Note: In the end of this post, I added context. This post is part of a research paper. (I got rid of the 1000, but there was a reason I included it in the original code.)
Suppose, the following values are either:
cV1[1, 2] = 1
cV1[2, 1] = 1
cV2[1, 2] = 1/4
cV2[2, 1] = 1/3
c[1, 2] = 1/4
c[2, 1] = 1/3
where:
cV1[x,y]==cV1[y,x] && cV1[y,x]>cV2[r,y,x] && cV2[y,x]==c[y,x] &&
c[y,x]>cV2[x,y] && cV2[x,y]==c[x,y]
or:
cV1[1, 2] = 1
cV1[2, 1 ] = 1
cV2[1, 2] = 1/3
cV2[2, 1] = 1/4
c[1, 2] = 1/3
c[2, 1] = 1/4
where:
cV1[x,y]==cV1[r,y,x] && cV1[y,x]>cV2[x,y] && cV2[x,y]==c[x,y] &&
c[x,y]>cV2[y,x] && cV2[y,x]==c[y,x]
Whenever:
cV1[1, 2] = 1
cV1[2, 1 ] = 1
cV2[1, 2] = 1/3
cV2[2, 1] = 1/4
c[1, 2] = 1/3
c[2, 1] = 1/4
and:
Q1[x_, y_] :=
Q1[x, y] =
cV1[x, y] >= cV1[y, x] && cV1[y, x] >= cV2[y, x] &&
cV2[y, x] >= c[y, x] && c[y, x] > c[x, y] &&
c[x, y] >= cV2[x, y]
Q2[x_, y_] :=
Q2[x, y] =
cV1[y, x] >= cV1[x, y] && cV1[x, y] >= cV2[x, y] &&
cV2[x, y] >= c[x, y] && c[x, y] >= c[y, x] &&
c[y, x] >= cV2[y, x]
Q3[x_, y_] :=
Q3[x, y] =
cV1[x, y] >= cV1[y, x] && cV1[y, x] >= cV2[x, y] &&
cV2[x, y] >= c[x, y] && c[x, y] >= cV2[y, x] &&
cV2[y, x] >= c[y, x]
Q4[x_, y_] :=
Q4[x, y] =
cV1[y, x] >= cV1[x, y] && cV1[x, y] >= cV2[y, x] &&
cV2[y, x] >= c[y, x] && c[y, x] >= cV2[x, y] &&
cV2[x, y] >= c[x, y]
we want
{Q1[1,2],Q2[1,2],Q3[1,2],Q4[1,2]}
to be the following:
{True, False, False, False}
and
{Q1[2,1],Q2[2,1],Q3[2,1],Q4[2,1]}
to be the following:
{False, True, False, False}
However, we get
{True, False, False, True}
{False, True, True, False}
Also, when:
cV1[1, 2] = 1
cV1[2, 1 ] = 1
cV2[1, 2] = 1/3
cV2[2, 1] = 1/4
c[1, 2] = 1/3
c[2, 1] = 1/4
we want
{Q1[1,2],Q2[1,2],Q3[1,2],Q4[1,2]}
to be the following:
{False, False, True, False}
and
{Q1[2,1],Q2[2,1],Q3[2,1],Q4[2,1]}
to be the following:
{False, False, False, True}
However, we get
{False, True, True, False}
{True, False, False, True}
Question: How do we add criteria to Q1, Q2, Q3, Q4 to get:
{True, False, False, False}
{False, True, False, False}
{False, False, True, False}
{False, False, False, True}
without “cheating”:
- we can’t use
cV1[1000, Min[1,2], Max[1,2]]: I cannot use this notation in my math paper - we can use the
Signfunction (e.g.,Sign[c[r,x,y]-c[r,y,x]]) but when I applied it toQ1,Q2,Q3, andQ4, I couldn’t get what I wanted.
Context:
I added context due to the comments.
Note the following:
cV1[1,2],cV1[2,1],cV2[1,2],cV2[2,1],c[1,2],c[2,1]are definitions.Q1,Q2,Q3, andQ4can be changed withSign,Min,Max,<,<=,>,>=,==,!=. If neither of those are possible, includePiecewise.- The
1000is not relevant for this post. I will get rid of it.