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C Integ Progr 58257 pair #381736330
details
property
value
status
complete
benchmark
PastaC3.c
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n006.star.cs.uiowa.edu
space
Stroeder_15
run statistics
property
value
solver
AProVE
configuration
c
runtime (wallclock)
2.27504992485 seconds
cpu usage
6.200259591
max memory
4.84175872E8
stage attributes
key
value
output-size
3951
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_c /export/starexec/sandbox/benchmark/theBenchmark.c /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.c # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given C Problem could be proven: (0) C Problem (1) CToIRSProof [EQUIVALENT, 0 ms] (2) IntTRS (3) TerminationGraphProcessor [SOUND, 43 ms] (4) IntTRS (5) IntTRSCompressionProof [EQUIVALENT, 31 ms] (6) IntTRS (7) PolynomialOrderProcessor [EQUIVALENT, 14 ms] (8) IntTRS (9) IntTRSCompressionProof [EQUIVALENT, 0 ms] (10) IntTRS (11) PolynomialOrderProcessor [EQUIVALENT, 0 ms] (12) YES ---------------------------------------- (0) Obligation: c file /export/starexec/sandbox/benchmark/theBenchmark.c ---------------------------------------- (1) CToIRSProof (EQUIVALENT) Parsed C Integer Program as IRS. ---------------------------------------- (2) Obligation: Rules: f1(x, y, z) -> f2(x_1, y, z) :|: TRUE f2(x1, x2, x3) -> f3(x1, x4, x3) :|: TRUE f3(x5, x6, x7) -> f4(x5, x6, x8) :|: TRUE f6(x9, x10, x11) -> f9(arith, x10, x11) :|: TRUE && arith = x9 + 1 f7(x36, x37, x38) -> f10(x36, x37, x39) :|: TRUE && x39 = x38 + 1 f5(x15, x16, x17) -> f6(x15, x16, x17) :|: x15 < x17 f5(x18, x19, x20) -> f7(x18, x19, x20) :|: x18 >= x20 f9(x21, x22, x23) -> f8(x21, x22, x23) :|: TRUE f10(x24, x25, x26) -> f8(x24, x25, x26) :|: TRUE f4(x27, x28, x29) -> f5(x27, x28, x29) :|: x27 < x28 f8(x30, x31, x32) -> f4(x30, x31, x32) :|: TRUE f4(x33, x34, x35) -> f11(x33, x34, x35) :|: x33 >= x34 Start term: f1(x, y, z) ---------------------------------------- (3) TerminationGraphProcessor (SOUND) Constructed the termination graph and obtained one non-trivial SCC. ---------------------------------------- (4) Obligation: Rules: f4(x27, x28, x29) -> f5(x27, x28, x29) :|: x27 < x28 f8(x30, x31, x32) -> f4(x30, x31, x32) :|: TRUE f9(x21, x22, x23) -> f8(x21, x22, x23) :|: TRUE f6(x9, x10, x11) -> f9(arith, x10, x11) :|: TRUE && arith = x9 + 1 f5(x15, x16, x17) -> f6(x15, x16, x17) :|: x15 < x17 f10(x24, x25, x26) -> f8(x24, x25, x26) :|: TRUE f7(x36, x37, x38) -> f10(x36, x37, x39) :|: TRUE && x39 = x38 + 1 f5(x18, x19, x20) -> f7(x18, x19, x20) :|: x18 >= x20 ---------------------------------------- (5) IntTRSCompressionProof (EQUIVALENT) Compressed rules. ---------------------------------------- (6) Obligation: Rules: f8(x30:0, x31:0, x32:0) -> f8(x30:0 + 1, x31:0, x32:0) :|: x31:0 > x30:0 && x32:0 > x30:0 f8(x, x1, x2) -> f8(x, x1, x2 + 1) :|: x1 > x && x2 <= x ---------------------------------------- (7) PolynomialOrderProcessor (EQUIVALENT) Found the following polynomial interpretation: [f8(x, x1, x2)] = -x + x1 The following rules are decreasing: f8(x30:0, x31:0, x32:0) -> f8(x30:0 + 1, x31:0, x32:0) :|: x31:0 > x30:0 && x32:0 > x30:0 The following rules are bounded: f8(x30:0, x31:0, x32:0) -> f8(x30:0 + 1, x31:0, x32:0) :|: x31:0 > x30:0 && x32:0 > x30:0 f8(x, x1, x2) -> f8(x, x1, x2 + 1) :|: x1 > x && x2 <= x ----------------------------------------
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