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Post by Admin on Jun 5, 2014 15:45:42 GMT
[a,b (1)(1) c,d]HC Is the next step and it is simplified as such
[a,b ([c,d]HC) c,d]HC
and to put it in a more general formula
[a,b (β)(1) c,d]HC=[a,b (β-1)([c,d]HC) c,d]HC
When you have a case where the second number is not 1, solve as if the first number was not there
[a,b (β)(γ) c,d]HC=[a,b (β)(γ-1) a,b (β)(γ-1)...(β)(γ-1) a,b]HC
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Post by Admin on Jun 5, 2014 15:50:49 GMT
When there are more than two numbers in parentheses such as the example here [a,b (β)(γ)(δ) c,d] HCFollow these set of rules - If γ=1 and δ=1 → [a,b (β-1)([a,b (1)(1) c,d]HC)HC
- If δ=1 → [a,b (β)(γ-1)([c,d]HC) c,d]HC
- If not then → [a,b (β)(γ)(δ-1) a,b (β)(γ)(δ-1)...(β)(γ)(δ-1) a,b]HC
In the case of n entries [a,b (β)(γ)(δ)...(ψ)(ω) c,d] HC- if all but β equal 1 → [a,b (β-1)([a,b (1)(1)...(1) c,d]HC)([a,b (1)...(1) c,d]HC)...([a,b (1)(1) c,d]HC) c,d]HC. The number of 1's in the second parentheses is equal to n-1, the number of 1's in the third parentheses is n-2 and so on until the last parentheses which only have 2 1's
- if all but β and γ are equal to 1 → [a,b (β)(γ-1)([a,b (1)(1)..(1)]HC)([a,b (1)...(1) c,d]HC)...([a,b (1)(1) c,d]HC) c,d]HC The number of parentheses in the third parentheses is n-2 and in the fourth n-3
- If all but the mth entry and the numbers leading up to the mth entryare equal to 1 → [a,b (β)(γ)...(mth entry-1)([a,b (1)(1)...(1) c,d]HC)(a,b (1)...(1) c,d]HC)...([a,b (1)(1) c,d]HC) c,d]HC Where the parentheses after (mth entry -1) will have n-m 1's and descending all the way to 2 1's in the last parentheses
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Post by Admin on Jun 5, 2014 17:11:42 GMT
[a,b ((1)) c,d]HC
it can also be written as
[a,b (1)2 c,d]HC
is simplified to [a,b ([a,b (1) c,d]HC)([a,b (1) c,d]HC-1)...(1) c,d]HC
These guys follow the same rules as the single parentheses
[a,b (1)3 c,d]HC=[a,b ([a,b (1)2 c,d]HC)2([a,b (1)2 c,d]HC-1)2...(1)2 c,d]HC
or in a more general manner
[a,b (1)n c,d]HC=[a,b ([a,b (1)n-1 c,d]HC)n-1([a,b (1)n-1 c,d]HC-1)n-1...(1)n c,d]HC
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Post by Admin on Jun 9, 2014 20:40:04 GMT
[a,b {1} c,d] HC has a different kind of 1, instead of being in a parentheses it is in a squiggly bracket. [a,b {1} c,d] HC=[a,b ([a,b (1) [c,d]HC c,d] HC) [c,d]HC c,d] HCThe "{}" follow the same fules as the "()" [a,b |1| c,d] HC=[a,b {[a,b {1} [c,d]HC c,d] HC} [c,d]HC c,d] HC
The order of the "Parentheses" are as follows |
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