Case Studies
C. W. Brown, U.S. Naval Academy
1
Epidemiological Example
Recall that the existence of an endemic equalibirium for the SEIT model was
equivalent to

−νS 2β12
2
2
2

+ β1νS β2 + dβ1Sr2 − d β2S + d β1S + β1Sr1r2 − dνSβ2
∃S  +νβ1Sqr2 − dβ2r2S + dνSβ1 − β1Sνβ2 + β1Sr1d + β2d2 + νβ2d 
+β2dr2 = 0 ∧ 0 < S < 1
... under the assumption that all parameters are positive.
C. W. Brown, U.S. Naval Academy
2
νβ1 − dr2 − d2 − νd > 0 ∧ ν 2β12β22 − 2νdr2β1β22 − 2νd2β1β22 − 2ν 2dβ1β22 + d2r22β22 + 2d3r2β22
+ 2νd2r2β22 + d4β22 + 2νd3β22 + ν 2d2β22 − 2νr1r2β12β2 + 2νdr2β12β2 − 2ν 2qr2β12β2 − 2νdr1β12β2
2 2
2
2
2
2 2
2
2
.+ 2νd β1 β2 + 2ν dβ1 β2 − 2dr1r2 β1β2 − 2d r2 β1β2 − 2νqdr2 β1β2 − 4d r1r2β1β2 − 2νdr1r2β1β2
− 4d3r2β1β2 − 2νqd2r2β1β2 − 4νd2r2β1β2 − 2ν 2qdr2β1β2 − 2d3r1β1β2 − 2νd2r1β1β2 − 2d4β1β2
− 4νd3β1β2 − 2ν 2d2β1β2 + r12r22β12 + 2dr1r22β12 + 2νqr1r22β12 + d2r22β12 + 2νqdr22β12 + ν 2q 2r22β12
+2dr12r2β12 +4d2r1r2β12 +2νqdr1r2β12 +2νdr1r2β12 +2d3r2β12 +2νqd2r2β12 +2νd2r2β12 +2ν 2qdr2β12
+d2r12β12 +2d3r1β12 +2νd2r1β12 +d4β12 +2νd3β12 +ν 2d2β12 >= 0∧[νβ1 −r1r2 −dr2 −νqr2 −dr1 −d2
−νd > 0∨β2 > root1ν 2β12β22 −2νdr2β1β22 −2νd2β1β22 −2ν 2dβ1β22 +d2r22β22 +2d3r2β22 +2νd2r2β22
+ d4β22 + 2νd3β22 + ν 2d2β22 − 2νr1r2β12β2 + 2νdr2β12β2 − 2ν 2qr2β12β2 − 2νdr1β12β2 + 2νd2β12β2
+ 2ν 2dβ12β2 − 2dr1r22β1β2 − 2d2r22β1β2 − 2νqdr22β1β2 − 4d2r1r2β1β2 − 2νdr1r2β1β2 − 4d3r2β1β2
− 2νqd2r2β1β2 − 4νd2r2β1β2 − 2ν 2qdr2β1β2 − 2d3r1β1β2 − 2νd2r1β1β2 − 2d4β1β2 − 4νd3β1β2
− 2ν 2d2β1β2 + r12r22β12 + 2dr1r22β12 + 2νqr1r22β12 + d2r22β12 + 2νqdr22β12 + ν 2q 2r22β12 + 2dr12r2β12
+4d2r1r2β12 +2νqdr1r2β12 +2νdr1r2β12 +2d3r2β12 +2νqd2r2β12 +2νd2r2β12 +2ν 2qdr2β12 +d2r12β12
+ 2d3r1β12 + 2νd2r1β12 + d4β12 + 2νd3β12 + ν 2d2β12]
C. W. Brown, U.S. Naval Academy
3
Epidemiological Example
In this model, the condition β1 > β2 should also hold, so ...

−νS 2β12
2
2
2

+ β1νS β2 + dβ1Sr2 − d β2S + d β1S + β1Sr1r2 − dνSβ2
∃S  +νβ1Sqr2 − dβ2r2S + dνSβ1 − β1Sνβ2 + β1Sr1d + β2d2 + νβ2d 
+β2dr2 = 0 ∧ 0 < S < 1
... under the assumption that all parameters are positive, and β1 > β2. This
produces the quantifier-free equivalent formula:
C. W. Brown, U.S. Naval Academy
4
Epidemiological Example
In this model, the condition β1 > β2 should also hold, so ...

