Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Integer_TRS_Innermost 2019-03-21 04.53 pair #429995139
details
property
value
status
complete
benchmark
eratosthenes.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n009.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
1.10763 seconds
cpu usage
1.1292
user time
0.603341
system time
0.52586
max virtual memory
125128.0
max residence set size
8664.0
stage attributes
key
value
starexec-result
MAYBE
output
1.08/1.10 MAYBE 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 isdiv#(x, y) -> isdiv#(0 - x, y) [0 > x] 1.08/1.10 isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1] 1.08/1.10 isdiv#(I2, I3) -> isdiv#(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0] 1.08/1.10 if_2#(false, I7, I8, zs) -> filter#(I7, zs) 1.08/1.10 filter#(I9, cons(I10, B0)) -> if_2#(isdiv(I9, I10), I9, I10, B0) 1.08/1.10 filter#(I9, cons(I10, B0)) -> isdiv#(I9, I10) 1.08/1.10 if_1#(true, I11, I12, B1) -> filter#(I11, B1) 1.08/1.10 filter#(I13, cons(I14, B2)) -> if_1#(isdiv(I13, I14), I13, I14, B2) 1.08/1.10 filter#(I13, cons(I14, B2)) -> isdiv#(I13, I14) 1.08/1.10 sieve#(cons(I16, ys)) -> sieve#(filter(I16, ys)) 1.08/1.10 sieve#(cons(I16, ys)) -> filter#(I16, ys) 1.08/1.10 nats#(I17, I18) -> nats#(I17 + 1, I18) [I18 > I17] 1.08/1.10 primes#(I23) -> sieve#(nats(2, I23)) 1.08/1.10 primes#(I23) -> nats#(2, I23) 1.08/1.10 mem#(I24, cons(I25, B3)) -> mem#(I24, B3) [I24 > I25] 1.08/1.10 mem#(I26, cons(I27, B4)) -> mem#(I26, B4) [I27 > I26] 1.08/1.10 isprime#(I31) -> mem#(I31, primes(I31)) 1.08/1.10 isprime#(I31) -> primes#(I31) 1.08/1.10 R = 1.08/1.10 isdiv(x, y) -> isdiv(0 - x, y) [0 > x] 1.08/1.10 isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1] 1.08/1.10 isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0] 1.08/1.10 isdiv(I4, I5) -> false [I4 > I5 && I5 > 0] 1.08/1.10 isdiv(I6, 0) -> true [I6 > 0] 1.08/1.10 if_2(false, I7, I8, zs) -> cons(I7, filter(I7, zs)) 1.08/1.10 filter(I9, cons(I10, B0)) -> if_2(isdiv(I9, I10), I9, I10, B0) 1.08/1.10 if_1(true, I11, I12, B1) -> filter(I11, B1) 1.08/1.10 filter(I13, cons(I14, B2)) -> if_1(isdiv(I13, I14), I13, I14, B2) 1.08/1.10 filter(I15, nil) -> nil 1.08/1.10 sieve(cons(I16, ys)) -> cons(I16, sieve(filter(I16, ys))) 1.08/1.10 sieve(nil) -> nil 1.08/1.10 nats(I17, I18) -> cons(I17, nats(I17 + 1, I18)) [I18 > I17] 1.08/1.10 nats(I19, I20) -> cons(I19, nil) [I19 = I20] 1.08/1.10 nats(I21, I22) -> nil [I21 > I22] 1.08/1.10 primes(I23) -> sieve(nats(2, I23)) 