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Integer_TRS_Innermost 2019-03-21 04.53 pair #429994987
details
property
value
status
complete
benchmark
poly3.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n178.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.978849 seconds
cpu usage
0.9893
user time
0.70918
system time
0.28012
max virtual memory
719376.0
max residence set size
12692.0
stage attributes
key
value
starexec-result
MAYBE
output
0.74/0.97 MAYBE 0.74/0.97 0.74/0.97 DP problem for innermost termination. 0.74/0.97 P = 0.74/0.97 eval#(x, y, z) -> eval#(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 eval#(I0, I1, I2) -> eval#(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] 0.74/0.97 R = 0.74/0.97 eval(x, y, z) -> eval(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 eval(I0, I1, I2) -> eval(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] 0.74/0.97 0.74/0.97 We use the reverse value criterion with the projection function NU: 0.74/0.97 NU[eval#(z1,z2,z3)] = z1 0.74/0.97 0.74/0.97 This gives the following inequalities: 0.74/0.97 x >= 0 && z * z * z >= y ==> x >= x 0.74/0.97 I0 >= 0 && I2 * I2 * I2 >= I1 ==> I0 > I0 - 1 with I0 >= 0 0.74/0.97 0.74/0.97 We remove all the strictly oriented dependency pairs. 0.74/0.97 0.74/0.97 DP problem for innermost termination. 0.74/0.97 P = 0.74/0.97 eval#(x, y, z) -> eval#(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 R = 0.74/0.97 eval(x, y, z) -> eval(x, y - 1, z + y) [x >= 0 && z * z * z >= y] 0.74/0.97 eval(I0, I1, I2) -> eval(I0 - 1, I1 - 1, I2) [I0 >= 0 && I2 * I2 * I2 >= I1] 0.74/0.97 0.74/3.95 EOF
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