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Integer TRS Innermost pair #487100559
details
property
value
status
complete
benchmark
gcd_minmax.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n151.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.586269 seconds
cpu usage
0.597174
user time
0.299318
system time
0.297856
max virtual memory
124240.0
max residence set size
8652.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = gcd#(x, y) -> gcd#(y - x, x) [y > x && x > 0] gcd#(I0, I1) -> gcd#(I0 - I1, I1) [I0 >= I1 && I1 > 0] R = gcd(x, y) -> gcd(y - x, x) [y > x && x > 0] gcd(I0, I1) -> gcd(I0 - I1, I1) [I0 >= I1 && I1 > 0] gcd(0, I2) -> I2 gcd(I3, 0) -> I3 We use the reverse value criterion with the projection function NU: NU[gcd#(z1,z2)] = z2 This gives the following inequalities: y > x && x > 0 ==> y > x with y >= 0 I0 >= I1 && I1 > 0 ==> I1 >= I1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = gcd#(I0, I1) -> gcd#(I0 - I1, I1) [I0 >= I1 && I1 > 0] R = gcd(x, y) -> gcd(y - x, x) [y > x && x > 0] gcd(I0, I1) -> gcd(I0 - I1, I1) [I0 >= I1 && I1 > 0] gcd(0, I2) -> I2 gcd(I3, 0) -> I3 We use the reverse value criterion with the projection function NU: NU[gcd#(z1,z2)] = z1 This gives the following inequalities: I0 >= I1 && I1 > 0 ==> I0 > I0 - I1 with I0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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