1.08/1.10 mem(I24, cons(I25, B3)) -> mem(I24, B3) [I24 > I25] 1.08/1.10 mem(I26, cons(I27, B4)) -> mem(I26, B4) [I27 > I26] 1.08/1.10 mem(I28, cons(I29, B5)) -> true [I28 = I29] 1.08/1.10 mem(I30, nil) -> false 1.08/1.10 isprime(I31) -> mem(I31, primes(I31)) 1.08/1.10 1.08/1.10 The dependency graph for this problem is: 1.08/1.10 0 -> 1, 2 1.08/1.10 1 -> 0, 2 1.08/1.10 2 -> 2 1.08/1.10 3 -> 4, 5, 7, 8 1.08/1.10 4 -> 3 1.08/1.10 5 -> 0, 1, 2 1.08/1.10 6 -> 4, 5, 7, 8 1.08/1.10 7 -> 6 1.08/1.10 8 -> 0, 1, 2 1.08/1.10 9 -> 9, 10 1.08/1.10 10 -> 4, 5, 7, 8 1.08/1.10 11 -> 11 1.08/1.10 12 -> 9, 10 1.08/1.10 13 -> 11 1.08/1.10 14 -> 14, 15 1.08/1.10 15 -> 14, 15 1.08/1.10 16 -> 14, 15 1.08/1.10 17 -> 12, 13 1.08/1.10 Where: 1.08/1.10 0) isdiv#(x, y) -> isdiv#(0 - x, y) [0 > x] 1.08/1.10 1) isdiv#(I0, I1) -> isdiv#(I0, 0 - I1) [0 > I1] 1.08/1.10 2) isdiv#(I2, I3) -> isdiv#(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0] 1.08/1.10 3) if_2#(false, I7, I8, zs) -> filter#(I7, zs) 1.08/1.10 4) filter#(I9, cons(I10, B0)) -> if_2#(isdiv(I9, I10), I9, I10, B0) 1.08/1.10 5) filter#(I9, cons(I10, B0)) -> isdiv#(I9, I10) 1.08/1.10 6) if_1#(true, I11, I12, B1) -> filter#(I11, B1) 1.08/1.10 7) filter#(I13, cons(I14, B2)) -> if_1#(isdiv(I13, I14), I13, I14, B2) 1.08/1.10 8) filter#(I13, cons(I14, B2)) -> isdiv#(I13, I14) 1.08/1.10 9) sieve#(cons(I16, ys)) -> sieve#(filter(I16, ys)) 1.08/1.10 10) sieve#(cons(I16, ys)) -> filter#(I16, ys) 1.08/1.10 11) nats#(I17, I18) -> nats#(I17 + 1, I18) [I18 > I17] 1.08/1.10 12) primes#(I23) -> sieve#(nats(2, I23)) 1.08/1.10 13) primes#(I23) -> nats#(2, I23) 1.08/1.10 14) mem#(I24, cons(I25, B3)) -> mem#(I24, B3) [I24 > I25] 1.08/1.10 15) mem#(I26, cons(I27, B4)) -> mem#(I26, B4) [I27 > I26] 1.08/1.10 16) isprime#(I31) -> mem#(I31, primes(I31)) 1.08/1.10 17) isprime#(I31) -> primes#(I31) 1.08/1.10 1.08/1.10 We have the following SCCs. 1.08/1.10 { 14, 15 } 1.08/1.10 { 11 } 1.08/1.10 { 9 } 1.08/1.10 { 3, 4, 6, 7 } 1.08/1.10 { 0, 1 } 1.08/1.10 { 2 } 1.08/1.10 1.08/1.10 DP problem for innermost termination. 1.08/1.10 P = 1.08/1.10 isdiv#(I2, I3) -> isdiv#(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0] 1.08/1.10 R = 1.08/1.10 isdiv(x, y) -> isdiv(0 - x, y) [0 > x] 1.08/1.10 isdiv(I0, I1) -> isdiv(I0, 0 - I1) [0 > I1] 1.08/1.10 isdiv(I2, I3) -> isdiv(I2, 0 - I2 + I3) [I3 >= I2 && I2 > 0] 1.08/1.10 isdiv(I4, I5) -> false [I4 > I5 && I5 > 0]
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Integer_TRS_Innermost 2019-03-21 04.53