In my previous blog post,I talked about the quickly-extending linear hydra!

Here is the second part,that ironicly contains different parts on it's own,but don't worry,they will come together in

an epic one thing in Part III!

New basics!:

Condition 1:If the hydra is \(0:[a]\) is equal to \(b\),then \(V_a = b\)

\(V_A\),for \(A\) is an arbitrary hydra,is the final result value of the hydra

Condition 2:If there is an X-hydra:\(0:[a]^X = 0:[[a-1]^X]^{X-1}\),where \(0:[a]^{X-a} = 0:[a]\) and \(0:[1]^X = 2\)

Condition 3:If there is a forward slash-hydra \(0:[a]/[b]\),it's solved like so:

\(0:[a]/[b] = 0:[a-1,b'^a]/[b-1] \leadsto 0:[1,c'^d]/[0] = 0[1,(c-1)'^{d+1}]^X/[1,(c-1)'^d]/[0]\),\(0:[1,0'^d] = 0:[d]\),and ultimately,

\(0:[0]/[0] =1\)

New basics for second subpart:

Condition 1b:If there is a normal bracket hydra:\((a,b) = (a+1,b-1))\)

Condition 2b:If there is a zero immediately after the first entry:\((a,0) = ((a))\)

Condition 3b:symplified hydra:\((_d a)_d = (+(_{d-2}a)_{d-2})\),where d denotes the pairs of brackets

Condition 4b:the hydra has a + root:\((+(a)) = (+(_a a-1)_a-)\)

Condition 5b:If there is a 1 in a + root hydra,reduse the number of hte brackets by 1 and turn the 1 into 2.

\((+(_a 1)_a) = (+(_{a-1}2)_{a-2})\)

Condition 6b:If there is a negative end,reduse the main entry by one and add an X,turning it into an X-hydra,which will be revealed in part III: \((+(_ab)_a-) = (+(_ab)X)_{a-1})\)

Condition 7b:If the hydra contains entries inbetween:\((+((a)b)) = (+(+((a-1)b-1)))\)

Condition 8b:any ones inside can be removed:\((+((a)1)) = (+((a)))\)

Condition 9b:when the hydra is compleately set to a 1:\((+(1)) = ((()))\)

Condition 10b:when the hydra has been simplyfied completely:\((_a)_a = (_{a-1}()()())())_{a-2}\),\(() = (((\)

The final value of the hydra \(V_a\) is the string of left brackets left.

Example:\((+(1)) = ((())) = ((()()())()) = (((((((((((()()) = (((((((((((((((() = (((((((((((((((((\) ,so \(V_1 = 17\)

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