base value).
2. It contains no triple-triple combination (hence easier).
3. It has greater GOE potential (hence a disincentive to perform the triple-triple).
I'm not going to bother to prove the case where n < 7 since it's even more trivial than the above.