2
−νS 2β12
2
2

+ β1νS β2 + dβ1Sr2 − d β2S + d β1S + β1Sr1r2 − dνSβ2
∃S  +νβ1Sqr2 − dβ2r2S + dνSβ1 − β1Sνβ2 + β1Sr1d + β2d2 + νβ2d 
+β2dr2 = 0 ∧ 0 < S < 1
... under the assumption that all parameters are positive, and β1 > β2. This
produces the quantifier-free equivalent formula:
νβ1 − r1r2 − dr2 − νqr2 − dr1 − d2 − νd > 0
C. W. Brown, U.S. Naval Academy
4
The Epidemiological Example Shows ...
• the importance of preparing input (recall the substitutions made to eliminate
E, I, adn T )
• Order of variables is crucial for this problem (the earlier simple heuristic was
used).
• you can only get a simple defining formula if the set your Q.E. problem
describes has a simple defining formula.
• sometimes CADs of high dimension (8 variables for this problem) are feasible.
C. W. Brown, U.S. Naval Academy
5
External Trisectors Problem
Recall: For triangle 4ABC the existence of the external trisector of B with
respect to A is equivalent to
2
2
2
2
2
2
2
2
a +b −c < ab∨a +b −c ≥ ab∧−c a + b − c
2 3
2 2
2
2
+3a b c a + b − c
2
under the assumption that a, b, c define a non-degenerate triangle, i.e. that
a>0∧b>0∧c>0∧a<b+c∧b<a+c∧c<a+b
Solution: The above is equivalent to c2 + bc − a2 > 0 under the assumptions.
C. W. Brown, U.S. Naval Academy
6
External Trisectors: a deeper look
Normalize a = 1 and view the CAD in the bc-plane:
condition
C. W. Brown, U.S. Naval Academy
7
External Trisectors: a deeper look
Normalize a = 1 and view the CAD in the bc-plane:
condition
C. W. Brown, U.S. Naval Academy
condition with assumptions
7
External Trisectors: a deeper look
Normalize a = 1 and view the CAD in the bc-plane:
condition
condition with assumptions
simplified CAD
From this it’s easy to read off the formula c2 + bc − a2 > 0.
C. W. Brown, U.S. Naval Academy
7
Realizable Sign Stack Sequences
• What sign stack sequences are realized by a generic, monic quartic polynomial?
• Consider f = x4 + ax3 + bx2 + cx + d.
• Construct CAD for f, f 0, f 00, f 000 with order a ≺ b ≺ c ≺ d ≺ x.
• Consider cell R in induced CAD of R4.
• Since f, f 0, f 00, f 000 are delineable over R, every (a, b, c, d) ∈ R defines a
polynomial with the same sign stack sequence, which can be read off of the
stack over R.
• So we can enumerate all realizable sign stack sequences.
C. W. Brown, U.S. Naval Academy
8
Realizable Sign Stack Sequences Cont.
• The problem is only interested in the “generic” situtation, i.e. no common
zeros between any pair of f, f 0, f 00, f 000. Thus it suffices to consider only full
dimensional cells.
• List sign stack sequence from each full dimensional cell in R4 (there are 72)
• Remove duplicate sequences (there are now 36, versus 42 legal sequences)
• Compare with list of all legal sign stack sequences to see which legal sequences
are not realizable. Here’s an example of an unrealizable sequence:
+
−
+
−
+
C. W. Brown, U.S. Naval Academy
−
−
+
−
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−
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+
−
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−
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−
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−
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9
Realizable Sign Stack Sequences: Wrapup
• A trivial (less than 1 second) calculation answers the question: there are
unrealizable sign stack sequences for 4th degree polynomials. In fact we can
list them all.
• Can easily determine that the thesis’ proposition that sequences containing
the subsequence (− + − − +), (+ − − − +), (+ − + + +), (− ∗ + + +) are
unrealizable still holds if (− ∗ + + +) is dropped.
• Degree 5 case is doable (516 realizable vs. 1000 legal).
• The Point: Make use of the properties of CADs. Don’t view CAD as
simply a black box for Q.E. or formula simplification.
C. W. Brown, U.S. Naval Academy
10