Difference between revisions of "User:Zhan"
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==DoA== | ==DoA== | ||
+ | === Discrete formulation === | ||
What are the next starts i should choose to reduce gemstones type difference, the discrete optimization point of view. | What are the next starts i should choose to reduce gemstones type difference, the discrete optimization point of view. | ||
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Let us set ''h<sub>(i,j)</sub>(y)'' = |g<sub>i</sub>(y) - g<sub>j</sub>(y)|. This is a polyhedral function with respect to ''n'': | Let us set ''h<sub>(i,j)</sub>(y)'' = |g<sub>i</sub>(y) - g<sub>j</sub>(y)|. This is a polyhedral function with respect to ''n'': | ||
− | ''h<sub>(i,j)</sub>(n, y) = |''y<sub>i</sub> - y<sub>j</sub> + k<D<sub>(i,j)</sub>,n>''| | + | ''h<sub>(i,j)</sub>(n, y) = |''y<sub>i</sub> - y<sub>j</sub> + k<D<sub>(i,j)</sub>,n>''| where <., .> is the usual scalar product. |
=== Continuous setting === | === Continuous setting === |
Revision as of 17:22, 8 June 2021
DoA
Discrete formulation
What are the next starts i should choose to reduce gemstones type difference, the discrete optimization point of view.
Let us consider the gemstones as a vector y = (y1, y2, y3, y4) in N4, where the components respectively respresent the number of margonite gemstones, stygian gemstones, torment gemstones, titan gemstones.
Then, we consider successful runs by the following functions going from N4 to N4:
f1(x) = x + k(1, 2, 3, 4)
f2(x) = x + k(4, 1, 2, 3)
f3(x) = x + k(3, 4, 1, 2)
f4(x) = x + k(2, 3, 4, 1)
Where k = 2 in Hard Mode (HM), and where the functions respectively represent start city, start veil, start gloom, start foundry.
Now, let g be a finite number ni of composition of functions fi, (i in {1,...,4}). We consider g to be solution if it is the minimal number of composition which gives the minimum possible difference of gemstones. This gives the following minimization problem:
argming (max(i, j) |gi(y) - gj(y)| + d||n||1), where d is small.
Let us set h(i,j)(y) = |gi(y) - gj(y)|. This is a polyhedral function with respect to n:
h(i,j)(n, y) = |yi - yj + k<D(i,j),n>| where <., .> is the usual scalar product.
Continuous setting
Let us now consider z in R. The discrete problem is equivalent to this one:
minz z + <n, 1>
under:
z >= yi - yj + k<D(i,j),n>
z >= yj - yi + k<D(j,i),n>
for all couple (i,j) in {(1, 2), (2, 3), (3, 4), (1, 3), (2, 4), (1, 4)}
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In game character name Kunvie Zhan